MATH 224 Sol.

245
Solutions Manual for A Course in Ordinary Differential Equations by Randall J. Swift Stephen A. Wirkus

Transcript of MATH 224 Sol.

Page 1: MATH 224 Sol.

Solutions Manual

forA Course in Ordinary Di!erential

Equations

byRandall J. Swift

Stephen A. Wirkus

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2

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Preface

This solutions manual is a guide for instructor’s using A Course in Ordinary Di!erential Equations.Many problems have their solution presented in its entirety while some merely have an answer andfew are skipped. This should provide su!cient guidance through the problems posed in the text.

As with the book, code for Matlab, Maple, or Mathematica is not given. It is our experience thatthe syntax given in the book is su!cient to learn the relevant commands used to obtain solutions tothe various problems in the book. Please give Appendix A a chance, if you have not done so already.

This solutions manual was put together by many people and we note a few of them here. We owea big thanks to our former students David Monarres, for help in preparing portions of this book,and Walter Sosa and Moore Chung, for their help in preparing solutions. More recently Scott Wildehas helped tremendously in shaping this manual. Jenny Switkes and other colleagues and studentshave also given feedback on various drafts of this manual and all have been helpful.

This book has evolved over the last few years and we have tried to make this solution manualstay in step. However, we realize that there are probably many typos throughout and we encourageyou to contact us with your corrections. Hopefully future printings of this manual will have anexponentially decreasing number of such errors.

We would appreciate any comments that you might have regarding the book and the manual.

Randall J. Swift (e-mail: [email protected])Stephen A. Wirkus (e-mail: [email protected])

URL for typos and errata: http://www.csupomona.edu/!swirkus/ACourseInODEs

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Contents

1 Traditional First-Order Di!erential Equations 11.1 Some Basic Terminology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Separable Di"erential Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.3 Physical Problems with Separable Equations . . . . . . . . . . . . . . . . . . . . . . 241.4 Exact Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 271.5 Linear Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 331.6 Chapter 1: Additional Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

2 Geometrical and Numerical Methods for First-Order Equations 512.1 Direction Fields—the Geometry of Di"erential Equations . . . . . . . . . . . . . . . 512.2 Existence and Uniqueness for First-Order Equations . . . . . . . . . . . . . . . . . . 562.3 First-Order Autonomous Equations—Geometrical Insight . . . . . . . . . . . . . . . 592.4 Population Modeling: An Application of Autonomous Equations . . . . . . . . . . . 822.5 Numerical Approximation with the Euler Method . . . . . . . . . . . . . . . . . . . . 832.6 Numerical Approximation with the Runge-Kutta Method . . . . . . . . . . . . . . . 872.7 An Introduction to Autonomous Second-Order Equations . . . . . . . . . . . . . . . 912.8 Chapter 2: Additional Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91

3 Elements of Higher-Order Linear Equations 933.1 Some Terminology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 933.2 Essential Topics from Linear Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . 943.3 Reduction of Order—The Case of n = 2 . . . . . . . . . . . . . . . . . . . . . . . . . 1013.4 Operator Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1043.5 Numerical Consideration for nth Order Equations . . . . . . . . . . . . . . . . . . . 1063.6 Chapter 3: Additional Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112

4 Techniques of Higher-Order Linear Equations 1154.1 Homogeneous Equations with Constant Coe!cients . . . . . . . . . . . . . . . . . . . 1154.2 A Mass on a Spring . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1234.3 Cauchy-Euler (Equidimensional) Equation . . . . . . . . . . . . . . . . . . . . . . . . 1294.4 Nonhomogeneous Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1334.5 Method of Undetermined Coe!cients via Tables . . . . . . . . . . . . . . . . . . . . 1384.6 Method of Undetermined Coe!cients via the Annihilator Method . . . . . . . . . . . 1464.7 Variation of Parameters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1544.8 Chapter 4: Additional Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165

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5 Fundamentals of Systems of Di!erential Equations 1675.1 Systems of Two Equations—Motivational Examples . . . . . . . . . . . . . . . . . . 1675.2 Useful Terminology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1715.3 Linear Transformations and the Fundamental Subspaces . . . . . . . . . . . . . . . . 1745.4 Eigenvalues and Eigenvectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1775.5 Matrix Exponentials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1805.6 Chapter 5: Additional Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182

6 Techniques of Systems of Di!erential Equations 1856.1 A General Method, Part I: Solving Systems with Real, Distinct Eigenvalues . . . . . 1856.2 A General Method, Part II: Solving Systems with Repeated Real or Complex Eigen-

values . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1886.3 Solving Linear Homogeneous and Nonhomogeneous Systems of Equations . . . . . . 1936.4 Nonlinear Equations and Phase Plane Analysis . . . . . . . . . . . . . . . . . . . . . 1956.5 Epidemiological Models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2046.6 Chapter 6: Additional Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 207

7 Laplace Transforms 2137.1 Fundamentals of the Laplace Transform . . . . . . . . . . . . . . . . . . . . . . . . . 2137.2 Properties of the Laplace Transform . . . . . . . . . . . . . . . . . . . . . . . . . . . 2147.3 Step Functions, Translated Functions, and Periodic Functions . . . . . . . . . . . . . 2147.4 The Inverse Laplace Transform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2157.5 Laplace Transform Solution of Linear Di"erential Equations . . . . . . . . . . . . . . 2167.6 Solving Linear Systems using Laplace Transforms . . . . . . . . . . . . . . . . . . . . 2177.7 The Convolution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2177.8 Chapter 7: Additional Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 218

8 Series Methods 2218.1 Power Series Representations of Functions . . . . . . . . . . . . . . . . . . . . . . . . 2218.2 The Power Series Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2238.3 Ordinary and Singular Points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2248.4 The Method of Frobenius . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2258.5 Bessel Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2268.6 Chapter 8: Additional Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 229

B Graphing Factored Polynomials 231

C Selected Topics from Linear Algebra 233C.1 A Primer on Matrix Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 233C.2 Gaussian Elimination, Matrix Inverses, and Cramer’s Rule . . . . . . . . . . . . . . . 235C.3 Coordinates and Change of Basis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 238

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Chapter 1

Traditional First-Order Di!erentialEquations

1.1 Some Basic Terminology

1. With y(x) = 2x3, we have y!(x) = 6x2. Substituting into the ODE gives x(6x2) = 3(2x3),which is true for x ! ("#,#).

2. y = 2,dy

dx=

d

dx(2) = 0. Substituting y and

dy

dxinto the ODE, we get 0 = x3(2 " 2)2, which

is true for all x. Thus y = 2 is a solution tody

dx= x3(y " 2)2 on ("#,#).

3. y(x) ="1

5x + 4= "(5x + 4)"1,

dy

dx= (5x + 4)"2(5) =

5(5x + 4)2

. y anddy

dxdo not exist when

5x + 4 = 0 $ x = "45. Substituting y and

dy

dxinto the ODE, we get

5(5x + 4)2

= 5!

"15x + 4

"2

=5("1)2

(5x + 4)2

=5

(5x + 4)2

which is true for {x | x %= " 45}. Thus y =

"15x + 4

is a solution to the ODE on (" 45 ,#).

4. Since y(x) = ex " x, we calculate y!(x) = ex " 1. Substituting into the ODE gives

(ex " 1) + (ex " x)2 = e2x + (1 " 2x)ex + x2 " 1$ ex " 1 + e2x " 2xex + x2 = e2x + ex " 2xex + x2 " 1

which is true for all x ! ("#,#).

5. y(x) = x3,dy

dx= 3x2. Substituting y and

dy

dxinto the ODE, we get

3x2 = 3(x3)2/3

= 3x2

1

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2 Section 1.1

which is true for all x. Thus y = x3 is a solution to the ODE on ("#,#).

6. y(x) ="1

x " 3,

dy

dx= (x " 3)"2 =

1(x " 3)2

. y anddy

dxdo not exist when x " 3 = 0 $ x = 3.

Substituting y anddy

dxinto the ODE, we get

1(x " 3)2

=!

"1x " 3

"2

=("1)2

(x " 3)2

=1

(x " 3)

which is true for {x | x %= 3}. Thus y ="1

x " 3is a solution to the ODE on ("#, 3).

7. Taking the derivative of y(x) = x2 " x"1 gives us y!(x) = 2x " x"2 and y!!(x) = 2 + x"3.Substitution gives

x2(2 + x"3) = 2(x2 " x"1)$ 2x2 + x"1 = 2x2 " x"1

which is true where both sides are defined, which occurs when x %= 0.

8. y(x) = sin x + 2 cosx,dy

dx= cosx " 2 sinx, and

d2y

dx2= " sinx " 2 cosx. Substituting y and

d2y

dx2into the ODE, we get

(" sinx " 2 cosx) + (sin x + 2 cosx) = 0

which is true for all x. Thus y = sinx + 2 cosx is a solution to the ODE on ("#,#).

9. y(x) = x,dy

dx= 1, and

d2y

dx2= 0. Substituting y and

d2y

dx2into the ODE, we get

(0) + (x) = x

which is true for all x. Thus y = x is a solution to the ODE on ("#,#).

10. y(x) = x + C sin x,dy

dx= 1 + C cosx, and

d2y

dx2= "C sin x. Substituting y and

d2y

dx2into the

ODE, we get

("C sin x) + (x + C sinx) = x

x = x

which is true for all x. Thus y = x + C sin x is a solution to the ODE on ("#,#) for anyconstant C.

11. (a) y = ex, y! = ex, y!! = ex $ y!! " 3y! + 2y = ex " 3ex + 2ex = 0(b) y = e2x, y! = 2e2x, y!! = 4e2x

$ y!! " 3y! + 2y = 4e2x " 6e2x + 2e2x = 0

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Section 1.1 3

12. (a) y = ex, y! = ex, y!! = ex $ y!! " 2y! + y = ex " 2ex + ex = 0(b) y = xex, y! = xex + ex, y!! = xex + 2ex

$ y!! " 2y! + y = xex + 2ex " 2xex " 2ex + ex = 0

13. (a) y = sin 3x, y! = 3 cos 3x, y!! = "9 sin 3x$ y!! + 9y = "9 sin3x + 9 sin 3x = 0 &YES

(b) y = sinx, y! = cosx, y!! = " sinx$ y!! + 9y = " sinx + 9 sinx %= 0 &NO

(c) y = cos 3x, y! = "3 sin3x, y!! = "9 cos 3x$ y!! + 9y = "9 cos 3x + 9 cos 3x = 0 &YES

(d) y = e3x, y! = 3e3x, y!! = 9e3x

$ y!! + 9y = 9e3x + 9e3x %= 0 &NO(e) y = x3, y! = 3x2, y!! = 6x

$ y!! + 9y = 6x + 9x3 %= 0(unless x = 0)&NO

14. y!! + 6y! + 9y = 0

(a) y = ex, y! = ex, y!! = ex $ex + 6ex + 9ex %= 0 &NO

(b) y = e"3x, y! = "3e"3x y!! = 9e"3x

$ 9e"3x " 18e"3x + 9e"3x = 0 &YES(c) y = xe"3x, y! = "3xe"3x + e"3x y!! = 9e"3x " 6e"3x

$ 9xe"3x " 6e"3x " 18xe"3x + 6e"3x + 9xe"3x = 0 &YES(d) y = 4e3x, y! = 12e3x, y!! = 36e3x

$ 36e3x + 72e3x + 36e3x %= 0 &NO(e) y = e"3x(x + 2), y! = e"3x("3x " 5), y!!e"3x(9x + 12)

$ e"3x(81x + 108 " 18x " 30 + 18 + 9x) %= 0 &NO

15. y!! " 7y! + 12y = 0

(a) y = e2x, y! = 2e2x, y!! = 4e2x

$ 4e2x " 14e2x + 12e2x %= 0 &NO(b) y = e3x, y! = 3e3x, y!! = 9e3x

$ 9e3x " 21e3x + 12e3x = 0 &YES(c) y = e4x, y! = 4e4x, y!! = 16e4x

$ 16e4x " 28e4x + 12e4x = 0 &YES(d) y = e5x, y! = 5e5x, y!! = 25e5x

$ 25e5x " 35e5x + 12e5x %= 0 &NO(e) y = e3x + 2e4x, y! = 3e3x + 8e4x, y!! = 9e3x + 32e4x

$ 32e4x + 9e3x " 21e3x " 56e3x + 12e3x + 24e4x = 0 &YES

16. y!! + 4y! + 5y = 0

(a) y = e"2x, y! = "2e"2x, y!! = 4e"2x

$ 4e"2x " 8e"2x + 5e"2x %= 0 &NO(b) y = e"2x sin 2x, y! = 2e"2x cos 2x " 2e"2x sin 2x, y!! = "8e"2x cos 2x

$ "8e"2x cos 2x + 8e"2x cos 2x " 8e"2x sin 2x + 5e"2x sin 2x %= 0&NO

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4 Section 1.1

(c) y = e"2x cos 2x, y! = "2e"2x cos 2x " 2e"2x sin 2x, y!! = 8e"2x sin 2x$ 8e"2x sin 2x " 8e"2x cos 2x " 8e"2x sin 2x + 5e"2x cos 2x %= 0&NO

(d) y = cos 2x, y! = "2 sin2x, y!! = "4 cos 2x$ "4 cos 2x " 8 sin 2x + 5 cos 2x %= 0 &NO

17. y = erx, y! = rerx, and y!! = r2erx. The ODE gives

r2erx + 3rerx + 2erx = 0$ erx(r2 + 3r + 2) = 0$ erx(r + 2)(r + 1) = 0

$ r = "2,"1

18. y!! + 4y! + 4y = 0y = xerx, y! = rxerx + erx, y!! = r2xerx + 2rerx $

erx#r2x + 2r + 4rx + 4 + 4x

$= 0

$ xr2 + (2 + 4x)r + (4 + 4x) = 0

$ r ="2 " 4x ±

'4

2x

="4x

2xor

"4 " 4x

2x

= "2 or"2 " 2x

x

We can disregard the “value” involving x, which leaves only r = "2.

19. (a) 3y!! + y = sinx (i) 2nd order (ii) linear (iii) N/A(b) y!! + sin y = 0 (i) 2nd order (ii) nonlinear (iii) N/A(c) y(3) + (sin x)y(2) + y = x, y(0) = 1, y!(0) = 0, y!!(0) = 2

(i) 3rd order (ii) linear (iii) IVP(d) y! + exy = y4, y(0) = 0

(i) 1st order (ii) nonlinear (iii) IVP(e) y!! + y! " y (i) 2nd order (ii) linear (iii) N/A(f) y!! + exy! + y2 = 0, y(0) = 1, y(!) = 0

(i) 2nd order (ii) nonlinear (iii) BVP

20. (a) y!! " 3yy! = x (i) 2nd order (ii) nonlinear (iii) N/A(b) y!! = sin x (i) 2nd order (ii) linear (iii) N/A(c) y!! + 3y! = 0, y(0) = 1, y!(1) = 0

(i) 2nd order (ii) linear (iii) BVP(d) y!! = 0, y(1) = 1, y!(1) = 2

(i) 2nd order (ii) linear (iii) IVP(e) y!! " 4y! + 4y = 0, y(0) = 1, y!(0) = 1

(i) 2nd order (ii) linear (iii) IVP(f) x2y!! + y! + (lnx)y = 0 (i) 2nd order (ii) linear (iii) N/A

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1.2. SEPARABLE DIFFERENTIAL EQUATIONS 5

1.2 Separable Di!erential Equations

1.

4xy dx + (x2 + 1) dy = 04x

x2 + 1dx = "1

ydy

"%

4x

x2 + 1dx =

%dy

y

Letting u = x2 + 1,

"2%

du

u=

%dy

y

"2 ln |u| + C1 = ln |y|"2 ln(x2 + 1) + C1 = ln |y|

e"2 ln(x2+1)+C1 = |y||y| = eC1(x2 + 1)"2, since eC1 > 0

y =C

(x2 + 1)2, where C = ±eC1

2.

tan xdy + 2y dx = 0

"%

dy

2y=

%dx

tan x

" ln |y| = 2 ln |sin x| + C1

1|y| = |sinx|2eC1

y = C · csc2 x, c = ±eC1

3. (ex + 1) cos y dy + ex(sin y + 1) dx = 0, y(0) = 3. Then

$%

cos y

sin y + 1dy =

% "ex

ex + 1dx

Substituteu1 = sin y + 1 u2 = ex + 1

du1 = cos y dy du2 = ex dx

Then

ln |sin y + 1| = " ln |ex + 1| + C $ sin y + 1 = A

&1

(ex + 1)

'

Apply IC: (sin 3) + 1 =A

(e0 + 1)$ A = 2(sin 3) + 2

Therefore,

sin y + 1 =2 + 2 sin 3

ex + 1

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6 Section 1.2

4. We have 2x(y2 + 1) dx + (x4 + 1) dy = 0, y(1) = 1. Then

(x4 + 1) ds = "2x(s2 + 1) dr%dy

y2 + 1=

% "2x

x4 + 1dx =

(u = x2

du = 2xdx

"%

1u2 + 1

du

Therefore,

arctan(y) = " arctan(u) + C

arctan(y) = " arctan(x2) + C

arctan(1) = " arctan(1) + C

+2 arctan(1)( )* +!/4

= C $ C =!

2

Finally,

arctan(y) = " arctan(x2) +!

2y = tan

,!2" arctan(x2)

-

5. xy dx + (x + 1) dy = 0 &%

dy

y=% "x

x + 1dx. Substituting u = x + 1, x = u" 1, du = dx, we

have

ln |y| =%

" (u " 1)u

du =% !

1u" 1"

du = ln |x + 1|" x + C

$ y = A(x + 1)e"x

6.

(y2 + 1)1/2 dx = xy dy

dx

x=

y.y2 + 1

dy

%dx

x=

%y.

y2 + 1dy

Let u = y2 + 1 and du = 2y dy, so du2 = y dy. Then we have%

dx

x=

12

%du'

u

ln |x| + C =(1/2)

'u

(1/2)

ln |x| + C =.

y2 + 1

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Section 1.2 7

7. We have (x2 " 1)y! + 2xy2 = 0 and y('

2) = 1. Then

(x2 " 1)dy

dx= "2xy2

dy

y2=

2xdx

x2 " 1

"%

dy

y2=

%2x

x2 " 1dx

1y

= ln |x2 " 1| + C

y =1

ln |x2 " 1| + C

IC:

11

= ln |2 " 1| + C

$ C = 1

y =1

ln |x2 " 1| + 1

8. We have y! cotx + y = 2 and y(0) = "1. Then

dy

dxcotx = "y + 2

%dy

y " 2= "

%tan xdx

ln |y " 2| = ln |cosx| + C

|y " 2| = eC |cos x|y = ±eC(cosx) + 2 = A(cosx) + 2

Since y(0) = "1, "1 = A + 2 $ A = "3. Thus

y = "3(cosx) + 2

9.

y! = 10x+y $ dy

dx= 10x10y

dy

10y= 10x dx &

%10"y dy =

%10x dx

Use au = eu ln a. Then

"10"y = 10x + C1

10y ="1

10x + C1

$ y = log10 |"10x + C|

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8 Section 1.2

10.

xdx

dt+ t = 1

xdx = (1 " t) dt%xdx =

%(1 " t) dt

12x2 = t " 1

2t2 + C1

x2 = 2t " t2 + 2C1

x = ±.

2t " t2 + C, C = 2C1

11. Let u = y " x

& du

dx=

dy

dx" 1 & du

dx+ 1 = cosu

du

dx= cosu " 1

du

cosu " 1= dx

x =1

tan(u/2)

=1

tan(y"x2 )

$ y = x + 2 tan"1(1/x)

12. Let u = 2x + y " 3 & du

dx= 2 +

dy

dx

u! " 2 = u

u! = u + 2du

u + 2= dx

ln(u + 2) = x

ex = u + 2 = 2x + y " 1y = ex " 2x + 1

13. We have (x + 2y)y! = 1, y(0) = "2. Let u = 2y + x. Then

du

dx= 2

dy

dx+ 1

Substituting into the ODE gives

u

!u! " 1

2

"= 1

This is separable:

u! " 1 =2u

$ u! =2u

+ 1 =2 + u

u

$%

u

2 + udu =

%dx

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Section 1.2 9

Note thatu

2 + u=

u + 2 " 22 + u

= 1 " 22 + u

Then % !1 " 2

2 + u

"du =

%dx $ u " 2 ln |2 + u| = x + C

Substitute back u = x + 2y:

(x + 2y) " 2 ln |2 + x + 2y| = x + C

Apply IC: 0 + 2("2) " 2 ln |2 + 0 + 2("2)| = 0 + C

"4 " 2 ln |"2| = C

The solution is given in implicit form as

2y " 2 ln |2 + x + 2y| + 4 + 2 ln 2 = 0

14. y! =.

4x + 2y " 1. Let u = 4x + 2y " 1 & dudx = 4 + 2 dy

dx $

u! = 2'

u + 4

dx =du

2'

u + 4x = "2

#ln#'

u + 2$$

"'

u

= "2,ln,.

4x + 2y " 1 + 2--

".

4x + 2y " 1

15. We have (y + 2) dx + y(x + 4) dy = 0, y("3) = "1. Then

(x + 4)y dy = "(y + 2) dx%y

y + 2dy = "

%dx

x + 4% !

1 " 2y + 2

"dy = "

%dx

x + 4y " 2 ln |y + 2| = " ln |x + 4| + C

ey"2 ln |y+2| = e" ln |x+4|+C

1|y + 2|2 · ey = eC · 1

|x + 4|ey

(y + 2)2= ± eC

(x + 4)=

A

(x + 4)where A = ±eC

Apply IC: y("3) = "1 $ e"1

("1 + 2)2=

A

("3 + 4)

$ A = e"1 $ ey

(y + 2)2=

e"1

(x + 4)

16. We have 8 cos2 y dx + csc2 xdy = 0, y(!/12) = !/4. Then

csc2 xdy = "8 cos2 y dx%sec2 y dy = "8

%sin2 xdx

tan y = "8!

12x " 1

4sin 2x + C1

"

tan y = 2 sin 2x " 4x + C, where C = "8C1

Page 16: MATH 224 Sol.

10 Section 1.2

Using y(!/12) = !/4, we solve for C:

tan,!

4

-= 2 sin

,!6

-" !

3+ C

C =!

3

Sotan y = 2 sin 2x " 4x +

!

3

17. We havedy

dx=

y3 + 2y

x2 + 3x, y(1) = 1. Separating variables and integrating gives

%1

y3 + 2ydy =

%1

x2 + 3xdx

$% ! "y

2(y2 + 2)+

12y

"dy =

% ! "13(x + 3)

+13x

"

$ "14

ln |y2 + 2| + 12

ln |y| = "13

ln |x + 3| + 13

ln |x| + C

Now apply IC:

"14

ln(12 + 2) +12

ln(1) = "13

ln(1 + 3) +13

ln(1) + C

$ C =13

ln 4 " 14

ln 3

The solution is then

"14

ln(y2 + 2) +12

ln(y) = "13

ln |x + 3| + 13

ln |x| + 13

ln 4 " 14

ln 3

18. We have y! = ex2, y(0) = 0. Then

dy

dx= ex2

%dy =

%ex2

dx

y =%

ex2dx

Apply IC:% x

0y dy =

% x

0et2 dt

y(x) " y(0) =% x

0et2 dt

y(x) =% x

0et2 dt

Page 17: MATH 224 Sol.

Section 1.2 11

19. We have y! = xyex2, y(0) = 1. Then

%dy

y=

%xex2

dx

ln |y| =12ex2

+ C

C = ln |y|" 12ex2

, when x = 0, y = 1

C = "12

Thus

ln |y| =12x2 " 1

2|y| = e"1/2e("1/2)x2

y = ±e1/2ex2"1

y is even since we have only an x2 term.

20. (a) We have x2y! " cos 2y = 1, limx#+$ y = 9!/4. Then

x2 dy

dx= cos 2y + 1

First we use a trigonometric identity as follows:

cos 2y + 1 = (cos2 y " sin2 y) + (cos2 y + sin2 y)= 2 cos2 y

Thus we obtain

dy

2 cos2 y=

dx

x2

2%

sec2 y dy =%

dx

x2

2 tan y = " 2x

+ C1

tan y = " 1x

+ C2

C2 = tan y +1x

limx#+$

C2 = tan!

9!4

"+ 0 = 1

Thustan y = 1 " 1

xor y = arctan

!1 " 1

x

"

Page 18: MATH 224 Sol.

12 Section 1.2

(b) We have 3y2y! + 16x = 2xy3. Then

3y2 dy

dx= 2x(y3 " 8)

%3y2

y3 " 8dy =

%2xdx

ln |y3 " 8| = x2 + C1

|y3 " 8| = eC1ex2

y3 = ±eC1ex2+ 8

y = [Cex2+ 8]1/3 where C = ±eC1

Notice that no value for C bounds y since eC1 %= 0.

21.

dN

dt= rN

1N

dN = r dt%

1N

dN =%

r dt

ln N = rt + C1

Note N is number of population, i.e. > 0.

elnN = ert+C1

N = eC1ert

N = C2ert

Note C2 is > 0, dealing with possible population. Thus N = C2ert is a solution to the ODE.Case 1. For values r ) 0 we have exponential growth as t & #.Case 2. For values r < 0 we have exponential decay as t & #.

Page 19: MATH 224 Sol.

Section 1.2 13

22.

dN

dt= r

!1 " N

K

"N

dN

rN#1 " N

K

$ = dt

(Let u = 1 " N/K)du

dt= " 1

K

dN

dt

"Kdu

dt= ruK(1 " u)

du

dt= ru(u " 1)

dt =du

ru(u " 1)

$ t =ln |u " 1|" ln |u|

r

=1r

ln

/1 " 1#

1 " NK

$0

$ etr = 1 " 1#1 " N

K

$

1 " N

K=

11 " etr

N = K " K

1 " etr

limt#$

N = K

23. We have (x + y) dx " xdy = 0. Then

(x + y)x

=dy

dx$ 1 +

y

x=

dy

dx

Letu =

y

x$ y! = u = xu!

Thenu + xu! = 1 + u $ u! =

1x$ u = ln |x| + C

Substituting for u givesy = x(ln |x| + C)

24. We havedy

dx=

y2 + 2xy

x2=

y2

x2+

2xy

x2

Letu =

y

x$ y! = u + xu!

Thenu + xu! = u2 + 2u $ u! =

u2 + u

x

Page 20: MATH 224 Sol.

14 Section 1.2

This equation is separable:%

du

u2 + u=%

1x

dx = ln |x| + C1

Use partial fractions to rewrite1

u2 + u.

1

223

1u(u + 1)

=A

u+

B

u + 1$ 1 = A(u + 1) + Bu = u(A + B) + A

$ A + B = 0A = 1

4$ B = "1

5

667

%1

u(u + 1)du =

% !1u" 1

u + 1

"du = ln |u|" ln |u + 1| = ln

8888u

u + 1

8888

Thus

ln8888

u

u + 1

8888 = ln |x| + C $8888

u

u + 1

8888 = eln |x|+C1 = eC1 · |x|

$ u

u + 1= C · x $ u = Cx(u + 1)

$ u(1 " Cx) = Cx

$ u =Cx

1 " Cx

Plugging in u = y/x then gives

y =Cx2

1 " Cx

25. We have2x2 dy

dx= x2 + y2 $ 2

dy

dx= 1 +

y2

x2

Let u = yx $ y! = u + xu!. Substituting gives

2(u + xu!) = 1 + u2 $ 2xu! = 1 + u2 " 2u = (u " 1)2

This is separable: %2 du

(u " 1)2=%

1x

dx

Letting w = u " 1, dw = du,

$ "2u " 1

= ln |x| + C $ u " 1 ="2

ln |x| + C

$ u = 1 " 2ln |x| + C

Plugging in u = y/x gives

y = x " 2x

ln |x| + C

Page 21: MATH 224 Sol.

Section 1.2 15

26.

xy! " y =.

x2 + y2

xy! =.

x2 + y2 + y

y! =.

x2 + y2

x+

y

x

=9

1 +, y

x

-2+, y

x

-(homogeneous)

Let y = vx, y! = v!x + v. Then

dv

dxx + v =

.1 + v2 + v

xdv

dx=

.1 + v2

1'1 + v2

dv =1x

dx

arcsinh v = ln |x| + C

Substituting back in, we have

y

x= sinh(ln |x| + C) $ y = x sinh(ln |x| + C)

27. We have (x + 2y) dx " xdy = 0. Then

x + 2y " xdy

dx= 0

x + 2y = xdy

dxdy

dx= 1 + 2

,y

x

-(homogeneous)

Let y = vx, y! = v!x + v. Then

dv

dxx + v = 1 + 2v

dv

dx= 1 + v

1v + 1

dv = dx

ln |v + 1| = x + C

v + 1 = Cex

y

x= Cex " 1

y = Cxex " x

28. We have(y2 " 2xy) dx + x2 dy = 0 $ dy

dx=

2xy " y2

x2= 2,y

x

-", y

x

-2

Page 22: MATH 224 Sol.

16 Section 1.2

Let u = yx , y! = u!x + u. Then

xu! + u = 2u " u2 $ u!x = u " u2

$ " ln |u " 1| + ln |u| = ln x + C1

$ u

u " 1= eln x+C1 = x · C

$ y/x

(y/x) " 1= xC

$ yx

y " x= Cx2

$ y

y " x= Cx

$ y = xCy " Cx2

$ y =Cx2

1 + Cx

29. We have 2x3y! = y(2x2 " y2). Then

dy

dx=

y3(2x2 " y2)2x3

$ dy

dx=, y

x

-!1 " 1

2

,y

x

-2"

Let u =y

x,

dy

dx= x

du

dx+ u. Then

xdu

dx+ u = u

!1 " 1

2u2

"$ du

dx· x = "1

2u3

$%

"2u"3 du =%

1x

dx

$ u2 = ln x + C

$ y2 = x2 ln x + Cx2

$ y = ±x'

ln x + C

Page 23: MATH 224 Sol.

Section 1.2 17

30. We have (x2 + y2)y! = 2xy. Then

dy

dx=

2xy

x2 + y2$ dv

dxx + v =

2x2v

x2 + v2x2

$ dv

dxx + v =

2v

1 + v2

$ dv

dxx =

2v

1 + v2" v

$ dv

dxx =

2v " v " v3

1 + v2

$ dv

dxx =

v " v3

1 + v2

$ 1 + v2

v " v3dv =

1x

dv

$ 1v(1 " v2)

dv +v

(1 " v2)dv =

1x

dx

$ ln |v|" ln |v2 " 1| = ln x + Cv

v2 " 1= Cx $ y

# yx

$2 " 1= Cx2

$ y = Cx2,y

x

-2" Cx2

y = Cy2 " Cx2 $ y = C(y2 " x2)

31.

xy! " y = x tan(y/x)

y! =y

x+ tan

, y

x

-(Homogeneous)

,y = vx, y! = v + xv!, v =

y

x

-

v + xv! = v + tan vdv

tan v=

dx

xln |x| = ln | sin v| + C

|x| = eC | sin v| = eC | sin y

x|

y

x= sin"1 | x

eC|

y = x sin"1 | x

eC|

32.

(2x + y) dx " (4x + 2y) dy = 0dy

dx=

12

y =x

2+ C

Page 24: MATH 224 Sol.

18 Section 1.2

33. We have y2 + x2y! = xyy!. Then

y2 = y!(xy " x2) $ y! =y2

xy " x2

$ y! =4

y " x

$ dy

dx=

y

(y " x)$ (y " x) dy = y dx

$ (y " x) dy + ("y) dx = 0

Let y! = v!x + v. Then

dv

dxx + v =

vx

vx " y=

v

v " 1$ dv

dxx =

v

v " 1" v

$ dv

dxx =

v " v(v " 1)v " 1

$ dv

dxx =

v " v2 + v

v " 1

$ dv

dxx =

2v " v2

v " 1v " 1

2x " v2dv =

1x

dx $ "12

ln(v " 2) " 12

ln |v| = ln x + C

$ ln8888

1'v " 2

8888+ ln8888

1'v

8888 = ln x + C

1.v(v " 2)

= Cx $ 1:#yx

$ #yx " 2

$ = Cx

1:1x2 (y)(y " 2x)

= Cx

=1.

y(y " 2x)= C

34. (x " y) + (y " x)y! = 0 $ y! = 1( for y %= x) $ y = x + C. If x = y, the equation is truetrivally.

Page 25: MATH 224 Sol.

Section 1.2 19

35.

(x + 4y)y! = 2x + 3y

y! =2x + 3y

x + 4y=

2 + 3(y/x)1 + 4(y/x)

,y = vx, y! = v + xv!, v =

y

x

-

v + xv! =2 + 3v

1 + 4vdx

x=

(1 + 4v)dv

2 + 2v " 4v2

(Partial fractions yields:)

ln |x| = "16;ln(1 + 2v)(2 " 2v)5

<

ln |x| = "16;ln(1 + 2(y/x))(2 " 2(y/x))5

<

36. We have(x " y) dx + (x + y) dy = 0 $ dy

dx=

y " x

x + y=

yx " 11 + y

x

Let u = yx , y! = xu! + u. Then

xu! + u =u " 11 + u

$ du

dx=

u " 11 + u

" u

$ du

dx=

"1 " u2

1 + u

$%

1 + u

1 + u2du =

%"dx

% !1

1 + u2+

u

1 + u2

"du =

%"dx $ arctanu +

12

ln |1 + u2| = "x

$ arctan, y

x

-+

12

ln!

1 +,y

x

-2"

= "x

Page 26: MATH 224 Sol.

20 Section 1.2

37.

y dx = (2x + y) dy

dy

dx=

(y/x)2 + (y/x)

(y = vx, y! = v + xv!)

v + xv! =v

2 + v

xv! = "v2 + v

v + 2

" 2 + v

v2 + vdv =

dx

x

ln!|v + 1|

v2

"= ln |x| + C

|v + 1|v2

= eC |x|

|(y/x) + 1|(y/x)2

= eC |x|

38.

y! = 2!

y

x + y

"2

= 2!

(y/x)1 + (y/x)

"2

,y = vx, y! = v + xv!, v =

y

x

-

v + xv! =2v2

(1 + v)2

xv! =v2 " v

(1 + v)2

(1 + v)2

v2 " vdv =

dx

x

ln!

(v " 1)4

v

"+ v = ln |x| + C

ln!

((y/x) " 1)4

(y/x)

"+ (y/x) = ln |x| + C

41. (a) M(1,y

x)

t= 1x= M(tx, ty)

Homog. deg. n= tnM(x, y)

=!

1x

"n

M(x, y) $ M(x, y) = xnM(1,y

x)

(b) Repeat above exchanging M for N .

(c) M(x, y)dx + N(x, y)dy = 0 $ dy

dx=

"M(x, y)N(x, y)

="M(1, y

x)N(1, y

x )

Define g1(t) = "M(1, t), g2(t) = N(1, t), g(t) ="g1(t)g2(t)

$ dy

dx=

"M(1, yx )

N(1, yx )

="g1( y

x)g2( y

x )= g(

y

x)

Page 27: MATH 224 Sol.

Section 1.2 21

42. In (23.)F (x, y) = (x + y)dx " xdy = 0 $F (tx, ty) = (tx + ty)t dx " tx t dy = t2((x + y)dx " xdy) = t2F (x, y) $Homogeneous of degree 2. The others are similar.

43. Given that M(x, y) dx + N(x, y) dy = 0 is homogeneous, thendy

dx=

"M(x, y)N(x, y)

="M(1, (y/x))N(1, (y/x))

and the substitution

x = r cos t

dx = "r sin t dt + cos t dr

y = r sin t

dy = r cos t dt + sin t dr

dy

dx=

"M(1, tan t)N(1, tan t)

="M

N

dy

dx=

r cos t dt + sin t dr

r sin t dt + cos t dr

$ "M

N=

r cos t dt + sin t dr

r sin t dt + cos t dr$ (r cos t dt + sin t dr)N = "M("r sin t dt + cos t dr)

N sin t dr + M cos t dr = "Nr cos t dt + Mr sin t dtdr

dt= r

!"N cos t + M sin t

N sin t + M cos t

"

Since the function M and N are functions of t, the right hand side is a function of r times afunction of t. Hence, it is seperable.

44. (a) From 43.

dr

dt= r

!r cos2 t + (cos t + sin t) sin t

" cos t sin t + (cos t + sin t) cos t

"

= r

!cos2 t + sin t cos t + sin2 t

cos2 t

"

= r

!1 + sin t cos t

cos2 t

"

dr

r=

!1 + sin t cos t

cos2 t

"dt

ln |r| = ln

/91

cos2 t

0+ tan t

|r| = exp

=ln

/91

cos2 t

0+ tan t + C

>

with r =.

x2 + y2 and t = tan"1,y

x

-, we finally have

.x2 + y2 = exp

=ln

/?1

cos2(tan"1# y

x

$)

0+ tan(tan"1

,y

x

-) + C

>

(b)

Page 28: MATH 224 Sol.

22 Section 1.2

45. (a)dy

dx=

y " x

y + x=

(y/x) " 1(y/x) + 1

which is homogeneous. As before, y = vx and

v + v!x =v " 1v + 1

v!x ="2

v + 1

(v + 1) dv ="2 dx

xv2

2+ v + C = = "2 ln |x|

|x| = Ke"(v2/4)"(v/2) = K exp(" y2

4x2" y

2x)

(b) i. (x " y " 1) dx + (y + x + 5) dy = 0 & M(x, y) = x " y " 1& M(tx, ty) = tx " ty " 1 %= tM(x, y). Hence, M is not homogeneous. Similarly forN , and so the equation is not homogeneous.

ii. x = X " 2, y = Y " 3, dx = dX , dy = dY and dydx = dY

dX = Y "XY +X

iii. |X | = K exp(" Y 2

4X2" Y

2X) $ |x + 2| =

K exp(" (y + 3)2

4(x + 2)2" y + 3

2(x + 2))

46. Lines are parallel, so there is no point of intersection.

47. Lines are parallel, so there is no point of intersection.

48.dy

dx=

2x + 3y " 5x + 4y

which gives point of intersection to be (4,"1). Thus, we let x = X " 4 and

y = Y + 1 and get to

dY

dX=

2X + 3Y

X + 4Y

=2 + 3(Y/X)1 + 4(Y/X)

Y = vX Y ! = v!X + v

v!X =1 " v

1 + 4vln |X | = "5 ln |1 " v|" 4X

ln |x + 4| = "5 ln88881 " y " 1

x + 4

8888" 4(x + 4)

Page 29: MATH 224 Sol.

Section 1.2 23

49. Point of intersection is (3,"2). Thus we let x = X + 3 and y = Y " 2 and get

dY

dX=

Y

2X + YdX

dY=

2X + Y

Y= 2(x/y) + 1

X = uYdX

dY= u!Y + u

u!Y + u = 2u + 1du

u + 1=

dY

Yln |u + 1| = ln |Y | + C

|Y | = K|u + 1| = K|(X/Y ) + 1|

y + 2 = ±K

!x " 3y + 2

+ 1"

50. y! = 2!

y + 2x + y " 1

"2

and as in 49.,

dY

dX= 2

!Y

X + Y

"2

dX

dY=

12

!X

Y+ 1"2

X = uYdX

dY= u!Y + u

u + u!X =12u2 + u +

12

u!X =12(u2 + 1)

du

u2 + 1=

12

dX

X

tan"1(u) =12

ln |X | + C

tan"1

!x " 3y + 2

"=

12

ln |x " 3| + C

Page 30: MATH 224 Sol.

24 Section 1.3

1.3 Physical Problems with Separable Equations

1. r = kv2, ve = 60m/s, v0 = 0, m = 75kg, s0 = 1800m, se = 500m

mdv

dt= mg " kv2

75v! = 75(9.8)" kv2

= 735 " kv2

75735 " kv2

dv = dt

"1.38 ln888%

kv"27.11%kv+27.11

888'

k= t + C

88888

'kv " 27.11'kv + 27.11

88888 = exp

='kt + C

"1.38

>

From the given information limt#$

v(t) = 60 $88888

'k60 " 27.11'k60 + 27.11

88888 = 0 & k = .20417

v0 = 0 & C = 0

Thus, we have.45v " 27.11.45v + 27.11

= e"(.326t)

and after some simplification

v(t) = 60!

1 + e".326t

1 " e".326t

"

2. r = kv2, r = .48N when v = 1 & k = .48

.4dv

dt= (.4)(9.8) " .48v2

dv

dt= 9.8 " 1.2v2

dv

9.8 " 1.2v2= dt

If air resistance is neglected, thens(t) = "4.9t2 + 20t, v(t) = "9.8t + 20 = 0& t = 2.04, s(2.04) = 20.4ft

4. v(t) = "gt + v0, v(3) = "3g + v0 = 0 &v0 = 3g = 3s(32ft/s2) = 96ft/s

5. (a) v0 = 0, v(t) = gt = 32t, v(4) = 128ft/s

(b) Ave velocity=14

% 4

032t dt = 64ft/s

(c) v(t) = 32t, s(t)16t2 + s0, s(4) = 0 & s(4) = 162 + s0 = 0& s0 = "256 & cli" was 256 ft tall

6. a(t) = 32, v(t) = 32t + v0 = 32t, s(t) = 16t2 + s0,s(5) = 16(25) + s0 = 0 & s0 = "400 & cli" was 400 ft tall

Page 31: MATH 224 Sol.

Section 1.3 25

7. s0 = "350, v0 = 0, s(t) = 16t2 " 350 = 0 & t ='

3504

& average velocity =2km'350/4

=8km'350s

= .428km/s

8. T (t) = (T0 " Ts)ekt + Ts, Ts = "2, T0 = 15, T (2) = 5

& T (2) = 17e2k " 2 = 5 & k =12

ln!

717

". Then to find eating time (t = eat),

T (eat) = 17ek(eat) " 2 = 0 & eat = 2ln(2/17)ln(7/17)

= 4.82hrs * 5pm

9. Let t = 0 be 3pm. T (0) = 79, T (3) = 68, Ts = 60.T (t) = 19ekt + 60, T (3) = 19e3k + 60 = 68

19e3k = 8

k =13

ln .421 = ".288

T (t) = 19e".288t + 60(Solve for t when T = 98.6) = 98.6

t = "2.46

Thus the person died about 12:30pm

10. T (t) = 80ekt + 20, T (10) = 80e10k + 20 = 60 & k = ".069

T (t) = 80e".069t + 20 = 25 & t =10 ln 16

ln 2= 40 min

13. N(t) = amount of N in tank after t seconds. N(0) = 16. dNdt = rate in - rate out = .1 " N

20(.1) &

200dN

20 " N= dt & N(t) = 20 " e("t/200)+C ,

N(0) = 16 = 20 " eC & C = ln 4 * 1.39 N(t) = 20 " e("t/200)+1.39. We want to solve N(t) =20(.99) = 19.8 for t. t = "200(ln(.2) " 1.39) * 599 seconds * 10 min.

14. s(t)=amount of salt at time t. s(0) = 10.ds

dt= 0 "

!s(t)kg

100L

"(5L/m) =

"s(t)20

& s(t) = e("t/20)+C & C = ln 10s(t) = 10e("t/20)+ln10 & s(60) = 4.97 * 5kg

15. (a)ds

dt= (4lb/gal)(2gal/m)" (s(t)lb/50gal)(2gal/min) = 8 " s(t)

25& s(t) = 16 + 9e("t/25) & concentration= s(t)/25.

(b)ds

dt= 8 " 3s

50& s(t) =

4003

" 3253

e("3t/50) & conc= s(t)/25 " t.

16. s(t) = amount of salt in pond, s0 = 0, v(t) =volume of pond, v0 = 10000, dvdt = "50 & v(t) =

10000" 50t.ds

dt= (1250m3/d)

!5kg

1000m3

""!

s(t)kg

(10000" 5t)m3

"(1300m3/d)

= 6.25 " 1300s(t)10000" 50t

Yes, after 200 days, there is no water in the pond which will result in no salt in the pond.

Page 32: MATH 224 Sol.

26 Section 1.3

17. C(t) = amount of CO2, C0 = (.0015)(200) = .3m3, dCdt =

,20m3

min

-(.0004)"

,Cm3

200m3

-(20m3/min) =

.008 " .1C This gives us that the concentration will be half of what it started (i.e. there willbe .15m3 of CO2) in 11.45 minutes.

18.dA

dt= kA,

dA

A= k dt & A(t) = ekt+C = A0e

kt.

A(30) = A0/2 = A0e30k

12

= e30k

k =ln(1/2)

30= ".023

A(t) = .01A0 = A0e".023t

t =ln(.01)".023

* 200 days

19. A(1) = .56 = ek & k = ".5798 & t = 1.195 years

20. p = amount of decayed tin, M=original amount of tin.

dp

dt= k(M " p)(p) where M " p is the amount of tin left

dP

p(M " p)= k dt

1M

ln!

p

M " p

"= kt + C

$ p(t) = M " M

1 + eMkt+C

where C would be determined by the original amount of decayed tin in the organ.

22. Let (x, y) be the point of tangency and c be the x-intercept of the tangent line. Then 12 |c "

x||y| = a2 & |c " x| = 2a2

|y| .

dy

dx=

y

|c " x| =y

2a2

|y|

=y|y|2a2

dy

y2=

12a2

dx

y = ±2a2

x

Because of symmetry, either graph will give the desired result.

23. Let (x, y) be the point of tangency and c be the x-intercept of the tangent line. Then |c"x|+|y| = b.

dy

dx=

y

|c " x| =y

b " |y|b ln |y|" |y| = x

This is an implicitly defined function.

Page 33: MATH 224 Sol.

1.4. EXACT EQUATIONS 27

33. Let A(t) be the amount of snow on the ground. Then A(t) = kt since it is falling at aconstant rate and there was none when it started. Let P (t) be the distance plowed. ThendPdt = m

A(t) = mkt . Thus dP = m

ktdt & P = mk ln t + C, P (1) = 2 = C & P (t) = m

k ln t + 2.P (2) = m

k ln 2 +2 = 1 & mk = "1.443. P (t) = "1.443 ln t + 2. Set P (t) = 0 to get t * 4 which

says it started snowing at about 8am.

1.4 Exact Equations

1. We have 2xy3 + (1 + 3x2y2) dydx = 0. Using M = 2xy3 and N = 1 + 3x2y2, we have

"M

"y= 6xy2,

"N

"x= 6xy2

so the equation is exact. We need to solve "f"x = M and "f

"y = N :

f =%"f

"xdx =

%2xy3 dx = x2y3 + g(y)

This f = x2y3 + g(y) must also satisfy "f"y = N .

1 + 3x2y2 ="

"y(x2y3 + g(y)) = 3x2y2 + g!(y)

$ g!(y) = 1 $ g(y) = y

Thus the solution is given by f = C:

x2y3 + y = C

2. We have (2xy + 1) + (x2 + 4y) dy = 0, y(0) = 1. Use M = 2xy + 1 and N = x2 + 4y. Then

"M

"y= 2x =

"N

"x

so the equation is exact. We set "f"x = M and "f

"y = N and solve.

f =%"f

"ydy =

%(x2 + 4y) dy = x2y + 2y2 + h(x)

This f must also satisfy "f"x = M :

2xy + 1 ="

"x(x2y + 2y2 + h(x)) = 2xy + h!(x)

$ h!(x) = 1 $ h(x) = x

The solution is given byx2y + 2y2 + x = C

Initial conditions yield C = 2, sox2y + 2y2 + x = 2

Page 34: MATH 224 Sol.

28 Section 1.4

3. We have (y sec2 x + secx tan x) + (tanx + 2y) dy = 0 with y(0) = 1. Setting M = (y sec2 x +secx tan x) and N = tan x + 2y leads to

"M

"y= sec2 x =

"N

"x

so the ODE is exact. Then,

F =%

(y sec2 x + secx tan x) dx

$ F = y tan x + secx + g(y)

tan x + 2y ="f

"y= tanx + g!(y)

$ 2y = g!(y) $ g(y) = y2

The solution is y tan x + secx + y2 = C. With initial conditions, 1(0) + 1 + 12 = C $ C = 2,so

y tan x + secx + y2 = 2

4. We have (2y sinx cos x + y2 sin x) + (sin2 x " 2y cosx) dy = 0 with y(0) = 3. Let M =2y sinx cos x + y2 sin x and N = sin2 x " 2y cosx. Then

"M

"y= 2 sinx cos x + 2y sinx

"N

"x= 2 sinx cos x " 2y(" sinx)

so the equation is exact. We set "f"x = M and "f

"y = N and solve.

f =%"f

"ydy =

%(sin2 x " 2y cosx) dy = y sin2 x " y2 cosx + h(x)

Then

2y sinx cos x + y2 sin x ="

"x(f) = 2y sin x cos x + y2 sin x + h!(x)

$ h!(x) = 0 $ h(x) = C1

The solution is given byy2 sin2 x " y2 cosx = C

Initial conditions give us 32 sin2 0 " 32 cos 0 = 0, so C = "9 and

y2 sin2 x " y2 cosx = "9

5. We have 2xy + (x2 " y2) dy = 0. Setting M = 2xy and N = x2 " y2,

"M

"y= 2x =

"N

"x

Then"f

"x= 2xy $ f(x, y) = x2y + g(y)

"f

"y= x2 + g!(y) = x2 " y2 $ g!(y) = "y2

$ g(y) = "13y3

Page 35: MATH 224 Sol.

Section 1.4 29

Thus the solution satisfiesx2y " 1

3y3 = C

6. We have (2 " 9xy2)x + (4y2 " 6x3)y dy = 0. Using M = (2 " 9xy2)x and N = (4y2 " 6x3)ygives

"M

"y= 18x2y =

"N

"x

Then

"f

"x= 2x " 9x2y2 $ f(x, y) = x2 " 3x3y2 + g(y)

"f

"y= "6x3y + g!(y) = 4y3 " 6x3y

$ g(y) = y4

Thus the solution satisfiesx2 " 3x3y2 + y4 = C

With initial conditions y(1) = 1, 12 " 3(1)3(1)2 + 14 = C gives C = "1, so

x2 " 3x3y2 + y4 = "1

7. We have e"y " (2y + xe"4)dy = 0 with y(1) = 3. Using M = e"y and N = "2y " xe"y wehave

"M

"y= "e"y =

"N

"x

We need to solvef =

%M dx =

%e"y dx = xe"y + g(y)

Substituting into "f"y = N ,

"

"y(xe"y + g(y)) = "xe"y + g!(y) = "2y " xe"y $ g!(y) = "2y

$ g(y) = "y2

The solution is given by "xe"y + y2 = C. Now apply initial conditions: "1e"3 " (3)2 = C, soC = "9 " e"3. The solution is then

"xe"y " y2 = "e"3 " 9

8. We have (1 + y2 sin 2x) + ("2y cos 2x) dy = 0. Then

"

"y[1 + y2 sin 2x] = 2y sin 2x

"

"x["2y cos 2x] = 4y sin 2x

so the equation is not exact.

Page 36: MATH 224 Sol.

30 Section 1.4

9. We have 3x2(1 " ln y) +!

x3

y" 2y

"dy = 0. Then

"

"y[3x2 + 3x2 ln y] =

3x2

y

"

"x

&x3

y" 2y

'=

3x2

y

so the equation is exact.

"f

"x= 3x2 + 3x2 ln y $ f(x, y) = x3 + x3 ln y + #(y)

$ "f

"y=

x3

y+ #!(y) =

x3

y" 2y $ #(y) = "y2 + C

Thenf(x, y) = x3 + x3 ln y " y2 + C

10.,2 +

y

x2

-dx +

!y " 1

x

"dy = 0 & "M

"y=

1x2

,"N

"x=

1x2

$ the equation is exact. Thus"F"x =

2 + yx2 & F = 2x " y

x + g(y). "F"y = "1

x + g!(y) = y " 1x $ g!(y) = y, g(y) = y2/2 + C $

F (x, y) = 2x " yx + y2

2 + C

11. We have!

x

sin y+ 2"

dx +(x2 + 1) cos y

cos 2y " 1dy = 0. Then

"

"y

&x

sin y+ 2'

= "x cot[y] csc[y]

"

"x

&(x2 + 1) cos y

cos 2y " 1

'=

2x cos y

cos 2y " 1

We find

cos2 x =12

+12

cos(2y) $ 12

cos 2y = cos2 x " 12

$ cos 2y = 2 cos2 y " 1

So"N

"x=

2x cos y

2(cos2 x " 1)= "x cos y

sin2 y= "x cot y csc y

so it is exact. Then"f

"x=

x

sin y= 2

$ f(x, y) =12

x2

sin y+ 2x + #(y)

$ "f

"y= "1

2x2 cot y csc y + #!(y)

=x2 cos y + cos y

cos 2y " 1

=x2 cos y

2 sin2 y+

cos y

2 sin 2y

=12x2 cot y csc y " 1

2cot y csc y

Page 37: MATH 224 Sol.

Section 1.4 31

So#!(y) = "1

2cot y csc y $ #(y) =

csc(y)2

+ C

andf(x, y) =

12x2 csc y + 2x +

12

csc(y) + C

12. Solve axa"1y1"a dx + ((1 " a)xay"a) dy = 0.

(i) As a separable equation,

axa"1y1"a dx = (a " 1)xay"a dy

axa"1

xadx =

(a " 1)y"a

y1"ady

a

xdx =

(a " 1)y

dy

a ln |x| + C = (a " 1) ln(y)y(a"1) = Cxa

(ii) As an exact equation,

"

"y[axa"1y1"a] = a(1 " a)xa"1y"a

"

"y[(1 " a)xay"a] = a(1 " a)xa"1y"a

"f

"x= aa"1y1"a $ f(x, y) = xay(1"a) + #(y)

$ "f

"y= (1 " a)xay"a + #!(y) = (1 " a)xay"a

#!(y) = C

f(x, y) = xay(1"a) + C

13. Determine A ! R such that the equation is exact. Then solve the resulting equation.

(a) (x2 + 3xy) dx + (Ax2 + 4y) dy = 0. Then

"

"y[x2 + 3xy] = 3x

"

"x[Ax2 + 4y] = 2Ax

so 2A = 3 and A = 32 . Then

"f

"x= x2 + 3xy $ f(x, y) =

13x3 +

32x2y + #(y)

$ "f

"y=

32x2 + #!(y) =

32x2 + 4y

$ #(y) = 2y2 + C

Sof(x, y) =

13x3 +

32x2y + 2y2 + C

Page 38: MATH 224 Sol.

32 Section 1.4

(b)!

Ay

x3+

y

x2

"dx +

!1x2

" 1x

"dy = 0. We have

"

"y

&Ay

x3+

y

x2

'=

A

x3+

1x2

"

"x

&1x2

" 1x

'= " 2

x3+

1x2

so A = "2. Then

"f

"x=

y

x2" 2y

x3$ f(x, y) = " y

x+

y

x2+ #(y)

"f

"y= " 1

x+

1x2

+ #!(y) =1x2

" 1x$ #(y) = C

Thereforef(x, y) =

y

x2" y

x+ C

14. Assume"M

"y="N

"x. We want to find F (x, y) such that

"F

"x= M,

"F

"y= N . As in the proof

of Thm. 1.4.1, proceed replacing the argument of"F

"x= M(x, y) with one for

"F

"y= N(x, y) and continuing in the same manner.

15. Determine the most general function N(x, y) so that

(x3 + xy2) dx + N(x, y) dy = 0

is exact. For the equation to be exact,

"

"y[x3 + xy2] = 2xy =

"N

"x

soN(x, y) =

%2xy dx = x2y + #(y)

16. Determine the most general function M(x, y) such that

M(x, y) dx + (2yex + y2e3x) dy = 0

is exact. We have"M

"y="

"x[2yex + y2e3x] = 2yex + 3y2e3x

SoM(x, y) =

%2yex + 3y2e3x dy = y2ex + y3e3x + #(x)

17. Show that(Ax + By) dx + (Cx + Dy) dy = 0

is exact if B = C. To prove this, assume that B = C. Then

"

"y[Ax + By] = B

"

"x[Cx + Dy] = C

Page 39: MATH 224 Sol.

1.5. LINEAR EQUATIONS 33

Then"

"y[Ax + By] =

"

"x[Cx + Dy]

which by Theorem 2.8.1 implies that the equation is exact.Now assume that the equation is exact. Then by Theorem 2.8.1 we know that

B ="

"y[Ax + By] =

"

"x[Cx + Dy] = C

or that B = C.

18. Show that the homogeneous equation

(Ax2 + Bxy + Cy2) dx + (Dx2 + Exy + Fy2) dy = 0

is exact if and only if B = 2D and E = 2C. To prove this, assume that B = 2D and E = 2C.Then

"

"y[Ax2 + Bxy + Cy2] = Bx + 2Cy = 2Dx + Ey =

"

"x[Dx2 + Exy + Fy2]

Therefore, by Theorem 2.8.1, the equation is exact. Now assume that the equation is exact.Then by Theorem 2.8.1,

Bx + 2Cy ="

"y[Ax2 + Bxy + Cy2]

="

"x[Dx2 + Exy + Fy2] = 2Dx + Ey

which implies that B = 2D and 2C = E.

1.5 Linear Equations

1. y! + 1xy = x: Linear equation with a(x) = 1

x , b(x) = x. Let A(x) =@

1x dx $ ln |x|. Then

eA(x) = |x| and e"A(x) = 1|x| . The solution is given by

y(x) =1|x|

&%x · |x| dx

'+

C

|x| =

ABBBC

BBBD

1"x

!"x3

3

"+

C

"xif x < 0

1x

!x3

3

"+

C

xif x ) 0

Theny(x) =

x2

3+

C

xfor all x.

2. y! " 2x1+x2 y = x2. Linear with a(x) = "2x

1+x2 , b(x) = x2. Then

A(x) =% "2x

1 + x2dx = " ln |1 + x2| = " ln(1 + x2)

because 1 + x2 is always positive. Substituting u = 1 + x2, du = 2xdx leads to

eA(x) = e" ln(1+x2) =1

1 + x2; e"A(x) = eln(1+x2) = 1 + x2

Page 40: MATH 224 Sol.

34 Section 1.5

Then%

b(x)eA(x) dx =%

x2

!1

1 + x2

"dx =

% !1 " 1

1 + x2

"dx = x " arctanx

The solution is therefore

y = (1 + x2)(x " arctanx) + (1 + x2) + C

3. (2x + 1)y! = 4x + 2y leads to

y! " 2y

2x + 1=

4x

2x + 1Using P = "2

2x+1 and Q = 4x2x+1 gives% "2

2x + 1dx = " ln |2x + 1|

e@

p(x) = e" ln |2x+1| =1

|2x + 1|e"@

p(x) = eln |2x+1| = |2x + 1|%4x

2x + 1|2x + 1| dx = 2x2 · sgn(2x + 1)

where

sgn(2x + 1) =

E"1 if 2x + 1 < 01 if 2x + 1 ) 0

The solution is y = |2x + 1|(2x2 · sgn(2x + 1) + C) or

y = (2x + 1)(2x2 + C1)

4. We have y! + y tan x = secx with y(!) = 1. Then

N(x) = secx $ d

dx[y secx] = sec2 x

so y secx = tanx + C.

y = sinx + C cosx

$ 1 = sin! + C cos! $ 1 = "C $ C = "1y = sinx " cosx

5. We have dy =!

2x +xy

x2 " 1

"dx. Then

dy

dx" x

x2 " 1y = 2x $ Q(x) = " x

x2 " 1

SoN(x) = e"

@(x/(x2"1))dx

Let u = x2 " 1 = du/2x = dx. Then

N(x) = e"(1/2)@

(1/u) du = eln |(1/%

x2"1)| =1'

x2 " 1

Page 41: MATH 224 Sol.

Section 1.5 35

Sod

dx

&y'

x2 " 1

'=

2x'x2 " 1

$ y'x2 " 1

=%

2x'x2 " 1

Let u = x2 " 1. Then du/2x = dx and%

2x'x2 " 1

dx =%

1'u

du =%

u"1/2 du = 2.

x2 " 1 + C

so y'x2 " 1

= 2.

x2 " 1 + C

andy = 2(x2 " 1) + C

.x2 " 1

6. y! " y = 4ex and y(0) = 4. Let P (x) = "1, Q(x) = 4ex. Then%

"dx = "x $ e@

p(x) = e"x; e"@

p(x) dx = ex

%4ex · e"x dx =

%4 dx = 4x

The solution isy = ex(4x + C)

With initial conditions, 4 = e0(4(0) + C) $ C = 4, thus

y = 4ex(x + 1)

7. y! + xy = 2x, P (x) = x, Q(x) = 2x, µ(x) = ex2/2 $%

2xex2/2 dx

= 2ex2/2 + C.Thus the general solution is y = e"x2/2

,2ex2/2 + C

-

8. P (x) =1x

, Q(x) = ex, µ(x) = eln x = x $%

xex dx = xex " ex.

Thus the general solution is y = e" ln x (xex " ex + C) =1x

(xex " ex + C)

9. P (x) = 2, Q(x) = "3x, µ(x) = e2x $%

"3xe2x dx

="3xe2x

2+

34e2x + C. Thus the general solution is

y = e"2x

!"3xe2x

2+

34e2x + C

"

10. We havedy

dx+

1x

y =cosx

xwith y

,!2

-=

4!

and x > 0. This time Q(x) = 1x so N(x) = x,

which implies thatd

dx[xy] = cos(x) $ xy = sin(x) + C

So with y(!2 ) = 4! , we have that

!

2· 4!

= sin,!

2

-+ C $ 2 = 1 + C

so C = 1 andy(x) =

sin(x)x

+ x"1

Page 42: MATH 224 Sol.

36 Section 1.5

11.

P (x) = "3x2

Q(x) = x2

µ(x) = e"x3

%Q(x)µ(x) dx = "e"x3

3+ C

y = ex3

/"e"x3

3+ C

0

12.

P (x) = "2x

Q(x) = x3

µ(x) = e"x2

%Q(x)µ(x) dx = " (x2 + 1)e"x2

2+ C

y =

13.

P (x) = 1Q(x) = cosx

µ(x) = ex

%Q(x)µ(x) dx =

ex

2(cos x + sinx) + C

y = e"x

!ex

2(cosx + sin x) + C

"

14.

P (x) = 1Q(x) = ex

µ(x) = ex

%Q(x)µ(x) dx =

e2x

2+ C

y = e"x

!e2x

2+ C

"

Page 43: MATH 224 Sol.

Section 1.5 37

15. y! + 2xy = 1 with y(2) = 1. Then we have

A(x) =%

2xdx = x2

eA(x) = ex2, e"A(x) = e"4x2

%b(x)eA(x) =

%ex2

dx% x

2y(t)eA(t) =

% x

2b(t)eAt

16. We have (x + y2) dy = y dx gives

x + y2 = ydx

dyor

dx

dy" 1

yx = y

Thus a(y) = " 1y , b(y) = y, A(y) =

@" 1

y dy = " ln y. Then

eA(y) = e" ln y =1y, e"A(y) = y

%b(y)eA(y) dy =

%y0

!1y

"dy =

%1 dy = y

Then x = y[y] + Cy andx = y2 + Cy

17. We have (2ey "x)y! = 1 or (2ey "x) dydx = 1. So 2ey"x = dx

dy and dxdy +x = 2ey. With Q(y) = 1

we have N(y) = ey.d

dy[xey] = 2e2y and xey = e2y + C

so the solution isx = ey + Ce"y

18. We have (sin 2y + x cot y)y! = 1. Rewriting, we have

(sin 2y + x cot y) =dx

dydx

dy" x cot y = sin 2y

With Q(x) = " cot y we have N(y) = e" ln(sin x) " csc y, so

d

dy[x csc y] =

sin 2y

sin y

$ d

dy[x csc y] =

2 sin y cos y

sin y

$ x csc y =%

2 cos y dy

$ x csc y = 2 sin y + C

$ x(y) = 2 sin2 y + C sin y

19. (a) y = 0, y! = 0,& y! + P (x)y = 0 + 0 = 0

Page 44: MATH 224 Sol.

38 Section 1.5

(b) y = y1 is a solution. y = ky1, y! = ky!1.

Then ky!1 + P (x)ky1 = k(y!

1 + P (x)y1) = k(0) = 0(c) y = y1 + y2, y! = y!

1 + y!2

Then y!1 + y!

2 + P (x)(y1 + y2) = (y!1 + P (x)y1) + (y!

2 + P (x)y2)= 0 + 0(Since y1 and y2 are solutions) = 0

20. (a) y = y1 + y2, y! = y!1 + y!

2

Then y!1 + y!

2 + P (x)(y1 + y2) = (y!1 + P (x)y1) + (y!

2 + P (x)y2)= 0 + r(x) (Since y1 and y2 are solutions) = r(x)

(b) y = y1 + y2, y! = y!1 + y!

2

Then y!1 + y!

2 + P (x)(y1 + y2) = (y!1 + P (x)y1) + (y!

2 + P (x)y2)= r(x) + q(x) (Since y1 and y2 are solutions)

(c) Solution to y! + 2y = e"x is y = e"2x(ex + C) and y! + 2y = cosx is

y = e"2x

!e2x

5(2 cosx + sin x) + C

".

Thus y = e"2x

!ex +

e2x

5(2 cosx + sin x) + C

"is a solution to the original equation.

21. (a)

y! = "P (x)ydy

y= "P (x) dx

ln |y| =%

"P (x) dx

yc = ±e@"P (x) dx

(b) y = A(x)yc, y! = A!(x)yc + A(x)y!c.

Then Q(x) = y! + P (x)y = A!(x)yc + A(x)y!c + P (x)A(x)yc =

A!(x)yc + A(x)(y!c + P (x)yc) = A!(x)yc

(c) A!(x)e@"P (x)dx = Q(x) & A!(x) = Q(x)e

@P (x) dx &

A(x) =%

Q(x)e@

P (x) dx dx

(d) yp = ycA(x), y = yc + yp = yc + ycA(x) = yc(1 + A(x)) =

e@"P (x) dx

!%Q(x)e

@P (x) dx dx

"

22. Let y be the amount of pollutant at time t. Then

y! = 12000(2)" 10000!

y

500000 + 2000t

"

= 24000" 10y

500 + 2t

Then y! +10y

500 + 2t= 24000 is linear with solution

y(t) =1

(t + 250)5#4000(t + 250)6 + C

$. C = "4000(250)6.

y(t) = 4000(t + 250) " 4000(250)6

(t + 250)5, y(10) = 218072 &

Concentration = 218072520000 = .419g/gal

Page 45: MATH 224 Sol.

Section 1.5 39

23. (a)dN

dt= rN + rA & dN

dt" rN = rA & N(t) = "a + Cert. N(0) = 0 &

A = C & N(t) = Aert " A

(b) r = ln 2. So, with A = 3, N(t) = 3ert " 3 = 3 · 2t " 3.N(24) = 50, 331, 645N(36) = 206, 158, 430, 205

27.dx

dt+ a(t)x = f(t), P (t) = a(t), µ(t) = e

@a(t) dt

$ x(t) = e@"a(t) dt

!%e@

a(t) dtf(t) dt + C

".

limt#$

x(t) = limt#$

!e@"a(t) dt

!%e@

a(t) dtf(t) dt + C

""= 0 since a(t) > 0, e

@"a(t) dt & 0

29. (a) See section 1.5.1. µ(x) = exp

!%1

x2y " x(1 " (2xy " 1)) dx

"=

1x2

which gives an exact

DE and F (x, y) = 2x " y

x+

y2

2+ C

(b)1N

!"M

"y" "N"x

"=

"1x

(1 " ("1)) ="2x

. So µ(x) =1x2

which gives an exact De and

F (x, y) =x3

3" ln |x|" 1

xy+ C

31. M and N change roles, but the argument is exactly the same as (1.31)

32."M

"y= 2y,

"N

"x= 0 & 1

N

!"M

"y" "N"x

"=

1y(2y) = 2 which a function only in x so µ(x) = e2x

33."M

"y=.

1 + x2,"N

"x=

x2

'1 + x2

+.

1 + x2 & 1N

!"M

"y" "N"x

"=

"x

1 + x2which a function

only in x so µ(x) =1'

1 + x2

34."M

"y= x + 2y,

"N

"x= y & 1

M

!"N

"x" "M"y

"=

"1y

which a function only in y so µ(y) =1y

35."M

"y= 3y2 + 1,

"N

"x= y2 " 2x + 1 Which does not yield anything of a function in x or y

36."M

"y= 1,

"N

"x= "1 & 1

N

!"M

"y" "N"x

"=

"2x

which a function only in x so µ(x) =1x2

37."M

"y= 1,

"N

"x= ex & 1

N

!"M

"y" "N"x

"= 1 which a function only in x so µ(x) = ex

38."M

"y= "2y + 1,

"N

"x= 2y " 1 & 1

N

!"M

"y" "N"x

"=

"2x

which a function only in x so µ(x) =1x2

40."M

"y= "2x cos y,

"N

"x= 2x cos y & 1

N

!"M

"y" "N"x

"=

"4x

(x2 + 1)2which a function only in x

so µ(x) =1

(x2 + 1)2

41. n = 2, v = y"1 $ v! " v = "x which is linear and gives solution

v = x + 1 + Cex $ y =1

x + 1 + Cex

Page 46: MATH 224 Sol.

40 Section 1.5

43. y! = y4 cosx + y tan x, sody

dx+ (y tan(x))y = y4 cosx

Let v = y"3. Thendv

dx= "3y"4 dy

dx= 7

dv

dx

&"1

3y4

'=

dy

dx

So

dv

dx

!"1

3y4

"+ (" tan(x))y = y4 cosx

dv

dx+ (3 tan(x))v = "3 cosx

N = e3@

tan x dx = eln(1/(cos3 x)) = sec3 xd

dx[v sec3 x] = "3 cos2 x

d

dx[v sec3 x] = "3

2+

32

cos 2x

v sec3 x =34

sin 2x " 32x + C

y"3 sec3 x =34

sin 2x " 32x + C

44. n = 4, v = y"3 $ v! " 9v = "3 which is linear and gives

v = Ce9x +13& y3 =

33Ce9x + 1

& y = 3

93

3Ce9x + 1

45. We have xy2y! = x2 + y3. Then

y! " y3

xy2=

x2

xy2$ y! " 1

xy3 = xy"2

This is a Bernoulli equation with r = "2, so 1 " r = 1 " ("2) = 3. Thus we need to solveu! + 3

#" 1

x

$u = 3(x). Take a(x) = " 3

x , b(x) = 3x, and A(x) =@" 3

x dx = "3 ln |x|. Then

eA(x) = e"3 ln |x| =1

|x|3 , e"A(x) = e3 ln |x| = |x|3

%b(x)eA(x) dx =

%3x

!1

|x|3

"dx

$x ) 0

%3x

!1x3

"dx =

%3!

1x2

"dx = " 3

x

x < 0%

3x

!" 1

x3

"dx =

%" 3

x2dx =

3x

FBBBG

BBBH= " 3

|x|

u = |x|3!" 3|x|

"+ |x|3 · C = "3x2 + C · |x|3

$ y3 = "3x2 + C · |x|3

$ y = ("3x2 + C|x|3)1/3

46. y = 0 is a trivial solution, so if y %= 0y!

y= 4 " 2x & dy

y= (4 " 2x) dx

ln |y| = 4x " x2 + C $ y = ±e4x"x2+C

Page 47: MATH 224 Sol.

Section 1.5 41

47. We have (x + 1)(y! + y2) = "y. Then

dy

dx+ y2 =

"y

x + 1$ dy

dx+

y

x + 1= y2

Let v = y"1, sodv

dx= "y"2 dy

dx$ dv

dx("y2) =

dy

dx

and thatdv

dx("y2) +

!1

x + 1

"y = y2

which implies that

dv

dx"!

1x + 1

"y"1 = "1 $ dv

dx"!

1x + 1

"v = "1

Q(x) = " 1x+1 , N(x) = e"

@(1/(x+1)) dx = 1

1+x , so

d

dx

&v · 1

1 + x

'=

11 + x

$ v

1 + x= ln

88881

1 + x

8888

v = (ln |1 + x|)(1 + x) $ 1y

= (ln |1 + x|)(1 + x)

$ y =1

(x + 1) ln |x + 1|

48. We have xy! " 2x2'y = 4y. Then

dy

dx" 2x

'y =

4x

y

dy

dx+!" 4

x

"y = 2xy1/2 (Bernoulli equation)

Let v = y1"(1/2) = y1/2, so dvdx = 1

2y"1/2 dydx . Then

dv

dx[2y1/2] +

!" 4

x

"y = 2xy1/2

dv

dx+!" 2

x

"v = x

!N(x) =

1x2

"

d

dx

&'y

x2

'=

1x

$'

y

x2= ln |x| + C

$ 'y = x2 ln |x| + Cx2

Page 48: MATH 224 Sol.

42 Section 1.5

49. We have xy! + 2y + x5y3ex = 0. Then

xdy

dx+ 2y + x5y3ex = 0

dy

dx+

2x

y = ("x5ex)y3 (Bernoulli equation)

v = y"2 $ dv

dx= "2y"3 dy

dx$ dv

dx

!"1

2y3

"=

dy

dx

dv

dx

!"1

2y3

"+

2x

y = ("x5e"x)y3

dv

dx" 4

xy = (2x5e"x)

Now use N(x) = 1x4 , so d

dx

;v · 1

x4

<= [2xe"x]. Then

v · 1x4

= 2e"x(x " 1) + C

$ v = 2x4e"x(x " 1) + Cx4

$ y =1

x2.

2e"x(x " 1) + C

50. We have xy dy = (y2 + x) dx. Then

dy

dx(xy) = y2 + x $ dy

dx= x"1y + y"1

dy

dx+!" 1

x

"y = y"1 (Bernoulli equation)

Let v = y1+1 = y2, so dvdx

#12y"1

$= dy

dx . Then

dv

dx

!12y"1

"+!" 1

x

"y = y"1 $ dv

dx+!" 2

x

"v = 2

Using N(x) = 1x2 we have d

dx

;vx2

<= 2

x2 , so

v

x2= " 2

x+ C

$ y2 = "2x + Cx2

51. Use the Bernoulli method to solve the logistic equation

dN

dt= rN

!1 " N

K

"

We havedN

dt= rN " r

KN2 $ dN

dt+ ("r)N = " r

KN2

Let v = N"1, so dvdt = "N"2 dN

dt and dvdt ("N2) = dN

dt . Then

dv

dt("N2) + ("r)N = " r

KN2

dv

dt+ rv =

r

K

Page 49: MATH 224 Sol.

1.6. CHAPTER 1: ADDITIONAL PROBLEMS 43

Using N = ert gives ddt [vert] = r

K ert, so

vert =1K

ert + C $ v =1k

+ Ce"rt

1N

=1K

+ ce"rt $ N(t) =1

1K + Ce"rt

1.6 Chapter 1: Additional Problems

1. False - It is linear, but no initial condition is given

2. True - y2 makes it non-linear

3. False - y(x) = f(x) is an explicit solution

4. The point is to get linear DE into an exact DE by multiplying by an integrating factor. Soeven if this were true, there would be no reason to do so.

5. False - Solutions can be defined on a restricted portion of the real line

6. True

7. True -dy

f(y)= dx

8. y! = C4x3ex4& 4x3(y " 1) = 4x3(1 + Cex4

" 1) = 4x3(Cex4) = y!

9. y! = Cex " 2x " 2 & x2 + y = x2 + Cex " x2 " 2x " 2= Cex " 2x " 2 = y!

10. y2 =1

C " 2xy3 = ±(C " 2x)"3/2, y = ±

91

C " 2x,

y! = ±(C " 2x)"3/2 = y3. Valid when C %= 2x

11. y! =1

(x + C)2, (y " 3)2 =

!"1

(x + C)

"2

=1

(x + C)2= y!

12. We have 2x2yy! + y2 = 2. Then

2x2ydy

dx= 2 " y2

2y dy

2 " y2=

dx

x2

$%

2y dy

y2 " 2=%

"dx

x2

ln |y2 " 2| =1x

+ C1

|y2 " 2| = eC1e1/x

y2 " 2 = Ce1/x, where C = ±C1

y = ±.

Ce1/x + 2

Page 50: MATH 224 Sol.

44 Chapter 1 Review

13. We have y! = 3 3.

y2 with y(2) = 0. Rewrite the equation as y! = 3y2/3. Separate variablesand integrate:

%y"2/3 dy =

%3 dx

3y1/3 = 3x + C

It is easiest to apply the initial conditions now: 3(0)1/3 = 3(2) + C leads to C = "6, so

3y1/3 = 3x " 6y1/3 = x " 2

y = (x " 2)3

14. We have y! " xy2 = 2xy. Then

dy

dx= x(y2 + 2y)

dy

y2 + 2y= xdx (by partial fractions)

% !"1/2y

+1/2

y " 2

"dy =

%xdx

"12

ln |y| + 12

ln |y " 2| =12x2 + C1

ln8888y " 2

y

8888 = x2 + 2C1

88881 " 2y

8888 = e2C1ex2

2y

= 1 ± e2C1ex2

y =2

1 ± e2C1x2

y =2

1 + Cx2where C = ±e2C1

15. We have e"x#1 + dx

dt

$= 1. Then

dx

dte"x = 1 " e"x

e"x

1 " e"xdx = dt

Letting v = 1 " e"x, du = e"x,%

du

u=%

dt

ln |u| = t + C1

|u| = eC1et

1 " e"x = ±eC1et

e"x = ±eC1et " 1"x = ln |±eC1et " 1|

x = " ln |Cet " 1| where C = ±eC1

Page 51: MATH 224 Sol.

Chapter 1 Review 45

16. We have xy! + y = y2 with y(1) = .5. Then

xy! = y2 " y $%

dy

y2 " y=%

dx

x

$% !

1y " 1

" 1y

"dy =

%dx

x

$ ln |y " 1|" ln |y| = ln |x| + C $ y " 1y

= Ax

Apply initial conditions:

12 " 1

12

= A(1) $ A = "1 $ 1 " 1y

= "x

Then we have1 + x =

1y$ 1

1 + x= y

17. We have xy! = y " xey/x. Rewriting we have y! = yx " ey/x. Let u = y

x , dydx = xdu

dx + u. Then

xdu

dx+ u = u " e"u

xdu

dx= "e"u $

%eu du =

%" 1

xdx

eu = " lnx + C $ eye1/x = " lnx + C

ey =" lnx + C

e1/x

y = ln!" lnx + C

e1/x

"

18. y! " y

x= (1 +

y

x) ln(1 +

y

x) which is Homogeneous. Let v =

y

x&

v + v!x " v = (1 + v) ln(1 + v)dv

(1 + v) ln(1 + v)=

dx

x

[u = ln(1 + v)]du

u=

dx

x|u| = |x|eC

| ln(1 +y

x)| = |x|eC

ln(1 +y

x) = ±xeC

Page 52: MATH 224 Sol.

46 Chapter 1 Review

19. We have xy! " y cos#ln# y

x

$$= 0. Then

yx! = y cos,ln, y

x

--

y! =y

xcos,ln,y

x

--(homogeneous)

dv

dxx + v = v cos(ln(v))

dv

dxx = v(cos(ln(v))) " v

dv

dxx = v(cos(ln(v)) " 1)

1v(cos(ln(v)) " 1)

dv =1x

dx

cot!

12

ln(v)"

= lnx + C

cot!

12

ln,y

x

-"= lnx + C

20. We have (y + 'xy) dx = xdy. We use M(x, y) = y + '

xy and N(x, y) = "x, so

M(tx, ty) = ty +.

t2xy = t(y +'

xy) + tM(x, y)N(tx, ty) = "tx = t("x) = tN(x, y)

We need to solve

dy

dx=

y

x+

'xy

x$ dv

dxx + v = v +

'v

$ dv

dxx =

'v

$ 1'v

dv =1x

dx

$ 2'

v = ln |x| + C

Then'

y = 2'

x ln x + 2C'

x

y = 4x(ln y + C)2

21. We have x dydx =

.x2 " y2 + y. Then

dy

dx=

91 ",y

x

-2+,y

x

-(homogeneous)

$ dv

dxx + v =

.1 " v2 + v

1'1 " v2

dv =1x

dx

$ arcsin(v) = lnx + C

$ arcsin, y

x

-= ln x + C

Page 53: MATH 224 Sol.

Chapter 1 Review 47

22.

dy

dx=

4y " 2x

x + y

=4(y/x) " 21 + (y/x)

Homogeneous (y = vx)

v + v!x =4v " 21 + v

1 + v

"v2 + 3v " 2dv =

dx

x

ln8888(1 " v)2

(2 " v)3

8888 = ln |x| + C

&(1 " (y/x))2

(2 " (y/x))3

'= ±eCx

23. We have (y/x) + (y3 + ln x) dy = 0. We find

"

"y

,y

x

-=

1x

"

"x[y3 + ln x] =

1x

so the equation is exact. Then,

"f

"x=

4x

$ f(x, y) = y ln x + #(y)

$ "f

"y= ln x + #!(y) = y3 + ln x

#(y) =14y4 + C

Sof(x, y) = y ln x +

14y4 + C

24. We have!

3x2 + y2

y2

"+!"2x3 " 5y

y3

"dy

dx= 0. Then

"

"y

&3x2 + y2

y2

'=

"6x2

y3

"

"x

&"2x3 " 5y

y3

'=

"6x2

y3

Page 54: MATH 224 Sol.

48 Chapter 1 Review

so the equation is exact. Solving, we get

"f

"x=

3x2

y2+ 1

$ f(x, y) =x3

y2+ x + #(y)

$ "f

"y=

"2x3

y3+ #!(y) =

"2x3

y3" 5

y2

$ #!(y) = " 5y2

#(y) =5y

+ C

f(x, y) =x3

y2+ x +

5y

+ C

25. We have 2x(1 +.

x2 " y) dx ".

x2 " y dy = 0. Then

"

"y

I2x + 2x

.x2 " y

J=

"x.x2 " y

"

"x

I".

x2 " yJ

= " 2x

2.

x2 " y=

"x.x2 " y

so the equation is exact. Solving,

"f

"x= 2x + 2x

.x2 " y

$ f(x, y) = x2 +%

2x.

x2 " y dx

$ f(x, y) = x2 +23(x2 " y)3/2 + #(y)

$ "f

"y= "

.x2 " y + #!(y)

= ".

x2 " y

#!(y) = 0 leads to #(y) = C, so

f(x, y) = x2 +23(x2 " y)3/2 + C

26. y! +1x

y =1x2

y"2 which is Bernoulli n = "2, v = y3,dv

dx= 3y2 dy

dx. Then we have

dv

dx+

3x

v =3x2

which is linear with solution v = y3 =32x

+C

x3

$ y = 3

932x

+C

x3

27."M

"y= 2y,

"N

"x= "y " 3x2 & 1

N

!"M

"y" "N"x

"=

3y + 3x2

"x(y + x2)which is only a function of x

which gives µ(x) = x"3 and F (x, y) ="y2

2x2" y + C

Page 55: MATH 224 Sol.

Chapter 1 Review 49

28.

dy

dx=

y

x" y2

(Homog) (y = vx)v + v!x = v " v2x2

dv

"v2= xdx

1v

=x2

2+ C

x

y=

x2

2+ C

29. Not linear, exact, homogeneous, Bernoulli, seperable

30. Not linear, exact, homogeneous, Bernoulli, seperable

31. y! +y

x= x3 Linear $ y =

x4

5+

C

x

32. Linear $ y = ex

&"e"x

2(cosx + sin x) + C

'

33. Linear $ y = e"2x2&58e2x2

(2x2 " 1) + C

'

34. Bernoulli n = 4, v = y"3, v! " 3x

v = "3 (linear) v = y"3 =3x

2+ Cx3

35. Bernoulli n = 3, v = y"2, v! + 2v = "2x (linear)

v = y"2 = e"2x

&"e2x

2(2x " 1) + C

'

36. Bernoulli n = 3, v = y"2, v! " xv = "4x (linear)v = y"2 = ex2/2

,4e"x2/2 + C

-

37. T (t) = (T0 " Ts)ekt + Ts = 95ekt + 5, k = ".055 $ soup will be edible in about 34 minutes.

38. T (t) = 50ekt + 70, k = ".105 $ co"ee is drinkable in about 5 minutes.

39. Let y(t) = amount of salt at time t, y(0) = 6

dy

dt= 2 " 6 + 2t

15 + 2t

=24 " 2t

15 + 2t

y =392

ln |2t + 15|" t + C

y(0) = 6 & C = "46.8The tank will fill in 15 minutes

y(15) = 12.43lbs of salt

Page 56: MATH 224 Sol.

50 Chapter 1 Review

40. y(t) lbs of sugar at time t, y(0) = 25

dy

dt= p " 25 + pt

100 " t

=100p" 25 " 2pt

100 " t

dy =100p" 25 " 2pt

100 " tdt

y = (25 " 100p) ln |100 " t|"200p ln |t " 100|" 2pt + C

C = 25 " 25 ln 100We want y(95) = 2.5 & p = .074lbs/min

41. y(t) amount of nitric acid, y(0) = 1

dy

dx= 1.2 " 8

!100 + 1.2t

200 " 2t

"

="560" 12t

200 " 2ty(t) = 880 ln |t " 100|+ 6t + 1 " 880 ln100

The solution reaches 10% when y(t) = 20" .2t which is when t = 138 minutes. Since the tankempties in 100 minutes, this will never happen.

42. If (x, y) is the point of tangency, then (x2 , 0) is the point of intersection and

dy

dx=

yx2

=2y

x.

Thendy

y=

2 dx

x$ y = x2 is such a curve.

Page 57: MATH 224 Sol.

Chapter 2

Geometrical and NumericalMethods for First-Order Equations

2.1 Direction Fields—the Geometry of Di!erential Equations

1. Looking at the point (2, 1), y! = 92 , which matches graph b.

2. Looking at the point (2, 1), y! = 18, which matches graph c.

3. Looking at the point (2, 1), y! = 29 , which matches graph a.

4. Looking at the point (2, 1), y! = 118 , which matches graph d.

5. Graph B

6. Graph C

7. Graph D

8. Graph A

9. y! = y4 #9 #10

10. y! = cos y

51

Page 58: MATH 224 Sol.

52 Section 2.1

11. y! = sin y

12. y! = e"y

13. y! = x4

Page 59: MATH 224 Sol.

Section 2.1 53

14. y! = cos x

15. y! = sin x

16. y! = e"x

Page 60: MATH 224 Sol.

54 Section 2.1

17. y! = xy

18. y! = x + y

19. y! = (x2 + 1)(y + 1)

Page 61: MATH 224 Sol.

Section 2.1 55

20. y! = ex2

21. y! =x " 1

y " 1

22. y! =x2 " 1

y2 + 1

Page 62: MATH 224 Sol.

56 Section 2.2

23. y! = y(y2 " 2)

24. xy(x2 + 2)

25. y! =x3(y2 + 1)

y2 + x2

2.2 Existence and Uniqueness for First-Order Equations

1. (a) y! = y2 " x2, y(0) = 0, f = y2 " x2,"f

"y= 2y. The theorem guarantees that a solution

exists and is unique on some interval.

Page 63: MATH 224 Sol.

Section 2.2 57

(b) y! = y"2 "x2, y(0) = 0, f =1y2

"x2,"f

"y= "y"3. The theorem does not guarantee that

a solution exists or is unique on some interval.

(c) y! = y +1

1 " x, y(1) = 0, f = y +

11 " x

,"f

"y= 1. The theorem does not guarantee that

a solution exists or is unique on some interval.

2. (a) f(x, y) = 3x(y + 2)2/3,"f

"y=

2x

(y + 2)1/3, which is discontinuous when y = "2, so a

solution exists and is unique everywhere except possibly along y = "2.

(b) f(x, y) = (x " y)1/5,"f

"y=

"15(x " y)4/5

, which is discontinuous when y = x, so solutions

exist and are unique everywhere except possibly along y = x. Alternatively, f(x, y) =(x " y)!

5,"f

"y= "1

5, so solutions exist and are unique everywhere.

(c) f(x, y) = x2y"1,"f

"y= "x2y"2, which are both discontinuous when y = 0, so solutions

exist and are unique everywhere except possibly along y = 0.

(d) f(x, y) = (x + y)"2,"f

"y=

"2(x + y)3

, which are both discontinuous when y = "x, so

solutions exist and are unique everywhere except possibly along y = "x.

3. (a)"f

"y=

43xy("1/3) which is not continuous at y = 0. Hence a solution exists and is unique

everywhere except possibly along y = 0.

Actual solution: y =x6

27. This solution crosses y = 0 when x = 0.

(b)"f

"y= 2x(2/3) which, along with f(x, y), are continuous everywhere. Hence solution exists

and is unique.Actual solution: y = 0.

(c)"f

"y= "2

3xy("5/3) which is not continuous at y = 0. Hence a solution exists and is unique

everywhere except possibly along y = 0.

Actual solution: y =!

56

"(3/5)

x(6/5) This solution crosses y = 0 when x = 0.

(d)"f

"y=

23xy("1/3) which is not continuous at y = 0. Hence a solution exists and is unique

everywhere except possibly along y = 0.

Actual solution: y =x6

216which crosses y = 0 when x = 0.

4. f(x, y) = 2'y is continuous for y ) 0 and"f

"y=

1'

yis continuous for y > 0. Hence, by the

theorem, a solution exists and is unique for y(1) = 3.Actual solution is given by: y = (x +

'3 " 1)2.

5. f(x, y) = 5(y " 2)3/5 and y(0) = 2."y

"y=

3(y " 2)2/5

is discontinuous when y = 2, so solutions

exist and are unique everywhere except possibly along y = 2. Solving for the initial value,

y! = 5(y " 2)3/5 $%

(y " 2)"3/5 dy =%

5 dx

$ 52(y " 2)2/5 = 5x + C

Page 64: MATH 224 Sol.

58 Section 2.2

Applying initial conditions,

52(2 " 2)2/5 = 5(0) + C $ C = 0

$ (y " 2)2/5 = 2x $ y " 2 = ±(2x)5/2

$ y = 2 ± (2x)5/2

Solutions passing through (0, 2) are NOT unique.

6."f

"y= x(1/3)("2y) which is continuous everywhere (as is f), therefore, unique solution will

exist everywhere.

7."f

"y= (y " 1)("2/3) which is continuous everywhere except y = 1, so solutions exist and are

unique everywhere except possibly along y = 1.Actual solution: 1 ± 2

'2x(3/2)

8."f

"y= "2xy"3 which is continuous everywhere except y = 0, so solutions exist and are unique

everywhere except possibly along y = 0.

Actual solution:!"32

+x2

2

"1/3

9. f(x, y) and"f

"yare not continuous at y = 1, so solutions exist and are unique everywhere

except possibly along y = 1.Actual solution: y = 1 + 3

'3 " 3 cosx which IS unique.

10."f

"yDNE at y = 0; if y > 0, then "f

"y = 1 and if y < 0, then "f"y = "1. Hence,

"f

"y= lim

h#0

f(x, y + h) " f(x, y)h

DNE. Therefore, the solutions will exist and be unique except

possibly at y = 0.If y ) 0, y = eCex which will never be 0. If y < 0, y = "eCe"x which will never be 0. Thus,the only solution to this DE is y = 0 which IS unique.

Page 65: MATH 224 Sol.

2.3. FIRST-ORDER AUTONOMOUS EQUATIONS—GEOMETRICAL INSIGHT 59

2.3 First-Order Autonomous Equations—Geometrical Insight

1. y! = 2y + 3, 2y + 3 = 0

Root y = " 32

Multiplicity 1

End behavior: +y

(i) y " y!

(ii) By the phase line diagram, y = " 32 is an unstable equilibrium point.

(iii) y > " 32 , y & # as x & #

y < " 32 , y & "# as x & #

(iv) xy-plane

Page 66: MATH 224 Sol.

60 Section 2.3

2. y! = y2 + 4y + 4, y2 + 4y + 4 = 0, (y + 2)2 = 0

Root y = "2Multiplicity 2

End behavior: y2

(i) y " y!

(ii) By the phase line diagram, y = "2 is a half-stable equilibrium point.(iii) y > "2, y & # as x & #

y < "2, y & "2 as x & #(iv) xy-plane

Page 67: MATH 224 Sol.

Section 2.3 61

3. x! = x2 " x " 6 = (x " 3)(x + 2)

(i) y " y!

(ii) x = "2 is a stable equilibriumx = 3 is an unstable equilibrium

(iii) If x0 > 3, then x(t) & # as t & #.If "2 < x0 < 3, then x(t) & "3 as t & #.If x0 < "2, then x(t) & "3 as t & #.

(iv) xy-plane

Page 68: MATH 224 Sol.

62 Section 2.3

4. x! = x(x + 2)(x " 3)

Equilibrium Multiplicityx = 0 1

x = "2 1x = 3 1

Highest power and coe!cient: x(+x)(+x) = +x3

(i) y " y!

(ii) x = "2, 3 unstable; x = 0 stable(iii) For x0 ! ("#,"2), x & "# as t & #

For x0 ! ("2, 3), x & 0 as t & #For x0 ! (3,#), x & # as t & #

(iv) xy-plane

Page 69: MATH 224 Sol.

Section 2.3 63

5. x! = (x " 2)3(x2 " 9) = (x " 2)3(x + 3)(x " 3)

Equilibrium 2 "3 3Multiplicity 3 1 1

Highest power coe!cient: (x)3(x2) = +x5

(i) y " y!

(ii) x = "3, 3 unstable, x = 2 stable(iii) For x0 ! ("#,"3), x & "# as t & #

For x0 ! ("3, 3), x & 2 as t & #For x0 ! (3,#), x & # as t & #

(iv) xy-plane

Page 70: MATH 224 Sol.

64 Section 2.3

6. y! = sin y, "2! < y < 2!sin y = 0, "2! < y < 2!

Roots y = "! y = 0 y = !Multiplicity 1 1 1

End behavior: trigonometric sine wave

(i) y " y!

(ii) By the phase line diagram, y = "! is a stable equilibrium point.y = 0 is an unstable equilibrium point.y = ! is a stable equilibrium point.

(iii) For y > !, y & ! as x & #For 0 < y < !, y & ! as x & #For "! < y < 0, y & "! as x & #For y < "!, y & "! as x & #

(iv) xy-plane

Page 71: MATH 224 Sol.

Section 2.3 65

7. y! = e"y2/2 " e"2, e"y2/2 " e"2 = 0. When

"y2

2= "2

"y2 = "4y2 = 4

(i) y " y!

(ii) By the phase line diagram, y = 2 is stabley = "2 is unstable

(iii) For y > 2, y &For "2 < y < 2, y &For y < "2, y &

(iv) xy-plane

Page 72: MATH 224 Sol.

66 Section 2.3

8. y! = "y2, "y2 = 0

Root y = 0Multiplicity 2

End behavior: "y2

(i) y " y!

(ii) By the phase line diagram, y = 0 is a half-stable equilibrium point.(iii) For y > 0, y & 0 as x & #.

For y < 0, y & "# as x & #.(iv) xy-plane

Page 73: MATH 224 Sol.

Section 2.3 67

9. x! = x2(2 " x)

Equilibrium Multiplicityx = 0 2x = 2 1

Highest power and coe!cient: x2("x) = "x3

(i) y " y!

(ii) x = 0 is a half-stable point; x = 2 is a stable point.(iii) For x ! ("#, 0), x & 0 as t & #.

For x ! (0,#), x & 2 as t & #(iv) xy-plane

Page 74: MATH 224 Sol.

68 Section 2.3

10. x! = (x " 1)2(x " 2)3(1 + x)

Equilibrium 1 2 "1Multiplicity 2 3 1

Highest power and coe!cient: (x)2(x)3(+x) = +x6

(i) y " y!

(ii) x = "1 is stable; x = 1 is half-stable; x = 2 is unstable.(iii) If x0 > 2, then x & # as t & #.

If 1 < x0 < 2, then x & 1 as t & #.If "1 < x0 < 1, then x & "1 as t & #.If x0 < "1, then x & "1 as t & #.

(iv) xy-plane

Page 75: MATH 224 Sol.

Section 2.3 69

11. y! = cos y + 1, "2! < y < 2!. cos y + 1 = 0, cos y = "1.

Roots y = "! y = !Multiplicity 1 1

End behavior: cosine wave

(i) y " y!

(ii) By the phase line diagram, y = "! is a half-stable equilibrium point. y = ! is a half-stableequilibrium point.

(iii) For y > !, y & # as x & #.For "! < y < !, y & ! as x & #.For y < "!, y & "! as x & #.

(iv) xy-plane

Page 76: MATH 224 Sol.

70 Section 2.3

12. x! = x3(2 + x)(5 + x)7(3 " x)(x " 2)4

Equilibrium 0 "2 "5 3 2Multiplicity 3 1 7 1 4

Highest power and coe!cient: x3(+x)(+x)7("x)(x)4 = "x16

(i) y " y!

!!

!!

!!

!

(ii) x = "5, 0 are unstable points; x = "2, 3 are stable points; x = 2 is a half-stable point.(iii) If x0 < "5, then x & "# as t & #.

If "5 < x0 < "2, then x & "2 as t & #.If "2 < x0 < 0, then x & "2 as t & #.If 0 < x0 < 2, then x & 2 as t & #.If 2 < x0 < 3, then x & 3 as t & #.If x0 > 3, then x & 3 as t & #.

(iv) xy-plane

Page 77: MATH 224 Sol.

Section 2.3 71

13. v! = g " kmv. Equilibrium satisfies g " (k/m)v = 0, so v =

gm

k

(i) v " v!

!

(ii) v =gm

kis a stable equilibrium point.

(iii) For v ! [0,#), v & gmk as t & #

(iv) Figures will vary. With g = 32, m = .25, k = 2 as in example 2 in 1.3 then

14. v! = g" kmv2. Equilibrium satisfies g" (k/m)v2 = 0, so v2 = gm/k yields v = ±

.gm/k. Note

that this is a free-fall problem where v > 0 in the downward direction. We thus ignore v < 0.

(i) Graphs will vary

(ii)9

gm

kis a stable equilibrium point

(iii) For v ! [0,#), v & +.

gm/k as t & #(iv) Graphs will vary

15. x! = x2(2 " x)(x + 3)2, x2(2 " x)(x + 3)2 = 0

Roots x = "3 x = 0 x = 2Multiplicity 2 2 1

End behavior: x2("x)(x)2 = "x5

Page 78: MATH 224 Sol.

72 Section 2.3

(i) x " x!

(ii) By the phase line diagram, x = 2 is a stable equilibrium point; x = 0 is a half-stableequilibrium point; x = "3 is a half-stable equilibrium point.

(iii) For x > 2, & 2 as t & #.For 0 < x < 2, & 2 as t & #.For "3 < x < 0, & 0 as t & #.For x < "3, & "3 as t & #.

(iv) tx-plane

Page 79: MATH 224 Sol.

Section 2.3 73

16. x! = (2 " x)3(x2 + 4), (2 " x)3(x2 + 4) = 0

Roots 2 ±2iMultiplicity 3 ignore

End behavior: ("x)3(x2) = "x5

(i) x " x!

(ii) By the phase line diagram, x = 2 is a stable equilibrium point.(iii) For x > 2, & 2 as t & #.

For x < "2, & 2 as t & #.(iv) tx-plane

Page 80: MATH 224 Sol.

74 Section 2.3

17. y! = (2 " y)3(y2 + 4)2

Root 2Multiplicity 3

Highest power and coe!cient: ("y)3(y2)2 = "y7

(i) x " x!

(ii) By the phase line diagram, y = 2 is a stable equilibrium point.(iii) If y0 < 2, then y & 2 as x & #.

If y0 > 2, then y & 2 as x & #.(iv) tx-plane

Page 81: MATH 224 Sol.

Section 2.3 75

18. y! = "y2(4 " y)(9 " y2)

Roots 0 4 3 "3Multiplicity 2 1 1 1

Highest power and coe!cient: "y2("y)("y2) = "y5

(i) x " x!

(ii) y = "3, 4 are stable; y = 0 is half-stable; y = 3 is unstable.(iii) If y0 < "3, y & "3 as x & #.

If "3 < y0 < 0, y & "3 as x & #.If 0 < x < 3, y & 0 as x & #.If 3 < x < 4, y & 4 as x & #.If x > 4, y & 4 as x & #.

(iv) tx-plane

Page 82: MATH 224 Sol.

76 Section 2.3

19. x! = x5(1 " x)(1 " x3), x5(1 " x)(1 " x3) = 0

Roots 0 1 two normalMultiplicity 5 2 ignore

End behavior: x5("x)("x3) = x9

(i) x " x!

(ii) By the phase line diagram, = 1 is a half-stable equilibrium point; = 0 is an unstableequilibrium point.

(iii) For x > 1, & # as t & #.For 0 < x < 1, & 1 as t & #.For x < 0, & "# as t & #.

(iv) tx-plane

Page 83: MATH 224 Sol.

Section 2.3 77

20. x! = x(x " 3)(1 + x3)(1 " x2)2, x(x " 3)(1 + x3)(1 " x2)2 = 0

Roots 0 3 "1 two non-real ±1Multiplicity 1 1 1 ignore 2

End behavior: x(x)(x3)("x2)2 = x9

(i) x " x!

(ii) By the phase line diagram, x = 3 is an unstable equilibrium point; x = 1 is a half-stableequilibrium point; x = 0 is a stable equilibrium point; x = "1 is an unstable equilibriumpoint.

(iii) If x > 3, & # as t & #.If 1 < x < 3, & 1 as t & #.If 0 < x < 1, & 0 as t & #.If "1 < x < 0, & 0 as t & #.If x < "1, & "# as t & #.

(iv) tx-plane

Page 84: MATH 224 Sol.

78 Section 2.3

21. x! = x2(1 " 2x)3(x2 " 1), x2(1 " 2x)3(x2 " 1) = 0

Roots 0 12 ±1

Multiplicity 2 3 1

End behavior: x2("x)3(x2) = "x7

(i) x " x!

(ii) By the phase line diagram, x = 1 is a stable equilibrium point; x = 12 is an unstable

equilibrium point; x = "1 is a stable equilibrium point.(iii) If x > 1, & 1 as t & #.

If 12 < x < 1, & 1 as t & #.

If "1 < x < 12 , & "1 as t & #.

If x < "1, & "1 as t & #.(iv) tx-plane

Page 85: MATH 224 Sol.

Section 2.3 79

22. x! = x3(x2 + 5)(x " 4)2(x + 5)

Roots 0 4 "5Multiplicity 3 2 1

End behavior: x3x2x2x = x8

(i) x " x!

(ii) By the phase line diagram, x = 1 is a stable equilibrium point; x = 12 is an unstable

equilibrium point; x = "1 is a stable equilibrium point.(iii) If x > 1, & 1 as t & #.

If 12 < x < 1, & 1 as t & #.

If "1 < x < 12 , & "1 as t & #.

If x < "1, & "1 as t & #.(iv) tx-plane

Page 86: MATH 224 Sol.

80 Section 2.3

23. y& = ±1, f !(y) = "2y, f !("1) = 2 $ Unstablef !(1) = "2 $ Stable

24. y& = ±1, f !(y) = "2y, f !("1) = 2 $ Unstablef !(1) = "2 $ Stable

25. y& = 0, ±!, f !(y) = cos y, f !(0) = 1 $ Unstablef !(±!) = "1 $ Stable

26. y& = "1, f !(y) = 3y2, f !("1) = 3 $ Unstable

27. y& = 0, f !(y) = "3y2, f !(0) = 0 $ Inconclusive. However, from the phase line diagram, y& = 0is stable.

28. If we Taylor expand the function about y& and keep the lowest order non-zero term, we seethat we have y! = f (3)(y&)y3 as the approximate solution near the equilibrium point. Phaseline analysis then shows the equilibrium point is stable.

29. (a) y! = ry " y3

"

"

Figure 2.1: Phase line for y! = ry " y3 : r < 0, r = 0, r > 0

Figure 2.2: Pitchfork bifurcation for y! = ry " y3

Page 87: MATH 224 Sol.

Section 2.3 81

(b) y! = ry + y3

"

"

Figure 2.3: Phaseline for y! = ry + y3 : r < 0, r = ", r > 0

Figure 2.4: Pitchfork bifurcation for y! = ry + y3

Page 88: MATH 224 Sol.

82 Section 2.4

2.4 Population Modeling: An Application of AutonomousEquations

1. (a) x = 0 is half-stable, x = 1 is stable. For logistic equation, x = 0 is unstable and x = k isstable. Yes, the are di"erent.

(b) For small x, logistic model growth is larger.

2. x = 0 - Stable, x = 1 - Unstable, x = 6 - Stable

3. x = 0 - Stable, x = 2 - Unstable, x = 10 - Stable

4. (a) x = 0 - Half-stable, x = 1 - Unstable, x = 4 - Stable(b) For small x, Allee e"ect model has larger growth rate.

5. (a) Exponential growth - Unlimited growth rate. No limitations placed on organisms.(b) Logistic model - Growth rate dependent on factors such as population amount or food

availability.(c) Allee e"ect - Growth rate dependent on factors such as population amount or food avail-

ability as well as a su!cient population to sustain itself.

6. (a) x = 0 - Unstable, x = a - Stable, x = 5 - Unstable(b) For x0 > 5, bacteria grows uninhibited.(c) The parameter a could represent the strength of the immune system or ability of the

body to fight o" the given bacteria. A healthy person would likely have an a-value that iscloser to 0 than to 5 because the lower value of a represents a lower value of the bacteria(that is stable).

7. (a) For a > 0,x = 0 - Unstablex = 5 "

'a - Stable

x = 5 +'

a - Unstablex = 10 - Stable

(b) For a = 0,x = 0 - Unstablex = 5 - Half-stablex = 10 - Stable

(c) For a < 0,x = 0 - Unstablex = 10 - Stable

(d) Saddle-node(e) Bacteria grows unchecked to a level of 10. No, the bacteria population eventually reaches

and levels o" at 10, which above the fatal level.(f) The parameter a could again represent the strength of the immune system or ability of

the body to fight o" the given bacteria.

8. x! = x(1 " x)(x " 6)(x " 10)

9. x! = x(2x " 1)(x " 1)2(x " 2)(x " 8) or x! = x(2x " 1)2(x " 1)(x " 2)(x " 8)

10. (a) x! = x2(2 " x)2(x " 4)

Page 89: MATH 224 Sol.

2.5. NUMERICAL APPROXIMATION WITH THE EULER METHOD 83

(b) x = 0 - Half-stable, x = 2 - Half-stable, x = 4 - Unstable

11. (a) x! = rx(x " a)(x " 1) r, a > 0(b) i. 0 < a < 1, x = 0 - Unstable, x = a - Stable, x = 1 - Unstable

ii. a = 1, x = 0 - Unstable, x = 1 - Half-stableiii. a > 1, x = 0 - Unstable, x = 1 - Stable, x = a - Unstable

(c) When a < 1, the bacteria goes to the stable level a if it starts with a level less than 1and grows without bound otherwise. When a = 1, the bacteria goes to the half-stablelevel a = 1 if it starts with a level less than 1 and grows without bound otherwise. Whena > 1, the bacteria goes to the stable level of 1 if it starts with a level less than a andgrows without bound otherwise.

2.5 Numerical Approximation with the Euler Method

1. dy/dx = x3, y(1) = 1; explicit solution: y = 14 (x4 + 3)

Euler Explicitxi yi y(xi)1.0 1 11.1 1.1 1.11601.2 1.2331 1.26841.3 1.4059 1.46401.4 1.6256 1.7104

2. dy/dx = x4y, y(1) = 1; explicit solution: y = e(x5"1)/5

Euler Explicitxi yi y(xi)1.0 1 11.1 1.1 1.12991.2 1.2611 1.34671.3 1.5225 1.72051.4 1.9574 2.4004

3.dy

dx= "y2 cosx, h = 0.1, y(0) = 1; explicit solution y =

11 + sinx

.

x0 = 0x1 = 0 + (0.1)(1) = 0.1x2 = 0 + (0.1)(2) = 0.2x3 = 0 + (0.1)(3) = 0.3x4 = 0 + (0.1)(4) = 0.4y0 = 1y1 = 1 + (0.1)("(1)2 cos(0)) = .9y2 = .9 + (0.1)("(.9)2 cos(0.1)) = .8194y3 = .8194 + (0.1)("(.8194)2 cos(0.2)) = .7536y4 = .7536 + (0.1)("(.7536)2 cos(0.3)) = .6993

Page 90: MATH 224 Sol.

84 Section 2.5

Explicit:

y(0) =1

1 + sin 0= 1

y(0.1) =1

1 + sin(0.1)= .909228

y(0.2) =1

1 + sin(0.2)= .834258

y(0.3) =1

1 + sin(0.3)= .7718907

y(0.4) =1

1 + sin(0.4)= .719725

Euler Explicitxi yi y(xi)0.0 1 10.1 .9 .9092280.2 .8194 .8342580.3 .7536 .77189070.4 .6993 .719725

4. y! =sin x

y3, y(!) = 2, h = .1.

x0 = !

x1 = ! + .1x2 = ! + .2x3 = ! + .3x4 = ! + .4y0 = 2

y1 = 2 + (.1)!

sin(!)23

"= 2

y2 = 2 + (.1)!

sin(! + .1)23

"= 1.99875

y3 = 1.995 + (.1)!

sin(! + .2)(1.995)3

"= 1.99626

y4 = 1.985 + (.1)!

sin(! + .3)(1.985)3

"= 1.99255

Explicit:

y = (12 " 4 cosx)1/4

y(!) = 2y(! + .1) = (12 " 4 cos(! + .1))1/4 = 1.99937y(! + .2) = (12 " 4 cos(! + .2))1/4 = 1.99750y(! + .3) = (12 " 4 cos(! + .3))1/4 = 1.99439y(! + .4) = (12 " 4 cos(! + .4))1/4 = 1.99006

Page 91: MATH 224 Sol.

Section 2.5 85

Euler Explicitxi yi y(xi)! 2 2

! + .1 2 1.99937! + .2 1.99875 1.99750! + .3 1.99626 1.99439! + .4 1.99255 1.99006

5. dy/dx = ye"x, y(0) = 1; explicit solution: y = exp(1 " e"x)Euler Explicit

xi yi y(xi)1.0 1 11.1 1.1 1.09981.2 1.1995 1.19871.3 1.2977 1.29591.4 1.3939 1.3905

6. dy/dx = xy2, y(0) = 1; explicit solution: y =2

2 " x2

Euler Explicitxi yi y(xi)0.0 1 10.1 1.1 1.00500.2 1.01 1.02040.3 1.0304 1.04710.4 1.0623 1.08700.5 1.1074 1.14290.6 1.1687 1.21950.7 1.2507 1.32450.8 1.3602 1.4706

7. dy/dx = y + cosx, y(0) = 0; explicit solution: y = 12 (sin x " cosx + ex)

Euler Explicitxi yi y(xi)0.0 0 00.1 0.1 0.10500.2 0.2095 0.22000.3 0.3285 0.34500.4 0.4568 0.48010.5 0.5946 0.62530.6 0.7418 0.78070.7 0.8986 0.94660.8 1.0649 1.1231

8. dy/dx = y + sin x, y(0) = 2; explicit solution: y = "12 (cosx + sin x " 5ex)

Page 92: MATH 224 Sol.

86 Section 2.5

Euler Explicitxi yi y(xi)0.0 2 20.1 2.2 2.21550.2 2.4300 2.46410.3 2.6929 2.74920.4 2.9917 3.07430.5 3.3298 3.44330.6 3.7107 3.86030.7 4.1383 4.32990.8 4.6165 4.8568

9. dy/dx = e"y, y(0) = 2; explicit solution: y = ln(x + e2)Euler Explicit

xi yi y(xi)0.0 2 20.1 2.0135 2.01340.2 2.0269 2.02670.3 2.0401 2.03980.4 2.0531 2.05270.5 2.0659 2.06550.6 2.0786 2.07810.7 2.0911 2.09050.8 2.1034 2.1028

10. dy/dx = x + y, y(0) = 0; explicit solution: y = "x " 1 + ex

Euler Explicitxi yi y(xi)0.0 0 00.1 0 0.00520.2 .01 0.02140.3 .031 0.04990.4 .0641 0.09180.5 .1105 0.14870.6 .1716 0.22210.7 .2487 0.31380.8 .3436 0.4255

11. dy/dx = (x + 1)(y2 + 1), y(0) = 0; explicit solution: y = tan(12x2 + x)

Euler Explicitxi yi y(xi)0.0 0 00.1 0.1 0.10540.2 0.2111 0.22360.3 0.3364 0.35940.4 0.4812 0.52060.5 0.6536 0.72150.6 0.8677 0.98930.7 1.1481 1.38370.8 1.5422 2.0660

Page 93: MATH 224 Sol.

2.6. NUMERICAL APPROXIMATION WITH THE RUNGE-KUTTA METHOD 87

12. a. x! = 1.7x#1 " x

1000

$, b. x! = 2.7x

#1 " x

1000

$

h (a.) x(1) (b.) x(1)1 100 100

0.5 100.425 100.6750.25 100.637 101.0120.1 100.765 101.214

2.6 Numerical Approximation with the Runge-Kutta Method

1. dy/dx = x3, y(1) = 1; explicit solution: y = 14 (x4 + 3)

Runge-Kutta Explicitxi yi y(xi)1.0 1 11.1 1.1160 1.11601.2 1.2684 1.2684

2. dy/dx = x4y, y(1) = 1; explicit solution: y = e(x5"1)/5

Runge-Kutta Explicitxi yi y(xi)1.0 1 11.1 1.1299 1.12991.2 1.3467 1.3467

3. y! = "y2 cosx, h = .1, y(0) = 1. We have x0 = 0, x1 = .1, x2 = .2.y1:

k1 = f(0, 1) = "(1)2 cos 0 = "1

k2 = f

!0 + .05, 1 +

(.1)("1)2

"= "(.95)2 cos(.05) = ".901372

k3 = f

!0 + .05, 1 +

(.1)2

(".902499)"

= ".9107543

k4 = f(0 + .1, 1 + (.1)(".911786)) = ".8220166

y1 = 1 +(.1)6

("1 + 2(".901372) + 2(".910754) + (".8220166))

= .9092288

y2:

k1 = ".822567k2 = ".745136k3 = ".751797k4 = ".68177

$ y2 = .8342587

4. y! =sin x

y3, h = .1, y(!) = 2

Page 94: MATH 224 Sol.

88 Section 2.6

y1:

k1 = 0k2 = ".049979k3 = ".0503557k4 = ".101356889

$ y1 = .994966

y2:

k1 = ".101356k2 = ".15405988k3 = ".1552968k4 = ".211447

$ y2 = .9794409

5. dy/dx = ye"x, y(0) = 1; explicit solution: y = exp(1 " e"x)Runge-Kutta Explicit

xi yi y(xi)1.0 1 11.1 1.0998 1.09981.2 1.1987 1.1987

6. dy/dx = xy2, y(0) = 1; explicit solution: y =2

2 " x2

Runge-Kutta Explicitxi yi y(xi)0.0 1 10.1 1.0050 1.00500.2 1.0204 1.02040.3 1.0471 1.04710.4 1.0870 1.08700.5 1.1429 1.14290.6 1.2195 1.21950.7 1.3245 1.32450.8 1.4706 1.4706

7. y! = y + cosx, y(0) = 0, y = 12 (sin x " cosx + ex)

x Runge-Kutta Explicit0 0 0.1 .1049999 .1050000.2 .22000 .22000.3 .34502 .34502.6 .78071 .78071.7 .946563 .946564.8 1.123094 1.123095.9 1.310658 1.3106601.0 1.509723 1.509725

8. dy/dx = y + sin x, y(0) = 2; explicit solution: y = "12 (cosx + sin x " 5ex)

Page 95: MATH 224 Sol.

Section 2.6 89

Runge-Kutta Explicitxi yi y(xi)0.0 2 20.1 2.2155 2.21550.2 2.4641 2.46410.3 2.7492 2.74920.4 3.0743 3.07430.5 3.4433 3.44330.6 3.8603 3.86030.7 4.3299 4.32990.8 4.8568 4.8568

9. dy/dx = e"y, y(0) = 2; explicit solution: y = ln(x + e2)Runge-Kutta Explicit

xi yi y(xi)0.0 2 20.1 2.0134 2.01340.2 2.0267 2.02670.3 2.0398 2.03980.4 2.0527 2.05270.5 2.0655 2.06550.6 2.0781 2.07810.7 2.0905 2.09050.8 2.1028 2.1028

10. y! = x + y, y(0) = 0, y = "x " 1 + ex

x Runge-Kutta Explicit.1 .00517 .00517.2 .02140 .02140.5 .14872 .14872.6 .22211 .22211.7 .31375 .31375.8 .42554 .42554.9 .55960 .559601.0 .71828 .71828

11. dy/dx = (x + 1)(y2 + 1), y(0) = 0; explicit solution: y = tan(12x2 + x)

Runge-Kutta Explicitxi yi y(xi)0.0 0 00.1 0.1054 0.10540.2 0.2236 0.22360.3 0.3594 0.35940.4 0.5206 0.52060.5 0.7215 0.72150.6 0.9893 0.98930.7 1.3837 1.38370.8 2.0660 2.0660

Page 96: MATH 224 Sol.

90 Section 2.6

12. y! = e"x2

13. y! = e"x2

14. y! = x3ey + 3x2 sin y

15. y! = e"x2

16. y! = e"x2

Page 97: MATH 224 Sol.

2.7. AN INTRODUCTION TO AUTONOMOUS SECOND-ORDER EQUATIONS 91

17.

h (a.) x(1) (b.) x(1)1 100.75 101.249

0.5 100.75 101.2490.25 100.75 101.2490.1 100.75 101.249

2.7 An Introduction to Autonomous Second-Order Equa-tions

1. y = cekx, y! = kcekx, y!! = k2cekx $yy!! = cekx + k2cekx = c2k2e2kx = (y!)2

2. If v = $!, then v =C + g cos $

mand

tanh"1

!(C"g) tan(#/2)'

g2"C2

"

.g2 " C2

=t

m+ K

4. (a) If u(x) =dy

dx, then we have u! =

.1 + u2.

(b) This is separable, so%

du'1 + u2

= sinh"1(u) = x " x0

$ u =e(x"x0) " e"(x"x0)

2for some constants c, d.

(d) Then, if u =dy

dx=

e(x"x0) " e"(x"x0)

2= sinh(x " x0),

then y = cosh(x " x0) + C.

2.8 Chapter 2: Additional Problems

1. False. It may not be unique.

2. False. RK use four function evaluations to calculate the next step.

3. False. Euler’s method is not superior.

4. False. Neither f nor "f"y is continuous everywhere.

5. True.

6. True. Phase line analysis will work and gives information on long-term behavior.

8. (a) (i) Solutions exist everywhere; (ii) solutions are unique everywhere except possibly alongy = 0.

(b) (i) Solutions exist everywhere; (ii) solutions are unique everywhere.(c) (i) Solutions exist everywhere except possibly when y = !

2 ± n! for n = 0, 1, 2, · · · ; (ii)solutions are unique everywhere except possibly when y = !

2 ± n! for n = 0, 1, 2, · · · .(d) (i) Solutions exist everywhere except possibly along x = "1; (ii) solutions are unique

except possibly along x = "1.

Page 98: MATH 224 Sol.

92 Chapter 2 Review

17. y! = 1"y2

2x , y(1) = !.xi RK:yi Euler:yx

1.0000 3.1416 3.14161.1000 2.7742 2.69811.2000 2.5144 2.41271.3000 2.3210 2.21181.4000 2.1713 2.06211.5000 2.0522 1.94591.6000 1.9550 1.85311.7000 1.8743 1.77701.8000 1.8061 1.7135

20. y! = y"y2

x+1 , y(2) = 0.All answers are 0 because y0 = 0 is an equilibrium point.

23. y! = 1"y2

2x , y(1) = !.xi RK:yi Euler:yx

1.4142 1.0000 1.00001.5142 1.1251 1.12071.6142 1.2592 1.24951.7142 1.4030 1.38681.8142 1.5571 1.53321.9142 1.7222 1.68912.0142 1.8992 1.85522.1142 2.0889 2.03212.2142 2.2924 2.2207

26. y = Cx2e"3/x

27. y = 0, (x±C)2

4x2

30. y = ln,

"1ln(x"2)+C

-

32. y = ±1'1+Cex2

33. y = x"lnx+C(x"1)2

34.

36.

38. y = ±'

6 sin3 x+9C3 sin x

41. y2/3 " x3 " 1

6 " Ce2x

42. y = ±'

e!2x(x2+C)e!2x

Page 99: MATH 224 Sol.

Chapter 3

Elements of Higher-Order LinearEquations

3.1 Some Terminology

1. By Theorem 3.1.1, a unique solution exists (a0(x) = 1 %= 0). y = sin x is the solution.

2. a0(x) = x2 = 0 when x = 0, so Thm 3.1.1 does not apply.For the given initial conditions, y = c2x

2 is a solution. (Notice it is not unique).

3. a0(x) = x2 = 0 when x = 0, so Thm 3.1.1 does not apply.y(0), y!(0) are undefined, so there is no solution.

4. a0(x) = x2 = 0 when x = 0, so Thm 3.1.1 does not apply.Any constants c1 and c2 will work.

5. a0(x) = 1 %= 0, so Thm 3.1.1 guarantees a unique solution does exist.y = 2e2x + 2e"2x is the solution.

6. (a) x ! ("#,#)

(b) x %= k!

2, k ! Z

(c) x %= 0(d) x %= 0

7. (a) x ! ("#,#)(b) x > 0(c) x ! ("#,#)(d) x %= 1

93

Page 100: MATH 224 Sol.

94 Section 3.2

3.2 Essential Topics from Linear Algebra

1. Any solution of y! + p(x)y = 0 looks like y = Ce"@

p(x) dx. If y1 and y2 are solutions, they areof the form y1 = c1e"

@p(x) dx and y2 = c2e"

@p(x) dx. Then

W (x) = det&

c1e"@

p(x) dx c2e"@

p(x) dx

"p(x)c1e"@

p(x) dx "p(x)c2e"@

p(x) dx

'

= e@

p(x) dx · det&

c1 c2

"c1p(x) "c2p(x)

'

= 0 $ linearly dependent

2. {x, 2x}. c1x + c2(2x) = 0. Choose c1 = "2, c2 = 1. This will hold for all x on [0, 1].

3. {e2x, xe2x}. By definition,

c1e2x + c2xe2x = 0 $ e2x(c1 + c2x) = 0

$ c1 + c2x = 0$ c1 = c2 = 0$ linearly independent

Using the Wronskian,

det&

e2x xe2x

2e2x e2x(1 + x)

'

x=0

= det&

1 02 1

'= 1 $ linearly independent

4. c1x + c2 sinx = 0. Then

d

dx(c1x + c2 sin x = 0) = c1 + c2 cosx = 0

We can solve for c1 and c2 as long as

W (x) =8888

x sin x1 cosx

8888

is not zero. Choose x = !4 . Then

W,!

4

-=

88888

!4 sin !

4

1 cos !4

88888 ='

22

,!4" 1-%= 0

Thus {x, sin x} are linearly independent.

5. By definition, c1ex + c2(x + 1) = 0. Then

x = 0 $ c1 + c2 = 0x = "1 $ c1e

"1 = 0

We then havec1 = 0 $ c2 = "c1 $ c2 = 0

Thus c1 = c2 = 0. Taking the determinant,

W (ex, x + 1) =8888

ex x + 1ex 1

8888 = ex + ex(x + 1) = ex(x + 2) %= 0 if x %= "2.

Therefore {ex, x + 1} is linearly independent.

Page 101: MATH 224 Sol.

Section 3.2 95

6. Using the Wronskian,

W (x) =8888

ex ex+5

ex ex+5

8888$ W (x) = e2x+5 " e2x+5 = 0

Therefore the set is linearly dependent. (Take c2 = "c1e5 )

7. Using the Wronskian,

W (x) = det&

sin 2x cos 2x2 cos 2x "2 sin2x

'$ W (0) = det

&0 12 0

'= "2 %= 0.

Therefore {sin 2x, cos 2x} is linearly independent.

8. c1(x3 " 4) + c2x + c33x = 0. Grouping like terms and choosing c1 = 0, c2 = "3, and c3 = 1shows the set is linearly dependent.

9. {x3 " 4x, x, 2x3}. By definition,

c1(x3 " 4x) + c2x + c3(2x3) = 0x3(c1 + 2c3) + x("4c1 + c2) = 0

Choosing c1 = 2, c3 = "1, and c2 = 8 shows the set is linearly dependent. Using the Wronskianwill not work because the set is linear dependent.

10. Letting c1 = 2, c2 = 8, c3 = "1 shows the set is linearly dependent.

11. Letting c1 = 1, c2 = "2, c3 = 2 shows the set is linearly dependent.

12. Letting c1 = 4, c2 = "1.5, c3 = "1 shows the set is linearly dependent.

13. Letting c1 = 1, c2 = "1, c3 = "3 shows the set is linearly dependent.

14. W (x) = 2 $ linearly independent.

15. W (x) = 2x3 " 3x2 & W (1) = "1 %= 0 $ linearly independent.

16. W (x) = 2 $ linearly independent.

17. Letting c1 = 2, c2 = "1, c3 = "2 shows the set is linearly dependent.

18. Letting c1 = 0, c2 = e"4, c3 = "1 shows the set is linearly dependent.

19. Letting c1 = 0, c2 = 1, c3 = e7 shows the set is linearly dependent.

20. W (x) = ex ("5 cosx cos 2x + 3 cos 2x sinx " 4 sinx sin 2x) & W (0) %= 0 $ linearly independent.

21. Letting c1 = 1, c2 = "1, c3 = "1, c4 = "1 shows the set is linearly dependent.

22. (a) On [0, 1], |x| = x. Then c1x+ c2x = 0 when c1 = "c2 shows the set is linearly dependent.(b) On ["1, 0], |x| = "x. Then c1x+c2("x) when c1 = c2 shows the set is linearly dependent.(c) On ["1, 1], we need c1x + c2|x| = 0. This can only happen when c1 = "c2 and when

c1 = c2. These are both true only if c1 = c2 = 0. Thus the set is linearly independent on["1, 1].

(d) W (x) = det&

x "x1 "1

'= 0 on ["1, 0]

Page 102: MATH 224 Sol.

96 Section 3.2

W (x) = det&

x x1 1

'= 0 on [0, 1]

23. (a) On [0, 1], |x|3 = x3. Then c1x3 + c2x3 = 0 when c1 = "c2 shows the set is linearlydependent.

(b) On ["1, 0], |x|3 = "x3. Then c1x3 + c2("x)3 when c1 = c2 shows the set is linearlydependent.

(c) On ["1, 1], we need c1x3 + c2|x|3 = 0. This can only happen when c1 = "c2 and whenc1 = c2. These are both true only if c1 = c2 = 0. Thus the set is linearly independent on["1, 1].

(d) W (x) = det&

x3 "x3

1 "1

'= 0 on ["1, 0]

(e) W (x) = det&

x3 x3

1 1

'= 0 on [0, 1]

24. y = c1x + c2x ln x, y! = c1 + c2(1 + lnx), y!! =c2

xx2y!! " xy! + y = c2x " c1x " c2x " c2x ln x + c1x + c2x ln x = 0Letting c1 = 3, c2 = "4 satifies the IC. This does not violate theorem since a0(x) %= 0 on thespecified interval.

25. y = c1 + c2x2, y! = 2c2x, y!! = 2c2

xy!! " y! = x(2c2) " 2c2x = 0Theorem DOES guarantee existance of unique solution on (0,#) since a0(x) %= 0 on this

inteval. y =x2

2is the solution that satisfies IC.

26. y = c1x2 + c2x, y! = 2c1x + c2, y!! = 2c1

x2y!! " 2xy! + 2y = 2x2(c1) " 2x(2c1x2 + c2) + 2(c1x

2 + c2x) = 0Theorem does NOT guarantee existance of unique solution on (0,#) since a0(x) = 0 at x = 0.y = c1x

2 + x is the general solution that satisfies IC (not unique).

27. There are two functions and the equation is 2nd order. Both functions are solutions andW (x) = "17 %= 0 $ functions are linearly independent. Hence, the set is a fundamental solu-tion set.

28. There are two functions and the equation is 2nd order. Both functions are solutions andW (x) = 4 %= 0 $ functions are linearly independent. Hence, the set is a fundamental solutionset.

29. There are two functions and the equation is 2nd order. We have

(e"2x)!! " (e"2x)! " 6(e"2x) = 4e"2x + 2e"2x " 6e"2x = 0

and(e3x)!! " (e3x)! " 6(e3x) = 0

so both are solutions to the ODE. Taking the determinant,

det&

e"2x e3x

"2e"2x 3e3x

'

x=0

= det&

1 1"2 3

'= 3 " ("2) = 5 %= 0

Therefore they are linearly independent, and {e"2x, e3x} forms a fundamental set of solutions.

Page 103: MATH 224 Sol.

Section 3.2 97

30. There are two functions and the equation is 2nd order. Both functions are solutions andW (x) = 3e7x %= 0 $ functions are linearly independent. Hence, the set is a fundamental solu-tion set.

31. Let y1 = sin(2x), y2 = cos(2x). Then

y!1 = 2 cos(2x), y!!

1 = "4 sin(2x), y!2 = "2 sin(2x), y!!

2 = "4 cos(2x)

There are 2 functions and the ODE is 2nd order. Substitution shows that y!!1 + 4y1 = 0 and

y!!2 + 4y2 = 0. Thus both are solutions to the ODE. Taking the determinant,

det&

sin 2x cos 2x2 cos 2x "2 sin 2x

'

x=0

= det&

0 12 0

'= "2 %= 0

Therefore they are linearly independent, and the equations form a fundamental set of solutions.

32. We have

y1 = ex $ y!1 = ex, y!!

1 " ex

y2 = xex, y!2 = (1 + x)ex, y!!

2 = (2 + x)ex

Then

y!!1 " 2y!

1 + y1 = ex " 2ex + ex = 0y!!2 " 2y!

2 + y2 = ex(2 + x) " 2(1 + x)ex + xex = 0

so both ex and xex are in the solution space of the ODE. Using the Wronskian,

W (x) = det&

ex xex

ex (1 + x)ex

'$ W (0) = det

&1 01 1

'= 1 %= 0

so they are linearly independent. Because we have two functions in a two-dimensional spaceand they are linearly independent, it follows that they span. Thus {ex, xex} is a basis.

33. We have

y1 = e"3x, y!1 = "3e"3x, y!!

1 = 9e"3x

y2 = ex/2, y!2 =

12ex/2, y!

3 =14ex/2

Then

2y!!1 + 5y!

1 " 3y1 = 2(9e"3x) + 5("3e"3x) " 3(e"3x)= (18 " 15 " 3)e"3x

= 0

2y!!2 + 5y!

2 " 3y2 = 2!

14ex/2

"+ 5!

12ex/2

"" 3(ex/2)

=!

12

+52" 3"

ex/2 = 0

Thus both e"3x and ex/2 are in the solution space of the ODE. Using the Wronskian,

W (x) = det&

e"3x ex/2

"3e"3x 12ex/2

'$ W (0) = det

&1 1"3 1

2

'%= 0

Thus they are linearly independent. Because we have two functions in a two-dimensional spaceand they are linearly independent, it follows that they span. Thus {e"3x, ex/2} is a basis.

Page 104: MATH 224 Sol.

98 Section 3.2

34. There are three functions and the equation is 3rd order. All functions are solutions andW (x) = 1 %= 0 $ functions are linearly independent. Hence, the set is a fundamental solutionset.

35. There are three functions and the equation is 3rd order. All functions are solutions andW (x) = "2e"3x %= 0 $ functions are linearly independent. Hence, the set is a fundamentalsolution set.

36. There are three functions and the equation is 3rd order. All functions are solutions andW (x) = 8 cos4 x + 8 cos2 x sin2 x %= 0 $ functions are linearly independent. Hence, the set is afundamental solution set.

37. No. y = x2 + 4 is not a solution.

38. There are two functions and the equation is 2nd order. All functions are solutions andW (x) = 3e3x %= 0 $ functions are linearly independent. Hence, the set is a fundamental solu-tion set.

39. NO. y = x"1 is not a solution.

40. There are two functions and the equation is 2nd order. All functions are solutions andW (x) = "15 %= 0 $ functions are linearly independent. Hence, the set is a fundamental solu-tion set.

41. There are two functions and the equation is 2nd order. All functions are solutions andW (x) = "10 %= 0 $ functions are linearly independent. Hence, the set is a fundamental solu-tion set.

42. No. cos 3x is not a solution.

43. No, because sinx is not a solution to y!! " y = 0.

44. There are two functions and the equation is 2nd order. All functions are solutions andW (x) = 6x %= 0 on interval $ functions are linearly independent. Hence, the set is a fun-damental solution set.

45. We have {x2, x, e"x} and the ODE y(4)(x)"y!!!(x) = 0. The number of possible solutions doesnot match the order of the ODE (4th order), thus {x2, x, e"x} does not form a fundamentalset of solutions for y(4)(x) " y!!!(x) = 0.

46. No. e"3x is not a solution.

47. There are two functions and the equation is 2nd order. All functions are solutions andW (x) = e"2x %= 0 $ functions are linearly independent. Hence, the set is a fundamental solu-tion set.

48. There are two functions and the equation is 2nd order. All functions are solutions andW (x) = e4x %= 0 $ functions are linearly independent. Hence, the set is a fundamental so-lution set.

49. There are three functions and the equation is 3rd order. We have

(x)!!! " (x)!! = 0, (3)!!! " (3)!! = 0, (ex)!!! " (ex)!! = 0

so all are solutions. Taking the determinant,

det

K

Lx 3 ex

1 0 ex

0 0 ex

M

N

x=0

= det

K

L0 3 11 0 10 0 1

M

N = "3 det&

1 10 1

'= 3 %= 0

Page 105: MATH 224 Sol.

Section 3.2 99

so they are linearly independent. Thus {x, 3, ex} form a fundamental set of solutions.

50. No. It is a 2nd degree ODE with 3 functions.

51. There are two functions and the equation is 2nd order. All functions are solutions andW (x) = 2e6x %= 0 $ functions are linearly independent. Hence, the set is a fundamental solu-tion set.

52. There are two functions and the equation is 2nd order. All functions are solutions andW (x) = "e"4x %= 0 $ functions are linearly independent. Hence, the set is a fundamentalsolution set.

53. No. W (x) = 0.

54. There are four functions and the equation is 4th order. All functions are solutions andW (x) = "16 %= 0 $ functions are linearly independent. Hence, the set is a fundamental solu-tion set.

55.&

xy

'= c1

&0"1

'+ c2

&11

'$&

xy

'=&

0 1"1 1

' &c1

c2

'. Because the determinant is

nonzero, we can always solve this system for c1, c2 given any x, y. Thus, the setO!

0"1

",

!11

"4

spans R2. The functions are linearly independent because c1

&0"1

'+ c2

&11

'=&

00

'only

when c1 = c2 = 0. Thus they form a basis for R2.For the solutions to 56-73, the reader is referred to Thm 5.4.1, which says (in ourneed) that if detA %= 0 then the column vectors of A form a basis.

56. det&

2 "1"1 1

'= 1

57. det& '

2 00

'5

'=

'10

58. det&

10 31 2

'= 17

59. det

K

L1 1 11 1 01 0 0

M

N = "1

60. det

K

L2 1 01 0 "2"1 1 0

M

N = 6

61. det

K

PPL

1 0 3 10 1 2 00 0 0 22 "1 1 0

M

QQN = 6

62. det

K

PPL

1 "1 1 "10 1 "1 00 0 1 "20 0 0 2

M

QQN = 6

Page 106: MATH 224 Sol.

100 Section 3.2

63. Yes,detA = 3

64. No, detA = 0

65. No, 3 vectors for R2

66. Yes, detA = "3

67. No, 3 vectors for R2

68. No, the vectors are not 3 dimensional.

69. No, only 3 vectors for R4

70. No, only 3 vectors for R4

71. Yes, detA = "3

72. No, detA = 0

73. Yes, detA = "2

74. We need to show that R2 satisfies Definition 3.5. So, let %u =!

u1

u2

", %v =

!v1

v2

"and

%w =!

w1

w2

"! R2, u1, u2, v1, v2, w1, w2 ! R

(a) %u + %v =!

u1 + v1

u2 + v2

"! R2

(b) c%v =!

cv1

cv2

"! R2

(c) %u + %v =!

u1 + v1

u2 + v2

"=!

v1 + u1

v2 + u2

"= %v + %u

(d) %u + (%v + %w) =!

u1 + (v1 + w1)u2 + (v2 + w2)

"=!

(u1 + v1) + w1

(u2 + v2) + w2

"= (%u + %v) + %w

(e) Let %0 =!

00

". Then %u +%0 =

!u1 + 0u2 + 0

"=!

0 + u1

0 + u2

"= %0 + %u = %u

(f) Let "%u =!

"u1

"u2

". Then %u + ("%u) =

!u1

u2

"+!

"u1

"u2

"

=!

u1 " u1

u2 " u2

"=!

00

"= %0

(g) c(%u + %v) =!

c(u1 + v1)c(u2 + v2)

"=!

cu1 + cv1

cu2 + cv2

"= c%u + c%v

(h) (c1 + c2)%u =!

(c1 + c2)u1

(c1 + c2)u2

"=!

cu1 + c2u1

c1u2 + c2u2

"

=!

c1u1

c2u2

"+!

c2u1

c2u2

"= c1%u + c2%u

(i) c1(c2%u) =!

c1(c2u1)c1(c2u2)

"=!

(c1c2)u1

(c1c2)u2

"= (c1c2)%u

(j) 1%u =!

1u1

1u2

"=!

u1

u2

"= %u

Page 107: MATH 224 Sol.

3.3. REDUCTION OF ORDER—THE CASE OF N = 2 101

75. Same as 74., except with 3D vectors instead of 2D.

76. Same as 74, except with 2 , 2 matrices.

77. Label the matrices A1 · · ·A4. It is clear that

c1A1 + c2A2 + c3A3 + c4A4 =!

c1 c2

c3 c4

"spans R2'2 and it is also clear that

c1A1 + c2A2 + c3A3 + c4A4 =!

c1 c2

c3 c4

"=!

0 00 0

"only if

c1, c2, c3, c4 all are 0. Hence, they are linearly independent and so they form a basis.

78. All vectors in this subset are of the form

1

3xy0

5

7. Set x = y = 0. Thus

1

3000

5

7 is in this

subset W = set of all vectors of the form

1

3xy0

5

7. Let

1

3x1

y1

0

5

7 and

1

3x2

y2

0

5

7 be in W . Then

1

3x1

y1

0

5

7+

1

3x2

y2

0

5

7 =

1

3x1 + x2

y1 + y2

0

5

7

is also in W . For any c ! R, c

1

3xy0

5

7 =

1

3cxcy0

5

7 is still in W . Thus

W is a subspace of R2.

79. Repeat argument for 78, changing all the vectors to

1

30yz

5

7.

80. Let W be the set of planes passing through the origin. Since each plane passes through the

origin,

1

3000

5

7 is a member of each element of W . Also, if %u and %v pass through the origin,

then %u + %v must pass through the origin. Hence, W is closed under addition. Finally, if %upasses through the origin, then any scalar multiple of %u will also pass through the origin, soc%u ! W . So W is a subspace of R3.

3.3 Reduction of Order—The Case of n = 2

FOR EXERCISES 1-21, THE GENERAL SOLUTION IS GIVEN BYy = v + p(x) where p(x) is the particular solution given in the exercises.

1. y = vex

y!! " 2y! + y = v!! = 0 & v = Cx + K

2. y = ve"x/2

4y!! + 4y! + y = 4v!! = 0 & v = Cx + K

3. y = v cos 5xy!! + 25y = v!! cos 5x " 10 sin 5xv! = 0, w = v!, w! = v!!

Page 108: MATH 224 Sol.

102 Section 3.3

& dw

w= 10 tan 5x & ln w = ln(cos"2 5x) + C & w = v! =

C

cos2 5xand

thus v =C

5tan 5x + K.

4. y = ve3x

y!! " 9y = v!! + 6v! = 0 $ v!!

v!= "6 & ln v! = "6x + C & v! = e"6x+C

& v ="16

e"6x+C + K

5. y = ve3x

y!! " 6y! + 9y = v!! = 0 & v! = C & v = Cx + K

6. y = ve4x

y!! " 8y! + 16y = v!! = 0 & v! = C & v = Cx + K

7. y = vex/2

2y!! + 5y! " 3y = 2v!! + 7v! = 0 & v!!

v!= "7

2& v! = e"

72 x+C

& v = "27e"

72 x+C + K

8. y = vx2

xy!! " y! = x3v!! + 3x2v! = 0 & v!!

v!=

"3x

& ln v! = ln x"3 + C

& v! =C

x3& v =

C

x2+ K

9. y = v ln x

xy!! + y! = v!! + v!!

2 + lnx

x ln x

"= 0 & v!!

v!= "

!2 + lnx

x ln x

"

& v! =C

x ln2 x& v =

C

ln x+ K

10. y = vx3

x2y!! " xy! " 3y = x5v!! + 5x4v! = 0 & v!!

v!=

"5x

& ln v! = ln x"5 + C

& v! =C

x5& v =

C

x4+ K

11. y = vx ln x

x2y!! " xy! + y = x3 ln xv!! + v!(2x2 + x2 ln x) = 0 & v!!

v!=

"(2x2 + x2 ln x)x3 ln x

& ln v! = ln x"1 + ln!

1ln2 x

"+ C & v! =

C

x ln2 x

& v =C

ln x+ K

12. y =v'x

2x2y!! + 5xy! + y = 2x3/2v!! + 3x1/2v! = 0 & v!!

v!=

"32x

& v! = Cx"3/2

& v =C'x

+ K

Page 109: MATH 224 Sol.

Section 3.3 103

13. y = v(x + 1)

(x + 1)2y!! " 3(x + 1)y! + 3y = (x + 1)3v!! " (x + 1)2v! = 0 & v!!

v!=

1x + 1

& v! = C(x + 1) & v =Cx2

2+ Cx + K

14. y = ve2x

(2x + 1)y!! " 4(x + 1)y! + 4y = e2x (4xv! + (1 + 2x)v!!) = 0 & v!!

v!=

"4x

1 + 2x& ln v! = "2x + ln(1 + 2x) + C & v! = ce"2x(1 + 2x)& v = ce"2x(1 " x) + K

15. y = v(x2 + 1)

(x2 " 1)y!! " 2xy! + 2y = v!!(x4 " 1) + v!(2x"6x) = 0 & v!!

v!=

6x " 2x3

x4 " 1

& ln v! = ln!

x2 " 1(x2 + 1)2

"+ C

& v! =C(x2 " 1)(x2 + 1)2

& v ="Cx

x2 + 1+ K

17. y = ve2x

y!!! " 2y!! " y! + 2y = v!!! + 4v!! + 3v! = 0 which is a 2nd degree ODE of v!. Hence, from thecharacteristic equation, v! = e"3x and v! = e"x are two solutions. So a general solution wouldbe v! = c1e"3x + c2e"x & v = k1e"3x + k2e"x + k3

18. y = vex

y!!! " 4y!! + 5y! " 2y = v!!! " v!! = 0 & v!!!

v!!= 1 & ln v!! = x + C

& v!! = Cex & v! = Cex + K& v = Cex + Kx + D

19. If xe2x is a solution, then so is e2x. So y = ve2x

y!!! " 5y!! + 8y! " 4y = e2x(v!!! + v!!) = 0 & v!!!

v!!= "1 & v!! = Ce"x

& v! = Ce"x + K & v = Ce"x + Kx + D

20. y!!! + 5y!! = 0 & y!!!

y!! = "5 & y!! = Ce"5x & y = Ce"5x + Kx + D is another solution.

21. The substitution y = 2v doesn’t simplify the ODE any. So just going ahead straightforward

gives: y!!! " 3y!! = 0 & y!!!

y!! = 3 & y!! = Ce3x

& y! = Ce3x + K, y = Ce3x + Kx + D

22. For simplicity, all functions depend on x. Thus we have:

Page 110: MATH 224 Sol.

104 Section 3.4

a0fw!

w= "2a0f

! " a1f

w!

w=

"2a0f !

a0f" a1f

a0f

= "2f !

f" a1

a0

ln w = "2 ln f "%

a1

a0+ C

= ln f"2 "%

a1

a0+ C

w = Cexp,"@

a1a0

-

f2

24. (a)d

dxcoshx =

d

dx

!ex + e"x

2

"=!

ex " e"x

2

"= sinhx

(b)d

dxsinh x =

d

dx

!ex " e"x

2

"=!

ex + e"x

2

"= coshx

(c) cosh2 x " sinh2 x =!

e2x

4+

12

+e"2x

4

""!

e2x

4" 1

2+

e"2x

4

"= 1

(d) det8888

coshx sinh xsinh x coshx

8888 = 1 %= 0 so they are linearly independent.

(e) y = c1 coshx + c2 sinh x, y! = c1 sinh x + c2 coshx,y!! = c1 coshx + c2 sinh x, & y!! " y = 0

3.4 Operator Notation

1. P (D) = D " 5

(a) P (D)(3x + 7) = 3 " 5(3x + 7) = "15x" 32(b) P (D)(cos x) = "5 sinx " 5 cosx

(c) P (D)(e5x) = 5e5x " 5e5x = 0

2. P (D) = D + 1

(a) P (D)(e"x) = "e"x + e"x = 0(b) P (D)(4 + sin x) = cosx + 4 + sin x

(c) P (D)(e"2x) = "2e"2x + e"2x = "e"2x

3. P (D) = D2 " 1

(a) P (D)(4e"x + e2x) = 3e2x

(b) P (D)(x3 + 4) = 6x " x3 " 4(c) P (D)(ex " 5e"x) = 0

4. P (D) = D2 + 1

Page 111: MATH 224 Sol.

Section 3.4 105

(a) P (D)(x3 + 2x) = x3 + 8x

(b) P (D)(cos x + sin x) = 0(c) P (D)(sin 2x) = "3 sin2x

5. P (D) = D2 " 4D + 5

(a) P (D)(e2x cosx) = e2x ("4 sinx + 3 cosx " 8 cosx + 4 sinx + 5 cosx)= 0

(b) P (D)(4x2 + ex) = 2ex + 20x2 " 32x + 8(c)

(D2 " 4D + 5)(x3 " cos 2x) = D2(x3 " cos 2x)"4D(x3 " cos 2x) + 5(x3 " cos 2x)

= (6x + 4 cos 2x) " 4(3x2 + 2 sin 2x)+(5x3 " 5 cos 2x)

6. P (D) = D2 + 2D + 1

(a) P (D)(ex) = 4ex

(b) P (D)(xe"x) = 0(c) P (D)(sin x) = 2 cosx

7.

[D3 + 2D](D2 " D)(x6 " 2x2) = [D3 + 2D](D6x5 " Dx4 " 6x5 + 4x)= (D3 + 2D)(30x4 " 4 " 6x5 + 4x)= 720x" 0 " 360x2 " 0

+240x3 " 0 " 60x4 + 8= "60x4 + 240x3 " 360x2 + 720x + 8

(D2 " D)(D3 + 2D)(x6 " 2x2) = (D2 " D)(120x3 " 0 + 12x5 " 8x)= (720x + 240x3 " 360x2 " 60x4 + 8)

Both P (D)Q(D)(y) and Q(D)P (D)(y) yield the same answer.

8. PQ(y) = QP (y) = 3x2 " 8x + 3 sin x " 5 cosx

9. PQ(y) = 12x2 + 3QP (y) = 6x

10. PQ(y) = 3x cosx + 2 sinxQP (y) = 3x cosx " sinx

11. PQ(y) = ex(2x + 7)QP (y) = 2ex(x + 3)

12. PQ(y) = 5x3 + 15x2 + 21x + 21QP (y) = 5x3 + 12x2 + 21x + 16

13. PQ(y) = e"x2(2x2 " 1)

QP (y) = 2e"x2(x2 " 1)

Page 112: MATH 224 Sol.

106 Section 3.5

14. PQ(y) = 1 " x sin xQP (y) = " cosx " x sin x

15. D3(D " 2)(D + 3)(x2 " 2x + 7) = (D " 2)(D + 3)D3(x2 " 2x + 7) = 0

16. 8e2x (The only function involved that has 3 derivatives is e2x)

17. (a) (D " 3)(D + 2)

(b)

/D "

/11 +

'73

2

00/D "

/11 "

'73

2

00

(c) (D2 + 1)2

18. (a) (D2 " 1)2

(b) D(D2 " D + 4)(c) D(D " 4)(D " 1)

19. (a) (D " 1)(D2 + D + 1)(b) (D + 2)(D2 " 2D + 4)

(c) (D " 1)

/D "

/1 +

'5

2

00/D "

/1 "

'5

2

00

20. (a) (D2 + 10D + 16)(y) = (D + 8)(D + 2)(y) = 0(b) (D5 + 4D3 + 4D)(y) = D(D2 + 2)2(y) = 0

21. (a) (D3 + 27)(y) = (D + 3)(D2 " 3D + 9)(y) = 0(b)

y!! + 3y! " 4y = 0 $ (D2 + 3D " 4)(y) = 0$ (D + 4)(D " 1)(y) = 0

22. (a) (D3 " 8)(y) = (D " 2)(D + 2D + 4)(y) = 0(b)

y!!! + 6y!! + 9y! = 0 $ (D3 + 6D2 + 9D)(y) = 0$ D(D + 3)2(y) = 0

23. See discussion on page 187.

3.5 Numerical Consideration for nth Order Equations

1. (a)u!

1 = u2

u!2 = "u2 + x(1 " 2u1)

andu1("1) = 1

u2("1) = 0

(b) h 1 .2 .01y(5) .434067 .433995 .433995

Page 113: MATH 224 Sol.

Section 3.5 107

(c) All plots on same graph

2. (a)u!

1 = u2

u!2 = " 4

7u2 + 37u1

andu1(0) = 0

u2(0) = 1

(b) h 1 .2 .01y(5) 5.96191 5.96191 5.96191

(c) All plots on same graph

3. (a)u!

1 = u2

u!2 = x"3u1

x+2

andu1(0) = 0

u2(0) = 4

(b) h 1 .2 .01y(5) -1.99941 -1.99941 -1.99941

(c) All plots on same graph

4. (a)u!

1 = u2

u!2 = "x2u2 " 12u1

andu1(0) = 0

u2(0) = 7

(b) h 1 .2 .01y(5) 1.76 , 1019 .0111645 .0111255

(c) All plots on same graph

Page 114: MATH 224 Sol.

108 Section 3.5

5. (a)u!

1 = u2

u!2 = 1 " 4u2 " 3 sinu1

andu1(0) = "1

u2(0) = !

(b) h 1 .2 .01y(5) .335637 .335636 .335636

(c) All plots on same graph

6. (a)u!

1 = u2

u!2 = ex"3u1

x2+2

andu1(0) = 1

u2(0) = 2

(b) h 1 .2 .01y(5) 5.73168 5.7317 5.7317

(c) All plots on same graph

7. (a)u!

1 = u2

u!2 = 3x"u2 sin x"u1

x3

andu1(1) = 1

u2(1) = 32

(b) h 1 .2 .01y(5) 8.90804 8.90804 8.90804

(c) All plots on same graph

Page 115: MATH 224 Sol.

Section 3.5 109

8. (a)u!

1 = u2

u!2 = sin x " 2u2 " 10u1

andu1(!) = e

u2(!) = 1

(b) h 1 .2 .01y(5) .0825329 .0825356 .0825358

(c) All plots on same graph

9. (a)u!

1 = u2

u!2 =u3

u!3 = ex " xu3u2 " u1

and

u1(0) = 0

u2(0) = 1

u3(1) = 1

(b) h 1 .2 .01y(5) "1.09 , 1029732975 "1.99, 102135498 19.5423

(c) All plots on same graph

!

!

!

!

10. (a)u!

1 = u2

u!2 =u3

u!3 = e"x " xu3u2 " sin u1

and

u1(1) = 1

u2(1) = 0

u3(1) = 1

(b) h 1 .2 .01y(5) 4.91114 4.91114 4.91114

(c) All plots on same graph

Page 116: MATH 224 Sol.

110 Section 3.5

11. (a)u!

1 = u2

u!2 =u3

u!3 = "u3

8

and

u1(0) = 1

u2(0) = 0

u3(0) = 2

(b) h 1 .2 .01y(5) 21.5135 21.5135 21.5135

(c) All plots on same graph

12. (a)u!

1 = u2

u!2 =u3

u!3 = ln x " xu3 " xu2 " u3

1

and

u1(1) = 0

u2(1) = 0

u3(1) = 1

(b) h 1 .2 .01y(5) 1.42093 1.42093 1.42093

(c) All plots on same graph

13. (a)u!

1 = u2

u!2 =u3

u!3 = e"x2

+ u23 " u2 " xu1

and

u1(0) = 1

u2(0) = 0

u3(0) = 0

(b) h 1 .2 .01y(5) "4.67 , 10588 -1.68357 -1.68355

(c) All plots on same graph

Page 117: MATH 224 Sol.

Section 3.5 111

14. (a)u!

1 = u2

u!2 =u3

u!3 = 8u1

x3

and

u1(1) = 1

u2(1) = 1

u3(1) = 0

(b) h 1 .2 .01y(5) 45.6552 45.6554 45.6554

(c) All plots on same graph

15. (a)u!

1 = u2

u!2 =u3

u!3 = u4

u!4 = "7u4 " 6u3 + 32u2 " 32u1

and

u1(1) = "1

u2(1) = 0

u3(1) = 0

u4(1) = 1

(b) h 1 .2 .01y(5) 66.4945 66.4945 66.4945

(c) All plots on same graph

16. (a)u!

1 = u2

u!2 =u3

u!3 = u4

u!4 = 1

x+2 (u4 " 6u3 " 3u2 + 2u1)

and

u1("1) = 3

u2("1) = "1

u3("1) = 0

u4("1) = 1

(b) h 1 .2 .01y(5) 7.33357 7.33351 7.33351

(c) All plots on same graph

Page 118: MATH 224 Sol.

112 Chapter 3 Review

17. (a)u!

1 = u2

u!2 =u3

u!3 = u4

u!4 = u5

u!5 = 9u2

and

u1(1) = 2

u2(1) = 0

u3(1) = 0

u4(1) = 1

u5(1) = 1

(b) h 1 .2 .01y(5) 79.3336 79.3335 79.3335

(c) All plots on same graph

3.6 Chapter 3: Additional Problems

1. False. We only can conclude that this is a possibility.

2. False. This equation has no problems.

3. True.

4. False. Only constant-coe!cient di"erential operators commute.

5. False.

7. It is guaranteed to have a unique solution through the given IC. The constants are c1 =1 " e"1,c2 = e"1.

9. Linearly independent.

10. Linearly independent.

11. Linearly independent.

12. Linearly independent.

13. Linearly dependent.

14. Linearly dependent.

15. Linearly dependent.

16. Not a fundamental set because x is not a solution.

17. Yes, it forms a fundamental set of solutions.

18. Not a fundamental set because neither is a solution.

19. Not a fundamental set because e"x is not a solution.

Page 119: MATH 224 Sol.

Chapter 3 Review 113

20. Yes, it forms a fundamental set of solutions.

21. Yes, it forms a fundamental set of solutions.

22. Not a fundamental set because only 3 is a solution.

23. Yes, it forms a basis.

24. Yes, it forms a basis.

25. No, it is not a basis because vectors are linearly independent.

26. Yes, it forms a basis.

27. Yes, it forms a basis.

28. No, it is not a basis because vectors are linearly independent.

29. No, it is not a basis because there are more than 3 vectors.

30. No, it is not a basis because the vectors are not in the space.

31. No, it is not a basis because vectors are linearly independent.

35. y = e5x

37. y = e"5x

40. (a) (D " 4)(D " 3)(y) = 0(b) D(D " 4)(D + 4)

Page 120: MATH 224 Sol.

114 Chapter 3 Review

Page 121: MATH 224 Sol.

Chapter 4

Techniques of Higher-Order LinearEquations

4.1 Homogeneous Equations with Constant Coe"cients

1. Assuming y = y(x) or y = y(t),

y!! + 8y! + 12y = 0 $ r2 + 8r + 12 = 0$ (r + 6)(r + 2) = 0$ r = "6,"2

Linearly independent solutions are e"6t, e"2t and the general solution is

y = C1e"6t + C2e

"2t

2.

7r2 + 4r " 3 = 0(7r " 3)(r + 1) = 0

r = "1,37

General solution $ y = c1e"x + c2e

3x7

3.

8y!!! + y!! = 0 $ 8r3 + r2 = 0$ r2(8r + 1) = 0

$ r = 0, 0,"18

We have three linearly independent solutions: e0·x, xe0·x, e"x/8 and the general solution is

y = C1 + C2x + C3e"x/8

4.

y!! " 2y! = 0 $ r2 " 2r = 0$ r(r " 2) = 0$ r = 0, 2

115

Page 122: MATH 224 Sol.

116 Section 4.1

The two linearly independent solutions are {1, e2x} and the general solution is

y = C1 + C2e2x

5. y!! + y! " 12y = 0, y(0) = 0, y!(0) = 7.

r2 + r " 12 = 0 $ (r + 4)(r " 3) = 0 $ r = "4, 3

Two linearly independent solutions are e"4x and e3x, and the general solution is

y = C1e"4x + C2e

3x

y! = "4C1e"4x + 3C2e

3x

Applying initial conditions,

0 = C1 + C2 $ C1 = "C2

7 = "4C1 + 3C2

7 = "4("C2) + 3C2 $ C2 = 1, C1 = "1

Therefore,y = "e"4x + e3x

! !

!

!

6. r = 1, "2 $ y = c1ex + c2e

"2x and IC give y = ex.

! !

7. y!! + 4y! + 4y = 0, y(0) = 0, y!(0) = 1. The characteristic equation is

r2 + 4r + 4 = 0 $ (r + 2)2 = 0$ r = "2,"2$ e"2x, xe"2x

Page 123: MATH 224 Sol.

Section 4.1 117

are two linearly independent solutions. The general solution is

y = C1e"2x + C2xe"2x

y! = "2C1e"2x + C2(e"2x " 2xe"2x)

by the product rule. Applying initial conditions, 0 = y(0) = C1, 1 = y!(0) = "2(0)+C2 yieldsC2 = 1 and

y = xe"2x

! !

!

!

!

!

!

!

!

8. y!! + 4y! + 3y = 0, y(1) = 1, y!(1) = 2. The characteristic equation is

r2 + 4r + 3 = 0 $ (r + 3)(r + 1) = 0$ r = "3,"1$ e"3x, e"x

are two linearly independent solutions. The general solution is then

y = C1e"3x + C2e

"x

Applying initial conditions,

1 = y(1) = C1e"3 + C2e

"1

2 = y!(1) = "3C1e"3 " C2e

"1

Adding the equations we get 3 = "2C1e"3, so C1 = " 32e3 and then

C2 = e1

!1 "!"3

2e3

"e"3

"

= e1

!1 +

32

"=

52e1

Theny = "3

2e"3x+3 +

52e"x+1

Page 124: MATH 224 Sol.

118 Section 4.1

!

!

!

9. 4y!! + 4y! + y = 0, so the characteristic equation is 4r2 + 4r + 1 = 0 or (2r + 1)2 = 0.Thus r = " 1

2 ," 12 . Two linearly independent solutions are e"x/2 and xe"x/2, and the general

solution isy = C1e

"x/2 + C2xe"x/2

10. 2y!!"5y!+2y = 0, so the characteristic equation is 2m2"5m+2 = 0. We have a = 2, b = "5,and c = 2. Roots are

m =5 ±

'25 " 164

$ m = 2,12

Two linearly independent solutions are e2x and ex/2, and the general solution is

y(x) = C1e2x + C2e

x/2

2y!! + 5y! + 2y = 0 $ 2r2 + 5r + 2 = 0 & r = "2, "12

$ y = c1e"2x + c2e

"x/2.

With IC given, y = e"2x

!"2 + !

3

"+ e"x/2

!2 + 4!

3

"

! !

!

!

!

!

11. (D2 " 4D + 5)(y) = 0; the characteristic equation is r2 " 4r + 5 = 0, or

r =4 ±.

16 " 4(5)(1)2

= 2 ± i

Two linearly independent solutions are e2x cosx and e2x sin x, and the general solution is

y = C1e2x cosx + C2e

2x sin x

Page 125: MATH 224 Sol.

Section 4.1 119

12. (D2 + 2D + 10)(y) = 0; the characteristic equation is m2 + 2m + 10. We have a = 1, b = 2,and c = 10, so the roots are

m ="2 ±

'4 " 40

2$ m = "1 ± 3i

Two linearly independent solutions are e"x cos 3x and e"x sin 3x, and the general solution is

y(x) = C1e"x cos 3x + C2e

"x sin 3x

IC give y = "e"!"x

3sin 3x

! ! ! ! !

!

!

!

13. (D2 + 1)2(y) = 0; the characteristic equation is (m2 + 1)2 = 0, the roots are m = ±i,±i, thelinearly independent solutions are cosx, sin x, x cos x, and x sin x, and the general solution is

y(x) = C1 cosx + C2 sinx + C3x cosx + C4x sin x

14. (D2 + 4)y = 0, y(!) = 1, y!(!) = 1; the characteristic equation is r2 + 4 = 0 so r = ±2i andthe two linearly independent solutions are cos 2x, sin 2x. The general solution is then

y = C1 cos 2x + C2 sin 2x

y! = "2C1 sin 2x + 2C2 cos 2x

Applying initial conditions,

1 = y(!) = C1 cos(2!) + C2 sin(2!)1 = y!(!) = "2C1 sin 2! + 2C2 cos 2!

$ C1 = 1, C2 =12

The solution isy = cos 2x +

12

sin 2x

! ! !

!

!

Page 126: MATH 224 Sol.

120 Section 4.1

15. y!!! " 8y = 0; the characteristic equation is r3 " 8 = 0, or (r " 2)(r2 + 2r + 4) = 0, which

leads to r = 2,"2 ±

.4 " 4(4)2

or 2, "1 +'

3 i, and "1 "'

3 i. Three linearly independent

solutions are e2x, e"x cos('

3x), e"x sin('

3x), and the general solution is

y = C1e2x + C2e

"x cos('

3x) + C3e"x sin(

'3 x)

Initial conditions are y(0) = 0, y!(0) = 1, y!!(0) = 0, and the solution is

y =16e2x " 1

6e"x cos(

'3x) +

16'

3 e"x sin('

3x)

! ! !

16. y(4) " y = 0; the characteristic equation is r4 " 1 = 0, which yields

r2 = +1 $ r = ±1r2 = "1 $ r = ±i

Four linearly independent solutions are {cosx, sin x, ex, e"x} and the general solution is

y = C1 cosx + C2 sin x + C3ex + C4e

"x

17. r4 " 16 = 0 & r = ±2, ±2i

y = c1e2x + c2e

"2x + c3 sin 2x + c4 cos 2x

18. (D2 " 2D + 1)(y) = 0; the characteristic equation is r2 " 2r + 1 = 0 or (r " 1)2 = 0, r = 1, 1.Two linearly independent solutions are ex and xex, and the general solution is

y = C1ex + C2xex

19. r4 + 7r3 + 6r2 " 32r " 32 = 0 & r = "1, 2, "4, "4

y = c1e"4x + c2xe"4x + c3e

"x + c4e2x

20. D3(D2"6D+9)y = 0; the characteristic equation is r3(r2 "6r+9) = 0, so r = 0, 0, 0,"3,"3.Five linearly independent solutions are e0x, xe0x, x2e0x, e"3x, and xe"3x. This simplifies as1, x, x2, e"3x, xe"3x. The general solution is then

y = C1 + C2x + C3x2 + C4e

"3x + C5xe"3x

Page 127: MATH 224 Sol.

Section 4.1 121

21. D2(D " 1)2(D2 + 1)(y) = 0; the characteristic equation is r2(r " 1)2(r2 + 1) = 0, which leadsto r = 0, 0, 1, 1,±i. We have 6 linearly independent solutions, 1, x, ex, xex, sinx, and cosx.The general solution is

y = C1 + C2x + C3ex + C4xex + C5 sin x + C6 cosx

22. y(5) " 10y!!! + 9y! = 0; the characteristic equation is r5 " 10r3 + 9r = r(r4 " 10r2 + 9) = 0 orr(r2 " 9)(r2 " 1) = 0, which leads to r = 0,±1,±3. Five linearly independent solutions are 1,ex, e"x, e3x, e"3x. The general solution is

y = C1 + C2ex + C3e

"x + C4e"3x + C5e

"3x

23. y(4) + 2y!! + y = 0; the characteristic equation is m4 + 2m2 + 1 = 0, so (m2)2 + 2(m2) + 1 = 0or (m2 +1)2 = 0. The roots are m = ±i,±i, which leads to four linearly independent solutionscosx, sin x, x cosx, and x sin x. The general solution is then

y(x) = C1 cosx + C2 sinx + C3x cosx + C4x sin x

24. (D2 + 4)2(y) = 0; the characteristic equation is (m2 + 4)2 = 0. The roots are m = ±2i,±2i,so the four linearly independent solutions are cos 2x, sin 2x, x cos 2x, x sin 2x, and the generalsolution is

y(x) = C1 cos 2x + C2 sin 2x + C3x cos 2x + C4x sin 2x

25. r3 " 3r2 + 3r " 1 = 0 & (r " 1)3 = 0 & r = 1, 1, 1

y = c1ex + c2xex + c3x

2ex

26. r3 " r2 " r + 1 = 0 & r = "1, 1, 1

y = c1e"x + c2e

x + c3xex

27. r4 " 5r2 + 4 = 0 & r = ±2, ±1

y = c1e2x + c2e

"2x + c3ex + c4e

"x

28. D(D2 + 4)(D2 " 2D + D)(y) = 0; the characteristic equation is r(r2 + 4)(r2 " r) = 0, sor = 0, 0, 1,±2x. Five linearly independent solutions are 1, x, ex, cos 2x, sin 2x, and the generalsolution is

y = C1 + C2x + C3ex + C4 cos 2x + C5 sin 2x

29. y(5) + 8y!!! + 16y! = 0; the characteristic equation is r5 + 8r3 + 16r = 0 or r(r4 + 8r2 + 16) = 0or r(r2 +4)2 = 0. The roots are r = 0 (once), r = ±2i (twice). Linearly independent solutionsare 1, cos 2t, sin 2t, t cos 2t, t sin 2t. The general solution is thus

y = C1 + C2 cos 2t + C3 sin 2t + C4t cos 2t + C5t sin 2t

30. r4 + 16 = 0 & r ='

2 ± i'

2, "'

2 ± i'

2

y = e%

2x(c1 sin'

2x + c2 cos'

2x) + e"%

2x(c1 sin'

2x + c2 cos'

2x)

Page 128: MATH 224 Sol.

122 Section 4.1

31. r4 + 64 = 0 & r = 2 ± 2i,"2± 2i

y = e2x(c1 sin 2x + c2 cos 2x) + e"2x(c1 sin 2x + c2 cos 2x)

32. r2(r + 2) = 0y!!! + 2y!! = 0

33. Since r1 = 3i is a root, then r4 = "3i must also be a root with multiplicity 2. Hence,(r " 3i)2(r + 3i)2(r " (1 + i))(r " (1 " i)) = 0

y(6) " 2y(5) + 20y(4) " 36y!!! + 117y!! " 162y! + 162y = 0

34. r4(r " (2 + 3i))3(r " (2 " 3i))3 = 0

y(10) " 12y(9) + 87y(8) " 376y(7) + 1131y(6) " 2028y(5) + 2197y(4) = 0

35. (r " (2 + 3i))2(r " (2 " 3i))2(r + 5)(r " 2)3 = 0

y(8) " 9y(7) + 32y(6) + 50y(5) " 939y(4) + 4207y!!! " 10131y!! + 12948y! " 6760y = 0

36. ar2 + br + c = 0 &

r ="2b ±

'4b2 " 4ac

2a=

"b ±'

b2 " ac

a

y = e"bxa

!c1e

x%

b2!aca + c2e

x%

b2!aca

"= e"

bxa#c1e

( + c2e"($

Now,

y = c1 cosh- + c2 sinh-

= c1e( + e"(

2+ c2 e( " e"(

2

= e(!

c1 + c2

2

"+ e"(

!c1 " c2

2

"

= k1e( + k2e

"(

which is the desired form.

37.

ex = 1 + x +x2

2!+

x3

3!+

x4

4!+ · · ·

ei# = 1 + (i$) +(i$)2

2!+

(i$)3

3!+

(i$)4

4!+ · · ·

= 1 + (i$) " $2

2!" i$3

3!+$4

4!+ · · ·

=!

1 " $2

2!+$4

4!+ · · ·

"+ i

!$ " $

3

3!+$5

5!+ · · ·

"

= cos $ + i sin $

38. z1 = ei#, z!1 = iei# = iz, z(0) = 1z2 = cos $ + i sin $, z!2 = "sin$ + i cos $ = iz, z(0) = 1Since this is a first order ODE, there must only be one linearly independent solution. Hence,z1 = Kz2.

Page 129: MATH 224 Sol.

4.2. A MASS ON A SPRING 123

4.2 A Mass on a Spring

1.1.512

k = 12

k = 160

m =12lb

32ft/s2=

38

x!! +160

38

x = 0

r2 +1280

3= 0

r = ±i

91280

3

x(t) = A sin

/t

91280

3

0+ B cos

/t

91280

3

0

x(0) =16, x!(0) = 0

x(t) =16

cos(t9

12803

)

Amplitude =16ft = 2in

Period =2!:1280

3

Frequency =

:1280

3

2!

2.

.5k = 16k = 32

m =16lb

32ft/s2=

12

x!! + 64x = 0r2 + 64 = 0

r = ±8i

x(t) = A sin (8t) + B cos (8t)

x(0) =13, x!(0) = 2

x(t) =14

sin 8t +13

cos 8t

Amplitude =

?!14

"2

+!

13

"2

=15ft

Period =!

4

Frequency =4!

Page 130: MATH 224 Sol.

124 Section 4.2

3.

.5k = 4k = 8

m =4lb

32ft/s2=

18

x!! + 64x = 0r2 + 64 = 0

r = ±8i

x(t) = A sin (8t) + B cos (8t)x(0) = 0, x!(0) = 2

x(t) =14

sin 8t

Amplitude =14ft = 3in

Period =!

4

Frequency =4!

6. 12!

:K( 1

I1+ 1

I2)

7. A = B1"m

k $2

8. I = VR (1 " e

RtL )

11. I = q$CLe"

Rt2L sin&t, CR2 < 4L

14. x!! + x! + x = 0 leads to

r2 + r + 1 = 0

r ="1 ±

.1 " 4(1)(1)2(1)

= "12± i

'3

2

which is underdamped.

!

Page 131: MATH 224 Sol.

Section 4.2 125

15. 4x!! + 12x! + 3x = 0 leads to

b2 " 4mk =14" 4(4)(3) < 0

which is underdamped.

!

16. x!! + 4x! + 3x = 0 leads to

r2 + 4r + 3 = 0(r + 3)(r + 1) = 0

r = "3,"1

which is overdamped.

17. !x!! + 4x! + 2x = 0 leads tob2 " 4mk = 16 " 4(2)(!) < 0

which is underdamped.

Page 132: MATH 224 Sol.

126 Section 4.2

18. x!! + 2x! + x = 0 leads to

r2 + 2r + 1 = 0(r + 1)2 = 0

r = "1,"1

which is critically damped.

19. 3x!! + 7x! + x = 0 leads tob2 " 4mk = 49 " 4(3)(1) = 37

which is overdamped.

20. x!! + 6x! + 9x = 0 leads tob2 " 4mk = 36 " 9(4)(1) = 0

which is critically damped.

Page 133: MATH 224 Sol.

Section 4.2 127

21. Overdamped: x = c1er1t + c2er2t, x! = c1r1er1t + c2r2er2t. Substitute t = 0 and solve forconstants to get a solution

x = ""v0 + x0r2

r1 " r2er1t +

r1x0 " v0

r1 " r2er2t.

Set this equal to 0 (corresponding to crossing through rest position) and solve for t:

t =ln,

r1x0"v0"v0+x0r2

-

r1 " r2

In order for this crossing to be possible, the argument inside the natural log must be nonneg-ative. This gives the desired condition

|v0| > |x0r2|, where r2 = ("b ".

b2 " 4mk)/2m

is the smaller of the characteristic roots and v0 and x0 must have opposite sign.Critically damped: the argument is similar to the above one to obtain |v0| > |bx0/(2m)|where v0 and x0 must again have opposite sign.

23. mx!! + bx! + kx = 0 leads to m = 1, b = 2, and

r2 + 2r + k = 0

r ="2 ±

.22 " 4(1)(k)2(1)

which is underdamped for 4 " 4k < 0, 4 < 4k, 1 < k.

24. m = 1, k = 3 leads to r2 + br + 3 = 0, or

r ="b ±

.b2 " 4(1)(3)2(1)

which is critically damped with b2 " 12 = 0 or b = 2'

3.

25. We need b2 " 4mk > 0 to be overdamped. Then

22 " 4m(1) > 0 $ 4 " 4m > 0$ 4 > 4m

$ 1 > m

26. We need b2 " 4mk < 0 for underdamped. Then

22 " 4m(2) < 0 $ 4 " 8m < 0 $ 4 < 8m $ 12

< m

27 .

!

Page 134: MATH 224 Sol.

128 Section 4.2

28 .

!

!

29 .

!

30. (a) & = 10

!

!

(b) & = 1

!

!

Page 135: MATH 224 Sol.

4.3. CAUCHY-EULER (EQUIDIMENSIONAL) EQUATION 129

(c) It seems & = 1 gives the largest oscillations.

31. (a) & = 10

!

!

(b) & = 1

!

!

(c) It seems & = 1 gives the largest oscillations. Amplitude * 10.

4.3 Cauchy-Euler (Equidimensional) Equation

1.

x2y!! + 4xy! + 2y = 0.

r2 + (4 " 1)r + 2 = 0(r + 2)(r + 1) = 0

r = "2 r = "1y = Ax"2 + Bx"1

2.

x2y!! + 3xy! + y = 0.

r2 + (3 " 1)r + 1 = 0(r + 1)(r + 1) = 0

r = "1 r = "1y = Ax"1 + Bx"1 ln x

Page 136: MATH 224 Sol.

130 Section 4.3

3.

x2y!! + xy! + 4y = 0.

r2 + 4 = 0(r + 2i)(r " 2i) = 0

r = ±2i

y = A sin(2 lnx) + B cos(2 lnx)

4.

x2y!! + 3xy! + 2y = 0.

r2 + 2r + 2 = 0

r ="2 ±

'4 " 8

2= "1 ± i

y = Asin(ln x)

x+ B

cos(ln x)x

5.

x2y!! + 11xy! + 21y = 0.

r2 + 10r + 21 = 0(r + 7)(r + 3) = 0

r = "7 r = "3y = Ax"7 + Bx"3

6.

2x2y!! + xy! " y = 0.

2r2 " r " 1 = 0(2r + 1)(r " 1) = 0

r = "12

r = 1

y = Ax + Bx" 12

7.

2x2y!! " 3xy! + 3y = 0.

2r2 " 5r + 3 = 0(2r " 31)(r " 1) = 0

r = "12

r = 1

y = Ax32 + Bx

Page 137: MATH 224 Sol.

Section 4.3 131

8.

x2y!! " 3xy! + 3y = 0.

r2 " 4r + 3 = 0(r " 3)(r " 1) = 0

r = 3 r = 1y = Ax + Bx3

9.

x2y!! " 3xy! + 4y = 0.

r2 " 4r + 4 = 0(r " 2)2 = 0

r = 2 r = 2y = Ax2 + Bx2 ln x

10.

x2y!! + 3xy! " 2y = 0.

r2 + 2r " 2 = 0

r ="2 ±

'12

2r = "1 ±

'3

y = Ax"1+%

3 + Bx"1"%

3

11.

x2y!! + 5xy! " 3y = 0.

r2 + 4r " 3 = 0

r ="4 ±

'28

2= "2 ±

'7

y = Ax"2+%

7 + Bx"2"%

7

12.

3x2y!! + 3xy! + 9y = 0.

3r2 + 9 = 0r = ±i

'3

y = A sin('

3 ln x) + B cos('

3 ln x)

Page 138: MATH 224 Sol.

132 Section 4.3

13.

7x2y!! + 5xy! + y = 0.

r2 " 2r + 1 = 0

r =2 ±

'"24

14

=1 ± i

'6

7

y = Ax17 sin

/'6

7ln x

0+ Bx

17 cos

/'6

7ln x

0

14.

x2y!! + 5xy! + 8y = 0.

r2 + 4r + 8 = 0

r ="4 ±

'"16

2= "2 ± 2i

y = Ax"2 sin(2 ln x) + Bx"2 cos(2 ln x)

15.

3x2y!! + 13xy! + 11y = 0.

3r2 + 10r + 11 = 0

r ="10 ±

'"32

6

="5 ± i

'8

3

y = Ax" 53 sin(

'8

3ln x) + Bx" 5

3 cos('

83

ln x)

16.

3x2y!! + 5xy! + y = 0.

3r2 + 2r + 1 = 0

r ="2 ±

'"8

6

="1 ± i

'2

3

y = Ax" 13 sin(

'2

3ln x) + Bx" 1

3 cos('

23

ln x)

Page 139: MATH 224 Sol.

4.4. NONHOMOGENEOUS EQUATIONS 133

17. Examine the Wronskian, W (x)

W (x) =8888

y1 ln x y2 ln xy1x + y!

1 ln x y2x + y!

2 ln x

8888

=y1y2 ln x

x+ y1y

!2(ln x)2

"y1y2 ln x

x" y!

1y2(ln x)2

= (lnx)2(y1y!2 " y!

1y2)%= 0

since y1 and y2 are linearly independent for all x %= 1.

4.4 Nonhomogeneous Equations

1. Since 3ex("53 ) = "5ex, we need to multiply Yp = yp("5

3 ) = ex

2

4. Since 2x(A2 ) = Ax, we need to multiply Yp = yp(A

2 ) = Ax2

2 " Ax

5. We multiply y1p by 2, y2p by "12 and y3p by 6 to get

Yp = 2y1p " 12y2p + 6y3p = 3ex " 2x " 43

6. We multiply y1p by 8 and y2p by "3 to get

Yp = 8y1p " 3y2p = 2x " 2 " 3x2

2e"2x

Page 140: MATH 224 Sol.

134 Section 4.4

7. & = 1

!

& = 2

!

& = 3

!

!

From the graphs, resonant frequency happens when & = 2 and the amplitude is about .22.

Page 141: MATH 224 Sol.

Section 4.4 135

8. & = 1

!

& = 2

!

& = 3

!

!

From the graphs, resonant frequency happens when & = 1 and the amplitude is about .4.

Page 142: MATH 224 Sol.

136 Section 4.4

9. & =.

1/18

!

!

& =.

5/18

!

& =.

1/2

!

From the graphs, resonant frequency happens when & =.

1/2 and the amplitude is about .6.

Page 143: MATH 224 Sol.

Section 4.4 137

10. & =.

5/9

!

!

& =.

7/9

!

!

& =.

10/9

!

!

From the graphs, resonant frequency happens when & =.

10/9 and the amplitude is about.45.

Page 144: MATH 224 Sol.

138 Section 4.5

11. & = 1

!

& ='

2

!

& ='

3

!

From the graphs, resonant frequency happens when & ='

3 and the amplitude is about .25.

4.5 Method of Undetermined Coe"cients via Tables

1. y!!+y = e"x+x2; homogeneous solution r2+1 = 0 $ r = ±i; fundamental set is {cosx, sin x}.Particular UC sets: {e"x}, {x2, x, 1}, so no modification is necessary.

yp = Ae"x + Bx2 + Cx + E

Page 145: MATH 224 Sol.

Section 4.5 139

2. y!! + 3y! = x2e2x + cosx; homogeneous solution r2 + 3r = 0 $ r = 0, "3; fundamental set is{1, e"3x}. Particular UC sets: {e2x, xe2x, x2e2x},{cosx, sin x}, so no modification is necessary.

yp = Ae2x + Bxe2x + Cx2e2x + D cosx + E sin x

3. y!! + y = 4 sinx + ex cosx; homogeneous solution r2 + 1 = 0; linearly independent solutions{cosx, sin x}, yc = C1 cosx + C2 sin x. Particular solution:

4 sinx $ {sinx, cosx} $ modify as {x cosx, x sin x}ex cosx $ {ex cosx, ex sin x} $ no modification necessary

Henceyp = A1x cosx + A2x sin x + B1e

x cosx + B2ex sinx

4. y!!+4y!+y = sin 2x+cosx; homogeneous solution r2+4r+1 = 0 $ r = "2±'

3; fundamentalset is {e("2+

%3)x, e("2"

%3)x}. Particular UC sets: {sin 2x, cos 2x}, {cosx, sin x}, so no

modification is necessary.

yp = A sin 2x + B cos 2x + C sinx + D cosx

5. y!!! + y = xe2x cosx + sinx; UC set is {sinx, cos x, xe2x cosx, xe2x sin x, e2x cosx, e2x sin x}.From the homogeneous part, r3 + 1 = (r + 1)(r2 " r + 1). The linearly independent solutionsare e"x, ex/2 cos

%3

2 x, ex/2 sin%

32 x. Thus

yp = A sinx + B sin x + Cxe2x cosx + Exe2x sin x + Fe2x cosx + Ge2x sinx

6. y!! " y = ex + xe"x; homogeneous solution r2 " 1 = 0 $ r = ±1; fundamental set is{ex, e"x}. Particular solution: UC sets are {ex}, {xe"x, e"x}, both need modification, to{xex}, {x2e"x, xe"x}. Thus

yp = Axex + Bx2e"x + Cxe"x

7. y!!! + 8y = 4 sinx + cosx; homogeneous solution r3 + 8 = 0 or (r + 2)(r2 + 2r + 4) = 0, sor = "2, 1 ±

'3 i. Linearly independent solutions are {e"2x, ex cos

'3x, ex sin

'3x}, and

yc = C1e"2x + C2e

x cos'

3 x + C3ex sin

'3 x

Particular solution: 4 sinx + cosx $ {sinx, cos x} (UC set), no modification necessary. Thus

yp = A sinx + B cosx

8. y(5) + y!! = x3 + xex cosx; homogeneous solution

r5 + r2 = 0r2(r3 + 1) = 0

r2(r + 1)(r2 " r + 1) = 0

r = 0, 0,"1,1 ±

'3 i

2

Page 146: MATH 224 Sol.

140 Section 4.5

Fundamental set is {1, x, e"x, ex/2 cos%

32 x, ex/2 sin

%3

2 x}. Particular solution: UC sets {x3, x2, x, 1},{xex cosx, xex sin x, ex cosx, ex sin x}; the first one needs modification. Multiplying by x2 willgive linearly independent solutions, so {x5, x4, x3, x2}. Then

yp = Ax5 + Bx4 + Cx3 + Ex2 + Gxex cosx

+ Hxex sin x + Iex cosx + Jex sin x

9. y!! + 4y! + 13y = e"2x. We have

yp = Ae"2x $ y!p = "2Ae"2x, y!!

p = 4Ae"2x

y!!p + 4y!

p + 13yp = (4Ae"2x) + 4("2Ae"2x) + 13(Ae"2x) = e"2x

$ 4A " 8A + 13A = 1 $ 9A = 1 $ A =19

10. y!! + 4y! = x2 " 3, yp = Ax3 + Bx2 + Cx. Substitution gives

12Ax2 + (8B + 6A)x + 4C + 2B = x2 " 3

Equating coe!cients of like terms,

12A = 18B + 6A = 04C + 2B = "3

so A = 112 , B = " 1

16 , C = " 2332 .

11. y!! " 4y! = cosx + 3 sinx, yp = A cosx + B sin x. Substitution gives

("A " 4B) cosx + ("B + 4A) sin x = cosx + 3 sinx

Equating coe!cients of like terms,

"A " 4B = 1"B + 4A = 3

so A = 1117 , B = " 7

17 .

12. A = 1/8, B = 0, C = 0, E = "1/35

13. A = 0, B = 8/441, C = 1/21, E = 0

14. y!! + 2y! + 17y = ex + 2, yp = Aex + B. Substituting into the ODE and equating coe!cientsgives 20Aex + 17B = ex + 2, so A = 1

20 , B = 217 .

15. A = "5/36, B = "1/6, C = 1/5, E = 3/50

16. yc = Ae2x + Bex

UC sets:{x2, x, 1} No modificationyp = Cx2 + Dx + E and equating coe!cients givesyp = 2x2 + 6x + 7

Page 147: MATH 224 Sol.

Section 4.5 141

17. y!! " 2y! " 8y = 4e2x " 21e"3x; homogeneous solution r2 " 2r " 8 = 0 or (r " 4)(r + 2) = 0, sor = 4,"2. The fundamental set is {e4x, e"2x}, so yh = C1e4x + C2e"2x. Particular solution:UC set is {e2x, e"3x}, which does not need modification, so yp = Ae2x + Be"3x. Substitutionand simplification gives

"8Ae2x + 7Be"3x = 4e2x " 21e"3x

Equating

"8A = 47B = "21

gives A = " 12 , B = "3, so

yp = "12e2x " 3e"3x

The general solution is

y = C1e4x + C2e

"2x " 12e2x " 3e"3x

18. yc = Ae"x sin 2x + Be"x cos 2xUC sets:{cos 2x, sin 2x} No modificationyp = C cos 2x + D sin 2x and equating coe!cients givesyp = " cos 2x + 2 sin 2x

19. yc = Ae"4x + Be"3x cosx + Ce"3x sin xUC sets:{xe"4x, e"4x} modify & {x2e"4x, xe"4x}{e"3x cosx, e"3x sinx} modify & {xe"3x cosx, xe"3x sin x}yp = Dx2e"4x + Exe"4x + Fxe"3x cosx + Gxe"3x sin x and equating coe!cients gives

yp =14x2e"4x +

12xe"4x " 1

2xe"3x cosx +

12xe"3x sin x

20. y!! " 4y = 32x, y(0) = 0, y!(0) = 6. Homogeneous solution: r2 " 4 = 0 so r = ±2. Thefundamental set is {e2x, e"2x}, yh = C1e2x + C2e"2x. Particular solution: UC set is {x, 1},and does not need modification, so yp = Ax + B. Substitution gives

"4Ax " 4B = 32x $ A = "8, B = 0$ yp = "8x

General solution:

y = C1e2x + C2e

"2x " 8x

y! = 2C1e2x " 2C2e

"2x " 8

Applying initial conditions,

0 = y(0) = C1 + C2

6 = y!(0) = 2C1 " 2C2 " 8

so C1 = 72 and C2 = " 7

2 . The solution is

y =72e2x " 7

2e"2x " 8x

Page 148: MATH 224 Sol.

142 Section 4.5

21. y!! " 2y! + 2y = ex + x cosx. Homogeneous solution:

m2 " 2m + 2 = 0

$ m =2 ±.

4 " 4(2)2

=2 ±

'"4

2= 1 ± i

yh = C1ex cosx + C2e

x sinx

Particular solution: UC set is {ex, x cosx, x sin x, cosx, sin x}, so

yp = Aex + Bx cosx + Cx sin x + D cosx + E sin x

Substituting into the ODE and equating coe!cients:

ex: A = 1sin x: 2B + E " 2C + 2D = 0cosx: 2C + D " 2B " 2E = 1

x sin x: C + 2B = 0x cosx: B " 2C = 1

Hence A = 1, B = 15 , C = " 2

5 , D = " 125 , and E = " 28

25 , so

y = C1ex cosx + C2e

x sin x + ex +15x cosx " 2

5x sin x " 1

25cosx " 28

25sin x

22. y!! +6y! +10y = 3xe"3x "2e3x cosx, y(0) = 1, y!(0) = "2. Homogeneous solution: m2 +6m+10 = 0, so

m ="6 ±

.36 " 4(10)2

="6 ± 2i

2= "3 ± i

soyh = C1e

"3x cosx + C2e"3x sin x

Particular solution: UC set is {xe"3x, e"3x, e3x cosx, e3x sin x}, so

yp = Bxe"3x + Ce3x cosx + De3x sin x

Substituting into the ODE and equating coe!cients:

xe"3x: B = 3e3x cosx: 36C + 12D = "2e3x sin x: "12C + 36D = 0

so B = 3, C = " 120 , D = " 1

60 , giving

y = C1e"3x cosx + C2e

"3x sinx + 3xe"3x " 120

e3x cosx " 160

e3x sin x

Applying initial conditions,

1 = C1 "120

"2 = "3C1 + C2 +176

so C1 = 2120 , C2 = " 101

60 , leading to

y =2120

e"3x cosx " 10160

e"3x sinx + 3xe"3x " 120

e3x cosx " 160

e3x sin x

Page 149: MATH 224 Sol.

Section 4.5 143

23. yc = Ae4x cos 2x + Be4x sin xUC sets:{xe4x sin 2x, e4x sin 2x, xe4x cos 2x, e4x cos 2x} modify& {x2e4x sin 2x, xe4x sin 2x, x2e4x cos 2x, xe4x cos 2x}yp =

516

xe4x sin x " 58x2e4x cos 2x

24. yc = Ae"5x + Be"2x

UC sets:{xe"2x cos 5x, e"2x cos 5x, xe"2x sin 5x, e"2x sin 5x} No modificationyp = Cxe"2x cos 5x + De"2x cos 5x + Exe"2x sin 5x + Fe"2x sin 5xyp = "1

34 xe"2x cos 5x + 637225e"2x cos 5x + 3

170xe"2x sin 5x + 5578e"2x sin 5x

25. yc = Aex cos 2x + Bex sin 2xUC sets:{xex, ex} No modification{ex cos 2x, ex sin 2x} modify & {xex cosx, xex sin x}yp = Cxex + Dex + Exex cos 2x + Fxex sin 2x

yp =12xex " 1

4xex sin 2x

26. yc = Aex + Bxex

UC sets:{xex, ex} modify & {x3ex, x2ex}{ex sin 2x, ex cos 2x} No modificationyp = Cx3ex + Dx2ex + Eex cos 2x + Fex sin 2x

y =x3

3ex " ex

4sin 2x

27. yc = Ae2x + Bex

UC sets:{ex} modify & {xex}yp = Cxex

yp = "xex

y = Ae2x + Bex " xex, y(0) = 1, y!(0) = 0 & A = 0, B = 1y = ex " xex

28. yc = Aex + Be"x

UC sets:{ex} modify & {xex}{e"x} modify & {xe"x}yp = Cxex + Dxe"x

yp = xex + xe"x

29. yc = Ae"3x + Be"x

UC sets:{ex} No modification{e"x} modify & {xe"x}yp = Cex + Dxe"x

yp =ex

16+

xe"x

430. yc = A cos 2x + B sin 2x

UC sets:

Page 150: MATH 224 Sol.

144 Section 4.5

{ex sin 2x, ex cos 2x} No modification{e"x sin 2x, e"x cos 2x} No modificationyp = Cex sin 2x + Dex cos 2x + Ee"x sin 2x + Fe"x cos 2xyp = 1

34ex sin 2x " 217ex cos 2x " 1

34e"x sin 2x " 217e"x cos 2x

31. yc = Ae"x sin x + Be"x cosxUC sets:{ex sin x, ex cosx} No modification{e"x sin x, e"x cosx} Modify & {xe"x sin x, xe"x cosx}yp = Cex sin x + Dex cosx + Ee"x sin x + Fe"x cosxyp = 1

16ex sin x " 116ex cosx " 1

4xe"x cosx

32. y(4) " 18y!! + 81y = e3x; Homogeneous solution: r4 " 18r2 + 81 = 0, or (r2 " 9)2 = 0, sor = 3, 3,"3,"3. The four linearly independent solutions are {e3x, xe3x, e"3x, xe"3x}, giving

yc = C1e3x + C2xe3x + C3e

"3x + C4xe"3x

Particular solution: e3x leads to the UC set {e3x}, which needs to be modified as {x2e3x}, so

yp = Ax2e3x

Substitution into the ODE gives

72Ae3x = e3x $ 72A = 1 $ A =172

Thus yp = 172x2e3x and the general solution is

y = yc + yp = C1e3x + C2xe3x + C3e

"3x + C4xe"3x +172

x2e3x

33. (a) yc = Ae4x, UC = {x2, x, 1}

yp = Bx2 + Cx + D & yp = "x2

4" x

8" 1

32(b) yc = Ae"x, UC = {cos 2x, sin 2x}

yp = B cos 2x + C sin 2x & yp =15

cos 2x +25

sin 2x

(c) yc = Aex, UC = {e4x}

yp = Be4x & yp =e4x

336.

A(w) =F0m.

4'2w2 + (w2 " w20)2

A!(w) = "F0

m

!8'2w + 4w(w2 " w2

0)4'2w2 + (w2 " w2

0)3

"

A!(w) = 0.

0 = 8'2w + 4w3 " 4ww20

= 4w3 + 4w(2'2 " w20)

= 4w(w2 + 2'2 " w20)

w = 0 or w2 = w20 " 2'2

w =:

w20 " 2'2

Page 151: MATH 224 Sol.

Section 4.5 145

37. (b) If b = 0, the terms of yc must involve sin $ and cos $ $ limx#$(y2 " y1) may not exist.(c) yc has terms involving e"Ax, UC= {1} & cyp = k & yp = k

c . Therefore y = yp + yc =kc +terms of e"Ax and limx#$ y = k

c

(d) r(ar + b) = 0 & yc = A + Be"ba x

UC= {1} modify & {x} & yp = Cx, bC = k & C = k/b & yp = kb x $ y =

A + Be"ba x + k

b x

(e) ay!! = k & y!! = ka & y! = k

ax + C & y = k2ax2 + Cx + D

limx#$ y = #

38. (a) y!!+ay!+by =Rn

i=0 cnxn. Homogeneous solution from r2+ar+b = 0 & r = "a±%

a2"4b2 =

±-. Thus yc = Ae(x + Be"(x. UC= {xn, xn"1, . . . , 1} $ yp =Rn

i=0 dnxn

(b) yc = Ae"2x + Be"x

UC= {x2, x, 1}yp = "x2 + 4x " 1

2

39. (a)

0 / x / 1 | x > 1yc = A cosx + B sin x | yc = A cosx + B sin x

UC = {x, 1} | UC = {1}yp = Cx + D | yp = C

0 | 0y!!

p = 0 | yp = 1yp = x | yp = 1

y = A cosx + B sin x + x | y = A cosx + B sin x + 1

y =O

A cosx + B sin x + x, 0 / x / 1A cosx + B sin x + 1, x > 1

(b)

y(0) = A = 0y!(0) = B + 1 = 1 & B = 0

y =O

x, 0 / x / 11, x > 1

Page 152: MATH 224 Sol.

146 Section 4.6

4.6 Method of Undetermined Coe"cients via the Annihila-tor Method

1.

y!! + y = e"x + x2

(D2 + 1)y =(Annihilator or rhs = (D + 1)(D3)

$

(D2 + 1)(D + 1)(D3)y = 0r3(r + 1)(r2 + 1) = 0

$ y = A + Bx + Cx2 + De"x + E cosx + F sin x

yc = a cosx + b sin x

$ yp = A + Bx + Cx2 + De"x

2.

y!! + 3y! = x2e2x + cosx

(D2 + 3D)y =(Annihilator or rhs = (D " 2)3(D2 + 1)

$

(D2 + 3D)(D " 2)3(D2 + 1)y = 0r(r + 3)(r " 2)3(r2 + 1) = 0

$ y = A + Be"3x + Ce2x + Dxe2x

+Ex2e2x + F cosx + G sin x

yc = a + be"3x

$ yp = Ce2x + Dxe2x + Ex2e2x + F cosx + G sin x

3.

y!! + y = 4 sinx + ex cosx

(D2 + 1)y =(Annihilator or rhs = (D2 " 2D + 2)(D2 + 1)

$

(D2 + 1)2(D2 " 2D + 2)y = 0(r2 + 1)2(r2 " 2r + 2) = 0

$ y = A sinx + B cosx + Cx sin x + Dx cos x

+Eex sin x + Fex cosx

yc = a cosx + b sinx

$ yp = Cx sin x + Dx cosx

+Eex sin x + Fex cosx

Page 153: MATH 224 Sol.

Section 4.6 147

4.

y!! + 4y! + y = sin 2x + cosx

(D2 + 4D + 1)y =(Annihilator or rhs = (D2 + 1)(D2 + 4)

$

(D2 + 1)(D2 + 4)(D2 + 4D + 1)y = 0(r2 + 1)(r2 + 4)(r2 + 4r + 1) = 0

$ y = A cosx + B sin x

+C cos 2x + D sin 2x

+Ee("2+%

3)x + Fe("2"%

3)x

yc = ae("2+%

3)x + be("2"%

3)x

$ yp = A cosx + B sin x

+C cos 2x + D sin 2x

5.

y!!! + y = xe2x cosx + sinx

(D3 + 1)y =(Annihilator or rhs = (D2 + 1)(D2 " 4D + 5)2)

$

(D2 + 1)(D2 " 4D + 5)2(D3 + 1)y = 0(r2 + 1)(r2 " 4r + 5)2(r3 + 1) = 0

$ y = Ae"x + B cosx + C sin x

+De2x cos 2x + Ee2x sin 2x

+Fex/2 cos('

32

)x

+Gex/2 sin('

32

)x

yc = ae"x + bex/2 cos('

32

)x

+cex/2 sin('

32

)x

$ yp = B cosx + C sin x

+De2x cos 2x + Ee2x sin 2x

6.

y!! " y = ex + xe"x

(D2 " 1)y =(Annihilator or rhs = (D " 1)(D + 1)2

$

(D2 " 1)(D " 1)(D + 1)2y = 0(r " 1)2(r + 1)3 = 0

$ y = Aex + Bxex

+Ce"x + Dxe"x + Ex2e"x

yc = aex + be"x

$ yp = Bxex + Dxe"x + Ex2e"x

Page 154: MATH 224 Sol.

148 Section 4.6

7.

y!!! + 8y = 4 sinx + cosx

(D3 + 8)y =(Annihilator or rhs = (D2 + 1)

$

(D2 + 1)(D3 + 8)y = 0(r3 + 8)(r2 + 1) = 0

$ y = A cosx + B sinx + Ce"2x

+Dex cos'

3x + Eex sin'

3xyc = ae"2x + bex cos

'3x + cex sin

'3x

$ yp = A cosx + B sinx

8.

y(5) + y!! = xex cosx + x3

(D5 + D2)y =(Annihilator or rhs = D4(D2 " 2D + 2)2

$

(D5 + D2)(D4)(D2 " 2D + 2)2y = 0(r5 + r2)(r4)(r2 " 2r + 2)2 = 0

$ y = A + Bx + Cx2 + Dx3 + Ex4 + Fx5

+Gex cosx + Hxex cosx

+Iex sin x + Jxex sin x

+Kex2 cos(

'3

2)x + Le

x2 sin(

'3

2)x

+Me"x

yc = a + bx + cex2 cos(

'3

2)x

+dex2 sin(

'3

2)x + fe"x

$ yp = Cx2 + Dx3 + Ex4 + Fx5

+Gex cosx + Hxex cosx

+Iex sin x + Jxex sin x

9. y!! + 4y! + 13y = e"2x. We have

yp = Ae"2x $ y!p = "2Ae"2x, y!!

p = 4Ae"2x

y!!p + 4y!

p + 13yp = (4Ae"2x) + 4("2Ae"2x) + 13(Ae"2x) = e"2x

$ 4A " 8A + 13A = 1 $ 9A = 1 $ A =19

10. y!! + 4y! = x2 " 3, yp = Ax3 + Bx2 + Cx. Substitution gives

12Ax2 + (8B + 6A)x + 4C + 2B = x2 " 3

Page 155: MATH 224 Sol.

Section 4.6 149

Equating coe!cients of like terms,

12A = 18B + 6A = 04C + 2B = "3

so A = 112 , B = " 1

16 , C = " 2332 .

11. y!! " 4y! = cosx + 3 sinx, yp = A cosx + B sin x. Substitution gives

("A " 4B) cosx + ("B + 4A) sin x = cosx + 3 sinx

Equating coe!cients of like terms,

"A " 4B = 1"B + 4A = 3

so A = 1117 , B = " 7

17 .

12. A = 1/8, B = 0, C = 0, E = "1/35

13. A = 0, B = 8/441, C = 1/21, E = 0

14. y!! + 2y! + 17y = ex + 2, yp = Aex + B. Substituting into the ODE and equating coe!cientsgives 20Aex + 17B = ex + 2, so A = 1

20 , B = 217 .

15. A = "5/36, B = "1/6, C = 1/5, E = 3/50

16. y!! " 3y! + 2y = 4x2; Homogeneous solution: r2 " 3r + 2 = 0 or (r " 2)(r " 1) = 0, so r = 2, 1.Two linearly independent solutions are e2x, ex. We have P (D) = (D " 2)(D " 1). What givesrise to x2 on the right-hand side?

r = 0 (3 times) $ Q(D) = D3

Q(D)P (D)(y) = 0 $ D3(D " 2)(D " 1)(y) = 0$ r3(r " 2)(r " 1) = 0

Roots are then r = 0 (3 times), 2, 1, which gives the linearly independent solutions 1, x, x2,e2x, ex. Ignoring those that were part of the homogeneous solution,

yp = a1 + a2x + a3x2

Then

y!p = a2 + 2a3x

y!!p = 2a3

Substituting into the ODE we get

(2a3) " 3(a2 + 2a3x) + 2(a1 + a2x + a3x2) = 4x2

Equating coe!cients gives

2a3 " 3a2 + 2a1 = 0"6a3 + 2a2 = 0

2a3 = 4

Thus a3 = 2, a2 = 6, and a1 = 7. The general solution is then

yh + yp = C1e2x + C2e

x + 7 + 6x + 2x2

Page 156: MATH 224 Sol.

150 Section 4.6

17. y!! " 2y! " 8y = 4e2x " 21e"3x; homogeneous solution r2 " 2r " 8 = 0 or (r " 4)(r + 2) = 0, sor = 4,"2. The fundamental set is {e4x, e"2x}, so yh = C1e4x + C2e"2x. Particular solution:UC set is {e2x, e"3x}, which does not need modification, so yp = Ae2x + Be"3x. Substitutionand simplification gives

"8Ae2x + 7Be"3x = 4e2x " 21e"3x

Equating

"8A = 47B = "21

gives A = " 12 , B = "3, so

yp = "12e2x " 3e"3x

The general solution is

y = C1e4x + C2e

"2x " 12e2x " 3e"3x

18. yc = Ae"x sin 2x + Be"x cos 2xyp = " cos 2x + 2 sin 2x

19. yc = Ae"4x + Be"3x cosx + Ce"3x sin x

yp =14x2e"4x +

12xe"4x " 1

2xe"3x cosx +

12xe"3x sin x

20. yc = C1e2x + C2e"2x

yp = "8x

y =72e2x " 7

2e"2x " 8x

21. y!! " 2y! + 2y = ex + x cosx; Homogeneous solution: r2 " 2r + 2 = 0, so

r =2 ±.

4 " 4(2)2

= 1 ± i

Linearly independent solutions are ex cosx, ex sin x. For particular solution, ex comes fromr = 1 $ (D " 1). x cos x comes from r = ±i (twice), which leads to

(r2 + 1)2 = 0 $ (D2 + 1)2

Q(D) = (D " 1)(D2 + 1)2

Q(D)P (D)(y) = 0

Linearly independent solutions that come from the characteristic equation are ex, cosx, sinx,x cosx, x sin x. Ignoring those from the homogeneous solution, ex cosx, ex sinx. This gives

yp = a1ex + a2 cosx + a3 sin x + a4x cosx + a5x sin x

y!p = a1e

x " a2 sinx + a3 cosx + a4(cosx " x sin x) + a5(sin x + x cosx)y!!

p = a1ex " a2 cosx " a3 sin x + a4(" sin x " sinx " x cos x)

+ a5(cosx + cosx " x sin x)

Substituting into the ODE gives

ex(a1 " 2a1 + 2a1) + (cosx)(a2 " 2a5 " 2a3 + 2a4)+ (sin x)(a3 " 2a4 + 2a2 " 2a5) + (x sin x)(a4 + 2a5)+ (x cos x)(a5 " 2a4) = ex + x cosx

Page 157: MATH 224 Sol.

Section 4.6 151

Equating coe!cients,

a1 = 1a2 " 2a3 + 2a4 " 2a5 = 02a2 + a3 " 2a4 " 2a5 = 0

a4 + 2a5 = 0"2a4 + a5 = 1

which gives a1 = 1, a2 = 225 , a3 = " 14

25 , a4 = 15 , a5 = " 2

5 , so

yp = ex +225

cosx " 1425

sin x " 25x sin x +

15x cos x

The general solution isC1e

x cosx + C2ex sinx + yp

22. y!! + 6y! + 10y = 3xe"3x " 2e3x cosx; Homogeneous solution: r2 + 6r + 10 = 0,

r ="6 ±

.36 " 4(10)2

= "3 ± i

Linearly independent solutions e"3x cosx, e"3x sin x. The annihilator of xe"3x is (D +3)2 andof e3x cosx is (D2 " 6D + 10). We calculate the characteristic equation for Q(D)P (D), and,ignoring the linearly independent solutions that occurred in the homogeneous part we havee"3x, xe"3x, e3x cosx, e3x sin x. Thus

yp = a1e"3x + a2xe"3x + a3e

3x cosx + a4e3x sin x

y!p = "3a1e

"3x + a2(e"3x " 3xe"3x) + a3(3e3x cosx " e3x sin x)

+ a4(3e3x sin x + e3x cosx)y!!

p = 9a1e"3x + a2("3e"3x " 3e"3x + 9xe"3x)

+ a3(9e3x cosx " e3x sin x " 3e3x sin x " e3x cosx)+ a4(9e3x sin x + 3e3x cosx + 3e3x cosx " e3x sin x)

Substituting into the ODE and equating coe!cients gives

e"3x: 10a1 " 18a1 + 6a2 + 9a1 " 6a2 = 0xe"3x: 10a2 " 18a2 + 9a2 = 3

e3x cosx: 10a3 + 18a3 + 6a4 + 8a3 + 6a4 = "2e3x sin x: 10a4 " 6a3 + 18a4 " 6a3 + 8a4 = 0

which gives a1 = 0, a2 = 3, a3 = " 120 , a4 = " 1

60 , so

yp = 3xe"3x " 120

e3x cosx " 160

e3x sin x

The general solution isC1e

"3x cosx + C2e"3x sinx + yp

23. yc = Ae4x cos 2x + Be4x sin x

yp =516

xe4x sin x " 58x2e4x cos 2x

Page 158: MATH 224 Sol.

152 Section 4.6

24. yc = Ae"5x + Be"2x

yp = "134 xe"2x cos 5x + 63

7225e"2x cos 5x + 3170xe"2x sin 5x + 5

578e"2x sin 5x

25. yc = Aex cos 2x + Bex sin 2x

yp =12xex " 1

4xex sin 2x

26. yc = Aex + Bxex

y =x3

3ex " ex

4sin 2x

27. yc = Ae2x + Bex

y = ex " xex

28. yc = Aex + Be"x

yp = xex + xe"x

29. yc = Ae"3x + Be"x

yp =ex

16+

xe"x

430. yc = A cos 2x + B sin 2x

yp = 134ex sin 2x " 2

17ex cos 2x " 134e"x sin 2x " 2

17e"x cos 2x

31. yc = Ae"x sin x + Be"x cosxyp = 1

16ex sin x " 116ex cosx " 1

4xe"x cosx

32. yc = C1e3x + C2xe3x + C3e"3x + C4xe"3x

yp = 172x2e3x

33. (a) yc = Ae4x

yp = "x2

4" x

8" 1

32(b) yc = Ae"x

yp =15

cos 2x +25

sin 2x

(c) yc = Aex

yp =e4x

336.

A(w) =F0m.

4'2w2 + (w2 " w20)2

A!(w) = "F0

m

!8'2w + 4w(w2 " w2

0)4'2w2 + (w2 " w2

0)3

"

A!(w) = 0.

0 = 8'2w + 4w3 " 4ww20

= 4w3 + 4w(2'2 " w20)

= 4w(w2 + 2'2 " w20)

w = 0 or w2 = w20 " 2'2

w =:

w20 " 2'2

Page 159: MATH 224 Sol.

Section 4.6 153

37. (a)(b) If b = 0, the terms of yc must involve sin $ and cos $

$ limx#$(y2 " y1) may not exist.(c) yc has terms involving e"Ax, UC= {1} & cyp = k & yp = k

c . Therefore y = yp + yc =kc +terms of e"Ax and limx#$ y = k

c

(d) r(ar + b) = 0 & yc = A + Be"ba x

UC= {1} modify & {x} & yp = Cx, bC = k & C = k/b & yp = kb x $ y =

A + Be"ba x + k

b x

(e) ay!! = k & y!! = ka & y! = k

ax + C & y = k2ax2 + Cx + D

limx#$ y = #

38. (a) y!!+ay!+by =Rn

i=0 cnxn. Homogeneous solution from r2+ar+b = 0 & r = "a±%

a2"4b2 =

±-. Thus yc = Ae(x + Be"(x. yp =Rn

i=0 dnxn

(b) yc = Ae"2x + Be"x

yp = "x2 + 4x " 12

39. (a)

0 / x / 1 | x > 1yc = A cosx + B sin x | yc = A cosx + B sin x

UC = {x, 1} | UC = {1}yp = Cx + D | yp = C

0 | 0y!!

p = 0 | yp = 1yp = x | yp = 1

y = A cosx + B sin x + x | y = A cosx + B sin x + 1

y =O

A cosx + B sin x + x, 0 / x / 1A cosx + B sin x + 1, x > 1

(b)

y(0) = A = 0y!(0) = B + 1 = 1 & B = 0

y =O

x, 0 / x / 11, x > 1

Page 160: MATH 224 Sol.

154 Section 4.7

4.7 Variation of Parameters

1.

yc = cx + d

yp = u1x + u2Ou!

1x + u!2 = 0

u!1 = 1

1+x2

u!1 =

11 + x2

u!2 = " x

1 + x2

u1 = tan"1 x

u2 = ln!

11 + x2

"

yp = x tan"1 x + ln!

11 + x2

"

2.

yc = Aex + Bxex

yp = u1ex + u2xex

Ou!

1ex + u!

2xex = 0u!

1ex + u!

2(ex(x + 1)) = ex'x

u!1 = "x3/2

u!2 =

'x

u1 = "25x5/2

u2 =23x3/2

yp =415

x5/2ex

3.

yc = A sinx + B cosx

yp = u1 sin x + u2 cosxO

u!1 sin x + u!

2 cosx = 0u!

1 cosx " u!2 sin x = sec x

u!1 = 1

u!2 = " tanx

u1 = x

u2 = ln(cosx)yp = x sin x + cosx ln(cosx)y = A sinx + B cosx + x sin x + cosx ln(cosx)

y(0) = B = 1y!(0) = A = 2

y = 2 sinx + cosx + x sin x + cosx ln(cosx)

Page 161: MATH 224 Sol.

Section 4.7 155

4.

yc = A sinx + B cosx

yp = u1 sin x + u2 cosxO

u!1 sin x + u!

2 cosx = 0u!

1 cosx " u!2 sin x = tan x

u!1 = sinx

u!2 = " sinx tan x

u1 = " cosx

u2 = "2 tanh"1,tan,x

2

--+ sin x

yp = "2 tanh"1,tan,x

2

--

y = A sinx + B cosx " 2 tanh"1,tan,x

2

--

y(0) = B = "1y!(0) = A " 1 = 1 & A = 2

y = 2 sinx " cosx " 2 tanh"1,tan,x

2

--cosx

5.

yc = A sin x + B cosx

yp = u1 sin x + u2 cosxO

u!1 sin x + u!

2 cosx = 0u!

1 cosx " u!2 sin x = sec2 x

u!1 = sec x

u!2 = " secx tan x

u1 = " cosx

u2 = 2 tanh"1,tan,x

2

--

yp = 2 tanh"1,tan,x

2

--sin x " 1

y = A sin x + B cosx " 2 tanh"1,tan,x

2

--

y(0) = B " 1 = 0 & B = 1y!(0) = A = 1

y = sin x + cosx + 2 tanh"1,tan,x

2

--sinx " 1

Page 162: MATH 224 Sol.

156 Section 4.7

6.

yc = A sin 2x + B cos 2x

yp = u1 sin 2x + u2 cos 2xO

u!1 sin 2x + u!

2 cos 2x = 02u!

1 cos 2x " 2u!2 sin 2x = sec x

u!1 =

cos 2x

2 sinx

u!2 = " sin 2x

2 sinx= " cosx

u1 = cosx +12

ln,tan

x

2

-

u2 = " sinx

yp = sin 2x cosx + sin x cosx ln,tan

x

2

-" cos 2x sin x

7.

yc = Aex + Bxex

yp = u1ex + u2xex

Ou!

1ex + u!

2xex = 0u!

1ex + u!

2(xex + ex) = ex

x2

u!1 = " 1

x

u!2 =

1x2

u1 = " lnx

u2 = " 1x

yp = "ex ln x " ex

8.

yc = Ae"x + Bxe"x

yp = u1e"x + u2xe"x

Ou!

1e"x + u!

2xe"x = 0"u!

1e"x + u!

2("xe"x + e"x) = e!x

x4

u!1 = " 1

x3

u!2 =

1x4

u1 =1

2x2

u2 = " 13x3

yp =e"x

6x2

Page 163: MATH 224 Sol.

Section 4.7 157

9. y!! " 7y! + 10y = e3x; Homogeneous part: r2 " 7r + 10 = 0, or (r " 2)(r " 5) = 0 and r = 2, 5.Linearly independent solutions are e2x, e5x. Let #1 = e2x, #2 = e5x. Then

u!1 =

det&

0 e5x

e3x 5e5x

'

det&

e2x e5x

2e2x 5e5x

' ="e8x

5e7x " 2e7x=

"ex

3$ u1 = "1

3ex

u!2 =

det&

e2x 02e2x e3x

'

3e7x=

e5x

3e7x=

13e"2x $ u2 = "1

6e"2x

This gives

yp = u1#1 + u2#2 = "13exe2x +

!"1

6

"e"2xe5x = "1

3e3x " 1

6e3x = "1

2e3x

Then the general solution is

y = C1e2x + C2e

5x " 12e3x

y(0) = C1 + C2 " 12 = 1, y!(0) = 5C1 + 2C2 " 3

2 = 2 $ C1 = 16 , C2 = 4

3

$ y =e5x

6+

4e2x

3" e3x

210.

yc = Ae"3x + Be"2x

yp = u1e"3x + u2e

"2x

Ou!

1e"3x + u!

2e"2x = 0

"3u!1e

"3x " 2u!2e

"2x = e"x

u!1 = "e2x

u!2 = ex

u1 = "12e2x

u2 = ex

yp =e"x

2

y = Ae"3x + Be"2x +12e"x

y(0) = A + B +12

= 1

y!(0) = "3A " 2B " 12

= 2

A = "72

B = 4

y = "72e"3x + 4e"2x +

e"x

2

Page 164: MATH 224 Sol.

158 Section 4.7

11.

yc = A sin x + B cosx

yp = u1 sinx + u2 cosxO

u!1 sin x + u!

2 cosx = 0u!

1 cosx " u!2 sin x = 6x

u!1 = 6x cosx

u!2 = "6x sinx

u1 = 6 cosx + 6x sin x

u2 = 6x cosx " 6 sinx

yp = 6x

y = A sin x + B cosx + 6x

y(0) = B = 1y!(0) = A + 6 = 1

A = "5 B = 1y = "5 sinx + cosx + 6x

12.

yc = A sin x + B cosx

yp = u1 sinx + u2 cosxO

u!1 sin x + u!

2 cosx = 0u!

1 cosx " u!2 sin x = sin2 x

u!1 = cosx sin2 x

u!2 = " sin3 x

u1 =sin3 x

3

u2 =cos 3x

12" 3 cosx

4

yp =sin4 x

3" cosx cos 3x

12+

3 cos2 x

4y = A sin x + B cosx + yp

y(0) = B " 112

+34

= 1

y!(0) = A = 0

A = 0 B =13

y =cosx

3+ yp

Page 165: MATH 224 Sol.

Section 4.7 159

13.

yc = Aex + Bxex + Cx2ex

yp = u1ex + u2xex + u3x

2ex

AC

D

u!1e

x + u!2xex + u!

3x2ex = 0

u!1e

x + u!2(xex + ex) + u!

3(x2ex + 2xex) = 0u!

1ex + u!

2(xex + 2ex) + u!3(x2ex + 4xex + 2ex) = ex

x

u!1 =

x

2u!

2 = "1

u!3 =

12x

u1 =x2

4u2 = "x

u3 =12

ln x

yp = "34x2ex +

12x2 ln xex

14.

yc = A + Be"x + Ce2x

yp = u1 + u2e"x + u3e

2x

AC

D

u!1 + u!

2e"x + u!

3e2x = 0

"u!2e

"x + 2u!3e

2x = 0u!

2e"x + 4u!

3e2x = x3

u!1 = "x3

2

u!2 =

exx3

3

u!3 =

e"2xx3

6

u1 = "x4

8

u2 =x3ex

3" x2ex + 2xex " 2ex

u3 = "x3e"2x

12" x2e"2x

8" xe"2x

8" e"2x

16

yp = "x4

8+

x3

4" 9x2

8+

15x

8" 17

16

Page 166: MATH 224 Sol.

160 Section 4.7

15.

yc = A sin 3'

3x + B cos 3'

3xyp = u1 sin 3

'3x + u2 cos 3

'3x

Ou!

1 sin 3'

3x + u!2 cos 3

'3x = 0

3'

3u!1 cos 3

'3x " 3

'3u!

2 sin 3'

3x = e3x

u!1 =

e3x cos 3'

3x

3'

3

u!2 = "e3x sin 3

'3x

3'

3

u1 =e3x

36'

3

,cos 3

'3x +

'3 sin 3

'3x-

u2 =e3x

36'

3

,'3 cos 3

'3x " sin 3

'3x-

yp =e3x

36

16.

yc = A sinx + B cosx + C

yp = u1 sin x + u2 cosx + u3AC

D

u!1 sin x + u!

2 cosx + u!3 = 0

u!1 cosx " u!

2 sin x = 0"u!

1 sinx " u!2 cosx = tanx

u!1 = " sinx tan x

u!2 = " sinx

u!3 = sinx tan x + tan2 x

u1 = sinx " 2 tanh("1),tan

x

2

-

u2 = cosx

u3 = sinx " 2 tanh("1),tan

x

2

-+ tanx " x

yp = 1 " 4 tanh("1),tan

x

2

-+ sinx + tanx " x

Page 167: MATH 224 Sol.

Section 4.7 161

17.

y(4) " 16y = e4x =$ y = C1 sin(2x) + C2 cos(2x) + C3e2x + C4e

"2x +1

240e4x

18. y!! +4y! +3y = 65 cos(2x). The characteristic equation is r2 +4r +3 = 0 or (r +1)(r +3) = 0,so r = "1,"3. This gives

yh = C1e"x + C2e

"3x

Variation of parameters: Let #1 = e"x, #2 = e"3x. Then

u!1 =

det&

0 #2

65 cos(2x) #!2

'

det&#1 #2

#!1 #!2

' ="65e"3x cos(2x)

"2e"4x

u!2 =

det&#1 0#!1 65 cos(2x)

'

det&#1 #2

#!1 #!2

' =65e"x cos(2x)

"2e"4x

u1 =%

u!1 dx =

132

ex (cos(2x) + 2 sin(2x))

u2 =%

u!2 dx = "15

2e3x cos(2x) " 5e3x sin(2x)

This givesyp = u1#1 + u2#2

The general solution is

y = C1e"x + C2e

"3x " cos(2x) + 8 sin(2x)

Undetermined coe!cients: Roots that gave cos(2x) are ±2i, which gives Q(D) = D2 + 4; so

Q(D)P (D)y = 0 $ (r2 + 4r + 3)(r2 + 4) = 0$ r = "1,"3,±2i

$ yp = a1 cos(2x) + a2 sin(2x)

Plug into the ODE and simplify:

("a1 + 8a2) cos(2x) + ("a2 " 8a1) sin(2x) = 65 cos(2x)

Equating coe!cients:

"a1 + 8a2 = 65"a2 " 8a1 = 0

so a2 = "8a1, "a1 = 8("8a1) = 65, so a1 = "1 and a2 = 8. Then

yp = " cos(2x) + 8 sin(2x)

Page 168: MATH 224 Sol.

162 Section 4.7

19. Plugging both functions into the left-hand side of the DE results in a solution to the homoge-neous DE.

yp = u1 + u2(xe"x + e"x)O

u!1 + u!

2(xe"x + e"x) = 0u!

2("xe"x) = ex

1+x

u!1 =

ex

x

u!2 = " e2x

x(1 + x)

u1 =%

ex

xdx

u2 =% "e2x

x(1 + x)yp = u1 + u2(xe"x + e"x)

20. Plugging both functions into the left-hand side of the DE results in a solution to the homoge-neous DE.

yp = u1x + u2x ln xO

u!1x + u!

2x ln x = 0u!

1 + u!2(1 + lnx) = 1

x

u!1 = " ln x

x

u!2 =

1x

u1 = "12

ln2 x

u2 = lnx

yp =x

2ln2 x

21. Plugging both functions into the left-hand side of the DE results in a solution to the homoge-neous DE.

Page 169: MATH 224 Sol.

Section 4.7 163

yp = u1 sin(ln x) + u2 cos(ln x)O

u!1 sin(ln x) + u!

2 cos(ln x) = 0u!

1cos(ln x)

x " u!2

sin(ln x)x = 1

x

u!1 = cos(ln x)

u!2 = " sin(ln x)

u1 =x

2cos(ln x) +

x

2sin(ln x)

u2 =x

2cos(ln x) " x

2sin(ln x)

yp =x

2

22. Plugging all three functions into the left-hand side of the DE results in a solution to the ho-mogeneous DE.

We need to rewrite the DE as:

y!!! +3x

y!! =1

2x2

yp = u1 + u2

!1x

"+ u3x

AC

D

u!1 + u!

2

#1x

$+ u!

3x = 0"u!

2

#1x2

$+ u!

3 = 02u!

2

#1x3

$= 1

2x2

u!1 =

12

u!2 =

x

4

u!3 =

14x

u1 =x

2

u2 =x2

8

u3 =14

ln x

yp =5x

8+

x

4ln x

23. As the exercise is written, u(x) = yc(x) and v(x) = yp(x)

24. Note that yc will come from Cauchy-Euler material earlier, but variation of parameters will

Page 170: MATH 224 Sol.

164 Section 4.7

require dividing everything by x2.

yc =A

x+

B

x2

yp =u1

x+

u2

x2E

u"1

x + u"2

x2 = 0"u"

1x2 " 2u"

2x3 = ex

x2

u!1 = ex

u!2 = "xex

u1 = ex

u2 = "xex + ex

yp =ex

x2

25. Note that yc will come from Cauchy-Euler material earlier, but variation of parameters willrequire dividing everything by x2.

yc =A

x+

B

xln x

yp =u1

x+

u2

xln x

Eu"1

x + u"2

x ln x = 0"u"

1x2 + u!

21"ln x

x2 = ln xx2

u!1 = " ln2 x

u!2 = lnx

u1 = "2x + 2x ln x " x ln2 x

u2 = x ln x " x

yp = lnx " 2

26. Note that yc will come from Cauchy-Euler material earlier, but variation of parameters willrequire dividing everything by x2.

yc = A sin(2 ln x) + B cos(2 ln x)yp = u1 sin(2 lnx) + u2 cos(2 ln x)

Ou!

1 sin(2 lnx) + u!2 cos(2 ln x) = 0

u!1 cos(2 lnx)( 2

x ) + u!2 sin(2 ln x)( 2

x ) = 2x2

u!1 =

cos(2 ln x)x

u!2 = " sin(2 lnx)

x

u1 =sin(2 ln x)

2

u2 =cos(2 ln x)

2

yp =12

Page 171: MATH 224 Sol.

4.8. CHAPTER 4: ADDITIONAL PROBLEMS 165

4.8 Chapter 4: Additional Problems

1. False, this is true with constant coe!cients.

2. True.

3. False, it corresponds to the particular solution.

4. False.

5. False.

6. y(x) = c1e2x + c2e3x

8. y(x) = (3x + 1)e"3x

9. y(x) =12e"4x(3e2x " 1)

11. y(x) =13e2(x"!/2)(5 cos 3x " 6 sin 3x)

13. y(x) = c1e"x + c2 + c3x + c4e4x

15. y(x) = c1e"x + c2xe"x + c3x2e"x + c4

19. y(x) = c1e"7x + c2e"2x + c3 + c4e2x

21. y(x) =c1

x+

c2'x

22. y =c1

x+ c2

'x

23. y = c1 sin#'

2 lnx$

+ c2 cos#'

2 lnx$)

25. y(x) = c1e4x + c2 "

117

(4 sinx + cosx)

27. y(x) = c1e2x + c2xe2x +

116

e"2x

29. y(x) = c1 sin(5x

2) + c2 cos(

5x

2) +

229

e"x

30. yh + yp = C1e"x cos 4x + C2e

"x sin 4x +18xe"x sin 4x

31. y(x) = c1e"5x + c2e

2x " 1100

(10 + 3e3x + 10xe3x)

33. Overdamped.

34. Underdamped.

35. Overdamped.

36. Overdamped.

37. Critically damped.

38. Critically damped.

Page 172: MATH 224 Sol.

166 Chapter 4 Review

39. b < 2'

3

40. b < 4

41. k <18

42. k >94

43. m =14

44. m <4912

45. (a) y(x) = (3/4)e"x + (1/2)xe"x + (1/4)ex

(b) y(4) = 13.70

47. (a) y = (7/6) sin(2x) + cos(2x) + (1/3) sin(x),(b) y(4) = .7565

50. (a) y(x) = c1x + c2%

x

52. y(x) = c1 sin#'

2 lnx$

+ c2 cos#'

2 lnx$

Page 173: MATH 224 Sol.

Chapter 5

Fundamentals of Systems ofDi!erential Equations

5.1 Systems of Two Equations—Motivational Examples

1. (a) ! = !3 " Saddle

(b) ! = 6, " = 5, "2 ! 4! = 1 " Unstable Spiral1.(a) xy-plane 1.(b) xy-plane

2. (a) ! = 5," = 0, " Center(b) ! = 1, " = 0 " Center

2.(a) xy-plane 3.(b) xy-plane

167

Page 174: MATH 224 Sol.

168 Section 5.1

3. (a) ! = !5 " Saddle

(b) ! = !3 " Saddle3.(a) xy-plane 3.(b) xy-plane

4. (a) ! = 5," = 6,"2 ! 4! = 16 " Unstable Node

(b) ! = 13," = 6, "2 ! 4! ! 16 " Unstable Spiral4.(a) xy-plane 4.(b) xy-plane

5. (a) ! = 11," = 4, "2 ! 4! = !28 " Unstable Spiral

(b) ! = 8," = 4,"2 ! 4! = !16 " Unstable Spiral5.(a) xy-plane 5.(b) xy-plane

Page 175: MATH 224 Sol.

Section 5.1 169

6. (a) ! = 8," = !4,"2 ! 4! = !16 " Stable Spiral

(b) ! = 26," = !10, "2 ! 4! = !4 " Stable Spiral6.(a) xy-plane 6.(b) xy-plane

7. (a) ! = 7," = 4,"2 ! 4! = !12 " Unstable Spiral

(b) ! = !15 " Saddle7.(a) xy-plane 7.(b) xy-plane

8. (a) ! = !40 " Saddle

(b) ! = 1," = 2,"2 ! 4! = 0 " Unstable Node8.(a) xy-plane 8.(b) xy-plane

Page 176: MATH 224 Sol.

170 Section 5.1

9. (a) ! = 1," = !2,"2 ! 4! = 0 " Stable Node

(b) ! = 10," = !2,"2 ! 4! = !44 " Stable Spiral9.(a) xy-plane 9.(b) xy-plane

10. (a) ! = 4," = !4,"2 ! 4! = 0 " Stable Node

(b) ! = !18 " Saddle10.(a) xy-plane 10.(b) xy-plane

11. (a) ! = 3," = !4,"2 ! 4! = 4 " Stable Node

(b) ! = 29, " = !10, "2 ! 4! = !16 " Stable Spiral11.(a) xy-plane 11.(b) xy-plane

Page 177: MATH 224 Sol.

5.2. USEFUL TERMINOLOGY 171

12. (a) ! = 19," = !8,"2 ! 4! = !12 " Stable Spiral

(b) ! = 7, " = 8, "2 ! 4! = 36 " Unstable Node12.(a) xy-plane 12.(b) xy-plane

13. u1 = x, u2 = x!, u3 = x!! " x!! = !bx! ! kx, u3 = !bu2 ! ku1 " u!1 = u2 = 0u1 + u2, u

!2 =

!ku1 ! bu2 " ! = k," = !b,"2 ! 4! = b2 ! 4k "k > 0, b < 0 " Stablek > 0, b > 0 " Unstable

14. x = 2u1 + u2, y = u1 + u2 " u!1 = 2u1, u

!2 = 3u2 " u1 = c1e

2t, u2 = c2e3t " x = 2c1e

2t + c2e3t, y =

c1e2t + c2e3t

15. x = u1 + u2, y = u1 + 2u2 " u!1 = !u1, u!

2 = !3u2 " u1 = c1e"t, u2 = c2e"3t " x = c1e"t +c2e

"3t, y = c1e"t + 2c2e

"3t

16. x = 4u1, y = u1 + u2 " u!1 = 5u1, u

!2 = !3u2 " u1 = c1e

5t, u2 = c2e"3t " x = 4c1e

"t, y =c1e

"t + c2e"3t

17. x = u1+3u2, y = u1+u2 " u!1 = !2u1, u

!2 = 2u2 " u1 = c1e

"2t, u2 = c2e3t " x = c1e

"2t+3c2e2t, y =

c1e"2t + c2e

2t

18. x = u1 + u2, y = u1 + 3u2 " u!1 = !u1, u

!2 = u2 " u1 = c1e

"t, u2 = c2et " x = c1e

"t + c2et, y =

c1e"t + 3c2et

19. See Section 5.1.2. This is done verbatim there.

5.2 Useful Terminology

1.dAdt

=

!cos t et

2t 3

"

dBdt

=

!2 sin t cos t !te"t + e"t

0 1

"

%A dt =

!! cos t et

t3

33t2

2

"+

!C11 C12

C21 C22

"

%B dt =

! t2 ! 1

4 sin(2t) e"t(!1 ! t)

0 3t + t2

2

"+

!C11 C12

C21 C22

"

2.dAdt

=

!cos t et

2t 3

"

dBdt

=

!2e2t cos t ! e2t sin t 2te"t ! t2e"t

0 1t

"

%A dt =

!! cos t et

t3

33t2

2

"+

!C11 C12

C21 C22

"

%B dt =

!2e2t cos t

5 + e2t sin t5 e"t(!2 ! 2t ! t2)

3t !t + t ln t

"+

!C11 C12

C21 C22

"

Page 178: MATH 224 Sol.

172 Section 5.2

3.dAdt

=

1

30 !e"t 3e3t

1 0 6e3t

et !e"t 9e3t

5

7

dBdt

=

1

3cos t !te"t + e"t !e"t

0 0 6e3t

et 2t ! e"t ! sin t

5

7

%A dt =

1

23t !e"t e3t

3t2

2 0 2e3t

3et !e"t e3t

5

67 +

1

3C11 C12 C13

C21 C22 C23

C31 C32 C33

5

7

%B dt =

1

23! cos t e"t(!1 ! t) !e"t

0 0 2e3t

3

et t3

3 ! e"t sin t

5

67 +

1

3C11 C12 C13

C21 C22 C23

C31 C32 C33

5

7

4.dAdt

=

1

3e"t(cos t ! sin t !3e3t !3e"3t

12#

t4t3 ! sin t

43 t1/3 0 3te3t + e3t

5

7

dBdt

=

1

3sec2 t !te"t + e"t !e"t

0 0 6e3t

et 2t ! e"t ! sin t

5

7

%A dt =

1

23t !e"t e3t

3t2

2 0 2e3t

3et !e"t e3t

5

67 +

1

3C11 C12 C13

C21 C22 C23

C31 C32 C33

5

7

%B dt =

1

23! e!t

2 (cos t + sin t) ! e3t

3 ! e!3t

323 t3/2 t5

5 sin t37 t7/3 0 e3t( t

3 ! 19 )

5

67 +

1

3C11 C12 C13

C21 C22 C23

C31 C32 C33

5

7

5.dx1

dt=

!!4e4t

!8e4t

"= Ax1

dx2

dt=

!et

3et

"= Ax2

6.dx1

dt=

!4e"t

!2e"t

"= Ax1

dx2

dt=

!6e2t

!2e2t

"= Ax2

7.dx1

dt=

!!4e"2t

!6e"2t

"= Ax1

dx2

dt=

!!3e"3t

!6e"3t

"= Ax2

8.dx1

dt=

!4e2t

10e2t

"= Ax1

dx2

dt=

!et

3et

"= Ax2

9.dx1

dt=

1

3et

00

5

7 = Ax1

dx2

dt=

1

30

!e"t

0

5

7 = Ax2

dx3

dt=

1

3!2e2t

02e2t

5

7 = Ax3

Page 179: MATH 224 Sol.

Section 5.2 173

10.dx1

dt=

1

32e2t

2e2t

2e2t

5

7 #= Ax1

dx2

dt=

1

3e"t

!2e"t

!e"t

5

7 #= Ax2

dx3

dt=

1

3!2e"2t

00

5

7 #= Ax3

11.dx1

dt=

1

3!2e"2t

6e"2t

!2e"2t

5

7 = Ax1

dx2

dt=

1

3e"t

!2e"t

0

5

7 = Ax2

dx3

dt=

1

3et

0!et

5

7 = Ax3

12. If c1x!1 = c1Ax1, c2x

!2 = c2Ax2 then c1x

!1 + c2x

!2 = c1Ax1 + c2Ax2 = A(c1x1 + c2x2) " c1x1 + c2x2 is

a solution.

13. (a) W (t) =

8888et et

0 et

8888 = e2t #= 0 " Linearly Independent

(b) W (t) =

88881 21 !1

8888 = !3 #= 0 " Linearly Independent

(c) W (t) =

88883 !6!1 2

8888 = 0 " Linearly Dependent

14. (a) W (t) =

88883et !6e"t

!et 2e"t

8888 = 0 " Linearly Dependent

(b) W (t) =

8888!1 21 !1

8888 = !1 #= 0 " Linearly Independent

(c) W (t) =

888888

3 1 0!1 1 00 0 1

888888= 4 #= 0 " Linearly Independent

15. (a) W (t) =

888888

3et !3e2t 0et !e2t 00 0 e"t

888888= 0 " Linearly Dependent

(b) W (t) =

888888

2 !2 0!1 1 00 1 1

888888= 0 " Linearly Dependent

16. x!1 =

!e3t cos t + 3e3t sin t!e3t sin t + 3e3t cos t

"#= Ax1 =

!3e3t sin t + 2e3t cos t!2e3t sin t + 3e3t cos t

"" NOT a fundamental solution

set.

17. x!1 =

!2e3t cos 2t + 3e3t sin 2t!2e3t sin 2t + 3e3t cos 2t

"= Ax1

x!2 =

!!2e3t sin 2t + 3e3t cos 2t!2e3t cos 2t ! 3e3t sin 2t

"= Ax2 "

W (t) = !e3t #= 0 " Linearly Independent " this set IS a fundamental solution set.

18. x!1 =

!!5e5t

5e5t

"= Ax1

Page 180: MATH 224 Sol.

174 Section 5.3

x!2 =

!et

et

"= Ax2 "

W (t) = !2e6t #= 0 " Linearly Independent " this set IS a fundamental solution set.

19. x!1 =

!!e"t

e"t

"= Ax1

x!2 =

!!2e"t

!2e"t

"#= Ax2 =

!6e"t

!2e"t

"" this set is NOT a fundamental solution set.

20. x!1 =

1

3et

00

5

7 = Ax1

x!2 =

1

3e"t

!e"t

0

5

7 = Ax2

x!3 =

1

300

!2e"2t

5

7 = Ax3

W (t) =

888888

et !e"t 00 e"t 00 0 e"2t

888888= e"2t #= 0 " Linearly Independent " this set IS a fundamental solution

set.

21. x!1 =

1

300

!e"t

5

7 #= Ax1 =

1

300

!2e"t

5

7 " this set is NOT a fundamental solution set.

5.3 Linear Transformations and the Fundamental Subspaces

1. (a)

!3/2 00 3/2

"!13

"=

!3/29/2

"

(b) A =

!cos #/2 ! sin #/2sin #/2 cos #/2

"!13

"=

!0 !11 0

"!13

"=

!!31

"

(c) A =

!cos2 #/2 cos#/2 sin #/2

cos #/2 sin #/2 sin2 #/2

"!13

"=

!0 00 1

"!13

"=

!03

"

(d) A =

!2 cos2 #/2 ! 1 2 cos #/2 sin #/2

2 cos #/2 sin #/2 2 sin2 #/2 ! 1

"!13

"=

!!1 00 1

"!13

"=

!!13

"

2. (a)

!!2 00 !2

"!!1!2

"=

!24

"

(b) A =

!cos # ! sin #sin # cos #

"!!1!2

"=

!!1 00 !1

"!!1!2

"=

!12

"

(c) A =

!cos2 # cos # sin #

cos # sin # sin2 #

"!!1!2

"=

!1 00 0

"!!1!2

"=

!!10

"

(d) A =

!2 cos2 # ! 1 2 cos# sin #2 cos# sin # 2 sin2 # ! 1

"!!1!2

"=

!1 00 !1

"!!1!2

"=

!!12

"

3. (a)

!1/2 00 1/2

"!1!3

"=

!1/2!3/2

"

(b) A =

!cos #/3 ! sin #/3sin #/3 cos #/3

"!1!3

"=

!1/2 !

$3/2$

3/2 1/2

"!1!3

"=

!1/2 + 3

$2/2$

3/2 ! 3/2

"

(c) A =

!cos2 #/3 cos#/3 sin #/3

cos #/3 sin #/3 sin2 #/3

"!1!3

"=

!1/4

$3/4$

3/4 3/4

"!1!3

"=

/1"3

#3

4#3"94

0

Page 181: MATH 224 Sol.

Section 5.3 175

(d) A =

!2 cos2 #/3 ! 1 2 cos #/3 sin #/3

2 cos #/3 sin #/3 2 sin2 #/3 ! 1

"!1!3

"=

!!1/2

$3/2$

3/2 1/2

"!1!3

"

=

/"1"3

#3

2#3"32

0

4. (a)

!3 00 3

"!!23

"=

!!69

"

(b) A =

!cos #/3 ! sin #/3sin #/3 cos #/3

"!!23

"=

!1/2 !

$3/2$

3/2 1/2

"!!23

"=

!!1 ! 3

$3/2

!2$

3/2 + 3/2

"

(c) A =

!cos2 #/3 cos#/3 sin #/3

cos #/3 sin #/3 sin2 #/3

"!!23

"=

!1/4

$3/4$

3/4 3/4

"!!23

"

=

!!1/2 + 3

$3/4

!2$

3/4 + 9/4

"

(d) A =

!2 cos2 #/3 ! 1 2 cos #/3 sin #/3

2 cos #/3 sin #/3 2 sin2 #/3 ! 1

"!!23

"=

!!1/2

$3/2$

3/2 1/2

"!!23

"

=

!1 + 3

$3/2

!$

3 + 3/2

"

5. (a)

!!1 00 !1

"!21

"=

!!2!1

"

(b) A =

!cos!#/4 ! sin!#/4sin!#/4 cos!#/4

"!21

"=

! $2/2

$2/2

!$

2/2$

2/2

"!21

"=

! $2 +

$2/2

!$

2 +$

2/2

"

(c) A =

!cos2 0 cos 0 sin 0

cos 0 sin 0 sin2 0

"!21

"=

!1 00 1

"!21

"=

!21

"

(d) A =

!2 cos2 0 ! 1 2 cos 0 sin 02 cos 0 sin 0 2 sin2 0 ! 1

"!21

"=

!1 00 !1

"!21

"=

!2!1

"

6. (a)

!2 00 2

"!!31

"=

!!62

"

(b) A =

!cos!#/2 ! sin!#/2sin!#/2 cos!#/2

"!13

"=

!0 1!1 0

"!!31

"=

!13

"

(c) A =

!cos2 #/3 cos#/3 sin #/3

cos #/3 sin #/3 sin2 #/3

"!!31

"=

!1/4

$3/4$

3/4 3/4

"!!31

"

=

!!3/4

$3/4

!3$

3/4 + 3/4

"

(d) A =

!2 cos2 #/6 ! 1 2 cos #/6 sin #/6

2 cos #/6 sin #/6 2 sin2 #/6 ! 1

"!!31

"=

!1/2

$3/2$

3/2 !1/2

"!!31

"

=

!!3/2 +

$3/2

!3$

3/2 ! 1/2

"

7. A =

!2 cos2 $ ! 1 2 cos $ sin $2 cos $ sin $ 2 sin2 $ ! 1

""

A2 =

!(2 cos2 $ ! 1)2 + (2 cos $ sin $)2 B

B 4 cos2 $ sin2 $ + (2 sin2 $ ! 1)2

"

where B = 4 cos3 $ sin $ ! 2 cos $ sin $ + 4 cos $ sin3 $ ! 2 cos $ sin $

=

!(cos2 2$)2 + (sin2 2$) 2 cos $ sin $(2 cos2 $ ! 1 + 2 sin2 $ ! 1)

2 cos $ sin $(2 cos2 $ ! 1 + 2 sin2 $ ! 1) (cos2 2$)2 + (sin2 2$)

"=

!1 00 1

"

8. Column space A =

O!10

",

!01

"4, Null space A = %

Column space B =

O!1!3

"4, Null space B =

O!31

"4

Page 182: MATH 224 Sol.

176 Section 5.3

9. Column space A =

O!10

",

!01

"4, Null space A =

AC

D

1

34!35

5

7

FG

H

Column space B =

O!10

",

!01

"4, Null space B =

AC

D

1

3!1!11

5

7

FG

H

10. Column space A =

O!10

",

!01

"4, Null space A =

ABBC

BBD

1

223

3!504

5

667 ,

1

223

!1120

5

667

FBBG

BBH

Column space B =

O!10

",

!01

"4, Null space B =

ABBC

BBD

1

223

7!202

5

667 ,

1

223

!2110

5

667

FBBG

BBH

11. Column space A =

AC

D

1

310

5/4

5

7 ,

1

301

1/8

5

7

FG

H, Null space A = %

Column space B =

AC

D

1

31!21

5

7

FG

H, Null space B =

O!31

"4

12. Answers will vary

13. Answers will vary

14. Answers will vary

15. Use code from Section to complete.

16. NullSpace(A) =

AC

D

1

31!21

5

7

FG

H

ColumnSpace(A) =

AC

D

1

310!1

5

7 ,

1

3012

5

7

FG

H

RowSpace(A) =

AC

D

1

310!1

5

7 ,

1

3012

5

7

FG

H

LeftNullSpace(A) =

AC

D

1

31!21

5

7

FG

H

17. NullSpace(A) =%

ColumnSpace(A) =

AC

D

1

3100

5

7 ,

1

3010

5

7 ,

1

3001

5

7

FG

H

RowSpace(A) =

AC

D

1

3100

5

7 ,

1

3010

5

7 ,

1

3001

5

7

FG

H

LeftNullSpace(A) =%18. See Linear Algebra Text

19. See Linear Algebra Text

Page 183: MATH 224 Sol.

5.4. EIGENVALUES AND EIGENVECTORS 177

20. m = tan%

1m2 + 1

!1 ! m2 2m

2m m2 ! 1

"=

1tan2 %+ 1

!1 ! tan2 % 2 tan%

2 tan% tan2 %! 1

"

=1

sec2 %

!1 ! tan2 % 2 tan%

2 tan% tan2 %! 1

"

=

!cos2 %! sin2 % 2 sin% cos%2 sin% cos% sin2 %! cos2 %

"

=

!2 cos2 %! 1 2 sin% cos%2 sin % cos% 2 sin2 %! 1

"

5.4 Eigenvalues and Eigenvectors

1. (a) det(A ! &I) = &2 ! &! 2 = 0 " &1 = 2, &2 = !1

&1 = 2; (A ! &I)v1 = 0 " v1 =

!52

"

&2 = !1; (A! &I)v2 = 0 " v2 =

!11

"

(b) det(A ! &I) = &2 + 3&! 18 = 0 " &1 = !6, &2 = 3

&1 = !6; (A! &I)v1 = 0 " v1 =

!29

"

&2 = 3; (A ! &I)v2 = 0 " v2 =

!00

"

(c) det(A ! &I) = &2 + 4&+ 3 = 0 " &1 = !3, &2 = !1

&1 = !3; (A! &I)v1 = 0 " v1 =

!12

"

&2 = !1; (A! &I)v2 = 0 " v2 =

!11

"

2. (a) det(A ! &I) = (!2 ! &)(!5! &) = 0 " &1 = !2, &2 = !5

&1 = !2; (A! &I)v1 = 0 " v1 =

!10

"

&2 = !5; (A! &I)v2 = 0 " v2 =

!1!1

"

(b) det(A ! &I) = (!4 ! &)(!1! &) = 0 " &1 = !4, &2 = !1

&1 = !4; (A! &I)v1 = 0 " v1 =

!10

"

&2 = 3; (A ! &I)v2 = 0 " v2 =

!73

"

(c) det(A ! &I) = (1 ! &)(7 ! &) = 0 " &1 = 1, &2 = 7

&1 = 1; (A ! &I)v1 = 0 " v1 =

!10

"

&2 = 7; (A ! &I)v2 = 0 " v2 =

!!12

"

3. (a) det(A ! &I) = &2 ! 4&+ 3 = 0 " &1 = 3, &2 = 1

&1 = 3; (A ! &I)v1 = 0 " v1 =

!11

"

&2 = 1; (A ! &I)v2 = 0 " v2 =

!1!1

"

(b) det(A ! &I) = &2 ! 4 = 0 " &1 = 2, &2 = !2

&1 = 2; (A ! &I)v1 = 0 " v1 =

!11

"

Page 184: MATH 224 Sol.

178 Section 5.4

&2 = !2; (A! &I)v2 = 0 " v2 =

!1!1

"

(c) det(A ! &I) = &2 ! 2&+ 8 = 0 " &1 = 4, &2 = !2

&1 = 4; (A ! &I)v1 = 0 " v1 =

!11

"

&2 = !2; (A! &I)v2 = 0 " v2 =

!1!1

"

4. (a) det(A ! &I) = &2 ! 2&! 3 = 0 " &1 = 3, &2 = !1

&1 = 3; (A ! &I)v1 = 0 " v1 =

!11

"

&2 = !1; (A! &I)v2 = 0 " v2 =

!31

"

(b) det(A ! &I) = &2 ! 5&+ 6 = 0 " &1 = 3, &2 = 2

&1 = 3; (A ! &I)v1 = 0 " v1 =

!11

"

&2 = 2; (A ! &I)v2 = 0 " v2 =

!21

"

(c) det(A ! &I) = &2 + 2&! 15 = 0 " &1 = !5, &2 = 3

&1 = !5; (A! &I)v1 = 0 " v1 =

!!31

"

&2 = 3; (A ! &I)v2 = 0 " v2 =

!11

"

5. (a) det(A ! &I) = &2 ! 2&! 3 = 0 " &1 = 3, &2 = !1

&1 = 3; (A ! &I)v1 = 0 " v1 =

!13

"

&2 = !1; (A! &I)v2 = 0 " v2 =

!1!1

"

(b) det(A ! &I) = &2 ! 6&+ 5 = 0 " &1 = 5, &2 = 1

&1 = 5; (A ! &I)v1 = 0 " v1 =

!11

"

&2 = 1; (A ! &I)v2 = 0 " v2 =

!!13

"

(c) det(A ! &I) = &2 + 2&! 5 = 0 " &1 = !1 +$

6, &2 = !1 !$

6

&1 = !1 +$

6; (A ! &I)v1 = 0 " v1 =

!1

2 +$

6

"

&2 = !1 !$

6; (A ! &I)v2 = 0 " v2 =

!!1

!2 +$

6

"

6. (a) det(A ! &I) = (5 ! &)(!3 ! &) = 0 " &1 = 5, &2 = !3

&1 = 5; (A ! &I)v1 = 0 " v1 =

!41

"

&2 = !3; (A! &I)v2 = 0 " v2 =

!01

"

(b) det(A ! &I) = &2 ! 3&! 40 = 0 " &1 = 8, &2 = !5

&1 = 8; (A ! &I)v1 = 0 " v1 =

!61

"

&2 = !5; (A! &I)v2 = 0 " v2 =

!1!2

"

(c) det(A ! &I) = &2 + 2&+ 1 = 0 " &1 = !1

&1 = !1; (A! &I)v2 = 0 " v2 =

!1!1

"

Page 185: MATH 224 Sol.

Section 5.4 179

7. (a) det(A ! &I) = &2 + 1 = 0 " &1 = i, &2 = !i

&1 = i; (A ! &I)v1 = 0 " v1 =

!1!i

"" v2 =

!1!i

"

(b) det(A ! &I) = &2 + 2&+ 5 = 0 " &1 = !1 + 2i, &2 = !1 ! 2i

&1 = !1 + 2i; (A ! &I)v1 = 0 " v1 =

!1i

"" v2 =

!1!i

"

8. (a) det(A ! &I) = &2 + 1 = 0 " &1 = i, &2 = !i

&1 = i; (A ! &I)v1 = 0 " v1 =

!2

3+i10

"v2 = 0 " v2 =

!2

3"i10

"

(b) det(A ! &I) = &2 ! 6&+ 13 = 0 " &1 = 3 + 2i, &2 = 3 ! 2i

&1 = 3 + 2i; (A! &I)v1 = 0 " v1 =

! "1+i21

"" v2 =

! "1"i21

"

9. (a) det(A ! &I) = &2 + 4&+ 8 = 0 " &1 = !2 + 2i, &2 = !2 ! 2i

&1 = !2 + 2i; (A ! &I)v1 = 0 " v1 =

!1

1"2i5

"" v2 =

!1

1+2i5

"

(b) det(A ! &I) = &2 + 4&+ 8 = 0 " &1 = !2 + 2i, &2 = !2 ! 2i

&1 = !2 + 2i; (A ! &I)v1 = 0 " v1 =

!1

! 5+2i29

"" v2 =

!1

! 5"2i29

"

10. (a) det(A ! &I) = &2 + 10& + 26 = 0 " &1 = !5 + i, &2 = !5 ! i

&1 = !5 + i; (A ! &I)v1 = 0 " v1 =

!1

2"i5

"" v2 =

!1

2+i5

"

(b) det(A ! &I) = &2 + 5 = 0 " &1 = i$

5, &2 = !i$

5

&1 = !8; (A! &I)v1 = 0 " v1 =

!1

1"i#

56

"" v2 =

!1

1+i#

56

"

11. (a) det(A ! &I) = &2 + 10& + 29 = 0 " &1 = !5 + 2i, &2 = !5 ! 2i

&1 = !5 + 2i; (A ! &I)v1 = 0 " v1 =

!1i

"" v2 =

!1!i

"

(b) det(A ! &I) = &2 ! 4&+ 11 = 0 " &1 = 2 + i$

7, &2 = 2 ! i$

7

&1 = 2 + i$

7; (A! &I)v1 = 0 " v1 =

!1

"1+i#

78

"" v2 =

!1

"1"i#

78

"

12. (a) &1 = 1; v1 =

1

3001

5

7

&2 = 2; v2 =

1

31!11

5

7

&3 = 5; v3 =

1

31/210

5

7

(b) &1 = !4; v1 =

1

31!11

5

7

&2 = 1; v2 =

1

3001

5

7

&3 = 2; v3 =

1

31/210

5

7

Page 186: MATH 224 Sol.

180 Section 5.5

13. (a) &1 = !1; v1 =

1

3!132

5

7

&2 = !1 ! 2i; v2 =

1

3!i10

5

7

&3 = !1 + 2i; v3 =

1

3i10

5

7

(b) &1 = 1; v1 =

1

3!3!15

5

7

&2 = 2 ! 2i; v2 =

1

3!1 ! 2i

10

5

7

&3 = 2 + 2i; v3 =

1

3!1 + 2i

10

5

7

14. (a) &1 = !1; v1 =

1

3001

5

7

&2 = 3; v2 =

1

33/201

5

7

&3 = 3; v3 =

1

31/210

5

7

(b) &1 = !2; v1 =

1

31/210

5

7

&2 = !1; v2 =

1

3001

5

7

&3 = 0; v3 =

1

31!11

5

7

5.5 Matrix Exponentials

1. (a) & = 3, !6; p(3) = 3t'1 + '2 = e3t; p(!6) = !6t'1 + '2 = e"6t "

'1 =e3t ! e"6t

9t, '2 =

e"6t + 2e3t

3"

eAt = '1At + '2I =

!e3t 2

9 (!e3t + e"6t)0 e"6t

"

(b) & = !3, !1; p(!3) = (!3t)'1 + '2 = e"3t; p(!1) = !t'1 + '2 = e"t "

'1 =e"t ! e"3t

2t, '2 =

3e"t ! e"3t

2"

eAt = '1At + '2I =

!2e"t ! e"3t e"3t ! e"t

2e"t ! 2e"3t !e"t + 2e"3t

"

2. (a) & = 3, 1; p(3) = 3t'1 + '2 = e3t; p(1) = t'1 + '2 = et "

'1 =e3t ! et

2t, '2 =

3et ! e3t

2"

Page 187: MATH 224 Sol.

Section 5.5 181

eAt = '1At + '2I =

/e3t+et

2e3t"et

2e3t"et

2e3t+et

2

0

(b) & = 1, !1; p(1) = (t)'1 + '2 = et; p(!1) = !t'1 + '2 = e"t "

'1 =et ! e"t

2t, '2 =

et + e"t

2"

eAt = '1At + '2I =

/et+e!t

2et"e!t

2et"e!t

2et+e!t

2

0

3. From Corollary 5.5.1,

(a) a = 1, b = 1 " eAt = et

!cos t ! sin tsin t cos t

"

(b) a = 3, b = !2 " eAt = e3t

!cos(2t) sin(2t)! sin(2t) cos(2t)

"

4. (a) & = 5, !3; p(5) = 5t'1 + '2 = e5t; p(3) = 3t'1 + '2 = e"3t "

'1 =e5t ! e"3t

8t, '2 =

3e5t + 5e"3t

8"

eAt = '1At + '2I =

!e5t 0

e5t"e!3t

4 e"3t

"

5. (a) eAt =

1

23

2e!2t+et

3"2e!2t+2et+3tet

9"2e!2t+2et

30 et 0

"e!2t+et

3e!2t"et+3tet

9e!2t+2et

3

5

67

(b) eAt =

1

23et 0 0

"5+4et+e4t

21+3e4t

43"3e4t

4"15+16et"e4t

61"e4t

43+e4t

4

5

67

6. eAt =

1

2223

e2t !e"t + e2t "e!3t"5e!t+6e2t

10 e"t ! e2t

e2t"e4t

22e!t+5e2t+3e4t

10"3e!3t+e!t+3e2t"e4t

10"2e!t"5e2t+7e4t

100 0 e"3t 0

e2t"e4t

2"8e!t+5e2t+3e4t

102e!3t"4e!t+3e2t"e4t

108e!t"5e2t+7e4t

10

5

6667

11. (a) i. det(A ! &I) = &2 + 3&! 18 " A2 + 3A ! 18I =

!9 60 36

"+

!9 !60 18

"!!

18 00 18

"

=

!0 00 0

"

ii. det(A ! &I) = &2 + 4&+ 3 " A2 + 4A + 3I =

!!7 8!16 17

"+

!4 !816 20

"+

!3 00 3

"

=

!0 00 0

"

(b) i. det(A ! &I) = &2 ! 4&+ 3 " A2 ! 4A + 3I =

!5 44 5

"!!

8 44 8

"+

!3 00 3

"

=

!0 00 0

"

ii. det(A ! &I) = &2 ! 1 " A2 ! I =

!1 00 1

"!!

1 00 1

"

=

!0 00 0

"

(c) i. det(A ! &I) = &2 ! 2&+ 2 " A2 ! 2A + 2I =

!0 !22 0

"!!

2 !22 2

"+

!2 00 2

"

=

!0 00 0

"

Page 188: MATH 224 Sol.

182 Chapter 5 Review

ii. det(A ! &I) = &2 ! 6&+ 13 " A2 ! 6A + 13I =

!5 12

!12 5

"!!

18 12!12 18

"+

!13 00 13

"

=

!0 00 0

"

(d) i. det(A ! &I) = &2 ! 2&! 15 " A2 ! 2A ! 15I =

!25 04 9

"!!

10 04 !6

"!!

15 00 15

"

=

!0 00 0

"

ii. det(A ! &I) = &2 ! 4&+ 5 " A2 ! 4A + 5I =

!!1 !84 7

"!!

4 !84 12

"+

!5 00 5

"

=

!0 00 0

"

5.6 Chapter 5: Additional Problems

1. False. Only those that have the proper form have those specific interpretations.

2. True.

3. True. See Definition 5.1.

4. True.

5. True. See Theorem 5.3.2.

6. False.

7. (a) unstable node, &1,2 = 4, 2

(b) unstable node, &1,2 = 3, 1

8. (a) unstable spiral, &1,2 = 5 ± i

(b) unstable spiral, &1,2 = 1 ± i

9. (a) unstable spiral, &1,2 = 1 ± i

(b) saddle, &1,2 = 1,!2

10. (a) saddle, &1,2 = 2,!1

(b) saddle, &1,2 = 2,!2

11. (a) unstable spiral, &1,2 = 1 ± i

(b) center, &1,2 = ±i

12. (a) unstable spiral, &1,2 = 2 ± i$

3

(b) unstable node, &1,2 = 2, 3

13. (a) unstable spiral, &1,2 = 2 ± i$

3

(b) stable spiral, &1,2 = !4 ± i$

3

17. &1,2,3 = !2, 2, 3, v1 =

1

32!12

5

7, v2 =

1

31!11

5

7, v3 =

1

32!11

5

7

18. &1,2,3 = 1, 1, 2, v1 =

1

3!210

5

7, v2 =

1

3001

5

7, v3 =

1

31!11

5

7

19. &1,2,3 = !1, 1, 5, v1 =

1

3111

5

7, v2 =

1

311!1

5

7, v3 =

1

31!11

5

7

20. &1,2,3 = !3, 5, 1, v1 =

1

3!1!11

5

7, v2 =

1

31!11

5

7, v3 =

1

3111

5

7

Page 189: MATH 224 Sol.

Chapter 5 Review 183

21. &1,2,3 = 5, 1, 3, v1 =

1

31!11

5

7, v2 =

1

3111

5

7, v3 =

1

311!1

5

7

22. &1,2,3 = !27,!18,!9, v1 =

1

31/3!4/3

1

5

7, v2 =

1

321!3

5

7, v3 =

1

3216

5

7

23. Yes, it’s a fund set of solutions

24. No, it’s NOT a fund set of solutions

25. Yes, it’s a fund set of solutions

26. Yes, it’s a fund set of solutions

27. No, it’s NOT a fund set of solutions

28. Yes, it’s a fund set of solutions

29. Yes, it’s a fund set of solutions

30. No, it’s NOT a fund set of solutions

Page 190: MATH 224 Sol.

184 Chapter 5 Review

Page 191: MATH 224 Sol.

Chapter 6

Techniques of Systems ofDi!erential Equations

6.1 A General Method, Part I: Solving Systems with Real,Distinct Eigenvalues

1. (a) Eigenvalues: &1 = !2, &2 = 1 " Saddle

Eigenvectors: v1 =

&21

', v2 =

&1!1

'

General Solution: x(t) = c1

&21

'e"2t + c2

&1!1

'et

(b) Eigenvalues: &1 = 1, &2 = !1 " Saddle

Eigenvectors: v1 =

&12

', v2 =

&11

'

General Solution: x(t) = c1

&12

'et + c2

&1!1

'e"t

2. (a) Eigenvalues: &1 = !2, &2 = !1 " Stable

Eigenvectors: v1 =

&!12

', v2 =

&1!3

'

General Solution: x(t) = c1

&!12

'e"2t + c2

&1!3

'e"t

(b) Eigenvalues: &1 = 3, &2 = 1 " Unstable

Eigenvectors: v1 =

&12

', v2 =

&13

'

General Solution: x(t) = c1

&12

'e3t + c2

&13

'et

3. (a) Eigenvalues: &1 = 4, &2 = 2 " Unstable

Eigenvectors: v1 =

&11

', v2 =

&13

'

General Solution: x(t) = c1

&11

'e4t + c2

&13

'e2t

(b) Eigenvalues: &1 = 5, &2 = !1 " Unstable

Eigenvectors: v1 =

&21

', v2 =

&!11

'

General Solution: x(t) = c1

&21

'e5t + c2

&!11

'e"t

185

Page 192: MATH 224 Sol.

186 Section 6.1

4. (a) Eigenvalues: &1 = !2, &2 = 3 " Saddle

Eigenvectors: v1 =

&11

', v2 =

&13

'

General Solution: x(t) = c1

&11

'e"2t + c2

&13

'e3t

(b) Eigenvalues: &1 = !4, &2 = 1 " Saddle

Eigenvectors: v1 =

&!21

', v2 =

&1!1

'

General Solution: x(t) = c1

&!21

'e"4t + c2

&1!1

'et

5. (a) Eigenvalues: &1 = 4, &2 = !2 " Saddle

Eigenvectors: v1 =

&11

', v2 =

&12

'

General Solution: x(t) = c1

&11

'e4t + c2

&12

'e"2t

(b) Eigenvalues: &1 = !3, &2 = !2 " Stable

Eigenvectors: v1 =

&11

', v2 =

&12

'

General Solution: x(t) = c1

&11

'e"3t + c2

&12

'e"2t

6. (a) Eigenvalues: &1 = !2, &2 = 1 " Saddle

Eigenvectors: v1 =

&21

', v2 =

&11

'

General Solution: x(t) = c1

&21

'e"2t + c2

&11

'et

(b) Eigenvalues: &1 =3 !

$5

2, &2 =

3 +$

52

" Unstable

Eigenvectors: v1 =

&1

!1.618

', v2 =

&1.618

1

'

General Solution: x(t) = c1

&1

!1.618

'e0.382t + c2

&1.618

1

'e2.618t

7. (a) Eigenvalues: &1 = 5, &2 = 1 " Unstable

Eigenvectors: v1 =

&11

', v2 =

&!13

'

General Solution: x(t) = c1

&11

'e5t + c2

&!13

'et

(b) Eigenvalues: &1 = 2 +$

3, &2 = 2 !$

3 " Unstable

Eigenvectors: v1 =

&1.732

1

', v2 =

&!1.732

1

'

General Solution: x(t) = c1

&1.732

1

'e3.732t + c2

&!1.732

1

'e0.268t

8. (a) Eigenvalues: &1 = !5, &2 = 1 " Saddle

Eigenvectors: v1 =

&31

', v2 =

&21

'

General Solution: x(t) = c1

&31

'e"5t + c2

&21

'et

(b) Eigenvalues: &1 = !2 +$

6, &2 = !2 !$

6 " Saddle

Eigenvectors: v1 =

&1

1.225

', v2 =

&1

!1.225

'

General Solution: x(t) = c1

&1

1.225

'e0.450t + c2

&1

!1.225

'e"4.450t

Page 193: MATH 224 Sol.

Section 6.1 187

9. (a) Eigenvalues: &1 = 3 + i, &2 = 3 ! i " Stable

(b) Eigenvalues: &1 = 0 + 2i, &2 = 0 ! 2i " Center

10. (a) Eigenvalues: &1 = 0 + i, &2 = 0 ! i " Center

(b) Eigenvalues: &1 = 0 + 2i, &2 = 0 ! 2i " Center

11. (a) Eigenvalues: &1 = 1 + 2i, &2 = 1 ! 2i " Unstable

(b) Eigenvalues: &1 = !1 + 2i, &2 = !1 ! 2i " Stable

12. (a) Eigenvalues: &1 = !3 + i, &2 = !3 ! i " Stable

(b) Eigenvalues: &1 = !3 + 2i, &2 = !3 ! 2i " Stable

13. (a) Eigenvalues: &1 = !6, &2 = !10 " Stable

(b) Eigenvalues: &1 = 3 + i, &2 = 3 ! i " Unstable

14. (a) Eigenvalues: &1 = 1, &2 = 5, &3 = 2 " Unstable

(b) Eigenvalues: &1 = 1, &2 = 2, &3 = !4 " Saddle

15. (a) Eigenvalues: &1 = !1 + 2i, &2 = !1 ! 2i, &3 = 1 " Stable

(b) Eigenvalues: &1 =3 +

$3

2, &2 =

3 !$

32

, &3 = 1 " Unstable

16. (a) Eigenvalues: &1 = !1, &2 = !1 + i, &3 = !1 ! i " Stable

(b) Eigenvalues: &1 = 11, &2 = 1, &3 = !1 " Saddle

17. (a) Eigenvalues: &1 =!3 +

$3

2, &2 =

!3 !$

32

, &3 = !1 " Stable

(b) Eigenvalues: &1 = !1 + i, &2 = !1 ! i, &3 = !1 " Stable

18. Eigenvalues are &1 =12

,(a + d) +

.(a ! d)2 + 4bc

-, &2 =

12

,(a + d) !

.(a ! d)2 + 4bc

-

(a) Must be real, so (a ! d)2 + 4bc & 0 and (a + d) ±.

(a ! d)2 + 4bc > 0

(b) Must be real, so (a ! d)2 + 4bc & 0 and (a + d) ±.

(a ! d)2 + 4bc < 0

(c) Must be complex, so (a ! d)2 + 4bc < 0 and a + d > 0

(d) Must be complex, so (a ! d)2 + 4bc < 0 and a + d < 0

(e) Must be real, so (a ! d)2 + 4bc & 0 and ad < bc

19. Eigenvalues: &1 = !1, &2 = !3 " Stable

Eigenvectors: v1 =

&1!1

', v2 =

&!13

'

General Solution: x(t) = c1

&1!1

'e"t + c2

&!13

'e"3t

Page 194: MATH 224 Sol.

188 Section 6.2

20. Eigenvalues: &1 = 2, &2 = !4 " Stable

Eigenvectors: v1 =

&12

', v2 =

&!14

'

General Solution: x(t) = c1

&12

'e2t + c2

&!14

'e"4t

6.2 A General Method, Part II: Solving Systems with Re-peated Real or Complex Eigenvalues

1. (a) Eigenvalues: &1 = 3 + i, &2 = 3 ! i

Eigenvectors: v1 =

&1!i

', v2 =

&1i

'"

a =

&10

', b =

&0!1

'

' = 3, " = 1

General Solution: x(t) = e3t

Oc1

!&10

'cos t !

&0!1

'sin t

"+ c2

!&10

'sin t +

&0!1

'cos t

"4

(b) Eigenvalues: &1 = 0 + 2i, &2 = 0 ! 2i

Eigenvectors: v1 =

&i1

', v2 =

&!i1

'"

a =

&01

', b =

&10

'

' = 0, " = 2

General Solution: x(t) =

Oc1

!&01

'cos 2t !

&10

'sin 2t

"+ c2

!&01

'sin 2t +

&10

'cos 2t

"4

2. (a) Eigenvalues: &1 = 0 + i, &2 = 0 ! i

Eigenvectors: v1 =

&2 ! i

5

', v2 =

&2 + i

5

'"

a =

&25

', b =

&!10

'

' = 0, " = 1

General Solution: x(t) =

Oc1

!&25

'cos t !

&!10

'sin t

"+ c2

!&25

'sin t +

&!10

'cos t

"4

(b) Eigenvalues: &1 = 0 + 2i, &2 = 0 ! 2i

Eigenvectors: v1 =

&i1

', v2 =

&!i1

'"

a =

&01

', b =

&10

'

' = 0, " = 2

General Solution: x(t) =

Oc1

!&01

'cos 2t !

&10

'sin 2t

"+ c2

!&01

'sin 2t +

&10

'cos 2t

"4

3. (a) Eigenvalues: &1 = 1 + 2i, &2 = 1 ! 2i

Eigenvectors: v1 =

&1

!3 ! i

', v2 =

&1

!3 + i

'"

a =

&1!3

', b =

&0!1

'

' = 1, " = 2

General Solution: x(t) = et

Oc1

!&1!3

'cos 2t !

&0!1

'sin 2t

"+ c2

!&1!3

'sin 2t +

&0!1

'cos 2t

"4

(b) Eigenvalues: &1 = !1 + 2i, &2 = !1 ! 2i

Eigenvectors: v1 =

&2

!3 ! i

', v2 =

&2

!3 + i

'"

Page 195: MATH 224 Sol.

Section 6.2 189

a =

&2!3

', b =

&0!1

'

' = !1, " = 2

General Solution: x(t) = e"t

Oc1

!&2!3

'cos 2t !

&0!1

'sin 2t

"+ c2

!&2!3

'sin 2t +

&0!1

'cos 2t

"4

4. (a) Eigenvalues: &1 = !3 + i, &2 = !3 ! i

Eigenvectors: v1 =

&!3 + i

5

', v2 =

&!3 ! i

5

'"

a =

&!35

', b =

&10

'

' = !3, " = 1

General Solution: x(t) = e"3t

Oc1

!&!35

'cos t !

&10

'sin t

"+ c2

!&!35

'sin t +

&10

'cos t

"4

(b) Eigenvalues: &1 = !3 + 2i, &2 = !3 ! 2i

Eigenvectors: v1 =

&2i1

', v2 =

&!2i1

'"

a =

&01

', b =

&20

'

' = !3, " = 2

General Solution: x(t) = e"3t

Oc1

!&01

'cos 2t !

&20

'sin 2t

"+ c2

!&01

'sin 2t +

&20

'cos 2t

"4

5. (a) Eigenvalues: &1 = !10, &2 = !6

Eigenvectors: v1 =

&12

', v2 =

&1!2

'"

a =

&12

', b =

&00

'

' = !10, " = 0

General Solution: x(t) = e"10t

&c1

2c1

'

(b) Eigenvalues: &1 = 3 + i, &2 = 3 ! i

Eigenvectors: v1 =

&1

3 + i

', v2 =

&1

3 ! i

'"

a =

&13

', b =

&01

'

' = 3, " = 1

General Solution: x(t) = e3t

Oc1

!&13

'cos t !

&01

'sin t

"+ c2

!&13

'sin t +

&01

'cos t

"4

6. (a) Eigenvalues: &1 = &2 = 1 "

Eigenvectors: v1 =

&10

', u1 =

&012

'

General Solution: x(t) = et

&c1 + c2t

c22

'

(b) Eigenvalues: &1 = &2 = !2 "

Eigenvectors: v1 =

&10

', u1 =

&013

'

General Solution: x(t) = e"2t

&c1 + c2t

c23

'

7. (a) Eigenvalues: &1 = &2 = 3 "

Eigenvectors: v1 =

&12

', u1 =

&0! 1

2

'

General Solution: x(t) = e3t

&c1 + c2t

2c1 + c2(2t ! 12 )

'

Page 196: MATH 224 Sol.

190 Section 6.2

(b) Eigenvalues: &1 = &2 = !1 "

Eigenvectors: v1 =

&12

', u1 =

&0!1

'

General Solution: x(t) = e"t

&c1 + c2t

2c1 + c2(2t ! 1)

'

8. (a) Eigenvalues: &1 = &2 = !1 "

Eigenvectors: v1 =

&11

', u1 =

&!10

'

General Solution: x(t) = e"t

&c1 + c2(t ! 1)

c1 + c2t

'

(b) Eigenvalues: &1 = &2 = 1 "

Eigenvectors: v1 =

&!11

', u1 =

&!10

'

General Solution: x(t) = et

&!c1 + c2(!t ! 1)

c1 + c2t

'

9. (a) Eigenvalues: &1 = &2 = !1 "

Eigenvectors: v1 =

&10

', u1 =

&014

'

General Solution: x(t) = e"t

&c1 + c2t

14c2

'

(b) Eigenvalues: &1 = &2 = !1 "

Eigenvectors: v1 =

&11

', u1 =

&120

'

General Solution: x(t) = e"t

&c1 + c2(t + 1

2 )c1 + c2t

'

10. (a) Eigenvalues: &1 = &2 = 1 "

Eigenvectors: v1 =

&!11

', u1 =

&120

'

General Solution: x(t) = et

&!c1 + c2(!t + 1

2 )c1 + c2t

'

(b) Eigenvalues: &1 = &2 = 2 "

Eigenvectors: v1 =

&!32

', u1 =

&! 1

20

'

General Solution: x(t) = e2t

&!3c1 + c2(!3t ! 1

2 )2c1 + 2c2t

'

11. Eigenvalues: &1 = !1 + 2i, &2 = !1 ! 2i, &3 = 1 " ' = !1, " = 2

Eigenvectors: v1 =

K

L!i!i0

M

N, v3 =

K

L012

M

N

a =

K

L0!10

M

N , b =

K

L!100

M

N

x1(t) = e"t

1

3

K

L0!10

M

N cos 2t !

K

L!100

M

N sin 2t

5

7

x2(t) = e"t

1

3

K

L0!10

M

N sin 2t +

K

L!100

M

N cos 2t

5

7

x3 = et

K

L012

M

N

12. Eigenvalues: &1 =3 +

$3

2, &2 =

3 !$

32

, &3 = 1 " ' = 1.5, " = 0.866

Page 197: MATH 224 Sol.

Section 6.2 191

Eigenvectors: v1 =

K

L0.3536 + .0.6124i

0.70710

M

N, v3 =

K

L!0.3015!0.90450.3015

M

N

a =

K

L0.35360.7071

0

M

N , b =

K

L0.6124

00

M

N

x1(t) = e1.5t

1

3

K

L0.35360.7071

0

M

N cos(0.866t) !

K

L0.6124

00

M

N sin(0.866t)

5

7

x2(t) = e1.5t

1

3

K

L0.35360.7071

0

M

N sin(0.866t) +

K

L0.6124

00

M

N cos(0.866t)

5

7

x3 = et

K

L!0.3015!0.90450.3015

M

N

13. Eigenvalues: &1 = !1 + i, &2 = !1 ! i, &3 = !1 " ' = !1, " = 1

Eigenvectors: v1 =

K

L10!i

M

N, v3 =

K

L011

M

N

a =

K

L100

M

N , b =

K

L00!1

M

N

x1(t) = e"t

1

3

K

L100

M

N cos(t) !

K

L00!1

M

N sin(t)

5

7

x2(t) = e"t

1

3

K

L100

M

N sin(t) +

K

L00!1

M

N cos(t)

5

7

x3 = e"t

K

L011

M

N

14. Eigenvalues: &1 = !1 + i, &2 = !1 ! i, &3 = !1 " ' = !1, " = 1

Eigenvectors: v1 =

K

L11i

M

N, v3 =

K

L100

M

N

a =

K

L110

M

N , b =

K

L001

M

N

x1(t) = e"t

1

3

K

L110

M

N cos(t) !

K

L001

M

N sin(t)

5

7

x2(t) = e"t

1

3

K

L110

M

N sin(t) +

K

L001

M

N cos(t)

5

7

x3 = e"t

K

L100

M

N

15. Eigenvalues: &1 = !1.5 + 0.866i, &2 = !1.5 ! 0.866i, &3 = !1 " ' = !1, " = 1

Eigenvectors: v1 =

K

L0.3536 + 0.6124i

00.7071

M

N, v3 =

K

L111

M

N

a =

K

L0.3536

00.7071

M

N , b =

K

L0.6124

00

M

N

Page 198: MATH 224 Sol.

192 Section 6.2

x1(t) = e"1.5t

1

3

K

L0.3536

00.7071

M

N cos(0.866t) !

K

L0.6124

00

M

N sin(0.866t)

5

7

x2(t) = e"t

1

3

K

L0.3536

00.7071

M

N sin(0.866t) +

K

L0.6124

00

M

N cos(0.866t)

5

7

x3 = e"t

K

L111

M

N

16. Eigenvalues: &1 = 2, &2 = 2, &3 = 1

Eigenvectors: v1 =

K

L!110

M

N, v2 =

K

L1!11

M

N

u1 =

K

L100

M

N

x(t) = c1(e2t ! te2t) ! c2e

2t + c3et

y(t) = c1te2t + c2e

2t ! c3et

z(t) = c3et

17. Eigenvalues: &1 = !1, &2 = 3, &3 = 3

Eigenvectors: v1 =

K

L001

M

N, v2 =

K

L1023

M

N v3 =

K

L01! 1

3

M

N

x(t) = c2e3t

y(t) = c3e3t

z(t) = c1e"t +

23c2e

3t ! 13c3e

3t

18. Eigenvalues: &1 = !1, &2 = 1, &3 = 1

Eigenvectors: v1 =

K

L100

M

N, v2 =

K

L! 1

21!2

M

N

u1 =

K

L3401

M

N

x(t) = c1e"t ! 1

2c2e

t + c3(34et ! 1

2tet)

y(t) = c2et + c3te

t

z(t) = !2c2et + c3(e

t ! 2tet)

19. & =

K

PPPPPPL

!1 + 2i!1 ! 2i

1222

M

QQQQQQN, v =

K

PPPPPPL

0 0 1 20 0 0 10 0 0 00 0 0 0!i i 0 01 1 0 0

M

QQQQQQN

x =

K

PPPPPPL

c3et + 2c4e2t + 2c5(te2t ! e2t) + c6(t2e2t ! 2te2t + 2e2t)c4e

2t + c5te2t + 1

2 c6t2e2t

c5e2t + c6te

2t

c6e2t

e"t(c1 sin(2t) + c2 cos(2t))e"t(c1 cos(2t) ! c2 sin(2t))

M

QQQQQQN

Page 199: MATH 224 Sol.

6.3. SOLVING LINEAR HOMOGENEOUS AND NONHOMOGENEOUS SYSTEMS OF EQUATIONS193

6.3 Solving Linear Homogeneous and Nonhomogeneous Sys-tems of Equations

1. Solution: &1 = !2, &2 = 3, v1 =

&!21

', v2 =

&1!3

'

x =

&!2c1e

"2t + c2e3t

c1e"2t ! 3c2e

3t

'

(a) !(t) =

&!2e"2t e3t

e"2t !3e3t

'

(b) " = V "1AV =

&3 00 !2

'" y! = "y " y =

&c1e

"2t

c2e3t

'" x = V y

(c) eAt = !(t)!"1(t0), !"1(0) =

15

&!3 !1!1 !2

'"

eAt =15

&6e"2t ! e3t 2e"2t ! 2e3t

3e3t ! 3e"2t !e"2t + 6e3t

'

2. Solution: &1 = !3, &2 = !4, v1 =

&!11

', v2 =

&!21

'

x =

&!c1e

"3t ! 2c2e"4t

c1e"3t + c2e

"4t

'

(a) !(t) =

&!e"3t !2e"4t

e"3t e"4t

'

(b) " = V "1AV =

&!3 00 !4

'" y! = "y " y =

&c1e

"3t

c2e"4t

'" x = V y

(c) eAt = !(t)!"1(t0) =

&2e"4t ! e"3t !2e"3t + 2e"4t

e"3t ! e"4t !e"4t + 2e"3t

'

3. Solution: &1 = 2, &2 = 4, v1 =

&01

', v2 =

&!21

'

x =

&!2c2e4t

c1e2t + c2e

4t

'

(a) !(t) =

&0 !2e4t

e2t e4t

'

(b) " = V "1AV =

&2 00 4

'" y! = "y " y =

&c1e2t

c2e4t

'" x = V y

(c) eAt = !(t)!"1(t0) =

&e4t 0

! 12e4t + 1

2e2t e2t

'

4. Solution: &1 = 1, &2 = 2, v1 =

&!11

', v2 =

&!21

'

x =

&!c1e

t ! 2c2e2t

c1et + c2e

t

'

(a) !(t) =

&!e"t !2e2t

et e2t

'

(b) " = V "1AV =

&1 00 2

'" y! = "y " y =

&c1e

t

c2e2t

'" x = V y

(c) eAt = !(t)!"1(t0) =

&!et + 2e2t 2e2t ! 2et

!e2t + et 2et ! e2t

'

Page 200: MATH 224 Sol.

194 Section 6.3

5. Solution: &1 = !1, &2 = 1, &3 = 2, v1 =

K

L1!21

M

N , v2 =

K

L!110

M

N , v3 =

K

L010

M

N

x =

K

Lc1e

"t ! c2et

!2c1e"t + c2e

t + c3e2t

c1e"t

M

N

(a) !(t) =

K

Le"t !et 0

!2e"t et e2t

e"t 0 0

M

N

(b) " = V "1AV =

K

L!1 0 00 1 00 0 2

M

N " y! = "y " y =

K

Lc1e

"t

c2et

c3e2t

M

N " x = V y

(c) eAt = !(t)!"1(t0) =

K

Let 0 e"t ! et

!et + e2t e2t e2t ! 2e"t + et

0 0 e"t

M

N

6. Solution: &1 = !1, &2 = 1, &3 = 3, v1 =

K

L!101

M

N , v2 =

K

L11!2

M

N , v3 =

K

L001

M

N

x =

K

L!c1e

"t + c2et

c2et

c1e"t ! 2c2e"t + c3e3t

M

N

(a) !(t) =

K

L!e"t et 0

0 et 0e"t !2et e3t

M

N

(b) " = V "1AV =

K

L!1 0 00 1 00 0 3

M

N " y! = "y " y =

K

Lc1e

"t

c2et

c3e3t

M

N " x = V y

(c) eAt = !(t)!"1(t0) =

K

Le"t !e"t + et 00 et 0

e3t ! e"t e3t + e"t ! 2et e3t

M

N

7. Solution: &1 = !1, &2 = 3, &3 = 1, v1 =

K

L!312

M

N , v2 =

K

L112

M

N , v3 =

K

L11!2

M

N

x =

K

L!3c1e

"t + c2e3t + c3e

t

c1e"t + c2e3t + c3et

2c1e"t + 2c2e

3t ! 2c3et

M

N

(a) !(t) =

K

L!3e"t e3t et

e"t e3t et

2e"t 2e3t !2et

M

N

(b) " = V "1AV =

K

L!1 0 00 3 00 0 1

M

N " y! = "y " y =

K

Lc1e

"t

c2e3t

c3et

M

N " x = V y

(c) eAt = !(t)!"1(t0) =14

K

L3e"t + e3t e3t ! 3e"t2et e3t ! et

e3t ! e"t 2et + e3t + e"t e3t ! et

2e3t ! 2e"t 2e3t + 2e"t ! 4et 2et + 2e3t

M

N

Page 201: MATH 224 Sol.

6.4. NONLINEAR EQUATIONS AND PHASE PLANE ANALYSIS 195

6.4 Nonlinear Equations and Phase Plane Analysis

1. Equilibria: (±2, 0)

J(!2, 0) =

&0 14 0

'" &1 = 2, &2 = !2 " Saddle

J(2, 0) =

&0 1!4 0

'" &1 = 2i, &2 = !2i " Linear center

2. Equilibria: (±1, 1)

J(1, 1) =

&0 12 !1

'" &1 = 1, &2 = !2 " Saddle

J(!1, 1) =

&0 1!2 !1

'" &1 =

!1 + i$

72

, &2 =!1 ! i

$7

2" Spiral

Page 202: MATH 224 Sol.

196 Section 6.4

3. Equilibria: (0, 0), (!2,!2)

J(0, 0) =

&!1 10 2

'" &1 = !1, &2 = 2 " Saddle

J(2, 0) =

&!1 1!4 2

'" &1 =

1 + i$

72

, &2 =1 ! i

$7

2" Unstable Spiral

4. Equilibria: (0, 0), (9, 3)

J(0, 0) =

&!1 01 !3

'" &1 = !1, &2 = !3 " Stable

J(9, 3) =

&!1 61 !3

'" &1 = !2 +

$7, &2 = 2 !

$7 " Saddle

Page 203: MATH 224 Sol.

Section 6.4 197

5. Equilibria: (0, 0), (1, 0), (!1, 0)

J(0, 0) =

&0 1!1 0

'" &1 = i, &2 = !i " Linear center

J(1, 0) =

&0 12 0

'" &1 =

$2, &2 = !

$2 " Saddle

J(!1, 0) =

&0 12 0

'" &1 =

$2, &2 = !

$2 " Saddle

6. Equilibria: (#6

,12), (!#

6,!1

2)

J(#6

,12) =

& #3

2 !10 1

'" &1 = 1, &2 =

$3

2" Unstable

J(!#6

,!12) =

& #3

2 !10 !1

'" &1 = !1, &2 =

$3

2" Saddle

Page 204: MATH 224 Sol.

198 Section 6.4

7. Equilibria: (±1, 1)

J(!1, 1) =

&0 2!2 !1

'" &1 = !1

2+

12

$15, &2 = !1

2!

$15 " Stable

J(1, 1) =

&0 22 !1

'" &1 = !1

2+

12

$17, &2 = !1

2!

$17 " Stable

8. Equilibria: (0, 0)

J(0, 0) =

&1 10 !8

'" &1 = 1, &2 = !8 " Saddle

Page 205: MATH 224 Sol.

Section 6.4 199

9. Equilibria: (0, 0), (0, 2), (0, 3)

J(0, 0) =

&3 00 2

'" &1 = 3, &2 = 2 " Unstable

J(0, 2) =

&1 0!2 !2

'" &1 = 1, &2 = !2 " Saddle J(0, 3) =

&!3 !30 !1

'" &1 = !3, &2 = !1 "

Stable

10. Equilibria: (0, 0), (0, 2), (3, 0), (1, 1)

J(0, 0) =

&3 00 2

'" &1 = 3, &2 = 2 " Unstable

J(0, 2) =

&!1 0!2 !2

'" &1 = !1, &2 = !2 " Stable

J(3, 0) =

&3 !60 !1

'" &1 = !3, &2 = !1 " Stable

J(1, 1) =

&!1 !2!1 !1

'" &1 = !1 +

$2, &2 = !1 !

$2 " Saddle

Page 206: MATH 224 Sol.

200 Section 6.4

11. Equilibria: (0, 0), (3, 0), (2, 1)

J(0, 0) =

&3 00 !2

'" &1 = 3, &2 = !2 " Saddle

J(3, 0) =

&!3 !30 1

'" &1 = !3, &2 = 1 " Saddle

J(2, 1) =

&!2 !21 0

'" &1 = !1 + i, &2 = !1 ! i " Stable spiral

12. u1 = x, u2 = x! " u!1 = u2, u!

2 = !(u2(u21 ! 1) ! u1 " x! = y, y! = !(y(x2 ! 1) ! x " Equilibria: (0, 0)

J(0, 0) =

&0 1!1 (

'" &1 =

(+$(2 ! 42

, &2 =(!

$(2 ! 42

"

Stable for ( < 0, spiral for !2 < ( < 0, node for ( ' !2Unstable for ( > 0, spiral for 0 < ( < 2, node for ( & 2Center for ( = 0Never a saddle.

(a) ! = 0.1 (b) ! = 10

Page 207: MATH 224 Sol.

Section 6.4 201

13. u1 = x, u2 = x! # u!1 = u2, u!

2 = "u1 " !u31 # x! = y, y! = "x " !x3 #

Equilibria: (0, 0),

!"1$"!

, 0

",

!1

$"!

, 0

"

J(0, 0) =

&0 1"1 0

'# "1 = i, "2 = "i # Center, %!

J

!"1$"!

, 0

"=

&0 12 0

'# "1 =

$2, "2 = "

$2 # Saddle, %!

J

!1

$"!

, 0

"=

&0 12 0

'# "1 =

$2, "2 = "

$2 # Saddle, %!

(a) # = 0.25 (b) # = 1

14. $! = v, v! ="g

msin $

Equilibria: centers at ($, $!) = (±2n%, 0), n = 0, 1, 2, · · ·saddles at ($, $!) = (±(2n + 1)%, 0), n = 0, 1, 2, · · ·

The closed orbits around the centers correspond to the typical back-and-forth pendulum motion. The sad-dles correspond to the pendulum standing on end. The trajectories outside the ones connecting the saddlescorrespond to a pendulum whirling over the top. These do not stop because there is no friction present.

Page 208: MATH 224 Sol.

202 Section 6.4

15. $! = v, v! = "0.3$! " $Equilibria: spiral sinks at ($, $!) = (±2n%, 0), n = 0, 1, 2, · · ·saddles at ($, $!) = (±(2n + 1)%, 0), n = 0, 1, 2, · · ·

Because friction is present, almost all trajectories eventually go to the rest position of the hanging pendulum.Some trajectories will end up with the pendulum standing on end.

16. r < 0 =# no equilibriar = 0 =# (0, 0) is degenerate node (“half-stable”)r > 0 =# ("

$r, 0) is unstable node, (

$r, 0) is stable node

This is a saddlenode bifurcation.

r < 0 r = 0 r > 0

17. r < 0 =# (r, 0) is unstable node, (0, 0) is stable noder = 0 =# (0, 0) is degenerate node (“half-stable”)r > 0 =# (0, 0) is unstable node, (r, 0) is stable nodeThis is a transcritical bifurcation.

r < 0 r = 0 r > 0

Page 209: MATH 224 Sol.

Section 6.4 203

18. r < 0 =# (0, 0) is stable noder = 0 =# (0, 0) is stable noder > 0 =# (0, 0) is a saddle, (±r, 0) is stable nodeThis is a supercritical pitchfork bifurcation.

r < 0 r = 0 r > 0

19. r < 0 =# (0, 0) is stable node, (±r, 0) are saddlesr = 0 =# (0, 0) is a saddler > 0 =# (0, 0) is a saddleThis is a subcritical pitchfork bifurcation.

r < 0 r = 0 r > 0

20. Equilibria: (0, 0)

J(0, 0) =

&r 2"2 r

'# "1 = r + 2i, "2 = r " 2i # (0, 0) is a stable spiral when r < 0 and an unstable spiral

when r > 0.(a) r = "0.25, no limit cycle (b) r = 0.25, stable limit cycle

Page 210: MATH 224 Sol.

204 Section 6.5

21. Equilibria: (0, 0)

J(0, 0) =

&r 2"2 r

'# "1 = r + 2i, "2 = r " 2i # (0, 0) is a stable spiral when r < 0 and an unstable spiral

when r > 0.(a) r = "0.25, unstable limit cycle (b) r = 0.25, no limit cycle

6.5 Epidemiological Models

1. (a)d)ds

=")s ! ')!"s) = !1 +

'"s

" )! '"

ln s + s = C

(b) IC " )! '"

ln s + s = )(0) ! '"

ln s(0) + s(0)

Evaluate at (s$, )$) " )$ ! '"

ln s$ + s$ = )(0) ! '"

ln s(0) + s(0)

(c) )$ ! )(0) +'"

ln

!s(0)s$

"+ s$ ! s(0) = 0

Substitution gives'"

ln

!s(0)s$

"+ s$ ! 1 = 0 " "

'=

ln,

s(0)s#

-

1 ! s$

2. Disease spreads when R0 > !" > 1. Thus we need 3.5" > 1 " " >

13.5

= .285.

3. If we let v = %vaccinated, then 1! v are not vaccinated. We replace s(0) and s$ with (1! v)s(0) and

(1 ! v)s$ respectively. This gives us 1 >

,(1"v)s(0)(1"v)s#

-

1 ! (1 ! v)s$" v >

,s(0)s#

-

s$= 1.94, which does not make

sense. If we go back to 1c, we would obtain 1 >"'

,(1"v)s(0)(1"v)s#

-

(1 ! v)(1 ! s$)" v < 1 !

,s(0)s#

-

1 ! s$= !.41, which

again doesn’t make sense. Our system is nonlinear and thus we cannot make the assumption statedin this problem because a percentage change in the initial condition does not necessarily correspondto the same change in the final solution.

4. (a) s = 0 " s! = 0)! = !')

4" s(t) = 0 (because s(0) = 0), and )(t) ( 0

(b) ) = 0 " s! = 0)! = 0

4" s(t) = s(0) and )(t) = 0

(c) If s + ) = 1, then

O) = 1 ! ss = 1 ! ) "

Os! = !"s(1 ! s) < 0)! = !")(1 ! )) ! ') < 0

Thus, solutions cannot leave the closed triangle, which makes the problem well-posed.

Page 211: MATH 224 Sol.

Section 6.5 205

#4 #5

5. (a) s = 0 # s! = µ&! = "(' + µ)&

4# s(t) increases (not always 0, but doesn’t leave set) and &(t) decreases

(b) & = 0 # s! = µ(1 " s)&! = 0

4# s(t) = increases and &(t) = 0

(c) If s + & = 1, then

O& = 1 " ss = 1 " &

#

ABBC

BBD

s! = µ(1 " s) " (s(1 " s)< 0 for s & 1> 0 for s '

&! = "(&(1 " &) " (' + µ)& < 0Thus, solutions cannot leave the closed triangle, which makes the problem well-posed.

6. Let s = S/N, & = I/N, r = R/N . Substitute

dSdt = "( I

N SdIdt = ( I

N S " 'IdRdt = 'I,

FG

H =#

ABC

BD

d(sN)dt = "( #N

N (sN)d(#N)

dt = ( #N (sN) " '(&N)

d(rN)dt = '(&N),

which gives the desired result after cancelling the constant N .

7. Again substituting s = S/N, & = I/N, r = R/N , we haved(sN)

dt = "( #NN (sN)

d(#N)dt = ( #

N (sN) " '(&N)d(rN)

dt = '(&N),

FBG

BH=#

dsdt = µ " µs " (&sd#dt = (&s " (' + µ)&drdt = '& " µr.

Now we let s+ &+r = 1 (and thus r = 1"s" &) to get the desired results, where the r! equation can be ignoredbecause knowing s(t) and &(t) gives r(t).

8. (a) N is time dependent; N is constant; s = SN , & = I

N give proportions of the students still in school while

r = R/N is the drop out proportion of the original population.

(b)N

N=

S + I

S + I + R=

S + I + R " R

S + I + R= 1 "

R

S + I + R= 1 " r

(c)

d(sN)dt = "(&sN

d(#N)dt = (&sN " '&N

d(rN)dt = '&N

FBG

BH=#

ABC

BD

dsdt = "(&sd#dt = (&s " '&drdt = '&N

N= '&(1 " r)

9. The equation satisfied by N !(t) isdN

dt= !" µN.

Solving this for N(t) gives

N(t) =

!N(0) "

!

µ

"e"µt +

!

µ.

Note thatlim

t%&e"µt = 0.

It is thus clear that N(t) ( !µ as t ( ).

10. (a) dNdt = 0 implies that the total population size remains constant for all time t * 0.

(b) It is su"cient to study the first two equations since they are independent of the third variable R andR = N " S " I. That is, there are 3 equations and one constraint, so there are 2 independent equations.

The DFE is given by (S$, I$) =,

µNµ+v , 0

-.

Page 212: MATH 224 Sol.

206 Section 6.5

(c) The Jacobian is

J(DFE) =

/"(µ + v) !µ

µ+v

0 !µµ+v " ' " µ

0

,

Because the matrix is upper triangular, the eigenvalues are immediate: "1 = "(µ + v), "2 =(µ

µ + v" ' " µ

Thus the DFE is locally asymptotically stable if !µµ+v " ' " µ < 0 and is unstable if "2 > 0

(d) Manipulating the only eigenvalue that can change sign, "2, we see that R0 =(µ

(µ + v)(' + µ), which can

be rewritten as

R0(v) =(

µ + '

µ

µ + v.

With no vaccination, this becomes R0(0) = !µ+" . R0 depends on two factors: the !

µ+" factor gives a

measure of those entering the I-class versus those leaving, while the µµ+v factor gives a factor that is

always less than or equal to one and is farther from one when v is bigger.

11. (a) Adding the equations, we see that

dN

dt=

dS

dt+

dE

dt+

dI

dt+

dR

dt= 0.

Thus N is a constant.

(b) The reduced 3-D system is thus

dS

dt= "(S

I

N+ µN " µS

dE

dt= (S

I

N" µE " #E

dI

dt= #E " µI " )I.

(6.1)

(c) At the DFE (N, 0, 0), the Jacobian is1

3"µ 0 " (0 "(µ + #) (0 # "(µ + ))

5

7

One of the eigenvalues, "1 = "µ (always less than zero) can be read immediately. Its location allows usto separate the matrix into it’s block form and now just consider

J(DFE) =

1

3"(µ + #) (# "(µ + ))

5

7

to determine the stability of the DFE. We can calculate the trace and determinant as

trJ = "(µ + # + µ + )) < 0,

DetJ = (µ + ))(µ + #) " #(.

We require Det J > 0 to ensure that the disease-free equilibrium is asymptotically stable, so that thedisease will die out under the condition of Det J > 0. Intuitively, the extinction of a disease meansthat new infections cannot replace the old/current ones. It turns out that the basic reproductive numbershould be less than 1. Therefore, because Det J > 0, we have

(#

(# + µ)(µ + ))< 1.

So,

R0 =(

(µ + ))

#

(# + µ).

We have thus shown that if R0 < 1, the disease-free equilibrium is asymptotically stable, i.e., that thedisease will die out. We know that the average life-span for the infectious class is 1

µ+$ , so !µ+$ represents

the total e#ective contact number of a typical infection. The quantity %%+µ is the fraction or probability

that an individual in the latent class becomes infected. Therefore,

R0 =(

(µ + ))

#

(µ + #)

represents the total number of new cases generated by a typical infection during its lifetime.

Page 213: MATH 224 Sol.

6.6. CHAPTER 6: ADDITIONAL PROBLEMS 207

6.6 Chapter 6: Additional Problems

1. False. The systems must also be homogeneous, constant coe#cient and with a full set of eigenvectors.

2. True.

3. False. The method of solution combines the complex (non-real) method with the repeated eigenvaluemethod.

4. True.

5. True. See a linear algebra text for more details.

6. (a) i. &1 = 2, &2 = 1, v1 =

&11

', v2 =

&01

'

ii. Unstable node

iii. x1 = e2tv1, x2 = etv2

iv. x =

&c1e

2t

c1e2t + c2e

t

'

(b) i. &1 = 4i, &2 = !4i, v1 =

&1i2

', v2 =

&1"i2

'

ii. Center

iii. x1 =

&10

'cos 4t !

&0

1/2

'sin 4t, x2 =

&10

'sin 4t +

&0

1/2

'cos 4t

iv. x =

&c1 cos 4t + c2 sin 4t

12c2 cos 4t ! 1

2 c1 sin 4t

'

7. (a) i. &1 = 2, &2 = !2, v1 =

&11

', v2 =

&1!3

'

ii. Saddle

iii. x1 = e2tv1, x2 = e"2tv2

iv. x =

&c1e2t + c2e"2t

c1e2t ! 3c2e

"2t

'

(b) iv. x(t) = 2c1 cos 4t + 2c2 sin 4t,y(t) = !c1 sin 4t + c2 cos 4t

8. (a) i. &1 = 1, &2 = 5, v1 =

&15

', v2 =

&11

'

ii. Unstable node

iii. x1 = etv1, x2 = e5tv2

iv. x =

&c1e

t + c2e5t

5c1et + c2e5t

'

(b) i. &1 = 7 + 4i, &2 = 7 ! 4i, v1 =

&1!i

', v2 =

&1i

'

ii. Unstable spiral

iii. x1 = e7t

!&10

'cos 4t !

&0!1

'sin 4t

", x2 = e7t

!&10

'sin 4t +

&01

'cos 4t

"

iv. x =

&e7t (c1 cos 4t + c2 sin 4t)e7t (c2 cos 4t + c1 sin 4t)

'

9. (a) i. &1 = !3, &2 = !2, v1 =

&1!2

', v2 =

&1! 3

2

'

ii. Stable node

iii. x1 = e"3tv1, x2 = e"2tv2

iv. x =

&c1e

"3t + c2e"2t

!2c1e"3t ! 32 c2e"2t

'

(b) i. &1 = 4i, &2 = !4i, v1 =

&1

!4i

', v2 =

&14i

'

Page 214: MATH 224 Sol.

208 Chapter 6 Review

ii. Center

iii. x1 =

&10

'cos 4t !

&0!4

'sin 4t, x2 =

&10

'sin 4t +

&04

'cos 4t

iv. x =

&c1 cos 4t + c2 sin 4t

4c1 sin 4t + 4c2 cos 4t

'

10. (a) i. &1 = !2, &2 = 1, v1 =

&132

', v2 =

&12

'

ii. Saddle

iii. x1 = e"2tv1, x2 = etv2

iv. x =

&c1e

"2t + c2et

32c1e

"2t + 2c2et

'

(b) i. &1 = 1 + 2i, &2 = 1 ! 2i, v1 =

&1

3 ! 2i

', v2 =

&1

!3 + 2i

'

ii. Unstable spiral

iii. x1 = et

!&13

'cos 2t !

&0!2

'sin 2t

", x2 = et

!&1!3

'sin 2t +

&02

'cos 2t

"

iv. x =

&et (c1 cos 2t + c2 sin 2t)

et (3c1 cos 2t + 2c2 sin 2t ! 3c1 sin 2t + 2c2 cos 2t)

'

11. (a) i. &1 = !1, &2 = !2, v1 =

&!11

', v2 =

&1!2

'

ii. Stable node

iii. x1 = e"tv1, x2 = e"2tv2

iv. x =

&c1e

"t + c2e"2t

c1e"t ! 2c2e

"2t

'

(b) i. &1 = 2 + i, &2 = 2 ! i, v1 =

&!3 ! i

1

', v2 =

&!3 + i

1

'

ii. Unstable spiral

iii. x1 = e2t

!&!31

'cos t !

&!10

'sin t

", x2 = e2t

!&!31

'sin t +

&10

'cos t

"

iv. x =

&e2t (!3c1 cos t + c1 sin t ! 3c2 sin t + c2 cos t)

e2t (c1 cos t + c2 sin t)

'

12. (a) i. &1 = !8, &2 = !12, v1 =

&1!2

', v2 =

&12

'

ii. Stable node

iii. x1 = e"8tv1, x2 = e"12tv2

iv. x =

&c1e

"8t + c2e"12t

!2c1e"8t + 2c2e

"12t

'

(b) i. &1 = 1 + 2i, &2 = 1 ! 2i, v1 =

&1 ! i

1

', v2 =

&1 + i

1

'

ii. Unstable spiral

iii. x1 = et

!&11

'cos 2t !

&!10

'sin 2t

", x2 = et

!&11

'sin 2t +

&10

'cos 2t

"

iv. x =

&et (c1 cos 2t + c2 sin 2t + c1 sin 2t + c2 cos 2t)

et (c1 cos 2t + c2 sin 2t)

'

13. (a) i. &1 = !1, &2 = !5, v1 =

&15

', v2 =

&11

'

ii. Stable node

iii. x1 = e"tv1, x2 = e"5tv2

iv. x =

&c1e

"t + c2e"5t

5c1e"t + c2e

"5t

'

Page 215: MATH 224 Sol.

Chapter 6 Review 209

(b) i. &1 = 4 +$

3i, &2 = 4 !$

3i, v1 =

&1

1+#

3i2

', v2 =

&1

1"#

3i2

'

ii. Unstable spiral

iii. x1 = e4t

!&1

1/2

'cos

$3t !

&0$3/2

'sin

$3t

", x2 = e4t

!&1

1/2

'sin

$3t +

&0

!$

3/2

'cos

$3t

"

iv. x =

=e4t#c1 cos

$3t + c1 sin

$3t + c2 cos

$3t$

e4t,

12 c1 cos

$3t !

#3

2 c2 sin$

3t + 12c1 sin

$3t !

#3

2 c2 cos$

3t->

14. (a) i. &1 = !1, v1 =

&10

'

ii. Stable node

iii. u1 =

&01

'

iv. x =

&c1e

"t + c2te"t

c2e"t

'

(b) i. &1 = !2, v1 =

&1!3

'

ii. Stable node

iii. u1 =

&01

'

iv. x =

&c1e

"2t + c2te"2t

!3c1e"2t + c2(!3te"2t + e"2t)

'

15. (a) i. &1 = 1, v1 =

&132

'

ii. Unstable node

iii. u1 =

&0

!1/4

'

iv. x =

&c1e

t + c2tet

32c1et + 3

2 c2tet ! 14c2et

'

(b) i. &1 = 2, v1 =

&21

'

ii. Unstable node

iii. u1 =

&!1/3

0

'

iv. x =

&2c1e

2t + c2(2te2t ! 13e2t)

c1e2t + c2te

2t

'

16. (a) i. &1 = 3, v1 =

&10

'

ii. Unstable node

iii. u1 =

&0

1/2

'

iv. x =

&c1e3t + c2te3t

12c2e

3t

'

(b) i. &1 = 3, v1 =

&21

'

ii. Stable node

iii. u1 =

&!10

'

iv. x =

&2c1e3t + c2(2te3t ! e3t)

c1e3t + c2te

3t

'

Page 216: MATH 224 Sol.

210 Chapter 6 Review

17. x =

&c1e

2t

c1e2t + c2et

'

18. x =

&c1e

"2t + c2e2t

!3c1e"2t + c2e2t

'

19. x =

&15c1et + c2e5t

c1et + c2e

5t

'

20. x =

K

Lc1e2t + c2e3t + c3e"2t

!c1e2t ! 1

2c2e3t ! 1

2c3e"2t

c1e2t + 1

2c2e3t + c3e

"2t

M

N

21. x =

K

Lc1e

5t + c2et ! c3e

"t

!c1e5t + c2e

t ! c3e"t

c1e5t + c2e

t + c3e"t

M

N

23. a. i. &1 = !4,&2,3 = 1,

v1 =

1

3!101

5

7 ,v2 =

1

3!213

5

7; ii. saddle;

b. i. &1 = 3, &2 = 3 + 6i, &3 = 3 ! 6i,

v1 =

1

31!2!3

5

7 ,v2 =

1

3!21

92 + 3

2 i

5

7 v3 =

1

3!21

92 ! 3

2 i

5

7; ii. unstable;

26. a. i. &1 = 1, &2 = !1, &3 = 5,

v1 =

1

3111

5

7 ,v2 =

1

311!1

5

7 ,v2 =

1

31!11

5

7; ii. saddle;

b. i. &1 = !3, &2 = 5, &3 = 1,

v1 =

1

3!1!11

5

7 ,v2 =

1

31!11

5

7 v3 =

1

3111

5

7; ii. saddle

28. (!1, 1) unstable node; (!1,!1) saddle

Page 217: MATH 224 Sol.

Chapter 6 Review 211

29. (0, 1) saddle; (0,!1) linear center

30. (0, 0) stable spiral; (!1,!1) saddle; (1, 1) saddle

Page 218: MATH 224 Sol.

212 Chapter 6 Review

31. i. (x$, y$) = (0, 0), (1, 1);ii. (1, 1) is a saddle, (0, 0) is an indeterminate form since linearization gives a zero eigenvalue; graph-ically, (0, 0) is stable from above but unstable from below.

Page 219: MATH 224 Sol.

Chapter 7

Laplace Transforms

7.1 Fundamentals of the Laplace Transform

1.2s3

, s > 0

2.1 ! 3s

s2, s > 0

3.1

(s + 1)2, s > !1

4.2 ! 2e"4s

s, s > 0

5.2s + 1

s2, s > 0

6.1 ! e"s ! se"2s

s2, s > 0

7.e"2s(se2s + es(2 + s) ! 2(1 + 2s + 2s2))

s3, s > 0

8.1

s2 ! 1, s > 1

9. Laplace transform does not exist (integral does not converge).

10.28

s ! 13, s > 13

11. L{t"} =

% &

0

e"stt" dt

!Let st = u " t" =

u"

s", dt =

dus

"=

% &

0

e"u u"

s"dus

=1

s"+1

% &

0

u" e"u du

=$('+ 1)

s"+1which converges only if ' > !1

12. (a) LO

1$t

4=$(1/2)s1/2

=

9#s

(b) LS$

tT

=$(3/2)s3/2

=

$#

2s3/2

(c) LS

t$

tT

=$(5/2)s5/2

=2$#

4s5/2

13. )',", M, N such that e""t |f(t)| < M, e"!t |g(t)| < N and let ! = max {', "} "M + N > e""t |f(t)| + e"!t |g(t)| > e"$t [|f(t)| + |g(t)|] > e"$t |f(t) + g(t)| " f(t) + g(t) is exponen-tial.

14. Let I1 be the partition of [a, b] for f(t) and I2 be the partition of [a, b] for g(t). With I = I1 * I2, bothf and g are piecewise continuous on [a, b] with partition I . Since f and g both continuous impliesf + g is continuous, then f + g is piecewise continous on [a, b] with partition I .

213

Page 220: MATH 224 Sol.

214 Section 7.3

15. L{f(t) + g(t)} =

% &

0

e"st [f(t) + g(t)] dt =

% &

0

e"stf(t) dt +

% &

0

e"stg(t) dt = L{f(t)} + L{g(t)} and

this transform exists for s > ! (see problem #13 for !) by theorem 7.1.3 and problems 13 and 14.

16. (a) e""t8883 sin(et2)

888 ' 3 for t > 0 and ' = 1 " f(t) is of exponential order.

(b) f !(t) = 6tet2 cos(et2) and limt%&

e""t6tet2 cos(et2) = + " f !(t) is not of exponential order.

7.2 Properties of the Laplace Transform

1.1 ! 5s

s2

2.1 + 5s

s2

3.1

(s ! 2)2

4.5s ! 14(s ! 3)2

5.12s

+s

s2 + 4

6.2a2 + s2

s(s2 + 4a2)

7.4s

(s2 + 9)(s2 + 1)

8.2abs

(s2 + (b + a)2)(s2 + (a ! b)2)

9.6

(s2 + 9)(s2 + 1)

10.2

s(s2 + 4)+

2s2 + 1

11.'

s2 + '2

12.s

s2 + '2

13.' cos%! s sin%

s2 + '2

14. (a)

% &

s

F (z) dz =

% &

s

!% &

0

e"ztf(t) dt

"dz =

% &

0

!% &

s

e"ztf(t) dz

"dt

=

% &

0

!!1

te"ztf(t)

8888z=&

z=s

"dt =

% &

0

1te"stf(t) dt

(b) Di#culty comes from t in the denominator, but limt%0

sin tt

= 1.

(c) LO

sin tt

4=

% &

s

!% &

0

e"zt sin(t) dt

"dz =

% &

s

1z2 + 1

dz = arctan z|&s =#2! arctan s

7.3 Step Functions, Translated Functions, and Periodic Func-tions

1.3e"4s

s

2.5e"6s

s

Page 221: MATH 224 Sol.

7.4. THE INVERSE LAPLACE TRANSFORM 215

3.4 ! 4e"10s

s

4.25

#e"5s ! e"7s$

5.6e"9s ! 6e"3s

s6. NOTE: t > 5 for piecewise definition of function

1 + e"2s + e"3s ! 3e"5s

s

7.1 ! e"2s + e"6s

s

8.1 ! e"3s

s2

9.2 ! 2e"5s

s2

10. e"s 1e + es

=e"s"1

s + 1

11.3e"7s

s! 4e"4s

s+

e"4s(1 + 4s)s2

12.6s

+2e"s

s! 2e"s(1 + s)

s2+

2e"3s(1 + 3s)s2

13.1 + e2

e2

14. 0

15. 1

16.1 ! 2e4

e4

17.

34

=

% 1

0

sin2[#(t ! t0)]*(t !12) dt

= sin2[#(12! t0)]

=12!

cos(2#( 12 ! t0))

2

!12

= cos(2#(12! t0))

= cos(# ! 2#t0)

= cos #(cos(2#t0)) + sin #(sin(2#t0))

12

= cos(2#t0)

t0 =cos"1( 1

2 )

2#

=16

18. Integrand only has value at t = 2 " n = 3

19. !1 + U0(t)

7.4 The Inverse Laplace Transform

1.32

sin(2t)

Page 222: MATH 224 Sol.

216 Section 7.5

2. 5 cosh(3t)

3. 5te3t

4. 5e"2t ! 10te"2t

5. e"at cos($

3t)

6. e"4t(cos(2t) + sin(2t))

7.12

,1 ! cos

$2 +

$2 sin(

$2t)-

8.t6

sin(3t) +154

sin(3t) ! t18

cos(3t)

9. F (s) =2s + 7

(s + 3)4=

2(s + 3)3

+1

(s + 3)4" f(t) = t2e"3t +

16t3e"3t

10. F (s) =8(s + 1)

(2s ! 1)3=

(s ! 12 )

(s ! 12 )3

+32

(s ! 12 )3

" f(t) = tet/2 +34t2et/2

11. F (s) =s ! 2

s2 + 5s + 6=

5s + 3

+!4

s + 2" f(t) = 5e"3t ! 4e"2t

12. F (s) =5s + 8

s2 + 3s ! 10=

177

s + 5+

187

s ! 2" f(t) =

177

e"5t +187

e2t

13. L"1 {F (s)} = U&(t)f(t ! #) where f(t ! #) = !5 cos(3t) ! 2 sin(3t) =

O0 0 < t < #

!5 cos(3t) ! 2 sin(3t) t > #

14. L"1 {F (s)} = U3(t)f(t ! 3) where f(t ! 3) = 2e"2t+6 cos(3t ! 9) +53e"2t+6 sin(3t ! 9)

=

O0 0 < t < 3

2e"2t+6 cos(3t ! 9) + 53e"2t+6 sin(3t ! 9) t > 3

15. L"1 {F (s)} = U4(t)f(t ! 4) ! U7(t)f(t ! 7) =

AC

D

0 0 < t < 4t ! 4 4 < t < 7

3 t > 7

16.13

[sin(9 ! 3t)U3(t ! 3) + 2 sin(3t)U0(t)]

7.5 Laplace Transform Solution of Linear Di!erential Equa-tions

1.12e3t +

32et

2.12

#3 cos t + 3 sin t ! 5e"t$

3. 2 cos(2t) +32

sin(2t)

4. 7 cos(4t)

5. 2e3t

6. e3t

7.137

e"4t +157

e3t

8.5 sin(2

$2t)$

2

9. 3e"t sin(2t) + 2e"t cos(2t)

10.74e"t +

9t4

e"t +14e"3t +

t4e"3t

11. !52e5t +

152

e3t ! 6te4t

Page 223: MATH 224 Sol.

7.6. SOLVING LINEAR SYSTEMS USING LAPLACE TRANSFORMS 217

12.3118

e2t ! 229

e"t +1318

e"t (cos(3t) ! sin(3t))

13. !e2t (! cos t + 7 sin t) +47

#e"5t ! e2t$

15.!20936

e4t +11436

te4t +112

e3t +174

e2t ! 8918

et

7.6 Solving Linear Systems using Laplace Transforms

1. x(t) = !e2t + 2et + 2y(t) = !e2t + et

2. x(t) = 2e2t +12e"t ! 1

2et

y(t) = !e2t +12e"t +

12et

3. x(t) = !3t ! e"t

y(t) = !1 + e"t + 2t

4. x(t) = 2t + 3 +43t3

y(t) =83t3 ! 2t2 + 5t + 5

5. !2et + 5e4t

!4et + 4e4t

6. x(t) = ! sin(2t) + 2 cos(2t) + e2t

y(t) = 5 sin(2t) + 2e2t

7. x(t) = 8 sin t + 2 cos t

y(t) = cos t ! 13 sin t +12et ! 1

2e"t

8. x(t) = !e"2t + 3e4t + 1 ! ty(t) = !e"2t ! 3e4t + t

9. x(t) = 2e"t + 2e"2t

y(t) = 2e"t + 3 ! 8e"2t

10. x(t) =47

,!1 + e

7t5

-

y(t) = !17! 16

49e

7t5 +

1649

11. x(t) = !3et + 2y(t) = !2et ! 8

12. x(t) = 4 sin(2t) ! 4 cos(2t)y(t) = cos(2t) + sin(2t)

13. x(t) = !4 +t3

3+ 5t ! 2t2 + 5e"t

y(t) = 5 ! 5t + 2t2 ! 5e"t

7.7 The Convolution

1.t3

3

2. !25

cos(2t) +15

sin(2t) +25e"t

3.3

128sin(4t) +

14t3 ! 3

32t

4.s cos(2t) + 2 sin(2t) ! se"st

s2 + 4

Page 224: MATH 224 Sol.

218 Chapter 7 Review

5. t

6. ! cos t + cos tU0(# ! 2t)

7. f(t) = e"2t ! e"3t

8. f(t) = !15e"4t +

15et

9. f(t) = !19

cos(3t) +19

10. f(t) =113

!1 ! e"2t cos(3t) ! 2

3e"2t sin(3t)

"

11. f(t) =127

! 19t +

16t2 ! 1

27e"3t

12. f(t) =153

!e"7t ! cos(2t) +

72

sin(2t)

"

13.

(f , g) , h =

% t

0

[(f , g)(t! + )] h(+ ) d+

=

% t

0

&% t"'

0

f((t ! + ) ! z)g(z) dz

'h(+ ) d+

f , (g , h) =

% t

0

f(t ! + )(g , h)(+ ) d+

=

% t

0

f(t ! + )&% '

0

g(+ ! z)h(z) dz

'd+

(interchange order of integration)

=

% t

0

h(z)

&% t

z

f(t ! + )g(+ ! z) d+

'dz

(let u = + ! z, du = d+ )

=

% t

0

h(z)

&% t"z

0

f(t ! (z + u))g(u) du

'dz

=

% t

0

h(z)

&% t"z

0

f((t ! z) ! u)g(u) du

'dz

=

% t

0

f , g(t ! z)h(z) dz

= (f , g) , h

16. LS

t"1/2T

=

9#s" L

St"1/2 , t"1/2

T=#s" L"1

SLS

t"1/2 , t"1/2TT

= #

17. (a) sin kt (cos kt sin t + sin kt ! cos t sin kt)

(b) sin kt (! cos kt + cos t cos kt + sin t sin kt)

(c) cos kt (cos kt sin t + sin kt ! cos t sin kt)

(d)12t sin kt

7.8 Chapter 7: Additional Problems

1. False, f(t) must be a real-valued function for which

% &

0

e"stf(t) dt exists.

2. True. Theorem 7.3.1

3. True. Problem 7.2.14

Page 225: MATH 224 Sol.

Chapter 7 Review 219

4. False. L{C1f1(t) + C2f2(t)} = C1L{f1(t)} + C2L{f2(t)}.5. True. Uniqueness of Laplace transforms.

6. !5e"7s

s+

5s

7.48s5

8.2

(s + 1)3

9.2s

+1s2

10.s ! 1

(s ! 1)2 + 1

11.1

s ! (a + bi)

16. 1 ! e"3

17. 0

18. 0

19. tan(#/4) = 1

20.53

sin 3t

21. 7 cos 2t

22. 6te2t

23. 5e"3t ! 15te"3t

24.115

e"5t/2

&15 cos

!t$

152

"! 11

$15 sin

!t$

152

"'

25.13e4t +

173

et

26. 2 cos 3t + sin 3t

27. cos 9t +29

sin 9t

28.37e4t +

47e"3t

29. 6e"t cos 2t + 3e"t sin 2t

30. 2e"2t + te"2t ! e"t + 2te"t

31.209

e2t +159

te2t ! 479

e5t + 8e3t

32.3825

e2t ! e"t ! 1325

cos 4t e"t ! 39100

sin 4t e"t

33. !16et ! 17

42e"5t ! 3

7e2t

34.9136

! 136

t ! 2320

e"2t ! 34e"t +

67180

e3t

37. x(t) = !119

e"t +5963

e"10t +167

e"3t

y(t) = !119

e"t ! 236315

e"10t ! 247

e"3t +125

38. x(t) =875

e2t ! 14e3t ! 125

cos t ! 245

sin t

y(t) = !875

e2t +563

e3t +245

sin t +125

cos t ! 23

Page 226: MATH 224 Sol.

220 Chapter 7 Review

39. x(t) = !52e2t +

32

y(t) = !2e2t ! 8

40. x(t) = !14e3t/2

&2 cosh(3t/2) ! 1

3sinh(3t/2)

'

y(t) = !10e3t/2 cosh(3t/2)

41.34t4

42.113

#!3 cos 3t + 2 sin 3t + 3e"2t$

43.!316

t +12t3 +

364

sin 4t

44.b cos at + a sin at ! be"bt

b2 + a2

45. sin t ! sin tU0(# ! 2t)

46. e"3t ! e"4t

47.113

#e5t ! e"8t$

48.!14

cos 2t +14

49.16t ! 1

36sin 6t

50.1

128

#e"8t ! cos 8t + sin 8t

$

Page 227: MATH 224 Sol.

Chapter 8

Series Methods

8.1 Power Series Representations of Functions

1. (a) limn%&

8888xn+1

xn

8888 = limn%&

|x| = |x| < 1 for convergence.

For x = 1,&U

n=0

1n diverges

For x = !1,&U

n=0

(!1)n diverges (the sequence of partial sums oscillates between 0 and 1...they

don’t converge)

(b)U

xn is a geometric series which converges to1

1 ! x.

To see this, observe Sk =kU

n=0

xn = 1 + x + x2 + · · · + xk and xSk = x + x2 + x3 + · · · + xk + xk+1 "

Sk ! xSk = 1 ! xk+1 " Sk =1 ! xk+1

1 ! x. From above, we know that |x| < 1, so

S(x) = limk%&

Sk = limk%&

1 ! xk+1

1 ! x=

11 ! x

.

2. (a) f(x) = sin x f(0) = 0f !(x) = cos x f !(0) = 1

f !!(x) = ! sin x f !!(0) = 0f !!!(x) = ! cos x f !!!(0) = !1

......

"

f(x) = sin x = 0 +1(x)1!

+0(x2)

2!+

(!1)(x3)3!

+0(x4)

4!+ · · · = x ! x3

3!+

x5

5!+ · · ·

(b) f(x) = cos x f(0) = 1f !(x) = ! sin x f !(0) = 0f !!(x) = ! cos x f !!(0) = !1f !!!(x) = sin x f !!!(0) = 0

......

"

f(x) = cos x = 1 +0(x)1!

+(!1)(x2)

2!+

(0)(x3)3!

+1(x4)

4!+ · · · = 1 ! x2

2!+

x4

4!+ · · ·

221

Page 228: MATH 224 Sol.

222 Section 8.1

(c) f(x) = sinh x f(0) = 0f !(x) = cosh x f !(0) = 1f !!(x) = sinhx f !!(0) = 0f !!!(x) = cosh x f !!!(0) = 1

......

"

f(x) = sinh x = 0 +1(x)1!

+0(x2)

2!+

(1)(x3)3!

+0(x4)

4!+ · · · = x +

x3

3!+

x5

5!+ · · ·

(d) f(x) = cosh x f(0) = 1f !(x) = sinhx f !(0) = 0f !!(x) = cosh x f !!(0) = 1f !!!(x) = sin x f !!!(0) = 0

......

"

f(x) = cosh x = 1 +0(x)1!

+(1)(x2)

2!+

(0)(x3)3!

+1(x4)

4!+ · · · = 1 +

x2

2!+

x4

4!+ · · ·

To show analytic, sin x =&U

n=0

(!1)nx2n+1

(2n + 1)!" lim

n%&

888 ("1)n+1x2(n+1)+1

(2(n+1)+1)!

888888 x2n+1

(2n+1)!

888= lim

n%&

888 ("1)n+1x2n+3

(2n+3)!

888888 x2n+1

(2n+1)!

888=

limn%&

8888x2

(2n + 3)(2n + 2)

8888 = 0, -x. The argument is similar for the other functions.

3. (a) f !(0) = limx%0

e"1/x2

x= lim

x%0

1x

e1/x2 = limx%0

"1x2

e1/x2 "2x3

= limx%0

x

2e1/x2 = 0

f !!(0) = limx%0

f !(x) ! f !(0)x

= limx%0

2

x4e1/x2 = 0

4. limn%&

88888

x2n+3

2n+3

x2n+1

2n+1

88888 = limn%&

8888x2(2n + 1)

2n + 3

8888 = |x2| < 1 for convergence " |x| < 1

For x = 1,&U

n=0

(!1)n(1)2n+1

2n + 1which is an alternating series which converges by the Alternating Series

Test.

For x = !1,&U

n=0

(!1)n(!1)2n+1

2n + 1=

&U

n=0

(!1)3n+1

2n + 1which is an alternating series which converges by

the Alternating Series Test.Therefore, the interval of convergence is |x| ' 1 and radius of convergence is R = 1.

5. ex = 1 + x + x2/2! + x3/3! + · · · , e"x = 1 ! x + x2/2! ! x3/3! + · · ·" exe"x = (1 + x + x2/2! + x3/3! + · · · )(1 ! x + x2/2! ! x3/3! + · · · )= (1 ! x + x2/2! ! x3/3! + · · · ) + x(1 ! x + x2/2! ! x3/3! + · · · )

+ (x2/2!)(1 ! x + x2/2! ! x3/3! + · · · ) + · · ·= 1

6. sin 2x =&U

n=0

(!1)n(2x)2n+1

(2n + 1)!= 2x ! (2x)3

3!+

(2x)5

5!+ · · ·

= 2x ! 8x3

3!+

32x5

5!+ · · ·

2 sin x cos x = 2(x ! x3/3! + x5/5! + · · · )(1 ! x2/2! + x4/4! + · · · )= 2(x ! x3/2! + x5/4! + · · ·! x3/3! + x5/(2!3!) + · · · + x5/5! + · · · )= 2(x ! 3x3/6 ! x3/6 + x5/24 + x5/12 + x5/120 + · · · )

= 2x ! 4x3

3+

32x5

120+ · · · = 2x ! 8x3

3!+

32x5

120+ · · · = sin 2x

Page 229: MATH 224 Sol.

8.2. THE POWER SERIES METHOD 223

7.

eix = 1 + (ix) + (ix)2/2! + (ix)3/3! + (ix)4/4! + · · ·= 1 + ix ! x2/2! ! ix3/3! + x4/4! + · · ·= (1 ! x2/2! + x4/4! + · · · ) + i(x ! x3/3! + x5/5! + · · · )= cos x + i sin x

8. limx%0

1 ! cos xx

= limx%0

1 ! (1 ! x2/2! + x4/4! + · · · )x

= limx%0

x/2! ! x3/4! + x5/6! + · · · = 0

9. (a)

W (x) =

8888f(x) g(x)f !(x) g!(x)

8888

= f(x)g!(x) ! f !(x)g(x)

= (a0 + a1(x ! c) + a2(x ! c)2 + · · · )(b1 + 2b2(x ! c) + · · · )!(a1 + 2a2(x ! c) + · · · )(b0 + b1(x ! c) + b2(x ! c)2 + · · · )

= a0b1 + 2a0b2(x ! c) + · · · + a1b1(x ! c) + 2a1b2(x ! c)2

+ · · · + a2b1(x ! c)2 + 2a2b2(x ! c)3 + · · ·!a1b0 ! aab1(x ! c) ! a1b2(x ! c)2 ! 2a2b0(x ! c) ! 2a2b1(x ! c)2 ! 2a2b2(x ! c)3

= a0b1 ! a1b0

(b) See Theorem 3.2.4

8.2 The Power Series Method

1. y = 1 + x +x2

2+

2x3

3+

7x4

12+ · · ·

2. y = 1 + 3(x ! 1) + 4(x ! 1)2 + 3(x ! 1)3 +32(x ! 1)4 + · · ·

3. y =x2

2+

x3

6+

x4

6+

x5

15+ · · ·

4. y = 1 + x + x2 +x3

3+

x4

12+ · · ·

5. y = 1 + x + x2 +4x3

3+

7x4

6+ · · ·

6. y =x2

2+

x3

6+

x4

24+

x5

120+ · · ·

7. y = x + x2 +x3

2+

x4

8! x5

15+ · · ·

8. y = 1 + 2x + 4x2 +25x3

3+

49x4

4+ · · ·

9. y = 1 + x +x3

3! x4

3+ · · ·

10. y = x +x3

3+

x5

30! x7

56+ · · ·

11. y = 1 + 2(x ! 1) + 4(x ! 1)2 +253

(x ! 1)3 +814

(x ! 1)4 + · · ·

12. y = 1 + 2(x ! 1) +72(x ! 1)2 +

143

(x ! 1)3 +7312

(x ! 1)4 + · · ·

13. y = 1 ! x +x3

3+

x4

12! x5

60+ · · ·

Page 230: MATH 224 Sol.

224 Section 8.3

14. y = 1 ! x ! x2

2+

x3

6! x4

12! x5

24+ · · ·

15. y = c0

!1 ! x3

6+

3x5

40

"+ c1

!x ! x3

6! x4

12+

3x5

40

"+ · · ·

16. y = x + x2 +x3

2+

x4

6+ · · ·

17. y = 1 ! x4

+ x2 +x4

2+ · · ·

18. y = c0

!1 +

x3

6

"+ c1

!x +

x4

12

"

19. c0

!1 ! px2 +

16(p ! 2)px4

"+ c1

!x +

x3(1 ! p)3

+130

(3 ! p)x5 +130

(p ! 3)px5

"

8.3 Ordinary and Singular Points

1.

x0 = Singular point p(x) q(x) (x ! x0)p(x) (x ! x0)2q(x) Classify

x0 = 0x + 2

x2 ! 3x1

x2 ! 3xx + 2x ! 3

xx ! 3

Regular

x0 = 3x + 2

xx ! 3

xRegular

2.

x0 = Singular point p(x) q(x) (x ! x0)p(x) (x ! x0)2q(x) Classify

x0 = 0x2 ! 2xx3 + x2

4x3 + x2

x2 ! 2xx2 + x

4x + 1

Irregular

x0 = !1x2 ! 2x

x2

4x + 4x2

Regular

3.

x0 = Singular point p(x) q(x) (x ! x0)p(x) (x ! x0)2q(x) Classify

x0 = 01

x(x + 2)2x ! 1

x2(x + 2)1

x + 22x ! 1x + 2

Regular

x0 = !21x

(2x ! 1)(x + 2)x2

Regular

4.

x0 = Singular point p(x) q(x) (x ! x0)p(x) (x ! x0)2q(x) Classify

x0 = 01

x3(x + 3)(x ! 2)1

x3(x + 3)1

x2(x + 3)(x ! 2)1

x(x + 3)Irregular

x0 = 21

x(x + 3)x ! 2

x3(x + 3)Regular

x0 = !31

x(x ! 2)x + 3x3

Regular

5.

x0 = Singular point p(x) q(x) (x ! x0)p(x) (x ! x0)2q(x) Classify

x0 = 0!(x + 2)

x2(2x + 1)2ex

x2(2x + 1)!(x + 2)x(2x + 1)

2ex

2x + 1Irregular

x0 = ! 12

!(x + 2)2x2

ex(2x + 1)2x2

Regular

6.

x0 = Singular point p(x) q(x) (x ! x0)p(x) (x ! x0)2q(x) Classify

x0 = 02

x2(x ! 1)1

(x ! 1)22

x(x ! 1)x2

(x ! 1)2Irregular

x0 = 12x2

1 Regular

7. T5(x) = 5x ! 20x3 + 16x5

T6(x) = !1 + 18x2 ! 48x4 + 32x6

8. (a) y = cos(n arccos x) " y! = n sin(n arccos x)

!1$

1 ! x2

""

y!! =nx sin(n arccos x)

(1 ! x2)$

1 ! x2! n2 cos(n arccos x)

1 ! x2"

Page 231: MATH 224 Sol.

8.4. THE METHOD OF FROBENIUS 225

(1 ! x2)y!! =nx sin(n arccos x)$

1 ! x2! n2 cos(n arccos x)

!xy! =!nx sin(n arccos x)$

1 ! x2, n2y = n2 cos(n arccos x) " (1 ! x2)y!! ! xy! + n2y = 0

(b) • n = 0; y(x) = cos 0 = 1 = T0(x)

• n = 1; y(x) = cos(arccos x) = x = T1(x)

• n = 2; y(x) = cos(2 arccos x) = 2 cos2(arccos x) ! 1 = 2x2 ! 1 = T2(x)

8.4 The Method of Frobenius

The Mathematica code may need to be modified as follows:

yr[x_] = c0*x^r + c1*x^(r + 1) + c2*x^(r + 2) + c3*x^(r + 3) + c4*x^(r + 4)eqODE[x_] = (x^2)*y’’[x] + (3 x - x^2)*y’[x] + (5 - x)*y[x]

(*your differential equation*)eq1 = Collect[ReplaceAll[eqODE[x], {y[x] -> yr[x], y’[x] -> yr’[x],

y’’[x] -> yr’’[x]}],x^(r - 3)]eq4a = Coefficient[Coefficient[eq1, x^(r - 0)], x, 0](*In the above equation, the x^(r - 0) term needs to be adjusted based

on the result from eq1. If eq1 has a x^(r - 1),then r - 1 needs to replace r - 0 in eq4a and eq5a, r needs toreplace r + 1, r + 1 needs to replace r + 2,and r + 2 needs to replace r + 3 in the equations eq5b - 5d below*)

eq4b = Solve[eq4a == 0, r]eq5a = Coefficient[Expand[eq1], x^(r - 0)]eq5b = Coefficient[Expand[eq1], x^(r + 1)]eq5c = Coefficient[Expand[eq1], x^(r + 2)]eq5d = Coefficient[Expand[eq1], x^(r + 3)]eqsoln1 = Solve[ReplaceAll[{eq5a == 0, eq5b == 0, eq5c == 0, eq5d == 0},

eq4b[[1]]], {c1, c2, c3, c4}]eqsoln2 = Solve[ReplaceAll[{eq5a == 0, eq5b == 0, eq5c == 0, eq5d == 0},

eq4b[[2]]], {c1, c2, c3, c4}]y1[x_]=Collect[ReplaceAll[ReplaceAll[ReplaceAll[yr[x], eqsoln1[[1]]],

c4 -> 0],eq4b[[1]]], c0]y2[x_]=Collect[ReplaceAll[ReplaceAll[ReplaceAll[yr[x], eqsoln2[[1]]],

c4 -> 0],eq4b[[2]]], c0]

1. y1(x) = c0

!1$x! x3/2

2

"

y2(x) = c0

!x ! x3

14

"

4. y1(x) = c0

!x1/3 ! 3

2x7/3

"

y2(x) = c0

!x5/3 ! 3

10x11/3

"

5. y = 3c1x12

&U

n=0

2n+1(n + 1)(2n + 3)!

xn + c2x"1

/1 !

&U

n=1

2n"1(n ! 1)!(2n ! 2)!

xn

0

7. y = 3c1x13

!1 +

2x9

!2&U

n=2

(!1)n 1 · 4 · 7 . . . (3n ! 5)9nn!

xn + c2x

0

10. y1(x) = c0

!!1 +

1x

"

y2(x) = c0

!x ! x2

3+

x3

12! x4

60

"

Page 232: MATH 224 Sol.

226 Section 8.5

13. y = c1x3

1 + x+ c2(1 ! x + x2)

16. y = c1

&U

n=0

xn

(n + 1)!

19. y1(x) = c0$

x + c1x3/2

y2(x) = c0x3/2

21. y1(x) = c1

&U

n=0

(!1)n xn

(n!)2,

y = c1y1(x) + c2

/y1(x) ln x ! 2

&U

n=2

(!1)n 1 + 12 + . . . + 1

n

(n!)2xn

0

22. y1(x) = y2(x) = c0

#x ! x2 + x3 ! x4$

23. y1(x) = x + x2,

y = c1y1(x) + c2

/y1(x) ln x ! 2x2 !

&U

n=2

(!1)n xn+1

n(n ! 1)

0

24. Same as #16

27. y1(x) = xex,

y = c1y1(x) + c2

/y1(x) ln x1 !

&U

n=2

1 + 12 + . . . + 1

n

(n ! 1)!xn

0

28. p(x) = 1 ! 35x +

15x2 + . . .,

q(x) =15x ! 3

20x2 + . . .,

y = c1x[p(x) cos ln x ! q(x) sin lnx] + c2x[q(x) cos lnx + p(x) sin ln x]

8.5 Bessel Functions

1. limn%&

8888888

( x2 )2n+2

((n+1)!)2

( x2 )2n

(n!)2

8888888= lim

n%&

88888

#x2

$2n!n!

(n + 1)!(n + 1)!

88888 = limn%&

88888

#x2

$2

(n + 1)2

88888 = 0, -x " converges.

2. (a) limn%&

8888888

( x2 )2n+3

(n+1)!(n+2)!

( x2 )2n+1

(n!)(n+1)!

8888888= lim

n%&

88888

#x2

$2n!(n + 1)!

(n + 1)!(n + 2)!

88888 = limn%&

88888

#x2

$2

(n + 1)(n + 2)

88888 = 0, -x " converges.

(b) J !0(x) =

&U

n=1

(!1)n(2n)x2n"1

(n!)2(22n)=

&U

n=0

(!1)n+1x2n+1

(n + 1)!n!(22n+1)= (!1)

&U

n=0

(!1)nx2n+1

(n + 1)!n!(22n+1)= !J1(x)

3. (a) kxpJp"1(kx) =&U

n=0

(!1)n

n!$(n + p)k2n+px2n+2p"1

22n+p"1

Page 233: MATH 224 Sol.

Section 8.5 227

ddx

(xpJp(kx)) =ddx

/ &U

n=0

(!1)n k2n+p x2n+2p

n!$(n + p + 1) 22n+p

0

=&U

n=0

(!1)n k2n+p(2n + 2p)x2n+2p"1

n!$(n + p + 1) 22n+p

=&U

n=0

(!1)n k2n+p 2(n + p)x2n+2p"1

n!$(n + p + 1) 22n+p

=&U

n=0

(!1)n k2n+p (n + p)x2n+2p"1

n! (n + p)$(n + p) 22n+p"1

=&U

n=0

(!1)n k2n+p x2n+2p"1

n!$(n + p) 22n+p"1

= kxp Jp"1(kx)

(b) !kx"pJp+1(kx) =&U

n=0

(!1)n+1

n!$(n + p + 2)k2n+p+2x2n+1

22n+p+1

ddx

#x"pJp(kx)

$=

ddx

/ &U

n=0

(!1)n k2n+p x2n

n!$(n + p + 1) 22n+p

0

=&U

n=0

(!1)n k2n+p(2n)x2n"1

n!$(n + p + 1) 22n+p

=&U

n=0

(!1)n k2n+p x2n"1

(n ! 1)!$(n + p + 1) 22n+p"1

(n = m + 1)

=&U

m="1

(!1)m+1 k2m+p+2 x2m+1

m!$(m + p + 2) 22m+p+1

=(!1)0x"1kp

(!1)!$(p + 1)2p"1

&U

m=0

(!1)m+1 k2m+p+2 x2m+1

m!$(m + p + 2) 22m+p+1

=&U

m=0

(!1)m+1 k2m+p+2 x2m+1

m!$(m + p + 2) 22m+p+1

= !kx"p Jp+1(kx)

To see this,(!1)0x"1kp

(!1)!$(p + 1)2p"1=

x"1kp

$(0)$(p + 1)2p+1= 0

because limx%0!

$(x) = !+and limx%0+

$(x) = +

Page 234: MATH 224 Sol.

228 Section 8.5

4.

J0(kx) =&U

n=0

(!1)n k2n x2n

(n!)2 22n

" k2x J0(kx) =&U

n=0

(!1)n k2n+2 x2n+1

(n!)2 22n

= k2x +&U

n=1

(!1)n k2n+2x2n+1

(n!)2 22n

J !0(kx) =

&U

n=1

(!1)n k2n 2n x2n"1

(n!)2 22n

J !!0 (kx) =

&U

n=1

(!1)n k2n 2n (2n ! 1) x2n"2

(n!)2 22n

xJ !!0 (kx) =

&U

n=1

(!1)n k2n 2n (2n ! 1) x2n"1

(n!)2 22n

" xy!! + y! + k2xy = k2x +&U

n=1

(!1)n k2n x2n"1

(n!)2 22n

;k2 x2 + 4n2<

= k2x + k2x&U

n=1

(!1)n k2nx2n

(n!)2 22n+

&U

n=1

(!1)n k2nx2n"1 4n2

(n!)2 22n

= k2x + k2x J0(kx) +&U

n=1

(!1)n k2nx2n"1

((n ! 1)!)2 22n"2

= k2x + k2x (J0(kx) ! 1) + k2x&U

n=1

(!1)n k2n"2x2n"2

((n ! 1)!)2 22n"2

= k2x + k2x (J0(kx) ! 1) + k2x&U

n=0

(!1)n"1 k2nx2n

(n!)2 22n

= k2x + k2xJ0(kx) ! k2x + k2x (!J0(kx))

= 0

5. (a) y =u(x)$

x" y! =

$xu! ! u(1/2)x"1/2

x

x2 y!! = x3/2u!! ! x1/2u! +34u x"1/2

xy! =$

xu! ! u

2$

x#x2 ! p2$ y = x3/2u ! p2x"1/2u

x2y!! + xy! + (x2 ! p2)y = x3/2u!! ! x1/2u! +34ux"1/2 + x1/2u! ! 1

2ux"1/2 + x3/2u ! p2x"1/2u

= x3/2u!! +

!14! p2

"ux"1/2 + ux3/2

= x3/2

&u!! +

!14! p2

"ux"2 + u

'= 0

(assuming x #= 0) " u!! + u

&1 +

!14! p2

"x"2

'= 0

(b) p =12, y1 =

cos x$x

" u = cos x " u!! = ! cos x "! cos x + cos x

&1 +

!14! 1

4

"x"2

'= 0 Check!

(c) p =12, y1 =

sin x$x

" u = sin x " u!! = ! sin x " ! sin x + sin x

&1 +

!14! 1

4

"x"2

'= 0 Check!

Page 235: MATH 224 Sol.

8.6. CHAPTER 8: ADDITIONAL PROBLEMS 229

6.

y =$

xf('x!)

y! = f !('x!)('"x!"1/2) +f('x!)

2$

x

y!! = f !('x!)('"(" ! 12)x!"3/2) + f !!('x!)('"x!"1)('"x!"1/2)

+2$

xf !('x!)('"x!"1) ! f('x!)( 1#x)

4x

x2y!! +

!'2"2x2! +

14! p2"2

"y = '"2f !('x!)x!

$x ! 1

2f !('x!)'"x!

$x + f !!('x!)('2"2x2!)

$x

+12f !('x!)('")x!

$x ! 1

4

$xf('x!) + f('x!)'2"2x2!$x

+14

$xf('x!) ! p2"2f('x!)

$x

=$

x"2If !!('x!)'2x2! + f !('x!)'x! + f('x!)

,'2x2! ! p2

-J

,if u = 'x! , u2 = '2x2!

-

=$

x"2 ;u2f !!(u) + uf !(u) + f(u)#u2 ! p2$<

= 0 since f('x!) is a solution to a Bessel function of order p

8. Using the chain rule (from calculus) to convert the variables, we see thatdydt

=dydx

dxdt

= !e"t dydx

and

d2ydt2

=ddt

!!e"t dy

dx

"= e"2t d2y

dx2+ e"t dy

dx.

Substitution of this (together with x = e"t) into the ODE gives the desired result.

8.6 Chapter 8: Additional Problems

1. False. See Theorem 8.1.1.

2. True, this is just one reason they are useful.

3. If you think this is true, then you have missed the point of this chapter.

4. False. See Definition 8.1.

5. False, check out Taylor’s theorem.

9. y = 3 + 12x + 12x2 + 8x3 + 4x4 + · · ·

10. y = 1 + 4(x ! 1) + 7(x ! 1)2 +203

(x ! 1)3 +236

(x ! 1)4 + · · ·

11. y = 6 ! 2x ! 4x2 + x3 +3x4

4+ · · ·

12. y = 3 + 6x +13x2

2+

13x3

3+

13x4

6+ · · ·

13. y = 5 + 3x +25x2

2+ 10x3 +

67x4

2+ · · ·

14. y = 1 + 2x + x2 +x3

6+

x4

24+ · · ·

15. y = x + x2 + x3 +7x4

12+

3x5

10+ · · ·

16. y = 6 + 30x + 75x2 +376x3

3+

470x4

3+ · · ·

17. y = 2 + 16x + 256x2 +14338

3x3 +

2867363

x4 + · · ·

Page 236: MATH 224 Sol.

230 Chapter 8 Review

18. y = 2 + x +cos(2)

2x2 ! sin(2)

3x3 + · · ·

19. y = 4 + 16x + 64x2 + 256x3 +4097

4x4 + · · ·

20. y = 6 + 432x + 46656x2 +16796164

3x3 + 705438792x4 + · · ·

22. y = c1x43

,1 ! 3x2

16 + 9x4

896 ! . . .-

+c2x23

,1 ! 3x2

8 + 9x4

320 ! . . .-

25. y = c1x13

,1 ! 3x2

16 + 9x4

896 ! . . .-

+c2x!13

,1 ! 3x2

8 + 9x4

320 ! . . .-

29. y = c1

,1 + x + 3x2

10 + . . .-

+c2x13

,1 + 7x

12 + 5x2

36 ! . . .-

Page 237: MATH 224 Sol.

Appendix B

Graphing Factored Polynomials

1. f(x) = x2(x ! 2)(4 ! 3x)

2. f(x) = (x2 ! 9)2(x ! 1)

3. g(x) = (1 + x)(3 + 2x)(5 ! x)7

231

Page 238: MATH 224 Sol.

232 Appendix B

4. h(x) = x4(2x ! 3)3(x2 + 4)(x2 + 2x + 1)

5. P (x) = (x4 ! 16)2

6. x(t) = (16 ! 3t2)(t2 ! 4)3

7. f(x) = (4 ! x)8(2 ! x)(4 ! 3x2)4(2x + 2)(x + 5)

8. f(x) = x(4! 3x)3(2x + 1)4(x3 ! 1)

Page 239: MATH 224 Sol.

Appendix C

Selected Topics from LinearAlgebra

C.1 A Primer on Matrix Algebra

1. (a)

!3 51 1

"!xy

"=

!7!1

"

(b)

1

32 5 !1!1 2 01 0 4

5

7

1

3xyz

5

7 =

1

32!10

5

7

(c)

!x!

y!

"=

!1 !52 1

"!xy

"

2.

AB =

!2 !3!1 4

"!!2 21 !5

"

=

!!4 ! 3 4 + 152 + 4 !2 ! 20

"

=

!!7 196 !22

"

BA =

!!2 21 !5

"!2 !3!1 4

"

=

!!4 ! 2 6 + 82 + 5 !3 ! 20

"

=

!!6 147 !23

"

3.

AB =

1

31 2 !30 !1 43 0 2

5

7

1

33 !2 2!1 1 !50 2 2

5

7

=

1

33 ! 2 + 0 !2 + 2 + 6 2 ! 10 ! 60 + 1 + 0 0 ! 1 ! 8 0 + 5 + 89 + 0 + 0 !6 + 0 + 4 6 + 0 + 4

5

7

=

1

31 !6 !141 7 139 !2 10

5

7

233

Page 240: MATH 224 Sol.

234 Appendix C.1

BA =

1

33 !2 2!1 1 !50 2 2

5

7

1

31 2 !30 !1 43 0 2

5

7

=

1

33 + 0 + 6 6 + 2 + 0 !9 ! 8 + 4

!1 + 0 ! 15 !2 ! 1 + 0 3 + 4 ! 100 + 0 + 6 0 ! 2 + 0 0 + 8 + 4

5

7

=

1

39 8 !13

!16 !3 !36 !2 12

5

7

4. (a)

1

32 2 64 1 47 4 2

5

7

(b)

1

30 8 !2!6 !1 21 0 0

5

7

(c)

1

35 1 1613 3 917 10 5

5

7

(d)

1

36 0!2 42 2

5

7

5. (a) Not possible: Matrices are of di%erent order.

(b)

1

32 !22 1222 4 !24 4 2

5

7

(c) 3

(d) 6

6. (a) 6

(b) Not possible: Matrices are of di%erent order.

(c)

!3 !1 10 2 1

"

(d)

1

31 !1 45 0 22 3 1

5

7

7. (a)

1

32 4 72 1 46 4 2

5

7

(b) Not possible: Number of columns of A (2) does not equal the number of rows of A (3).

(c)

1

33 12 !611 2 47 7 !1

5

7

(d)

1

332 6 118 9 !117 !8 19

5

7

8. (a)

1

320 13 !38 27 145 17 13

5

7

Page 241: MATH 224 Sol.

C.2. GAUSSIAN ELIMINATION, MATRIX INVERSES, AND CRAMER’S RULE 235

(b)

1

312 !3!4 54 1

5

7

(c) Not possible: Number of columns of B (2) does not equal the number of rows of A (3).

(d)

!12 !6 30 4 2

"

9. (a)

1

310 !215 38 5

5

7

(b) Not possible: Number of columns of A (2) does not equal the number of rows of E (3).

(c)

!1 !8 1213 4 3

"

(d)

1

3!6 39 !2726 !1 1310 19 !7

5

7

10. (a) 8

(b) 55

(c) 33

(d) 64

11. (a) 1815

(b) 1815

(c) 1815

(d) 55

C.2 Gaussian Elimination, Matrix Inverses, and Cramer’sRule

1. (a) Yes

(b) Yes

(c) No

2. (a) Yes

(b) Yes

(c) No

3. (a)

x1 =18

x2 =34

(b)

x1 = 3

x2 = 1

x3 = 2

Page 242: MATH 224 Sol.

236 Appendix C.2

4. (a) There are an infinite number of solutions.

x1 = !17(1 + 3x3)

x2 =17(1 ! 4x3)

(b) No solution.

5. (a)

x = 6

y = 8

z = 3

(b)

x = 5

y = !6

z = 5

6. (a)

x =

8888!2 !31 1

88888888

2 !32 1

8888

=!2 ! (!3)2 ! (!6)

=18

y =

88882 !22 1

88888888

2 !32 1

8888

=!2 ! (!4)2 ! (!6)

=34

(b)

x =

888888

2 0 43 !3 00 !4 6

888888888888

1 0 42 !3 03 !4 6

888888

=!84!14

= 6

Page 243: MATH 224 Sol.

Appendix C.2 237

y =

888888

1 2 42 3 03 0 6

888888888888

1 0 42 !3 03 !4 6

888888

=!42!14

= 3

z =

888888

1 0 22 !3 33 !4 0

888888888888

1 0 42 !3 03 !4 6

888888

=14!14

= 1

(c) Since

888888

2 3 !21 !2 34 !1 4

888888= 0, there is no solution.

7. (a)

x1 = 1

x2 = 2

(b)

x =311

y =211

z = ! 111

8. (a)

x = !14455

y = !6155

z =4611

(b)

x = 1

y = 0

z = 2

w = 0

9.

x! =

8888x ! sin $y cos $

88888888

cos $ ! sin $sin $ cos $

8888

= x cos $ + y sin $

Page 244: MATH 224 Sol.

238 Appendix C.3

y! =

8888cos $ xsin $ y

88888888

cos $ ! sin $sin $ cos $

8888

= y cos $ ! x sin $

10. The formula for the 2 . 2 inverse:

A =

!a bc d

"then A! =

1detA

!d !b!c a

".

A!1 =

!!1 !21 1

"

A!2 =

119

!5 4!1 3

"

11. A!1 =

13

!7 !2!2 1

"

A!2 =

121

!6 53 !1

"

12. A! =12

1

3!32 !22 67 5 !1!5 !3 1

5

7

B! =134

1

317 34 !17!9 !6 7!11 4 1

5

7

C.3 Coordinates and Change of Basis

1. (a)111

!4 3!1 2

"

(b)

!2 !31 4

"

(c) [w]B =

!3!5

"

(d) [w]B" =

!! 3

11! 13

11

"

2. (a)12

!1 !5!1 !13

"

(b)19

!13 !5!1 !1

"

(c) [w]B =

!! 17

9119

"

(d) [w]B" =

!!4!7

"

3. (a)

1

334

34 0

! 34 ! 17

12 ! 32

0 23 1

5

7

(b) [w]B =

1

3! 31

95! 1

3

5

7

4. (a)

1

372 2 5

2! 5

2 !3 ! 32

6 1 5

5

7

Page 245: MATH 224 Sol.

Appendix C.3 239

(b) [w]B =

1

37!6!6

5

7

5. (a) Matrix of eigenvectors P =

!0 10 0

"which is not invertible. Hence P"1 does not exist.

(b) Matrix of eigenvectors P =

!0 21 7

", P"1 =

!! 7

2 112 0

"

P"1AP =

!!1 00 1

"

6. (a) Matrix of eigenvectors P =

!3+i

#3

23"i

#3

21 1

",

P"1 =

/! i#

31+i

#3

2i#3

1"i#

32

0

P"1AP =

/1+i

#3

2 0

0 1"i#

32

0

(b) Matrix of eigenvectors P =

1

31 0 00 0 00 1 0

5

7, which is not invertible. Hence P"1 does not exist.

7. (a) Matrix of eigenvectors P =

1

3!2 0 10 1 03 0 0

5

7, P"1 =

1

30 0 1

30 1 01 0 2

3

5

7

P"1AP =

1

35 0 00 3 00 0 2

5

7

(b) Matrix of eigenvectors P =

1

3!1 0 !20 1 11 0 1

5

7,

P"1 =

1

31 0 21 1 1!1 0 !1

5

7

P"1AP =

1

32 0 00 2 00 0 1

5

7