Lecture V0305

Introduction to Ordinary Differential Equations

December 5, 2013

Contents

1 Differential Equations and Their Solutions 5

1.1 Classification of Differential Equations; Their Origin and Application . . . . . . . . . 5

1.2 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

1.3 Initial-Value Problems, Boundary-Value Problems, and Existence of Solutions . . . . 13

2 First-Order Equations with Exact Solutions 19

2.1 Exact Differential Equations and Integrating Factors . . . . . . . . . . . . . . . . . . 19

2.2 Separable Equations and Equations Reducible to This Form . . . . . . . . . . . . . . 27

2.3 Linear Equations and Bernoulli Equations . . . . . . . . . . . . . . . . . . . . . . . . 32

2.4 Special Integrating Factors and Transformations . . . . . . . . . . . . . . . . . . . . . 38

3 Applications of First-Order Equations 49

3.1 Orthogonal Trajectories and Oblique Trajectories . . . . . . . . . . . . . . . . . . . . 49

3.2 Problems in Mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61

3.3 Rate Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67

4 Explicit Methods of Solving High-Order ODE 75

4.1 Basic Theory of Linear Differential Equations . . . . . . . . . . . . . . . . . . . . . . 75

4.2 The Homogeneous Linear Equation with Constant Coefficients . . . . . . . . . . . . . 83

4.3 The Method of Undetermined Coefficients . . . . . . . . . . . . . . . . . . . . . . . . 87

4.4 Variation of Parameters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90

4.5 The Cauchy-Euler Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93

4.6 Linear Differential Equations and Difference Equations . . . . . . . . . . . . . . . . . 95

4 CONTENTS

5 Applications of Second-Order Linear ODE 105

5.1 The Differential Equation of Vibrations of a Mass on a Spring . . . . . . . . . . . . . 105

5.2 Free, Undamped Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107

5.3 Free, Damped Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110

5.4 Forced Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115

5.5 Resonance Phenomena . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117

5.6 Electric Circuit Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 130

6 Systems of Linear Differential Equations 145

6.1 Basic Theory of Linear Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145

6.2 Matrices and Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151

6.3 Homogeneous Linear Systems with Constant Coefficients: Two Unknowns . . . . . . . 157

Chapter 1

Differential Equations and Their

Solutions

1.1 Classification of Differential Equations; Their Origin

and Application

A. Classification

1. Definition (Differential Equation)

An equation involving derivatives of one or more dependent variables with respect to one or

more independent variables is called a differential equation.

2. Example

(1.1.1) xy(dy

dx)2 = 0

(1.1.2)d4x

dt4+ 5

d2x

dt2+ 3x = sin t

(1.1.3)∂v

∂s+

∂v

∂t= 0

(1.1.4)∂2u

∂x2+

∂2u

∂y2+

∂2u

∂z2= 0

3. Definition (Ordinary Differential Equation, ODE)

A differential equation involving ordinary derivatives of one or more dependent variables with

respect to a single independent variable is called an ordinary differential equation.

4. Example

6 CHAPTER 1. DIFFERENTIAL EQUATIONS AND THEIR SOLUTIONS

(1.1.1) xy(dy

dx)2 = 0: x is the single independent variable. y is dependent.

(1.1.2)d4x

dt4+ 5

d2x

dt2+ 3x = sin t: t is the single independent variable. x is dependent.

5. Definition (Partial Differential Equation, PDE)

A differential equation involving partial derivatives of one or more dependent variables with

respect to more than one independent variable is called a partial differential equation.

6. Example

(1.1.3)∂v

∂s+

∂v

∂t= 0: s and t are independent. v is dependent.

(1.1.4)∂2u

∂x2+

∂2u

∂y2+

∂2u

∂z2= 0: x, y and z are independent. u is dependent.

7. Definition (Order)

The order of the highest ordered derivative involved in a differential equation is called the order

of the differential equation.

8. Example

(1.1.1): xy(dy

dx)2 = 0. It is the first order.

(1.1.2):d4x

dt4+ 5

d2x

dt2+ 3x = sin t. It is the fourth order.

(1.1.3):∂v

∂s+

∂v

∂t= 0. It is the first order.

(1.1.4):∂2u

∂x2+

∂u2

∂y2+

∂u2

∂z2= 0. It is the second order.

9. Definition (Linear)

A linear ordinary differential equation of order n, in the dependent variable y and the indepen-

dent variable x, is an equation that is in, or can be expressed in, the form

a0(x)dny

dxn+ a1(x)

dn−1y

dxn−1+ · · ·+ an−1(x)

dy

dx+ an(x)y = b(x), (1.1.5)

where a0 is not identically zero.

Observe

(1) that the dependent variable y and its various derivatives occur to the first degree only,

(2) that no products of y and / or any its derivatives are present, and

(3) that no transcendental functions of y and / or its derivatives occur.

10. Example

1.1 Classification of Differential Equations; Their Origin and Application 7

(1.1.6)d2y

dx2+ 5

dy

dx+ 6y = 0

(1.1.7)d4y

dx4+ x2 d

3y

dx3+ x3 dy

dx= xex.

11. Definition (Nonlinear)

A nonlinear ordinary differential equation is an ordinary differential that is not linear.

12. Example

(1.1.8)d2y

dx2+ 5

dy

dx+ 6y2 = 0, (nonlinear: 6y2)

(1.1.9)d2y

dx2+ 5(

dy

dx)3 + 6y = 0, (nonlinear: 5(

dy

dx)3)

(1.1.10)d2y

dx2+ 5y

dy

dx+ 6y = 0. (nonlinear: 5y

dy

dx)

13. Definition (Constant Coefficients)

Linear ordinary differential equations are further classified according to the nature of the coef-

ficients of the dependent variables and their derivatives.

Example:

Equation (1.1.6) is said to be linear with constant coefficients.

Equation (1.1.7) is said to be linear with variable coefficients.

B. Origin and Application of Differential Equations

1. Applicable Areas

Differential equations occur in connection with numerous problems that are encountered in various

branches of science and engineering. We indicate a few such problems in the following list, which

could easily be extended to fill many pages.

(1) The problem of determining the motion of a projectile, rocked, satellite, or planet.

(2) The problem of determining the change or current in an electric circuit.

(3) The problem of conduction of heat in a rod or in a slab.

2. Example (Free, Undamped Motion; See also Chapter 5)

(1) Hooke’s law

The magnitude of the force needed to produce a certain elongation, provided this elongation is

not too great. In mathematical form,

|F | = ks,


where F is the magnitude of the force, s is the amount of elongation, and k is a constant of

proportionality which we shall call the spring constant.

When a mass is hung upon a spring of spring constant k and thus produces an elongation of

amount s, the force F of the mass upon the spring therefore has magnitude ks. The spring at

the same time exerts a force upon the mass called the restoring force of the spring. This force

is equal in magnitude but opposite in sign to F and hence has magnitude −ks

L

Fig. (a)

L

l

m

Fig. (b)

L

l

O

x

m

Fig. (c)

(a) natural length L

(b) mass in equilibrium position; spring has stretched length L+ l.

(c) mass distance x below equilibrium position; spring has stretched length L+ l + x.

(2) Forces Acting Upon the Mass

(a) F1, the force of gravity of magnitude mg.

F1 = mg ↓ positive (1.1.11)

(b) F2, the restoring force of the spring.

F2 = −k(x+ l) ↑ negative (1.1.12)

When x = 0, F1 + F2 = 0, thus we have mg = kl.

Replacing kl by mg in (1.1.12) we see that

F2 = −kx−mg. (1.1.13)

(c) F3, the resisting force of the medium, called the damping force. For small velocities it is

approximately proportional to the magnitude of the velocity:

|F3| = a

∣

∣

∣

∣

dx

dt

∣

∣

∣

∣

(1.1.15)

1.1 Classification of Differential Equations; Their Origin and Application 9

where a > 0 is called the damping constant. When the mass is moving downward,F3 acts

in the upward direction and so F3 < 0.

F3 = −adx

dt, (a > 0) ↑ negative (1.1.16)

(d) F4, any external impressed forces that act upon the mass.

F4 = F (t) (1.1.17)

We now apply Newton’s second law, F = ma, where F = F1 + F2 + F3 + F4. We find

md2x

dt2= mg − kx−mg − a

dx

dt+ F (t) (1.1.18)

or

md2x

dt2+ a

dx

dt+ kx = F (t) (1.1.19)

(e) It is a nonhomogeneous second-order linear differential equation with constant coefficients.

(f) a = 0: If a = 0, the motion is called undamped; otherwise (a 6= 0)it is called damped.

(g) F (t) = 0: If F (t) = 0, the motion is called free; otherwise it is called forced.

(3) Free, Undamped Motion

The different equation (1.1.19) then reduces to

md2x

dt2+ kx = 0 (1.1.20)

where m(> 0) is the mass and k(> 0) is the spring constant. Let λ2 =k

m, we rewrite (1.1.20)

in the formd2x

dt2+ λ2x = 0 (1.1.21)

Hence the general solution of (1.1.21) can be written

x = c1 sinλt + c2 cosλt (1.1.22)

where c1 and c2 are arbitrary constants.

Initial conditions:

x(0) = x0, (initial displacement) (1.1.23)

x′(0) = v0, (initial velocity) (1.1.24)

Therefore

dx/dt|t=0 = x′(0) = v0 =⇒ [c1λ cosλt− c2λ sinλt]|t=0 = v0 (1.1.25)


We have

c2 = x0 and c1 =v0λ

The solution is

x(t) =v0λsin λt+ x0 cosλt (1.1.26)

An alternative form

x(t) = c[v0/λ

csin λt+

x0

ccos λt]

where c =√

(v0/λ)2 + x02. Let φ be defined by satisfying

v0/λ

c= − sin φ and

x0

c= cosφ. So

x(t) = c cos(λt + φ) (1.1.27)

=√

(v0/λ)2 + x02 cos(

√

k

m+ φ) (1.1.28)

(4) Simple Harmonic Motion

(a) The free, undamped motion of the mass is a simple harmonic motion.

(b) The constant c is called the amplitude of the motion and gives the maximum (positive)

displacement of the mass from its equilibrium position.

(c) The motion is a periodic motion.

x = c⇔√

k

mt+ φ = ±2nπ, n ∈ N ∪ 0

Therefore

t =

√

m

k(±2nπ − φ) > 0 (1.1.29)

The time interval between two successive maxima is called the period of the motion and it

is given by2π

λ=

2π√

k/m=

√

m

k· 2π.

(d) The reciprocal of the period, which gives the number of oscillations per second, is called

natural frequence (or simply frequence) of the motion and it is given by

λ

2π=

√

k/m

2π=

√

k

m· 1

2π.

(e) The number of φ is called the phase constant (or phase angle). See Fig. 1.1.

(5) Another View Point (Constant Cancellation)

Simple Harmonic Motion ⇒ x(t) = A cos(pt + α), so x′′(t) = −p2A cos(pt + α), and then

x′′(t) + x · p2 = 0. Then we have the ODE:

d2x

dt2+ p2x = 0.

1.2 Solutions 11

0 0.5 1 1.5 2

−0.4

−0.3

−0.2

−0.1

0

0.1

0.2

0.3

0.4

π /8 2π/8 3π/8 4π/8

Period π /4

Amplitude 0.28

t = −φ/8

X

t

Figure 1.1: x(t) = 0.28 cos(8t + φ) where φ=-0.46

1.2 Solutions

A. Nature of Solutions

1. Definition (Explicit and Implicit Solution)

Consider the nth-order ordinary differential equation

F

[

x, y,dy

dx, .....,

dny

dxn

]

= 0, (1.2.1)

where F is a real function of its (n+ 2) arguments x, y, ....., dny/dxn.

(1) Let f be a real function of defined ∀ x ∈ I = [a, b], a, b ∈ R and having an nth derivative (and

hence also all lower derivatives) ∀ x ∈ I. The function f is called an explicit solution of (1.2.1)

on I if it is fulfilled the following two requirements:

F [x, f(x), f ′(x), .....f (n)(x)] (A)

is defined ∀ x ∈ I, and

F [x, f(x), f ′(x), .....f (n)(x)] = 0 (B)

∀ x ∈ I.

(2) A relation g(x, y) = 0 is called an implicit solution of(1.2.1) on I if this relation defines at least

one real function f(x) on I such that f(x) is an explicit solutions of (1.10) on I.


(3) Both explicit solutions and implicit solutions will usually be called simply solutions.

2. Example 1.2.1d2y

dx2+ y = 0

Answer:

f(x) = 2 sin x+ 3 cosx.

3. Example 1.2.2

x+ ydy

dx= 0

Answer:

(1) Explicit Solutions:

f1(x) =√25− x2, ∀x ∈ I = (−5, 5)

f2(x) = −√25− x2, ∀x ∈ I = (−5, 5)

(2) Implicit Solutions:

(a) g(x, y) = x2 + y2 + 25 = 0→ No!

2x+ 2ydy

dx= 0. so

dy

dx= −x

y.

We obtain the formal identity: x + y(−xy) = 0. Thus g = 0 formally satisfies the differential

equation. We have no assurance that g = 0 defines any function that is an explicit solution

of the ODE. Moreover, y = ±√−25− x2 /∈ R. So g = 0 is not truly an implicit solution but

merely a formal solution.

(b) h(x, y) = x2 + y2 − 25 = 0→ Yes!

B. Methods of Solution

1. Closed-form Solution: Chap. 2 & 4.

2. Approximate Solution: series methods, numerical methods and graphical methods. Chap. 6 & 8.

3. Chap. 7: Systems of linear ODE.

4. Chap. 9: The Laplace transform.

C. Family of Solutions

1.3 Initial-Value Problems, Boundary-Value Problems, and Existence of Solutions 13

1. Consider the ODEdy

dx= 2x. (1.2.2)

The function fc defined on R by

fc(x) = x2 + c (1.2.3)

is a solution of the differential equation (1.2.2). We call the constant c in (1.2.3) an arbitrary

constant or parameter and refer to the family of functions defined by (1.2.3) as a one-parameter

family of solutions of (1.2.2). We write this one-parameter family of solutions as

y = x2 + c.

2. We must not conclude that every 1st-order ODE has a so-called one-parameter family of

solutions which contains all solutions of the differential equation. For example, the first-order

differential equation∣

∣

∣

∣

dy

dx

∣

∣

∣

∣

+ |y|+ 1 = 0

has no (real) solutions.

3. The general 1st-order ODEdy

dx= f(x, y), (1.2.4)

where f is a real function, may be interpreted geometrically as defining a slop f(x, y) at every

point (x, y) at which the function f is defined.

4. Assume that the ODE (1.2.4) has a so-called one-parameter family of solutions that can be

written in the form

y = F (x, c), (1.2.5)

where c is the arbitrary constant. (1.2.5) is represented geometrically by a so-called one-parameter

family of curves in xy-plane. These curves are also called the integral curves.

1.3 Initial-Value Problems, Boundary-Value Problems, and

Existence of Solutions

A. IVP and BVP

1. Definition (IVP and BVP)

In an ODE problem, if all of the associated supplementary conditions relate to one x value, the prob-

lem is called an initial-value problem ( or one-point boundary-value problem, IVP). If the conditions

relate to two different x values, the problem is called a two-point boundary-value problem ( or simply


a boundary-value problem, BVP).

2. Example (IVP)

d2y

dx2+ y = 0, y(1) = 3, y′(1) = −4.

3. Example (BVP)

d2y

dx2+ y = 0, y(0) = 1, y′(

π

2) = −4.

4. Definition (1st-order IVP)

Consider the 1st-order differential equation

dy

dx= f(x, y), (1.3.1)

where f is a continuous function of x and y in some domain D of the xy-plane; and let (x0, y0) be a

point of D. The initial-value problem associated with (1.3.1) is to find a solution φ of (1.3.1), defined

on some interval I containing x0, and satisfying the initial condition

φ(x0) = y0.

In the customary abbreviated notation, this IVP may be written

dy

dx= f(x, y),

y(x0) = y0.

5. Example (Separation of Variables)

dy

dx= −x

y,

y(3) = 4.

Answer.

ydy = −xdx,∫

ydy = −∫

xdx, y2 = −x2 + c.

y(3) = 4, 42 = −32 + c, c = 25.

y2 = 25− x2. Thus, y =√25− x2.


B. Existence of Solution

1. Example (No solution)

d2y

dx2+ y = 0,

y(0) = 1, y(π) = 5.

Answer. The auxiliary equation is

m2 + 1 = 0, m = ±i.

The corresponding part of the general solution is

c1 exp(ix) + c2 exp(−ix).

From Euler’s formula,

exp(iθ) = cos(θ) + i sin(θ),

we know the general solution of the ODE is

y = φ(x) = c1 cos(x) + c2 sin(x).

y(0) = 1

y(π) = 5⇒

c1 · 0 + c2 · 1 = 1

c1 · 0 + c2 · (−1) = 5⇒

c2 = 1

−c2 = 5⇒ No solution!

2. Example (Many solutions, no uniqueness)

dy

dx= y1/3,

y(0) = 0.

Answer. y−1/3dy = dx,3

2y2/3 = x+ c1 y = (

2

3x+ c2)

3/2.

We can find many solutions for this problem.

y = φ1(x) = 0, ∀x ∈ R.

y = φ2(x) =

(23x)3/2 if x ≥ 0;

0 if x ≤ 0.y = φ3(x) =

(23x− 2)3/2 if x ≥ 3;

0 if x ≤ 3.· · ·

This problem has infinitely many solutions. Notice that y1/3 is not differentiable ar x = 0. See Fig

1.2.


0 2 4 6 8 10 12

0

2

4

6

8

10

12

14

16

18

20

← φ2(x) ← φ

3(x)

Figure 1.2: Many solutions.

3. Theorem 1.3.1 Basic Existence and Uniqueness Theorem

Hypothesis. Consider the differential equation

dy

dx= f(x, y), (1.3.1)

where

(1) the function f is continuous in the domain D and

(2) the partial derivative ∂f/∂y is also continuous in D.

Let (x0, y0) ∈ D.

Conclusion. ∃! φ, φ is also the solution of (1.3.1), defined on some interval |x− x0| ≤ h, where

h is sufficiently small, that satisfies

φ(x0) = y0.

4. Example

dy

dx= x2 + y2,

y(1) = 3.

Answer. f(x, y) = x2 + y2 and ∂f/∂y = 2y are continuous in R2. (1, 3) ∈ R

2.

Thus, there is a unique solution φ defined on |x− 1| ≤ h such that φ(1) = 3.


5. Example

(1)dy

dx=

y√x, y(1) = 2.

(2)dy

dx=

y√x, y(0) = 2.

Answer. f(x, y) =y

x1/2and

∂f

∂y=

1

x1/2are continuous in R

2\x ≤ 0.(1) Theorem 1.3.1 says there is a unique solution since x0 = 1 > 0.

(2) Theorem 1.3.1 says nothing since x0 = 0.

1

ydy =

1√xdx, ln y = 2

√x+ c1, y = c · exp(2

√x).

h(x) := exp(2√x) > 0, (positive)

h′(x) = exp(2√x) · 1√

x> 0, (increasing),

h′′

(x) =

exp(2√x)√

x· √x− exp(2

√x) · 1

2√x

x,

h′′

(x) = 0 ⇒ 1 =1

2√x⇒ x =

1

4. (inflection point)

0 ∼ 1/4 1/4 ∼ր ր

concave down concave up

See Fig. 1.3. Notice that h(x) is not differentiable at x = 0 even if we define h(x) as follows,

h(x) :=

exp(2√x) if x > 0

1 if x ≤ 0.

0 0.1 0.2 0.3 0.4 0.5 0.6

0

1

2

3

4

5

6

← exp(2 sqrt(x))

Figure 1.3: The graph of e2√x.

Chapter 2

First-Order Equations with Exact

Solutions

2.1 Exact Differential Equations and Integrating Factors

A. Standard Forms of First-Order Differential Equations

1. Definition

The derivative form:dy

dx= f(x, y). (2.1.1)

The differential form: M(x, y)dx+N(x, y)dy = 0. (2.1.2)

Remark.

Newton and Leibniz each used a different notation when they published their discoveries of

calculus, thereby creating a notational divide between Britain and the European continent

that lasted for more than 50 years. The Leibniz notation dy/dx eventually prevailed because

it suggests correct formulas in a natural way, the chain rule

dy

dx=

dy

du· dudx

being a good example. The symbols ”dy” and ”dx” are called differentials. Assume that f

is differentiable at a point x, define dx to be an independent variable that can have any real

value, and define dy by the formula

dy = f ′(x) dx.

If dx 6= 0, then we can divide both sides by dx to obtain

dy

dx= f ′(x).

20 CHAPTER 2. FIRST-ORDER EQUATIONS WITH EXACT SOLUTIONS

Thus, they are called the differential and derivative forms respectively.

The symbol df is another common notation for the differential of a function y = f(x). For

example, if f(x) = sin x, then we can write df = cosx dx. All of the general rules of differenti-

ation then have corresponding differential versions:

Derivative Formula Differential Formulad

dx[c] = 0 d[c] = 0

d

dx[cf ] = c

df

dxd[cf ] = c df

d

dx[f + g] =

df

dx+

dg

dxd[f + g] = df + dg

d

dx[fg] = f

dg

dx+ g

df

dxd[f + g] = fdg + gdf

d

dx

[

f

g

]

=g dfdx− f dg

dx

g2d

[

f

g

]

=g df − f dg

g2

2. Example

(1)dy

dx=

x2 + y2

x− y(x2 + y2)dx− (x− y)dy = 0.

(2)dy

dx= −sin x+ y

x+ 3y(sin x+ y)dx+ (x+ 3y)dy = 0.

B. Exact Differential Equations

1. Definition (Total Differential)

Let F be a function of two variables such that F has continuous first partial derivatives in a domain

D. The total differential dF of the function F is defined by

dF (x, y) =∂F

∂xdx+

∂F

∂xdy

for all (x, y) ∈ D. Note that : F (x1, x2 · · ·xn), n variables

dF =∂F

∂x1dx1 + · · ·+

∂F

∂xndxn =

n∑

i=1

∂F (x1, · · · , xn)

∂xidxi.

2. Example

Find dF where F (x, y) = xy2 + 2x3y.

Answer. dF = (y2 + 6x2y)dx+ (2xy + 2x3)dy

3. Definition (Exact Differential)

The expression

M(x, y)dx+N(x, y)dy (2.1.3)

2.1 Exact Differential Equations and Integrating Factors 21

is called an exact differential in a domain D if there exists a function F of two real variables such

that this expression equals to the total differential dF (x, y), ∀(x, y) ∈ D

That is, (2.1.3) is an exact differential in D if ∃ F such that

∂F

∂x(x, y) = M(x, y) and

∂F

∂y(x, y) = N(x, y) ∀(x, y) ∈ D

If M(x, y)dx+N(x, y)dy is an exact differential, then the differential equation

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

is called an exact differential equation.

4. Example

(1) y2dx+ 2xydy = 0 (2) ydx+ 2xdy = 0

Answer. (1)F = xy2 exact! (2) not exact!

5. Theorem 2.1.1

Consider

M(x, y)dx+N(x, y)dy = 0 (2.1.4)

where M and N have continuous first partial derivatives at all (x, y) ∈ D

(1) If (2.1.4) is exact in D, then

∂M(x, y)

∂y=

∂N(x, y)

∂x

(2) Conversely, if∂M

∂y=

∂N

∂x∀(x, y) ∈ D, then (2.1.4) is exact.

Proof.

Part 1. ”Exact” ⇒ ∂M

∂y=

∂N

∂x

If (2.1.4) is exact in D, then ∃F such that

∂F

∂x(x, y) = M(x, y) and

∂F

∂y(x, y) = N(x, y).

Therefore∂M

∂y=

∂2F

∂y∂x=

∂2F

∂x∂y=

∂N

∂x.


Part 2. ”Exact” ⇐ ∂M

∂y=

∂N

∂x

Let us assume that∂F

∂x= M . Then

F (x, y) =

∫

M(x, y)dx+ φ(y)

∂F

∂y=

∂

∂y

∫

M(x, y)dx+dφ

dy(y)

In order to prove the differential equation is exact, we must solve φ(y) to satisfy

N(x, y) =∂

∂y

∫

M(x, y)dx+dφ

dy(y).

Hencedφ

dy= N(x, y)− ∂

∂y

∫

M(x, y)dx.

(1) We prove that N(x, y)− ∂∂y

∫

M(x, y)dx must be independent of x.

Claim : ∂∂x[N(x, y)− ∂

∂y

∫

M(x, y)dx] = 0

LHS =∂

∂xN(x, y)− ∂2

∂x∂y

∫

M(x, y)dx

=∂

∂xN − ∂2

∂x∂y[F (x, y)− φ(y)]

=∂

∂xN − ∂2

∂y∂x[F (x, y)− φ(y)]

=∂

∂xN − ∂2

∂y∂x

∫

M(x, y)dx

=∂

∂xN − ∂

∂yM

= 0

Therefore, dφdy

= N − ∂∂y

∫

M(x, y)dx is independent of x.

(2) We solve φ(y).

φ(y) =

∫ [

N(x, y)− ∂

∂y

∫

M(x, y) dx

]

dy.

Therefore,

F (x, y) =

∫

M(x, y)dx+

∫[

N(x, y)− ∂

∂y

∫

M(x, y) dx

]

dy.

6. Example

(1) y2dx+ 2xydy = 0 (2) ydx+ 2xdy = 0


Answer. (1)∂M

∂y= 2y =

∂N

∂xExact!

F =∫

y2dx+∫

[2xy − ∂∂y

∫

y2dx]dy

= xy2 +∫

[2xy − 2xy]dy

= xy2 + 0

dF = y2dx+ 2xydy. (Solutions : xy2 = c.)

(2)∂M

∂y= 1 6= 2 =

∂N

∂xNot exact!

7. Example

(2x sin y + y3ex)dx+ (x2 cos y + 3y2ex)dy = 0

Answer.∂M

∂y= 2x cos y + 3y2ex =

∂N

∂xExact!

F =

∫

(2x sin y + y3ex)dx+

∫[

x2 cos y + 3y2ex − ∂

∂y

∫

(2x sin y + y3ex)dx

]

dy

= x2 sin y + y3ex + 0

(Solutions :x2 sin y + y3ex = c.)

8. Theorem 2.1.2 (The Solution of Exact Differential Equations)

Suppose the differential equationM(x, y)dx+N(x, y)dy = 0 satisfies the differentiability requirements

of Theorem 2.1.1 and is exact in D. Then a one-parameter family of solutions of this ODE is given

by

F (x, y) = c

where F is the function such that

∂F

∂x= M and

∂F

∂y= N ∀(x, y) ∈ D

and c is an arbitrary constant.

Proof. Mdx+Ndy = 0 is exact. ∃ F such that∂F

∂x= M and

∂F

∂y= N

∂F

∂xdx+

∂F

∂ydy = 0 or simply dF (x, y) = 0. The relation

F (x, y) = c

is obviously a solution of this, where c is an arbitrary constant.

9. Example

(3x2 + 4xy)dx+ (2x2 + 2y)dy = 0


Answer.∂M

∂y= 4x =

∂N

∂x

F =

∫

(3x2 + 4xy)dx+ φ(y) = x3 + 2x2y + φ(y)

∂F

∂y= 2x2 +

dφ

dy= N = 2x2 + y

dφ

dy= 2y, φ(y) = y2 + c

We may write the solution as x3 + 2x2y + y2 = c.

Method of Grouping

(3x2 + 4xy)dx+ (2x2 + 2y)dy = 0

3x2dx+ (4xydx+ 2x2dy) + 2ydy = 0

d(x3) + d(2x2y) + d(y2) = d(c)

x3 + 2x2y + y2 = c.

Notice that it requires a good ”working knowledge”!

Example: (x3 + xy2 − y)dx+ (x2y + y3 + x)dy = 0.

Answer. x(x2 + y2)dx− ydx+ y(x2 + y2)dy + xdy = 0.

(x2 + y2)(xdx+ ydy) + (xdy − ydx) = 0

xdx+ ydy +xdy − ydx

x2 + y2= 0

d(x2

2) + d(

y2

2) +

d(y/x)

1 + (y/x)2= 0

x2

2+

y2

2+ tan−1 y

x= c,

where we use the formula

∫

1

1 + u2du = tan−1 u+ c.


The proof is

tan(tan−1 u) = u

d tan(tan−1 u)

du=

du

du= 1

sec2(tan−1 u)d tan−1 u

du= 1

(1 + u2)d tan−1 u

du= 1

d tan−1 u

du=

1

1 + u2.

10. Example

Solve the IVP: (2x cos y + 3x2y)dx+ (x3 − x2 sin y − y)dy = 0, y(0) = 2.

Answer.∂M

∂y= −2x sin y + 3x2 =

∂N

∂x.

(1) Standard method

F =

∫

Mdx+ φ(y)

= x2 cos y + x3y + φ(y)

∂F

∂y= −x2 sin y + x3 +

dφ(y)

dy= N(x, y) = x3 − x2 sin y − y

dφ

dy= −y ⇒ φ(y) = −y

2

2+ c0

F (x, y) = x2 cos y + x3y − y2

2+ c0

y(0) = 2 ⇒ 0 + 0− 4/2 + c0 = 0 ⇒ c0 = 2

x2 cos y + x3y − y2

2= −2.

(2) Method of grouping

(2x cos ydx− x2 sin ydy) + (3x2ydx+ x3dy)− ydy = 0

d(x2 cos y) + d(x3y)− d(y2/2) = d(c)

x2 cos y + x3y − y2

2= c.

Of course y(0) = 2 again yields the particular solution already obtained.

11. Definition (Integrating Factor)

If the differential equation

M(x, y)dx+N(x, y)dy = 0 (2.1.5)


is not exact in the domain, D, but the differential equation

µ(x, y)M(x, y)dx+ µ(x, y)N(x, y)dy = 0 (2.1.6)

is exact in D, then µ(x, y) is called an integrating factor of the differential equation (2.1.5).

12. Example

ydx− xdy = 0.

Answer.

(1) µ =1

x2:

y

x2dx− 1

xdy = 0

∂M

∂y=

1

x2=

∂N

∂x

(2) µ =1

y2:

1

ydx− x

y2dy = 0

∂M

∂y= − 1

y2=

∂N

∂x

Notice that µ = xy and µ = 1/(xy) are also integrating factors of the ODE. Integrating factor is not

unique.

13. Example

(3y + 4xy2)dx+ (2x+ 3x2y)dy = 0.

Answer.∂M

∂y= 3 + 8xy 6= 2 + 6xy =

∂N

∂x. Consider the integrating factor µ = x2y.

(3x2y2 + 4x3y3)dx + (2x3y + 3x4y2)dy = 0

∂[µM ]

∂y= 6x2y + 12x3y2 =

∂[µN ]

∂x

14. Remark

We referred to this resulting exact equation as ”essentially equivalent” to the original. However, the

multiplication may result in either

(1) the loss of (one or more) solutions of the original, or

(2) the gain of (one or more) functions which are solutions of the ”new” equation but not of the

original, or

(3) both of these phenomena.

The question now arises: how is an integrating factor found? We will answer this question in Sec.

2.4.

2.2 Separable Equations and Equations Reducible to This Form 27

2.2 Separable Equations and Equations Reducible to This

Form

A. Separable Equations

1. Definition (Separable equations)

An equation of the form

F (x)G(y)dx+ f(x)g(y)dy = 0

is called an equation with variables separable or simply a separable equation.

2. Example

(x− 4)y4dx− x3(y2 − 3)dy = 0.

3. Theorem 2.2.1

The separable differential equation

F (x)G(y)dx+ f(x)g(y)dy = 0

has an integrating factor

µ(x, y) =1

f(x)G(y).

Proof.

µF (x)G(y)dx+ µf(x)g(y)dy = 0

F (x)

f(x)dx+

g(y)

G(y)dy = 0

∂M

∂y=

∂[F (x)/f(x)]

∂y= 0 =

∂N

∂x=

∂[g(y)/G(y)]

∂x

4. Remark

If y0 ∈ R such that G(y0) = 0, then y = y0 is a (constant) solution of the original differential

equation; and this solution may (or may not) have been lost in the formal separation process.

5. Example

(x− 4)y4dx− x3(y2 − 3)dy = 0


Answer.

(x− 4)

x3dx− (y2 − 3)

y4dy = 0

(x−2 − 4x−3)dx− (y−2 − 3y−4)dy = 0

We have the one-parameter family of solutions

−1

x+

2

x2+

1

y− 1

y3= c.

y = 0 is a solution of the original equation. We lose it in the separation process.

6. Example

x sin ydx+ (x2 + 1) cos ydy = 0

y(1) = π/2.

Answer.

x

x2 + 1dx+

cos y

sin ydy = 0

1

2

∫

1

x2 + 1d(x2 + 1) +

∫

1

sin yd(sin y) = 0

ln |x2 + 1|+ 2 ln | sin y| = c0.

More neater,

ln[

|x2 + 1| · | sin y|2]

= ln c

(x2 + 1) sin2 y = c.

Now consider the solutions of sin y = 0. These are given by

y = nπ, n ∈ Z.

Each of these constant solutions y = nπ is a member of the solution

(x2 + 1) sin2 y = c,

for c = 0. Thus none of these solutions was lost in the separation process.

y(1) = π/2 ⇒ (12 + 1) · sin2 π/2 = c. We have the solution (x2 + 1) sin2 y = 2.

B. Homogeneous Equations

1. Definition (Homogeneous Equations)


The first-order differential equation M(x, y)dx + N(x, y)dy = 0 is said to be homogeneous if, when

written is the derivative formdy

dx= f(x, y),

there exists a function g such that f(x, y) can be expressed in the form g(y/x).

2. Example

(x2 − 3y2)dx+ 2xydy = 0

Answer.dy

dx=

3y2 − x2

2xy=

3y

2x− x

2y=

3

2

(y

x

)

− 1

2

(

1

y/x

)

.

3. Example

(y +√

x2 + y2)dx− xdy = 0

Answer.dy

dx=

y +√

x2 + y2

x=

y

x±√

x2 + y2√x2

=y

x±√

1 +(y

x

)2

.

depending on the sign of x.

4. Definition (Homogeneous Functions)

A function F is called homogeneous of degree n if

F (tx, ty) = tnF (x, y).

Example

(1) F (x, y) = x2 + 2xy + y2, F (tx, ty) = t2F (x, y).

(2) F (x, y) = x3 + 5x2y, F (tx, ty) = t3F (x, y).

5. Theorem 2.2.2

If M and N in M(x, y)dx + N(x, y)dy = 0 are both homogeneous functions of the same degree n,

then the differential equation is a homogeneous differential equation.

Proof. Let t = 1/x. Then

M(1

x· x, 1

x· y) =

(

1

x

)n

M(x, y)

M(1,y

x) =

(

1

x

)n

M(x, y)

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

x), N(x, y) = xnN(1,

y

x).


Thus we havedy

dx= −M(x, y)

N(x, y)= −x

nM(1, y/x)

xnN(1, y/x)= −M(1, y/x)

N(1, y/x).

6. Theorem 2.2.3

If

M(x, y)dx+N(x, y)dy = 0 (2.24)

is a homogeneous differential equation, then the change of variables y = vx transforms (2.24) into a

separable equation in the variables v and x.

Proof. Since Mdx+Ndy = 0 is homogeneous, it may be written in the form

dy

dx= g(

y

x).

Let y = vx (i.e., v =y

x). We have

dy

dx= v + x

dv

dx

v + xdv

dx= g(v)

[v − g(v)] dx+ xdv = 01

v − g(v)dv +

1

xdx = 0

∫

1

v − g(v)dv +

∫

1

xdx = c,

where c is an arbitrary constant. Let F (v) denote

∫

1

v − g(v)dv. Thus

F (v) + ln |x| = c

F (y

x) + ln |x| = c.

7. Example

(x2 − 3y2)dx+ 2xydy = 0

Answer. Since M and N are homogeneous of the same degree 2, the ODE is homogeneous.

dy

dx= − x

2y+

3y

2x.


Let y = vx. We obtain

v + xdv

dx= − 1

2v+

3v

2

xdv

dx= − 1

2v+

v

22v

v2 − 1dv =

1

xdx

ln |v2 − 1| = ln |x|+ c1

|v2 − 1| = |cx|

|y2

x2− 1| = |cx|

|y2 − x2| = |cx3|.

If y ≥ x > 0, then we have y2 − x2 = cx3.

8. Example

(y +√

x2 + y2)dx− xdy = 0, y(1) = 0.

Answer. Since M and N are homogeneous of the same degree 1, the ODE is homogeneous.

dy

dx=

y +√

x2 + y2

x.

Since the initial x value is 1, we consider only x > 0 and take x =√x2.

dy

dx=

y

x+

√

1 +(y

x

)2

.

Let y = vx and obtain

v + xdv

dx= v +

√1 + v2

xdv

dx=√1 + v2

1√v2 + 1

dv =1

xdx

∫

1√v2 + 1

dv =

∫

1

xdx+ c1

ln |v +√v2 + 1| = ln |x|+ ln |c|,

where we use the formula∫

1√u2 + 1

dv = sinh−1 u+ c = ln |u+√u2 + 1|+ c.


The proof is as follows. Let y = sinh−1 u. We have

u = sinh y =ey − e−y

22 · ey · u = (ey)2 − 1

ey =2u+

√4u2 + 4

2= u+

√u2 + 1

y = sinh−1 u = ln |u+√u2 + 1|

d sinh−1 u

du=

1 + u/√u2 + 1

u+√u2 + 1

=1√

u2 + 1.

Another viewpoint:

d(v +√v2 + 1) = (1 +

2v

2√v2 + 1

)dv =v +√v2 + 1√

v2 + 1dv

1√v2 + 1

dv =1

v +√v2 + 1

d(v +√v2 + 1).

Therefor we obtain

v +√v2 + 1 = cx

y

x+

√

y2

x2+ 1 = cx

y +√

x2 + y2 = cx2 y(1) = 0 ⇒ c = 1

y +√

x2 + y2 = x2 y =1

2(x2 − 1).

2.3 Linear Equations and Bernoulli Equations

A. Linear Equations

1. Definition (Linear ODE)

A first-order ordinary differential equation is linear in the dependent variable y and the independent

variable x if it is, or can be, written in the form

dy

dx+ P (x)y = Q(x). (2.3.1)

Example: xdy

dx+ (x+ 1)y = x3 or

dy

dx+ (1 +

1

x)y = x2

2. Theorem 2.3.1

The linear differential equationdy

dx+ P (x)y = Q(x). (2.3.1)

2.3 Linear Equations and Bernoulli Equations 33

has an integrating factor of the form

exp[

∫

P (x)dx].

A one-parameter family of solutions of this equation is

y · exp[∫

P (x)dx] =

∫

exp

(∫

P (x)dx

)

·Q(x)dx+ c;

that is,

y = exp[−∫

P (x)dx] ·[∫

exp

(∫

P (x)dx

)

·Q(x)dx+ c

]

.

Furthermore, it can be shown that this one-parameter family of solutions of the equation (2.3.1)

includes of all solutions of (2.3.1)

Proof. Let us write (2.3.1) in the form

[P (x)y −Q(x)]dx+ dy = 0 (2.3.2)

We know that∂

∂y[P (x)y −Q(x)] = P (x) and

∂

∂x[1] = 0.

Equation (2.3.2) is not exact unless P (x) = 0. Let us multiply (2.3.2) by µ(x), obtaining

[µ(x)P (x)y − µ(x)Q(x)]dx+ µ(x)dy = 0 (2.3.3)

∂

∂y[µ(x)P (x)y − µ(x)Q(x)] =

∂

∂x[N(x, y) ≡ µ(x)]

µ(x)P (x) =d

dx[µ(x)]

P (x)dx =1

µdµ

∫

P (x)dx = ln |µ| or µ = exp[

∫

P (x)dx], (2.3.4)

where it is clear that µ > 0. Multiplying (2.3.1) by (2.3.4) gives

exp[

∫

P (x)dx] · dydx

+ exp[

∫

P (x)dx] · P (x)y = exp[

∫

P (x)dx] ·Q(x)

which is preciselyd

dx[exp(

∫

P (x)dx) · y] = Q(x) · exp[∫

P (x)dx].

Integrating this we obtain

exp(

∫

P (x)dx) · y =

∫

exp

[∫

P (x)dx

]

·Q(x)dx+ c,

y = exp

[

−∫

P (x)dx

]

·[∫

exp

(∫

P (x)dx

)

·Q(x)dx+ c

]

.


3. Example

dy

dx+ (

2x+ 1

x)y = e−2x (2.3.5)

Answer. An integrating factor is

exp[

∫

p(x)dx] = exp[

∫

(2x+ 1

x)dx]

= exp(2x+ ln | x |)= exp(2x) · x.

Multiplying (2.31) by this, we obtain

xe2xdy

dx+ e2x(2x+ 1)y = x

d

dx(xe2xy) = x

xe2xy =x2

2+ c

y =1

2xe−2x + c · 1

xe−2x

4. Example

(x2 + 1)dy

dx+ 4xy = x (2.3.6)

y(2) = 1 (2.3.7)

Answer.

dy

dx+

4x

x2 + 1y =

x

x2 + 1(2.3.8)

An integrating factor is

exp[

∫

4x

x2 + 1dx] = exp[ln(x2 + 1) · 2] = (x2 + 1)2.


Multiplying (2.3.8) by this, we obtain

(x2 + 1)2dy

dx+ 4x(x2 + 1)y = x(x2 + 1)

d

dx[(x2 + 1)2y] = x3 + x

(x2 + 1)2y =1

4x4 +

x2

2+ c

y(2) = 1 ⇒ c = 19

(x2 + 1)2y =1

4x4 +

x2

2+ 19

y = (1

4x4 +

x2

2+ 19)/(x2 + 1)2.

5. Example (not linear in y but linear in x)

y2dx+ (3xy − 1)dy = 0. (2.3.9)

Answer.dy

dx=

y2

1− 3xy

It is not linear in y (not exact, not separable and not homogeneous). The roles of x and y are

interchangeable.

dx

dy=

1− 3xy

y2

dx

dy+ (

3

y)x =

1

y2. (2.3.10)

It is linear in x. An integrating factor is

exp[

∫

p(y)dx] = exp(

∫

3

ydy) = exp(ln |y|3) = y3.


y3dx

dy+ 3y2x = y,

d

dy[y3x] = y

y3x =1

2y2 + c, x =

1

2y+

c

y3.

B. Bernoulli Equations


1. Definition (Bernoulli Equations)

An equation of the formdy

dx+ P (x)y = Q(x)yn (2.3.11)

is called a Bernoulli differential equation.

2. Theorem 2.3.2

Suppose n 6= 0 and n 6= 1. Then the transformation v = y1−n reduces the Bernoulli equation (2.3.11)

to a linear equation in v.

Proof. First, we multiply (2.3.11) by y−n. thereby expressing it in the form

y−n dy

dx+ P (x)y1−n = Q(x). (2.3.12)

Let v = y1−n.dv

dx= (1− n)y−n dy

dx. From (2.38) we have

1

1− n

dv

dx+ P (x) · v = Q(x).

dv

dx+ (1− n)P (x) · v = (1− n)Q(x)

P1(x) := (1− n)P (x), Q1(x) := (1− n)Q(x)

dv

dx+ P1(x) · v = Q1(x).

This is a linear equation in v.

3. Example

dy

dx+ y = xy3

Answer. y−3 dy

dx+ y−2 = x. Let v = y1−3 = y−2. We have

dv

dx= −2y−3 dy

dx. Therefore

−12

dv

dx+ v = x

dv

dx− 2v = −2x. (2.3.13)


exp[

∫

−2dx] = e−2x.

Multiplying (2.40) by e−2x, we find

e−2x dv

dx− 2e−2xv = −2xe−2x

d

dx(e−2xv) = −2xe−2x

e−2xv =

∫

−2xe−2xdx+ c


u = x dv = −2e−2xdx

du = dx v = e−2x∫

−2xe−2x = xe−2x −∫

e−2x

e−2xv = (x+1

2)e−2x + c

v = x+1

2+ ce2x.

But v =1

y2. Therefore

1

y2= x+

1

2+ ce2x.

4. General Form of the Bernoulli Differential Equation

Consider the equationdf(x)

dy· dydx

+ P (x) · f(y) = Q(x) (2.3.14)

where f is a known function of y. Then

(1) The Bernoulli differential equation (2.3.11) is a special case of (2.3.14).

(2) The transformation v = f(y) reduces (2.3.14) to a linear equation in v.

Proof.

(1) Let f(y) = y1−n.

(1− n)y−n · dydx

+ P1(x) · y1−n = Q1(x)⇒dy

dx+ P (x)y = Q(x)yn

where P1(x) = (1− n)P (x) and Q1(x) = (1− n)Q(x). Therefore, (2.3.14) ⇒ (2.3.11)

(2) Let v = f(y). We have

dv

dx=

dv

dy

dy

dx=

df(y)

dy· dydx

dv

dx+ P (x) · v = Q(x)

which is linear in v.

5. Example

cos ydy

dx+

1

xsin y = 1

Answer. Let f(y) = sin y. Therefore, df(y)dy

= cos y

df(x)

dy· dydx

+1

x· f(y) = 1.


Let v = sin y. We havedv

dx+

1

xv = 1 (2.3.15)


exp[

∫

1

xdx] = exp[ln |x|] = x

Multiply (2.3.15) by x ,we find

xdv

dx+ v = x

d(xv) = x

xv =

∫

xdx+ c =1

2x2 + c

But v = sin y. Therefore, x sin y = 12x2 + c ( or sin y = 1

2x+ c

x).

Check :d

dx(x sin y) =

d

dx(1

2x2 + c)

d

dx(sin y) =

d

dx(1

2x+

c

x)

sin y + cos ydy

dxx = x+ 0 cos y

dy

dx=

1

2+−cx2

=1

2− 1

x2(sin y − x

2) · x

=1

2− sin y

x+

1

2

cos ydy

dx+

1

xsin y = 1

2.4 Special Integrating Factors and Transformations

A. Finding Integrating Factors

1. Theorem 2.4.1

Consider the differential equation

M(x, y)dx+N(x, y)dy = 0 (2.4.1)

(1) If1

N(x, y)

[

∂M(x, y)

∂y− ∂N(x, y)

∂x

]

(2.4.2)

depends upon x only, then

exp

∫

1

N(x, y)

[

∂M(x, y)

∂y− ∂N(x, y)

∂x

]

dx

(2.4.3)

2.4 Special Integrating Factors and Transformations 39

is an integrating factor of Equation (2.4.1).

(2) If1

M(x, y)

[

∂N(x, y)

∂x− ∂M(x, y)

∂y

]

(2.4.4)

depends upon y only, then

exp

∫

1

M(x, y)

[

∂N(x, y)

∂x− ∂M(x, y)

∂y

]

dy

(2.4.5)

is an integrating factor of Equation (2.4.1).

Proof.

(1) An integrating factor µ depends upon x alone.

µMdx+ µNdy = 0∂

∂y[µ(x)M(x, y)] =

∂

∂x[µ(x)N(x, y)]

µ(x) · ∂M∂y

= µ(x) · ∂N∂x

+N · dµ∂x

dµ

dx= µ(x)

[

1

N(∂M

∂y− ∂N

∂x)

]

(depends upon x only)

dµ

µ=

1

N

[

∂M

∂y− ∂N

∂x

]

dx

µ(x) = exp

∫

1

N

[

∂M

∂y− ∂N

∂x

]

dx

.

(2) An integrating factor µ depends upon y alone.

µMdx+ µNdy = 0∂

∂y· [µ(y)M(x, y)] =

∂

∂x· [µ(y)N(x, y)]

dµ

dy·M + µ(y) · ∂M

∂y= µ(y) · ∂N

∂x

dµ

dy= µ(y)

[

1

M(∂N

∂x− ∂M

∂y)

]

(depends upon y only)

dµ

µ=

1

M

[

∂N

∂x− ∂M

∂y

]

dy

µ(y) = exp

∫

1

M

[

∂N

∂x− ∂M

∂y

]

dy

.


2. Example

(2x2 + y)dx+ (x2y − x)dy = 0 (2.4.6)

It is not exact, not separable, not homogeneous, not linear, and not Bernoulli.

Answer.1

N

[

∂M

∂y− ∂N

∂x

]

=1

x2y − x[1− (2xy − 1)] = −2

x.

This depends upon x only, and so

µ(x) = exp

(

−∫

2

xdx

)

= exp(−2 ln |x|) = 1

x2


(2 +y

x2)dx+ (y − 1

x)dy = 0 (2.4.7)

By the method for solving exact ODE, we have

F (x, y) =

∫

(2 +y

x2)dx+ φ(y) = 2x− y

x+ φ(y)

∂F

∂y= −1

x+ φ′(y) = y − 1

x

φ(y) =y2

2+ c.

The solution is

2x− y

x+

y2

2= c

B. Special Transformations

1. Theorem 2.4.2

Consider the equation

(a1x+ b1y + c1)dx+ (a2x+ b2y + c2)dy = 0 (2.4.8)

where a1, b1, a2, b2, c1 and c2 are constants.

Case 1. Ifa2a16= b2

b1, then the transformation

x = X + h,

y = Y + k,


where (h, k) is the solution of the system

a1h + b1k + c1 = 0,

a2h + b2k + c2 = 0,

reduces Equation (2.4.8) to the homogeneous equation

(a1X + b1Y )dX + (a2X + b2Y )dY = 0

in the variables X and Y .

Case 2. Ifa2a1

=b2b1

= k , then the transformation

z = a1x+ b1y

reduces Equation (2.4.8) to a separable equation in x and z.

Proof.

Case 1.

(a1X + a1h+ b1Y + b1k + c1)dX + (a2x+ a2h+ b2Y + b2k + c2)dY = 0

(a1X + b1Y )dX + (a2X + b2Y )dY = 0

It is homogeneous. We can take Y = vX to obtain

(a1 + b1Y

X)dX + (a2 + b2

Y

X)dY = 0

dY

dX= −a1 + b1(

YX)

a2 + b2(YX)

v +Xdv

dX= −a1 + b1v

a2 + b2v

Then it is a separable equation.

Case 2. dz = a1dx+ b1dy

(z + c1)dx+ (kz + c2)(1

b1)(dz − a1dx) = 0

(b1z + b1c1)dx+ (kz + c2)dz − (a1kz + a1c2)dx = 0

(b1z + a1kz + b1c1 + a1c2)dx+ (kz + c2)dz = 0

dx+kz + c2

b1z + a1kz + b1c1 + a2c2dz = 0

It is separable.

2. Example

(x− 2y + 1)dx+ (4x− 3y − 6)dy = 0 (2.4.9)


Answer.

h− 2k + 1 = 0

4h− 3k + (−6) = 0

We have h=3, k=2.

Let

x = X + 3

y = Y + 2This reduce (2.53) to

(X − 2Y )dX + (4X − 3Y )dY = 0

dY

dX=

1− 2( YX)

3( YX− 4)

Let Y = vX to obtain

v +Xdv

dX=

1− 2v

3v − 43v − 4

3v2 − 2v − 1dv = − 1

XdX

∫

3v − 4

3v2 − 2v − 1dv =

1

2ln |3v2 − 2v − 1| − 3

4ln

∣

∣

∣

∣

3v − 3

3v + 1

∣

∣

∣

∣

(1) The Use of Tables.

∫

1

ax2 + bx+ cdx =

2√4ac− b2

tan−1 2ax+ b√4ac− b2

if b2 − 4ac < 0

1√b2 − 4ac

ln

(

2ax+ b−√b2 − 4ac

2ax+ b+√b2 − 4ac

)

if b2 − 4ac > 0

∫

x

ax2 + bx+ cdx =

1

2aln(ax2 + bx+ c)− b

2a

∫

1

ax2 + bx+ cdx

Then we have

∫

3v − 4

3v2 − 2v − 1dv = 3

∫

v

3v2 − 2v − 1dv − 4

∫

1

3v2 − 2v − 1dv

= 3 ·[

1

6ln |3v2 − 2v − 1| − −2

6

∫

1

3v2 − 2v − 1dv

]

− 4

∫

1

3v2 − 2v − 1dv

=1

2ln |3v2 − 2v − 1| − 3 ·

∫

1

3v2 − 2v − 1dv

=1

2ln |3v2 − 2v − 1| − 3 · 1√

4 + 12ln

∣

∣

∣

∣

6x− 2− 4

6x− 2 + 4

∣

∣

∣

∣

=1

2ln |3v2 − 2v − 1| − 3

4ln

∣

∣

∣

∣

3v − 3

3v + 1

∣

∣

∣

∣


Moreover, multiplying 4

→ ln(3v2 − 2v − 1)2 − 3

[

ln | v − 1

3v + 1|+ ln 3

]

→ ln(3v2 − 2v − 1)2 − ln

∣

∣

∣

∣

v − 1

3v + 1

∣

∣

∣

∣

3

= ln

∣

∣

∣

∣

(3v + 1)5

v − 1

∣

∣

∣

∣

since 3v2 − 2v − 1 = (3v + 1)(v − 1)

(2) Partial Fraction

3v − 4

3v2 − 2v − 1=

a

3v + 1+

b

v − 1where a =

15

4and b =

−14

∫

3v − 4

3v2 − 2v − 1dv =

∫ 154

3v + 1dv +

∫ −14

v − 1dv

=5

4

∫

1

3v + 1d(3v + 1) + (

−14)

∫

1

v − 1d(v − 1)

=5

4ln |3v + 1| − 1

4ln |v − 1|

→ multiplying 4

5 ln |3v + 1| − ln |v − 1|

= ln

∣

∣

∣

∣

(3v + 1)5

v − 1

∣

∣

∣

∣

1

2ln |3v2 − 2v − 1| − 1

4ln |3v − 3

3v + 1|3 = − ln |X|+ C2

ln |3v2 − 2v − 1|2 − ln | v − 1

3v + 1|3 = − ln |X|4 + ln(C1)

4

ln

∣

∣

∣

∣

(3v + 1)5

v − 1

∣

∣

∣

∣

= ln((C1)

4

X4)

Finally, X4 · |(3v + 1)5| = C · |v − 1| where C = (C1)4. Now replacing v by Y/X , we obtain

|3Y +X|5 = C|Y −X|

Moreover, replacing X by x− 3 and Y by y − 2, we obtain

|3(y − 2) + (x− 3)|5 = C · |y − 2− x+ 3||x+ 3y − 9|5 = C · |y − x+ 1|


3. Example

(x+ 2y + 3)dx+ (2x+ 4y − 1)dy = 0 (2.4.10)

Answer. Let z = x+ 2y. Then we can transform the Equation (2.4.10) into

(z + 3)dx+ (2z − 1)(dz − dx

2) = 0

7dx+ (2z − 1)dz = 0

7x+ z2 − z = c.

Replacing z by x+ 2y we obtain

7x+ (x+ 2y)2 − (x+ 2y) = c

x2 + 4xy + 4y2 + 6x− 2y = c

4. Special Transformation I: v = ln y

Example.

x · dydx− y ln y = x2y (y > 0) (2.4.11)

Answer. Let v = ln y.dv

dx=

1

y· dydx

. Then we can transform (2.1) into

x · y dvdx− y · v = x2y

dv

dx− 1

x· v = x.


exp

(

−∫

1

xdx

)

= exp (− ln |x|) = 1

x.

Then we have

1

x· dvdx− 1

x2· v = 1

d(1

x· v) = 1,

v

x= x+ c

v = x · (x+ c)

Replacing v by ln y we obtain

ln y = x · (x+ c), y = ex(x+c).


5. Special Transformation II: u = x+ y, v = x/y

Example.x

y(dx+ dy) + (x+ y)(ydx− xdy) = 0 (2.4.12)

Answer. Let u = x+ y and v =x

y. We can transform (2.4.12) into

v · du+ u · ydx− xdy

y2· y2 = 0

v · du+ u · dv · 1

(v + 1)2· u2 = 0

1

u3du+

1

v(v + 1)2dv = 0

∫

1

v(v + 1)2dv = ln |v| − ln |v + 1|+ 1

v + 1

1

v(v + 1)2=

1

v+−1v + 1

+−1

(v + 1)2∫

1

v(v + 1)2dv = ln |v| − ln |v + 1|+ 1

v + 1

Then we have

−u−2 · 12+ ln |v| − ln |v + 1|+ 1

v + 1= c

Replacing u by x+ y and v by x/y, we obtain

− 1

2(x+ y)2+ ln | x/yx

y+ 1|+ 1

xy+ 1

= c

− 1

2(x+ y)2+ ln | x

x+ y|+ y

x+ y= c

6. Special Transformation III: v = xy

Example.

xdy + ydx = (x3y2 − x)dx (2.4.13)


Answer. Let v = xy. dv = ydx+ xdy. We can transform (2.4.13) into

dv = (x · v2 − x)dx,1

v2 − 1dv = xdx

∫

1

v2 − 1dv = [

∫

1

v − 1− 1

v + 1dv] · 1

2=

1

2· ln |v − 1

v + 1|

Therefore,

1

2· ln |v − 1

v + 1| =

1

2x2 + c1

ln |v − 1

v + 1| = x2 + c2 (c2 = 2c1)

v − 1

v + 1= exp[x2 + c2] (c = exp[c2])

xy − 1

xy + 1= c · ex2

7. Special Transformation IV: x = r cos θ, y = r sin θ

Example.

y2(xdx+ ydy) + x(ydx− xdy) = 0 (2.4.14)

Answer. Let

x = r cos θ therefore x2 + y2 = r2

y = r sin θ y/x = tan θ x/y = cot θ.

xdx+ ydy =1

2d(x2 + y2) =

1

2d(r2)

ydx− xdy = y2d(x

y) = y2 · d(cot θ)

Then we can transform (5.4) into

y2 · (12dr2) + x · (y2 · d cot θ) = 0

1

2r · 2 · dr + r cos θ · (− csc2 θ)dθ = 0

dr + (− cos θ csc2 θ)dθ = 0

dr + (− cot θ csc θ)dθ = 0

dr + d(csc θ) = 0


Therefore,

r + csc θ = c, r + 1sin θ

= c

r sin θ + 1 = c · sin θ, y + 1 = c · sin θr(y + 1) = c · r sin θ, r(y + 1) = c · yr2(y + 1)2 = c2y2, (x2 + y2)(y + 1)2 = c2y2.

Check:

(2x+ 2ydy

dx)(y + 1)2 + (x2 + y2) · 2(y + 1)

dy

dx= 2c2y

dy

dx(xdx+ ydy)(y + 1)2 + (x2 + y2) · (y + 1)dy = c2ydy

(xdx+ ydy)(y + 1)2 + (x2 + y2) · (y + 1)dy =1

y2· (x2 + y2)(y + 1)2ydy

Divide it by y + 1 and we have

(xdx+ ydy)(y + 1) + (x2 + y2)dy =1

y· (x2 + y2)(y + 1)dy

xy(y + 1)dx+ [y3 + y2 + y(x2 + y2)− (x2 + y2)(y + 1)]dy = 0

(xy2 + xy)dx+ (y3 − x2)dy = 0.

Chapter 3

Applications of First-Order Equations

3.1 Orthogonal Trajectories and Oblique Trajectories

A. Orthogonal Trajectories

1. Definition (Angle of Intersection)

By the angle of intersection of two coplanar curves at a point which they have in common is meant

−1.5 −1 −0.5 0 0.5 1 1.50.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

1.1

θ

Figure 3.1: Angle of intersection

the angles between the tangents to the curves at the common point. If the angles of intersection are

50 CHAPTER 3. APPLICATIONS OF FIRST-ORDER EQUATIONS

right angles, the two curves are said to be orthogonal. See Fig. 3.1.

2. Definition (Orthogonal Trajectories)

Let

F (x, y, c) = 0 (3.1.1)

be a given one-parameter family of curves in the xy-plane. A curve that intersects the curve of family

(3.1.1) at right angles is called an orthogonal trajectory of the given family.

3. Example 3.1.1

Consider the family of circles

F (x, y, c) = x2 + y2 − c2 = 0 (3.1.2)

with center at the origin and radius c. Each straight line through the origin,

y = kx, (3.1.3)

is an orthogonal trajectory of (3.1.2). Conversely, each circle of (3.1.2) is an orthogonal trajectory of

(3.1.3). See Fig. 3.2.

Electric Lines Equipotential Lines

Figure 3.2: The orthogonal trajectory of (3.1.2) and (3.1.3)

4. Procedure for Finding the Orthogonal Trajectories of a Given Family of Curves

3.1 Orthogonal Trajectories and Oblique Trajectories 51

Step1. From the equation

F (x, y, c) = 0 (3.1.1)

of the given family of curves, find the differential equation

dy

dx= f(x, y). (3.1.4)

Step2. Solvedy

dx= − 1

f(x, y).

Step3. Obtain a one-parameter family

G(x, y, c) = 0 or y = F (x, c).

Notice that

(1) For certain trajectories that are vertical lines and must be determined separately.

(2) In Step1, in finding the differential equation (3.1.4), be sure to eliminate the parameter c during

the process.

5. Example 3.1.2

F (x, y, c) = x2 + y2 − c2 = 0

Answer.

Step 1. Rewrite

2x+ 2ydy

dx= 0. So

dy

dx= −x

y.

Step 2. Solvedy

dx=

y

x.

1

ydy =

1

xdx.

ln |y| = ln |x|+ k. So y = cx where c = ek.

Step 3. Obtain G(x, y, c) = y − cx = 0. See Fig. 3.2.

6. Example 3.1.3

F (x, y, c) = y − cx2 = 0 (3.1.5)


Answer.

Step1.dy

dx− 2cx = 0 (3.1.6)

Eliminating the parameter c between (3.1.5) and (3.1.6),we have

dy

dx= 2(

y

x2)x

Step2. Solvedy

dx= − x

2y2ydy = −xdx

y2 = −12x2 + k

2y2 + x2 = c2 where c2 = 2k

Step3. G(x, y, c) = x2 + 2y2 − c2 = 0. See Fig. 3.3.

−2 −1.5 −1 −0.5 0 0.5 1 1.5 2

−2

−1.5

−1

−0.5

0

0.5

1

1.5

2

Figure 3.3: The orthogonal trajectories of (3.1.5) and (3.1.6)

7. Example 3.1.4

F (x, y, c) = x2 + (y − c)2 = c2 (3.1.7)


Answer.

Step1.

2x+ 2(y − c)dy

dx= 0 ,

dy

dx=−xy − c

(3.1.8)

Eliminating the parameter c between (3.1.7) and (3.1.8), we have

dy

dx=

−xy − x2+y2

2y

=2xy

x2 − y2

Step2. Solvedy

dx= −x

2 − y2

2xy= −1− ( y

x)2

2( yx)

Let y = vx.dy

dx= v + x

dv

dx. So we have v + x

dv

dx= −1 − v2

2v, x

dv

dx= −1 + v2

2v,

2v

1 + v2dv = −1

xdx

ln |1 + v2| = − ln |x|+ c1

1 + v2 = c2 ·1

x, x2 + y2 = c2x := +2cx (c :=

c22)

(x− c)2 + y2 = c2

Step3. G(x, y, c) = (x− c)2 + y2 − c2 = 0. See Fig. 3.4.

−8 −6 −4 −2 0 2 4 6 8−6

−4

−2

0

2

4

6

Figure 3.4: The orthogonal trajectories of (3.1.7) and (3.1.8)

B. Oblique Trajectories


1. Definition (Oblique or Isogonal Trajectories)

Let

F (x, y, c) = 0 (3.1.1)

be a one-parameter family of curves. A curve that intersects the curves of the family (3.1.1) at a

constant angle α 6= π2is called an oblique trajectories of the given family.

2. Theorem

dy

dx= f(x, y) and

dy

dx=

f(x, y) + tanα

1− f(x, y) tanα

ordy

dx= f(x, y) and

dy

dx=

f(x, y)− tanα

1 + f(x, y) tanα

have a constant angle of intersection, α.

Proof. Let θ1 = tan−1[f(x, y)]. Then we have

Case 1. θ2 = tan−1[f(x, y)] + α (i.e. θ2 − θ1 = α).

tan θ2 = tan[tan−1(f(x, y)) + α]

=tan[tan−1(f(x, y))] + tanα

1− tan[tan−1(f(x, y))] tanα

=f(x, y) + tanα

1− f(x, y) tanα

Hence the slope of this oblique trajectory is given by

dy

dx=

f(x, y) + tanα

1− f(x, y) tanα.

Case 2. θ2 = tan−1[f(x, y)]− α (i.e. θ1 − θ2 = α).

tan θ2 = tan[tan−1(f(x, y))− α]

=tan[tan−1(f(x, y))]− tanα

1 + tan[tan−1(f(x, y))] tanα

=f(x, y)− tanα

1 + f(x, y) tanα

Hence the slope of this oblique trajectory is given by

dy

dx=

f(x, y)− tanα

1 + f(x, y) tanα.

3. Example 3.1.5


Find a family of oblique trajectories that intersect the family of straight line y = cx at angle 45.

Answer.

Case 1.dy

dx=

f(x, y) + tanα

1− f(x, y) tanα

dy

dx= c =

y

x

Solvedy

dx=

yx+ tanπ/4

1− yxtan π/4

=x+ y

x− y

It is homogeneous. Let y = vx.

v + xdv

dx=

1 + v

1− v,

v − 1

1 + v2dv = −1

xdx

v

1 + v2dv − 1

1 + v2dv = −1

xdx

1

2ln |1 + v2| − tan−1 v = − ln |x| − ln |c|

⇒ ln c2x2(v2 + 1)− 2 tan−1 v = 0

Replacing v by y/x, we obtain

ln[(c2)(x2 + y2)]− 2 tan−1 y

x= 0

We have

G(x, y, c) = ln[c2(x2 + y2)]− 2 tan−1 y

x= 0

Case 2.dy

dx=

f(x, y)− tanα

1 + f(x, y) tanαSolve

dy

dx=

yx− tan π/4

1 + yxtan π/4

=y − x

x+ y

It is homogeneous. Let y = vx.

v + xdv

dx=

v − 1

1 + v,

v + 1

1 + v2dv = −1

xdx

v

1 + v2dv +

1

1 + v2dv = −1

xdx

1

2ln |1 + v2|+ tan−1 v = − ln |x| − ln |c|

⇒ ln c2x2(v2 + 1) + 2 tan−1 v = 0

Replacing v by y/x, we obtain

ln[(c2)(x2 + y2)] + 2 tan−1 y

x= 0


We have

G(x, y, c) = ln[c2(x2 + y2)] + 2 tan−1 y

x= 0

4. Self-Orthogonal

A given family of curves is said to be self-orthogonal if its family or orthogonal trajectories is the

same as the given family.

5. Example 3.1.6

y2 = 2cx+ c2

Proof.

−10 −8 −6 −4 −2 0 2 4 6 8 10−10

−8

−6

−4

−2

0

2

4

6

8

10

Figure 3.5: The self-orthogonal trajectory of Example 3.1.6.

2 · dydx· y = 2c+ 0 ⇒ dy

dx· y = c

Solve c2 + 2cx− y2 = 0 and we have c =−2x±

√

4x2 + 4y2

2= −x±

√

x2 + y2


Case (1) c > 0 : c = −x+√

x2 + y2 > 0

dy

dx· y = −x+

√

x2 + y2

dy

dx=−x+

√

x2 + y2

y

We must solvedy

dx= − y

−x+√

x2 + y2

to obtain the solution curves like y2 = 2cx + c2. Let r =√

x2 + y2 ≥ 0. So x2 + y2 = r2,

2x+ 2y · dydx

= 2rdr

dxand x+ y · dy

dx= r

dr

dx. So

dy

dx=

r drdx− x

y= − y

−x+√

x2 + y2=−y−x + r

(−x+ r)

(

rdr

dx− x

)

= −y2 = −r2 + x2 = (−r + x)(r + x)

rdr

dx− x = −r − x or r = x = r cos θ

dr

dx= −1 or cos θ = 1

r = −x+ c or θ = 2nπ

r2 = x2 + y2 = x2 − 2cx+ c2 or y = 0

Thus, we have G(x, y, c) = y2 + 2cx− c2 = 0.

Check the differential equation for the original curves.

x+ c = rdr

dx, r

dr

dx= x− x+

√

x2 + y2 = r

dr

dx= 1, r = x+ c1, (

√

x2 + y2)2 = (x+ c1)2

y2 = 2c1x+ c21

G(x, y, c) = y2 − 2cx− c2 = 0.

Case (2) c < 0 : c = −x−√

x2 + y2 < 0. Similarly we have

G(x, y, c) = y2 − 2cx− c2 = 0.


See Fig. 3.5.

6. Electric Field (Advanced Engineering Mathematics)

If an electrical current is flowing in a wire along the z-axis, the resulting equipotential lines in the

xy-plane are concentric circles about the origin, and the electric line of force are the orthogonal

trajectories of these circles. Find the differential equation of the these trajectories and solve it.

Answer. See Example 3.1.1 and Example 3.1.2.

Step 1. Rewrite

2x+ 2ydy

dx= 0. So

dy

dx= −x

y.

Step 2. Solvedy

dx=

y

x.

1

ydy =

1

xdx.

ln |y| = ln |x|+ k. So y = cx where c = ek.

Step 3. Obtain G(x, y, c) = y − cx = 0 or x = 0. See Fig. 3.2.

7. Electric Field (Advanced Engineering Mathematics)

Experiments show that the electric lines of force of two opposite charges of the same strength at (-1,

0) and (1, 0) are the circles through (-1, 0) and (1,0). Show that these circles can be represented by

the equation x2 + (y − c)2 = 1 + c2. Show also that the equipotential lines (orthogonal trajectories)

are the circles (x+ c)2 + y2 = c2 − 1.

Answer. See Example 3.1.4.

Step1. Solve the equations of the circles through (-1, 0) and (1,0). We can assume that the center is

(0, c). Thus the radius can be represented by r =√

(0− 1)2 + (c− 0)2 =√1 + c2. So these circles

can be represented by the equation

x2 + (y − c)2 = 1 + c2, or x2 + y2 − 2cy = 1.

For solving the orthogonal trajectories, we consider

2x+ 2(y − c)dy

dx= 0 ,

dy

dx=−xy − c

Eliminating the parameter c, we have

dy

dx=

−xy − x2+y2−1

2y

=2xy

x2 − y2 − 1

Step2. Solve

dy

dx= −x

2 − y2 − 1

2xy

2xydy = (y2 + 1− x2)dx


Consider the transformation u = y2 and we have

xdu = (u+ 1− x2)dx xdu− udx = (1− x2)dx

xdu− udx

x2= (

1

x2− 1)dx d(

u

x) = (

1

x2− 1)dx

u

x= −1

x− x− 2c

y2

x= −1

x− x− 2c

y2 = −1− x2 − 2cx (x+ c)2 + y2 = c2 − 1

Step3. G(x, y, c) = (x+ c)2 + y2 − c2 + 1 = 0. See Fig. 3.6.

Positive Point Charge Negative Point Charge

Electric Lines

Equipotential Lines

Figure 3.6: The orthogonal trajectories.

8. Fluid Flow (Advanced Engineering Mathematics)

Another area in which orthogonal trajectories play a role is fluid flow. Here the path of a par-

ticle of fluid is called a streamline, and the orthogonal trajectories of the streamlines are called

equipotential lines. Suppose that the streamlines are xy = c. Show that the walls of the channel in

Fig. 3.6 are streamlines, so that the flow may be regarded as a flow around a corner. Find and graph

the equipotential lines.

Answer.

Step 1. xy = c. y + xdy

dx= 0.

dy

dx= −y

x.

Step 2. Solvedy

dx=

x

y. ydy = xdx.


Step 3. G(x, y, c) = x2 − y2 + c = 0.

See Fig. 3.7.

9. Temperature Field

StreamLines

Equipotential Lines

Wall

Wall

Corner

Figure 3.7: The orthogonal trajectories of the fluid flow

Curves of constant temperature T (x, y) = c in a temperature field are called isotherms. Their orthog-

onal trajectories are the curves along which heat will flow (in regions that are free of sources or sinks

of heat and are filled with a homogeneous medium). If the isotherms are given by y2+2xy− x2 = c,

what are the curves of heat flow?

Answer.

Step 1. y2 + 2xy − x2 = c. 2ydy

dx+ 2y + 2x

dy

dx− 2x = 0.

dy

dx=

x− y

x+ y.

Step 2. Solvedy

dx= −x+ y

x− y. Let u = y/x.

u+ xdu

dx= −1 + u

1− u

∫ −2u+ 2

−u2 + 2u+ 1du =

∫ −2x

dx

ln | − u2 + 2u+ 1| = ln |x−2|+ c1 −(yx)2 + 2(

y

x) + 1 =

c

x2

x2 + 2xy − y2 = c.

Step 3. G(x, y, c) = x2 + 2xy − y2 − c = 0.

10. Cauchy-Riemann Equations (Complex Analysis)

3.2 Problems in Mechanics 61

Show that if the equation u(x, y) = c represents a family of curves, a representation of its orthogonal

trajectories of the form v(x, y) = c∗ can be obtained from

∂u

∂x=

∂v

∂y,

∂u

∂y= −∂v

∂x.

These so-called Cauchy-Riemann differential equations are basic in complex analysis and will be

considered below.

Answer.

u(x, y) = c ⇒ ∂u

∂x+

∂u

∂y· dydx

= 0dy

dx= −∂u/∂x

∂u/∂y

v(x, y) = c ⇒ ∂v

∂x+

∂v

∂y· dydx

= 0dy

dx= −∂v/∂x

∂v/∂y

Now we have(

−∂u/∂x∂u/∂y

)

·(

−∂v/∂x∂v/∂y

)

=

(

− ∂v/∂y

−∂v/∂x

)

·(

−∂v/∂x∂v/∂y

)

= −1.

Thus, v(x, y) = c∗ are the orthogonal trajectories of u(x, y) = c.

11. Example

Using the Cauchy-Riemann equations, find the orthogonal trajectories of ex cos y = c.

Answer. Let u(x, y) = ex cos y = c.∂u

∂x= ex cos y =

∂v

∂y,

∂u

∂y= −ex sin y =

∂v

∂x.

∂v

∂y= ex cos y

∂v

∂x= ex sin y

v =∫

ex cos ydy + φ(x) = ex sin y + φ(x). ∂v/∂x = ex sin y + φ′(x) = ex sin y ⇒ φ′(x) = 0.

v(x, y) = ex sin y = c.

3.2 Problems in Mechanics

A. Introduction

1. Definition (Mechanics)

Mechanics may be defined as that science which describes and predicts the conditions of rest or

motion of bodies under the action of forces. It is divided into three parts: mechanics of rigid bodies,

mechanics of deformable bodies, and mechanics of fluids.


(1) The mechanics of rigid bodies is divided into statics and dynamics.

(2) As far as the resistance of the structure to failure is concerned, the mechanics of deformable

bodies are studied.

(3) The mechanics of fluids is subdivided into the study of incompressible fluids and compressible

fluids. An important subdivision of the study of incompressible fluids is hydraulics, which deals with

problems involving liquids.

2. Definition

The momentum of a body is defined to be mv.

3. Newton’s Second Law

d

dt(mv) = k · F → m

dv

dt= k · F → F =

1

kma.

The simplest systems of units are those for which k = 1.

−F = m−a .

4. Acceleration of gravity w = mg, where w is the weight of the body and g is the acceleration of gravity.

5. Instantaneous velocity v =dx

dt, where v is the instantaneous velocity.

6. Instantaneous acceleration a =dv

dt=

d2x

dt, where a is the instantaneous acceleration.

7. Three forms We can express Newton’s second law in the following three forms:

F = mdv

dt(3.2.1)

F = md2x

dt2(3.2.2)

F = mvdv

dx(3.2.3)

wheredv

dt=

dv

dx

dx

dt=

dv

dx· v.

8. Table 3.2.1

British System CGS System MKS System

force pound dyne newton

mass slug g (gram) kg (kilogram)

distance foot cm (centimeter) m (meter)

time second second second

acceleration ft/sec2 cm/sec2 m/sec2


lb., libra(e): pound(s) in weight, g: the acceleration of gravity, g = 9.8 m/sec2 = 980 cm/sec2 =

32 ft/sec2.

B. Falling Body Problems

1. Definition (Modelling)

Modelling means setting up mathematical models of physics or other systems. In this section we

consider systems that can be modelled in terms of a 1st order ODE.

2. Example (A Falling Body with Air Resistance)

A body weighting 8 lb falls from rest toward the earth from a great height. As it falls, air resistance

acts upon it, and we shall assume that this resistance (in pounds) is numerically equal to 2v, where

v is the velocity (in feet per second). Find the velocity and distance fallen at time t seconds.

Answer. (1) Modelling (Formulation)

m =w

g=

8

32=

1

4(slug). The gravity force F1 = +8(pound). The air resistance F2 = −2v (pound).

Newton’s second law becomes F1 + F2 = mdv

dt.

Earth

O

B

+

x

F1 = +8

F2 = −2v

8− 2v =1

4

dv

dtv(0) = 0. The body was initially at rest.

(2) Solution

1

8− 2vdv = 4dt, 8− 2v = c · e−8t

v(0) = 0 ⇒ c = 8, v(t) = 4(1− e−8t).

dx

dt= 4(1− e−8t), x(t) = 4(t+

1

8e−8t) + c

x(0) = 0 ⇒ c = −1/2, x(t) = 4(t+1

8e−8t)− 1

2

(3) Interpretation of Results

(a) t→∞, v → the limiting velocity 4 (ft/sec)

(b) The limiting velocity is attained in a very short time.


(c) The body will not plow through the earth. When the body reaches the earth’s surface,

the differential equation no longer apply!

3. Example (Skydiver)

A skydiver equipped with parachute and other essential equipment falls from rest toward the earth.

The total weight of the man plus the equipment is 160 lb. Before the parachute opens, the air

resistance (in pounds) is numerically equal to 1/2 v, where v is the velocity (in feet per second). The

parachute opens 5 seconds after the fall begins; after it opens, the air resistance is numerically equal

to 5/8 v2. Find the velocity of the skydiver

(A) before the parachute opens, and

(B) after the parachute opens.


m =w

g=

160

32= 5, F1 + F2 = m

dv

dt

(A) 160− 1

2v = 5

dv

dt, v(0) = 0

(B) 160− 5

8v2 = 5

dv

dt, v(5) = v1 where v1 is the velocity when t = 5

(2) Solution

(A)1

v − 320dv = − 1

10dt, v = 320 + ce−t/10

v(0) = 0⇒ c = −320, v(t) = 320(1− e−t/10) ∀ 0 ≤ t ≤ 5

t = 5 v1 = v(5) = 320 · (1− e−5/10).= 126

(B) 160− 5

8v2 = 5

dv

dt,

1

v2 − 256dv = −1

8dt

1

32ln

∣

∣

∣

∣

v − 16

v + 16

∣

∣

∣

∣

= − t

8+ c2, ln

∣

∣

∣

∣

v − 16

v + 16

∣

∣

∣

∣

= −4t + c1∣

∣

∣

∣

v − 16

v + 16

∣

∣

∣

∣

= ce−4t, v(t) =16(ce−4t + 1)

1− ce−4t

v(5) = 126⇒ c =110

142e20, v(t) =

16(110142

e20−4t + 1)

1− 110142

e20−4t, ∀ t ≥ 5.

(3) Interpretation of Results

(a) The limit velocity = 320 ft/sec if the parachute never opened.

(b) The limit velocity = 16 ft/sec if the parachute opens.

(c) The well-known fact is that the velocity of impact with the open parachute is a small

fraction of the impact velocity that would have occurred if the parachute had not opened.

C. Frictional Forces

1. Friction


If a body moves on a rough surface, it will encounter not only air resistance but also another resis-

tance force due to the roughness of the surface. This additional force is called friction. It is shown

in physics that the friction is given by µ ·N , where

(1) µ is a constant of proportionality called the coefficient of friction, which depends upon the rough-

ness of the given surface; and

(2) N is the normal (that is, perpendicular) force which the surface exerts on the body.

2. Example

An object weighting 48lb is released from rest at the top of a plane metal slide that is inclined 30

to the horizontal. Air resistance (in pounds) is numerically equal to one-half the velocity (in feet per

second), and the coefficient of friction is one-quarter.

(A) What is the velocity of the object 2 second after it is released?

(B) If the slide is 24 feet long, what is the velocity when the object reaches the bottom?

Answer. (1) Modelling (Formulation) m = w/g = 48/32 = 3/2.

30° 48

48 sin 30°

Figure 3.8: The diagram to explain the gravitational, the frictional and the normal forces.

(A) F1, the component on the weight parallel to the plane, has value F1 = 48 sin 30 = 24.

(B) F2, the frictional force, has value F2 = −µN = −(1/4) · (48 cos 30) = −6√3.

(C) F3, the air resistance, has numerical value F3 = −(1/2)v.We apply Newton’s second law F = ma. Here F = F1 + F2 + F3.

24− 6√3− (1/2)v = (3/2)

dv

dt, v(0) = 0.

(2) Solution

(A)3

2

dv

dt= −1

2v + 24− 6

√3,

dv

48− 12√3− v

=dt

3, v(t) = 48− 12

√3− ce−t/3.

Since v(0) = 0, we have c = 48− 12√3. Therefore

v(t) = (48− 12√3)(1− e−t/3).


t = 2⇒ v(2) = (48− 12√3)(1− e−2/3)

.= 10.2 (ft/sec).

(B) x(t) =∫

v(t) dt = (48− 12√3)(t + 3e−t/3) + c. Since x(0) = 0 we have c = −(48− 12

√3) · 3.

x(t) = (48− 12√3)(t+ 3e−t/3 − 3).

The slide is 24 feet long. The object reaches the bottom at time T determined from

24 = (48− 12√3)(T + 3e−T/3 − 3).

T.= 2.6, v

.= (48− 12

√3)(1− e−0.9)

.= 12.3 (ft/sec).

D. Velocity of Escape from the Earth

1. Definition (Escape velocity)

A body of constant mass m is projected away from the earth in a direction perpendicular to the

earth’s surface with an initial velocity v0. Assuming that there is no air resistance, but taking into

account the variation of the earth’s gravitational field with distance. The least initial velocity v0 that

is required to lift the body such that the body will not return to the earth is called the escape velocity.

2. Example

Find the minimum initial velocity of a projectile that is fired in radial direction from the earth and

is supposed to escape from the earth. Neglect the air resistance and the gravitational pull of other

celestial bodies. The radius of the earth is R, the mass of projectile is m, the gravitational force is

w(x), the origin is at the surface of the earth and the position of the projectile is x.

R

x = 0 xw(x)


According to Newton’s law of gravitation, the gravitational force, and therefore also the acceleration,

a, of the projectile, is proportional to1

(x+R)2.

w(x) = − k

(x+R)2

where k is a constant. w(0) = −mg ⇒ −mg = −k/R2. Therefore k = mgR2.

w(x) = − mgR2

(x+R)2.

Since there are no other forces acting in the body, the equation of motion is

ma = mdv

dt= − mgR2

(x+R)2, v(0) = v0.

3.3 Rate Problems 67

(2) Solution

Expressdv

dtin terms of

dv

dx, hence

dv

dt=

dv

dx· dxdt

= v · dvdx

. Take it into the motion equation and we

have

mvdv

dx= − mgR2

(x+R)2, vdv = − gR2

(x+R)2dx,

v2

2=

gR2

x+R+ c.

t = 0⇒ x = 0; v(0) = v0. Thus we havev202

=gR2

0 +R+ c, c =

v202− gR.

v2

2=

gR2

x+R+

v202− gR, v = ±

√2 ·

√

v202− gR+

gR2

x+R.

The plus sign must be chosen if the body is rising, and the minus sign if it is falling back to the

earth. To determine the maximum altitude ξ that the body reaches, we set v = 0 and x = ξ.

v20 − 2gR+2gR2

ξ +R= 0

2gR2 = (R + ξ)(−v20 + 2gR) = −Rv20 + 2gR2 + ξ(−v20 + 2gR)

ξ =Rv20

−v20 + 2gR.

Solving the equation for v0, we find the initial velocity required to lift the body to the altitude ξ,

namely,

v0 =

√

2gR · ξ

R + ξ.

The escape velocity ve is then found by letting ξ →∞.

ve =√

2gR.

The numerical value of ve is approximately 6.9 miles/sec = 11.1 km/sec = 39960 km/hr.

3.3 Rate Problems

In certain problems the rate at which a quantity changes is a known function of the amount present

and/or the time, and it is desired to find the quantity itself. If x denotes the amount of the quantity

present at time t, then dx/dt denotes the rate at which the quantity changes and we are at once led

to a differential equation.

A. Rate of Growth and Decay

1. Example (Radioactive nuclei decay)

The rate at which radioactive nuclei decay is proportional to the number of such nuclei that are

present in a given sample. Half of the original number of radioactive nuclei have undergone disinte-

gration in a period of 1500 years.


(A) What percentage of the original radioactive nuclei will remain after 4500 years ?

(B) In how many years will only one-tenth of the original number remain ?


x0: the amount of radioactive nuclei initially present.

x: the amount of radioactive nuclei present after t years.dx

dt: the rate at which the nuclei decay.

The nuclei decay at a rate proportional to the amount present.

dx

dt= −kx, k > 0 (minus sign means decreasing in x)

x(0) = x0.

Half of the original number of radioactive nuclei have undergone disintegration in 1500 years.

x(1500) =1

2x0.

(2) Solution

(A)1

xdx = −kdx x = ce−kt

x(0) = x0 ⇒ c = x0 x(t) = x0 · e−kt

x(1500) =1

2x0 ⇒

1

2x0 = x0 · e−k·1500 e−k = (

1

2)

t1500

x(t) = x0(1

2)

t1500 x(4500) = x0(

1

2)45001500 =

1

8x0.

One-eighth (i.e. 12.5 %) of the original number remain after 4500 years.

(B)1

10x0 = x0(

1

2)

t1500 , ln(

1

10) =

t

1500ln(

1

2), t =

1500 ln 10

ln 2.= 4983 years.

B. Population Growth

1. Malthusian law

Assume that the rate of change of the population is proportional to the number of individuals in it

at any time.

x0: the number of individuals in a population at time t0.

x: the number of individuals in a population at time t.dx

dt: the rate at which the population grows.

dx

dt= +rx, r > 0 (plus sign means increasing in x)

x(t0) = x0.

⇒ x(t) = x0er(t−t0)


It can be shown that the Malthusian law turns out to be unreasonable when applied to the distance

future.

2. Logistic law

The logistic law of growth

dx

dt= rx− λx2

where r > 0 and λ > 0 are constants. The additional term −λx2 is the result of some cause that

tends to limit the ultimate growth of the population. For example, such a cause could be insufficient

living space or food supply, when the population becomes sufficiently large.

3. Example

The population x of a certain city satisfies the logistic law

dx

dt=

1

100x− 1

108x2

where time t is measured in years. Given that the population of this city is 100,000 in 1980, deter-

mine the population as a function of time t > 1980. In particular, answer the following questions:

(A) what will be the population in 2000?

(B) in what year does the 1980 population double?

(C) assuming the population model can be applied for all t > 0, how large will the population ulti-

mately be?


dx

dt=

1

100x− 1

108x2

x(1980) = 100, 000

(2) Solution

dx

10−2x− 10−8x2= dt, 100

(

1

x+

10−6

1− 10−6x

)

dx = dt.

Assuming that 0 < x < 106, we obtain

100[ln x− ln(1− 10−6x)] = t + c1x

1− 10−6x= c · exp( t

100)

x(t) =c · exp(t/100)

1 + 10−6c · exp(t/100)x(1980) = 100, 000 105 =

c · exp(1980/100)1 + 10−6c · exp(1980/100)

c =106

9 exp(1980/100)x(t) =

106

1 + 9 exp(19.8− t/100)


(A)

x(2000) =106

1 + 9 exp(19.8− 2000/100)=

106

1 + 9 exp(−0.2).= 119, 495

(B)

200, 000 =106

1 + 9 exp(19.8− t/100)⇒ exp(19.8− t/100) =

4

9⇒ t

.= 2061

(C)

limt→∞

x(t) = limt→∞

106

1 + 9 exp(19.8− t/100)=

106

1 + 0= 1, 000, 000

Remark:

(a) Consider the logistic equation modified from the Malthusian model.

dx

dt= h(x) · x.

We want to choose h(x) so that

• h(x) ≃ r > 0 when x is small,

• h(x) decreases as x grows larger, and

• h(x) < 0 when x is sufficiently large.

The simplest function that has these properties is

h(x) = r − λx,

where λ > 0. Then the following equation

dx

dt= r · x− λx2,

is known as the logistic equation (or the Verhulst equation).

(b) The equivalent form of the logistic equation is

dx

dt= r(1− x

k)x,

where k = r/λ. The constant r is called the intrinsic growth rate, i.e., the growth rate in the absence

of any limiting factors.

(c) Solve the ODE exactly.

dx

(1− x/k)x= r dt

(

1

x+

1/k

1− x/k

)

dx = r dt

ln |x| − ln |1− x/k| = rt+ c1


We need the assumption: 0 < x < k (and 0 < x0 < k). So we have

x

(1− x/k)= cert

x(0) = x0 ⇒ c = x0/[1− (x0/k)]

x(t) =x0k

x0 + (k − x0)e−rt

Then we can see that limt→∞ x(t) = k. It is called the saturation level.

• y = k: It is one of the equilibrium points and an asympotically stable solution.

• y = 0: It is one of the equilibrium points and an unstable solution.

C. Mixture Problems

1. Basic concepts

A substance S is allowed to flow into a certain mixture in a container at a certain rate, and the mixture

is kept uniform by stirring. Further, in one such situation, this uniform mixture simultaneously flows

out of the container at another (generally different) rate; in another situation this may not be the

case. In either case, we seek to determine the quantity of the substance S present in the mixture at

time t.

INOUT

Tank X

Figure: the mixture brine tank

x : the amount of S present at time tdx

dt: the rate of change of x with respect to t

IN : the rate of S enters the mixture

OUT : the rate of S at which it leavesdx

dt= IN−OUT

2. Example

A tank initially contains 50 gal. of pure water. Starting at time t = 0 a brine containing 2 lb of

dissolved salt per gallon flows into the tank at the rate of 3 gal/min. The mixture is kept uniform

by stirring and the well-stirred mixture simultaneously flows out of the tank at the same rate.

(A) How much salt is in the tank at any time t > 0?


(B) How much salt is present at the end of 25 min?

(C) How much salt is present after a long time?


dx

dt= IN−OUT

IN = ( 2 lb/gal) · (3 gal/min) = 6 lb/min.

OUT : the rate of outflow = the rate of inflow

⇒ the tank contains 50 gal of the mixture at any time t.

⇒ the concentration of salt at time t is (x/50) lb/gal.

OUT = [ (x/50) lb/gal] · (3 gal/min) = (3x/50) lb/min.

Initially there was no salt in the tank (pure water).

⇒ x(0) = 0 .

dx

dt= IN−OUT = 6− 3x

50x(0) = 0.

(2) Solution

(A)

dx

dt= 6− 3x

50dx

100− x=

3

50dt

x(t) = 100 + c · exp(− 3t

50)

x(0) = 0 ⇒ c = −100

x(t) = 100

[

1− exp(− 3t

50)

]

.

(B) x(25) = 100[1− exp(−3 · 25/50)] .= 78 (lb).

(C) limt→∞

100[1− exp(− 3t

50)] = 100 (lb).

3. Example

A tank initially contains 50 gal. of brine in which there is dissolved 10 lb of salt. Starting at time

t = 0 a brine containing 2 lb of dissolved salt per gallon flows into the tank at the rate of 5 gal/min.

The mixture is kept uniform by stirring and the well-stirred mixture simultaneously flows out of the

tank at the slower rate 3 gal/min. How much salt is in the tank at any time t > 0?


dx

dt= IN−OUT


IN = (2 lb/gal) · (5 gal/min) = 10 lb/min.

OUT : At time t = 0, the tank contains 50 gal of brine.

inflow: 5 gal/min, outflow: 3 gal/min, net gain: 5-3=2 gal/min

At the end of t minutes the amount of brine in the tank is (50 + 2t) gal.

The concentration of salt at time t is

(

x

50 + 2t

)

lb/gal.

OUT = (x

50 + 2tlb/gal ) · (3 gal/min) =

(

3x

50 + 2t

)

lb/min.

Initially there was 10 lb of salt in the tank, we have

⇒ x(0) = 10 .

dx

dt= IN−OUT = 10− 3x

50 + 2tx(0) = 10.

(2) Solution

dx

dt= 10− 3x

50 + 2tdx

dt+

3x

50 + 2t= 10

µ(t) = exp

(∫

3

50 + 2tdt

)

= (t+ 25)3/2

23/2µ(t) = 23/2 · (t+ 25)3/2 = (2t+ 50)3/2

d

dt

[

(2t+ 50)3/2x]

= 10 · (2t+ 50)3/2

x(t) = 2(2t+ 50) +c

(2t+ 50)3/2

x(0) = 10 ⇒ c = −(90)(50)3/2 = −22500√2

x(t) = 4t+ 100− 22500√2

(2t+ 50)3/2.

Chapter 4

Explicit Methods of Solving High-Order

ODE

4.1 Basic Theory of Linear Differential Equations

A. Definition and Basic Existence Theorem

1. Definition

A linear ordinary differential equation of order n in the dependent variable y and the independent

variable x is an equation that is in, or can be expressed in, the form

an(x)dny

dxn+ an−1(x)

dn−1y

dxn−1+ · · ·+ a1(x)

dy

dx+ a0(x)y = F (x) (4.1.1)

where an(x) is not identically zero. We shall assume that a0(x), a1(x), . . . , an(x) and F (x) are con-

tinuous real functions on a ≤ x ≤ b and that an(x) 6= 0 ∀ x ∈ [a, b]. The right-hand member F (x) is

called the nonhomogeneous term. If F is identically zero, Eqn. (4.1.1) reduces to

an(x)dny

dxn+ an−1(x)

dn−1y

dxn−1+ · · ·+ a1(x)

dy

dx+ a0(x)y = 0 (4.1.2)

and is then called homogeneous. For example, n = 2, the second-order nonhomogeneous linear ODE

a2(x)d2y

dx2+ a1(x)

dy

dx+ a0(x)y = F (x) (4.1.3)

and (4.1.2) reduces to the second-order homogeneous linear ODE

a2(x)d2y

dx2+ a1(x)

dy

dx+ a0(x)y = 0 (4.1.4)

2. Examples

(1) 2nd-order:d2y

dx2+ 3x

dy

dx+ x3y = ex.

(2) 3rd-order:d3y

dx3+ x

d2y

dx2+ 3x2 dy

dx− 5y = sin x.

76 CHAPTER 4. EXPLICIT METHODS OF SOLVING HIGH-ORDER ODE

3. Theorem 4.1.1

Hypothesis

(1) Consider the nth-order linear ODE (4.1.1) where a0(x), a1(x), . . . , an(x) and F (x) are continuous

real functions on a ≤ x ≤ b and that an(x) 6= 0 ∀ x ∈ [a, b]

(2) Let x0 ∈ [a, b] and c0, c1, . . . , cn−1 be n arbitrary real constants.

Conclusion

∃! the solution f of (4.1.1) such that

f(x0) = c0, f′(x0) = c1, . . . , f

(n−1)(x0) = cn−1

and the solution is defined over [a, b].

4. Exampled2y

dx2+ 3x

dy

dx+ x3y = ex, y(1) = 2, y′(1) = −5.

Theorem 4.1.1 assures us that there exist a unique solution of the given problem.

5. Corollary 4.1.2

Hypothesis

Let f be a solution of nth-order homogeneous ODE

an(x)dny

dxn+ an−1(x)

dn−1y

dxn−1+ · · ·+ a1(x)

dy

dx+ a0(x)y = 0 (4.1.2)

such that

f(x0) = 0, f ′(x0) = 0, . . . , f (n−1)(x0) = 0,

where x0 ∈ [a, b].

Conclusion

Then f(x) = 0 ∀ x ∈ [a, b].

6. Example

d3y

dx3+ 2

d2y

dx2+ 4x

dy

dx+ x2y = 0, f(2) = f ′(2) = f ′′(2) = 0.

Cor. 4.1.2 assures us that f(x) = 0 ∀ x ∈ R.

B. Homogeneous Equations

1. Theorem 4.1.3 (Basic Theorem on Linear Homogeneous Differential Equations)

Hypothesis

Let f1(x), f2(x), · · · , fm(x) be any m solutions of the homogeneous ODE

an(x)dny

dxn+ an−1(x)

dn−1y

dxn−1+ · · ·+ a1(x)

dy

dx+ a0(x)y = 0 (4.1.2)

4.1 Basic Theory of Linear Differential Equations 77

Conclusion

Then c1f1(x)+c2f2(x)+ · · ·+cmfm(x) is also a solution of (4.1.2) where c0, c1, . . . , cm are m arbitrary

real constants.

2. Definition (Linear Combination)

If f1(x), f2(x), · · · , fm(x) are m given functions, and c0, c1, . . . , cm are m constants, then the expres-

sion

c1f1(x) + c2f2(x) + · · ·+ cmfm(x)

is called a linear combination of f1(x), f2(x), · · · , fm(x).

3. Theorem 4.1.3 (Restated)

Any linear combination of solutions of the homogeneous linear ODE (4.1.2) is also a solution of

(4.1.2).

4. Exampled2y

dx2+ y = 0

Answer. c1 sin x+ c2 cosx.

5. Exampled3y

dx3− 2

d2y

dx2− dy

dx+ 2y = 0

Answer. c1ex + c2e

−x + c3e2x.

6. Definition (Linear Dependence)

The n functions f1(x), f2(x), · · · , fn(x) are called linearly dependent on a ≤ x ≤ b if there exist

constants c0, c1, . . . , cn, not all zeros, such that

c1f1(x) + c2f2(x) + · · ·+ cnfn(x) = 0 ∀ x ∈ [a, b].

7. Example

f1(x) = x, f2(x) = 2x

Answer. Let c1 = 2, c2 = −1. Therefore c1f1(x) + c2f2(x) = 0 ∀x ∈ R. f1(x) and f2(x) are linearly

dependent.

8. Definition (Linear Independence)

The n functions f1(x), f2(x), · · · , fn(x) are called linearly independent on a ≤ x ≤ b if they are not

linear dependent there. That is, the n functions f1(x), f2(x), · · · , fn(x) are linearly independent if

c1f1(x) + c2f2(x) + · · ·+ cnfn(x) = 0⇒ c0 = c1 = · · · = cn = 0.


In other words, the only linear combination of f1(x), f2(x), · · · , fn(x) that is identically zero on [a, b]

is the trivial linear combination

0 · f1(x) + 0 · f2(x) + · · ·+ 0 · fn(x).

9. Example

f1(x) = x, f2(x) = x2, [a, b] = [0, 1]

Answer. c1x+ c2x2 = 0, ∀x ∈ [0, 1].

Let x = 1/2 ⇒ 1/2 · c1 + 1/4 · c2 = 0 Therefore c1 = c2 = 0

x = 1 ⇒ c1 + c2 = 0

f1(x) and f2(x) are linearly independent.

10. Theorem 4.1.4

The nth-order homogeneous linear ODE (4.1.2) always possesses n solutions that are linearly inde-

pendent. Further, if f1(x), f2(x), · · · , fn(x) are n linearly independent solutions of (4.1.2), then every

solution of (4.1.2) can be expressed as a linear combination

c1f1(x) + c2f2(x) + · · ·+ cnfn(x).

11. Example

d2y

dx2+ y = 0, g(x) = sin(x+ π/6), f1(x) = sin x, f2(x) = cosx.

Answer.g(x) = sin(x+ π/6) = sin x cosπ/6 + cosx sin π/6

=

√3

2sin x+

1

2cosx =

√3

2f1(x) +

1

2f2(x)

12. Definition (Fundamental Set of Solutions)

If f1(x), f2(x), · · · , fn(x) are n linearly independent solutions of (4.1.2), then the set

f1(x), f2(x), · · · , fn(x) is called a fundamental set of solutions of (4.1.2) and the function f(x)

defined by

f(x) = c1f1(x) + c2f2(x) + · · ·+ cnfn(x)

where c1, c2, . . . , cn are arbitrary constants, is called a general solution of (4.1.2).

13. Example

(1)d2y

dx2+ y = 0. (2)

d3y

dx3− 2

d2y

dx2− dy

dx+ 2y = 0


Answer. (1) sin x, cos x, (2) ex, e−x, e2x.

14. Definition (Wronskian)

Let f1(x), f2(x), · · · , fn(x) be n real functions each of which has an (n−1)st derivative on [a, b]. The

determinant

W (f1, f2, . . . , fn) =

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

f1 f2 · · · fn

f ′1 f ′

2 · · · f ′n

· · ·f(n−1)1 f

(n−1)2 · · · f

(n−1)n

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

is called the Wronskian determinant, or simply the Wronskian of these n functions. We observe that

W (f1, . . . , fn) is itself a real function defined on [a, b]. Its values at x is denoted by W (f1, . . . , fn)(x)

or by W [f1(x), . . . , fn(x)].

15. Theorem 4.1.5

Suppose that f1, . . . , fn are n solutions of (4.1.2).

Then the n solutions f1, . . . , fn of (4.1.2) are linear independent on [a, b], i.e.,

the family of solutions

y(x) = c1f1(x) + · · ·+ cnfn(x)

with arbitrary coefficients c1, . . . , cn includes every solution of (4.1.2)

⇔ W (f1, . . . , fn) 6= 0 for some x ∈ [a, b].

16. Remark (Abel’s Theorem)

The Wronskian of n solutions f1, . . . , fn of (4.1.2) is either identically zero on [a, b] or else is never

zero on [a, b]. In particular,

if f1 and f2 are solutions of the differential equation (4.1.4) where a0(x), a1(x), and a2(x) are

continuous on an open interval I, then the Wronskian W (f1, f2)(x) is given by

W (f1, f2)(x) = c · exp[

−∫

a1(x)

a2(x)dx

]

,

where c is a certain constant that depends on f1 and f2, but not on x. Further, W (f1, f2)(x) either

is zero for all x in I (if c = 0) or else is never zero in I (if c 6= 0).

17. Example

(1)d2y

dx2+ y = 0 (2)

d3y

dx3− 2

d2y

dx2− d2y

dx2+ 2y = 0

Answer.

(1)

∣

∣

∣

∣

∣

sin x cosx

cosx − sin x

∣

∣

∣

∣

∣

= − sin2 x− cos2 x = −1 6= 0 (2)

∣

∣

∣

∣

∣

∣

∣

ex e−x e2x

ex −e−x 2e2x

ex e−x 4e2x

∣

∣

∣

∣

∣

∣

∣

= −6e2x 6= 0.


C. Reduction of Order

1. Theorem 4.1.6

Let f(x) be a nontrivial solution of the nth-order homogeneous ODE (4.1.2).

Then the transformation

y = f(x)v

reduces (4.1.2) to an (n-1)st-order homogeneous linear ODE in the dependent variable

w =dv

dx.

2. Theorem 4.1.7 (Reduction in 2nd-order ODE)

Hypothesis

Let f(x) be a nontrivial solution of the second-order homogeneous linear differential equation

a2(x)d2y

dx2+ a1(x)

dy

dx+ a0(x)y = 0 (4.1.4)

Conclusion 1.

The transformation y = f(x) · v reduces (4.1.4) to an 1st-order homogeneous linear ODE

a2(x)f(x)dw

dx+ [2a2(x)f

′(x) + a1(x)f(x)]w(x) = 0 (4.1.5)

in the dependent variable w(x), where w =dv

dx.

Conclusion 2.

The particular solution

w(x) =exp[−

∫ a1(x)a2(x)

dx]

[f(x)]2

of (4.1.5) give rise to the function v(x), where

v(x) =

∫

w(x)dx =

∫ exp[−∫ a1(x)

a2(x)dx]

[f(x)]2dx.

The function g(x) defined by g(x) = f(x)v(x) is then a solution of the 2nd-order equation (4.1.4).

Conclusion 3.

The original known solution f(x) and the ”new” solution g(x) are linearly independent solutions of

(4.1.4), and hence the general solution of (4.1.4) may be expressed as

c1f(x) + c2g(x).

3. Exampled2y

dx2+ y = 0, f(x) = sin x.


Find a linearly independent solution by reduction of order.

Answer. (1) Let y = f(x) · v = sin x · v.

dy

dx= cosx · v + sin x · v

d2y

dx2= − sin x · v + 2 cosx · dv

dx+ sin x · d

2v

dx2

d2y

dx2+ y = 0 ⇒ sin x · d

2v

dx2+ 2 cosx · dv

dx= 0.

(2) Let w = dv/dx.

sin x · dwdx

+ 2 cosx · w = 01

wdw = −2 cosx

sin xdx

ln |w| = −2∫

1

sin xd(sin x) = −2 ln | sin x| w =

1

sin2 xdv

dx= w =

1

sin2 xv(x) =

∫

1

sin2 xdx =

∫

csc2 xdx = − cot x.

g(x) = f(x) · v(x) = sin x · (− cotx) = − cosx.

(3) The general solution is y(x) = c1 sin x+ c2 cos x.

4. Example

(x2 + 1)d2y

dx2− 2x

dy

dx+ 2y = 0, f(x) = x.

Find a linearly independent solution by reduction of order.

Answer. (1) Let y = f(x) · v = x · v.

dy

dx= x

dv

dx+ v

d2y

dx2= x

d2v

dx2+ 2

dv

dx

x(x2 + 1)d2v

dx2+ 2

dv

dx= 0.

(2) Let w = dv/dx.

x(x2 + 1)dw

dx+ 2w = 0

1

wdw = − 2

x(x2 + 1)dx = (−2

x+

2x

x2 + 1)dx

ln |w| = −2 ln |x|+ ln(x2 + 2) + c1 w =x2 + 1

x2· c

dv

dx= w =

x2 + 1

x2dv = (1 +

1

x2)dx

Thus v(x) = x− 1

x. Therefore g(x) = f(x) · v(x) = x · (x− 1

x) = x2 − 1. (3) The general solution is

y = c1x+ c2(x2 − 1).


D. The Nonhomogeneous Equations

1. Theorem 4.1.8

Hypothesis

(1) Let v(x) be any solution of the given (nonhomogeneous) nth-order linear ODE (4.1.1).

(2) Let u(x) be any solution of the corresponding (homogeneous) nth-order linear ODE (4.1.2).

Conclusion

Then u(x) + v(x) is also a solution of the given (nonhomogeneous) equation (4.1.1).

2. Example

y = x is a solution of the eqn.d2y

dx2+ y = x (nonhomogeneous). y = sin x is a solution of the eqn.

d2y

dx2+ y = 0 (homogeneous).

Then

y = x+ sin x is also a solution ofd2y

dx2+ y = x (nonhomogeneous).

3. Theorem 4.1.9

Hypothesis

(1) Let yp(x) be a given solution of the nth-order nonhomogeneous linear equation (4.1.1) involving

no arbitrary constants.

(2) Let

yc(x) = c1y1(x) + c2y2(x) + · · ·+ cnyn(x)

be the general solution of the corresponding homogeneous equation (4.1.2).

Conclusion

Then every solution of the nth-order nonhomogeneous equation (4.1.1) can be expressed in the form

yc + yp

That is, c1y1(x) + c2y2(x) + · · · + cnyn(x) + yp(x) for suitable choice of the n arbitrary constants

c1, c2, . . . , cn.

4. Definition

Consider the nth-order nonhomogeneous linear ODE (4.1.1) and the corresponding homogeneous

equation (4.1.2).

(1) The general solution of (4.1.2) is called the complementary function of (4.1.2). We shall denote

this by yc.

(2) Any particular solution of (4.1.1) involving no arbitrary constant is called a particular integral

of (4.1.1). We shall denote this by yp.

(3) The solution yc + yp of (4.1.1) is called the general solution of (4.1.1).

5. Example

4.2 The Homogeneous Linear Equation with Constant Coefficients 83

d2y

dx2+ y = x

Answer. yc = c1 sin x+ c2 cosx. yp = x. Then the general solution is

y = yc + yp = c1 sin x+ c2 cosx+ x.

6. Theorem 4.1.10

Hypothesis

(1) Let f1(x) be a particular integral of

an(x)dny

dxn+ an−1(x)

dn−1y

dxn−1+ · · ·+ a1(x)

dy

dx+ a0(x)y = F1(x) (4.1.6)

(2) Let f2(x) be a particular integral of

an(x)dny

dxn+ an−1(x)

dn−1y

dxn−1+ · · ·+ a1(x)

dy

dx+ a0(x)y = F2(x) (4.1.7)

Conclusion

Then k1f1(x) + k2f2(x) is a particular integral of (1) Let f1(x) be a particular integral of

an(x)dny

dxn+ an−1(x)

dn−1y

dxn−1+ · · ·+ a1(x)

dy

dx+ a0(x)y = k1F1(x) + k2F2(x) (4.1.8)

where k1 and k2 are constants.

7. Example

d2y

dx2+ y = 3x+ 5 tanx

Answer.d2y

dx2+ y = x f1(x) = x (Sec. 4.3)

d2y

dx2+ y = tan x f2(x) = − cosx ln | sec x+ tanx| (Sec. 4.4)

Thus y = 3x− 5 cosx ln | sec x+ tan x| is a particular integral.

4.2 The Homogeneous Linear Equation with Constant Co-

efficients

A. Introduction

1. Definition (Characteristic Equation)


Consider

andny

dxn+ an−1

dn−1y

dxn−1+ · · ·+ a1

dy

dx+ a0y = 0 (4.2.1)

where an, an−1, . . . , a0 are real constants. The polynomial equation in the unknown m:

anmn + an−1m

n−1 + · · ·+ a1m+ a0 = 0 (4.2.2)

is called the characteristic equation or the auxiliary equation of the equation (4.2.1).

B. Case 1. Distinct Real Roots

1. Theorem 4.2.1

Consider the equation (4.2.1) with constant coefficients. If the characteristic equation (4.2.2) has the

n distinct real roots m1, m2, . . . , mn, then the general solution of (4.2.1) is

y(x) = c1em1x + c2e

m2x + · · ·+ cnemnx

where c1, c2, . . . , cn are arbitrary constants.

2. Exampled2y

dx2− 3

dy

dx+ 2y = 0

Answer. The characteristic equation is m2 − 3m+ 2 = 0. We have (m− 1)(m− 2) = 0.

y(x) = c1ex + c2e

2x

C. Case 2. Repeated Real Roots

1. Theorem 4.2.2

(1) Consider the nth-order ODE (4.2.1) with constant coefficients. If the characteristic equation

(4.2.2) has the real root m occurring k times, then the part of the general solution of (4.2.1) corre-

sponding to this k-fold repeated root is

(c1 + c2x+ · · ·+ ckxk−1)emx

(2) If, further, the remaining roots of the characteristic equation (4.2.2) are distinct real number

mk+1, . . . , mn, then the general solution is

y = (c1 + c2x+ · · ·+ ckxk−1)emx + ck+1e

mk+1x + · · ·+ cnemnx

(3) If, however, the remaining roots of the characteristic equation (4.2.2) are also repeated, then

the parts of the general solution (4.2.1) corresponding to each of these other repeated roots are

expressions similar to that corresponding to m in part (1).

4.2 The Homogeneous Linear Equation with Constant Coefficients 85

2. Exampled2y

dx2− 6

dy

dx+ 9y = 0

Answer. The characteristic equation is m2 − 6m+ 9 = 0, (m− 3)2 = 0.

y = (c1 + c2x)e3x

3. Exampled3y

dx3− 4

d2y

dx2− 3

dy

dx+ 18y = 0

Answer. The characteristic equation is m3 − 4m2 − 3m+ 18 = 0, (m− 3)2(m+ 2) = 0.

y = (c1 + c2x)e3x + c3e

−2x

D. Case 3. Conjugate Complex Roots

1. Euler’s Formula

(1) eiθ = cos θ + i sin θ

(2) eax · eibx = e(a+ib)x

(3) eiπ + 1 = 0

2. Theorem 4.2.3

(1) Consider the nth-order ODE (4.2.1) with constant coefficients. If the characteristic equation

(4.2.2) has the conjugate roots a + bi and a − bi, neither repeated, then the part of the general

solution of (4.2.1) corresponding to these two conjugate roots may be written

eax(c1 sin bx+ c2 cos bx)

(2) If, however, a + bi and a − bi are each k-fold roots of the characteristic equation (4.2.2), then

the part of the general solution of (4.2.1) corresponding to these repeated conjugate roots may be

written

eax[(c1 + c2x+ · · ·+ ckxk−1) sin bx+ (ck+1 + ck+2x+ · · ·+ c2kx

k−1) cos bx]

Note that

k1e(a+bi)x + k2e

(a−bi)x = eax[k1ebi + k2e

−bi]

= eax[k1(cos bx+ i sin bx) + k2(cos bx− i sin bx)]

= eax[(k1 + k2) cos bx+ i(k1 − k2) sin bx]

= eax[c1 cos bx+ c2 sin bx]


where c1 = k1 + k2, c2 = i(k1 − k2).

3. Example

(1)d2y

dx2+ y = 0 (2)

d2y

dx2− 6

dy

dx+ 25y = 0

Answer.m2 + 1 = 0 m2 − 6m+ 25 = 0

m = 0± i m = 3± 4i

y = c1 sin x+ c2 cos x y = e3x(c1 sin 4x+ c2 cos 4x)

4. Example

d4y

dx4− 4

d3y

dx3+ 14

d2y

dx2− 20

dy

dx+ 25y = 0

Answer.

m4 − 4m3 + 14m2 − 20m+ 25 = 0

(m2 − 2m+ 5)2 = 0

m = 1 + 2i, 1 + 2i, 1− 2i, 1− 2i

y = ex[(c1 + c2x) sin 2x+ (c3 + c4x) cos 2x]

5. Example

d2y

dx2− 6

dy

dx+ 25y = 0 y(0) = −3, y′(0) = −1

Answer.

m2 − 6m+ 25 = 0, m = 3± 4i

y = e3x(c1 sin 4x+ c2 cos 4x)

y(0) = −3 ⇒ −3 = e0(c1 sin 0 + c2 cos 0) c2 = −3y′(0) = −1 ⇒ −1 = e0[(3c1 − 4c2) sin 0 + (4c1 + 3c2) cos 0) c1 = 2

y(x) = e3x(2 sin 4x− 3 cos 4x)

=√13e3x(

2√13

sin 4x− 3√13

cos 4x)

=√13e3x sin(4x+ φ) where sin φ = − 3√

13, cosφ =

2√13

4.3 The Method of Undetermined Coefficients 87

4.3 The Method of Undetermined Coefficients

A. Introduction; An Illustrative Example

1. Example

d2y

dx2− 2

dy

dx− 3y = 2e4x

Answer. Let yp = Ae4x, y′p = 4Ae4x, y′′p = 16Ae4x.

16Ae4x − 2(4Ae4x)− 3Ae4x = 2e4x

A =2

5, yp(x) =

2

5e4x

2. Example

d2y

dx2− 2

dy

dx− 3y = 2e3x

Answer. Let yp = Ae3x. No! It is because yc = c1e3x + c2e

−x. Now we consider that

yp = Axe3x

y′p = 3Axe3x + Ae3x

y′′p = 9Axe3x + 6Ae3x

Solve the undetermined coefficient A.

(9Axe3x + 6Ae3x)− 2(3Axe3x + Ae3x)− 3(Axe3x) = 2e3x

A =1

2, yp(x) =

1

2xe3x

B. The Method

1. Definition (UC Function)

We shall call a function a UC function if it is

either (1) a function defined by one of the following:

(i) xn, n ∈ N ∪ 0,(ii) eax, a is a constant, a 6= 0,

(iii) sin(bx+ c), b and c are constants, b 6= 0,

(iv) cos(bx+ c), b and c are constants, b 6= 0,

or, (2) a function defined as a finite product of two or more functions of these four types.


2. Definition (UC Set)

Consider a UC function f(x). The set of functions consisting of f itself and all linearly indepen-

dent UC functions of which the successive derivatives of f are either constant multiples or linear

combinations will be called the UC set of f .

3. Table 4.3.1

UC function UC set

1. xn xn, xn−1, . . . , x, 1 2. eax eax 3. sin(bx+ c) or cos(bx+ c) sin(bx+ c), cos(bx+ c) 4. xneax xneax, xn−1eax, . . . , xeax, eax 5. xn sin(bx+ c) or xn cos(bx+ c) xn sin(bx+ c), xn cos(bx+ c),

xn−1 sin(bx+ c), xn−1 cos(bx+ c),

· · ·x sin(bx+ c), x cos(bx+ c),

sin(bx+ c), cos(bx+ c) 6. eax sin(bx+ c) or eax cos(bx+ c) eax sin(bx+ c), eax cos(bx+ c) 7. xneax sin(bx+ c) or xneax cos(bx+ c) xneax sin(bx+ c), xneax cos(bx+ c),

xn−1eax sin(bx+ c), xn−1eax cos(bx+ c),

· · ·xeax sin(bx+ c), xeax cos(bx+ c),

eax sin(bx+ c), eax cos(bx+ c)

4. Outline of the UC method

Outline the UC method for finding a particular integral yp of the equation (4.1.1) with constant

coefficients

andny

dxn+ an−1

dn−1y

dxn−1+ · · ·+ a1

dy

dx+ a0y = F (x)

where F (x) is a linear combination,

F (x) = A1u1(x) + A2u2(x) + · · ·+ Amum(x)

of UC functions u1, u2, . . . , um, the Ai’s being known constants.

Step 1. UC functions: u1, u2, . . . , um → UC set: S1, S2, . . . Sm.

Step 2. Si → S ′i → S

(2)i → · · · to make sure that yc 6∈ S

(n)i .

Step 3. If S(ki)i ⊆ S

(kj)j , then omit the (identical or smaller) S

(ki)i .

Find the union of S(k1)1 , S

(k2)2 , . . . , S

(km)m .

4.3 The Method of Undetermined Coefficients 89

Step 4. Form a linear combination of the set, with unknown constant coefficients

(undetermined coefficients).

Step 5. Determine these unknown coefficients by substituting the linear combination

formed in Step 4 into the differential equation and demanding that is identically

satisfies the differential equation.

5. Example

(1)d2y

dx2− 3

dy

dx+ 2y = x2ex (2)

d2y

dx2− 2

dy

dx+ y = x2ex

Answer. The UC set of x2ex is S = x2ex, xex, ex.

(1) yc = c1ex + c2e

2x (2) yc = c1ex + c2xe

x

ex ∈ S double root ex ∈ S

S ′ = x3ex, x2ex, xex, yc 6∈ S ′ S ′′ = x4ex, x3ex, x2ex, yc 6∈ S ′′

6. Example

d2y

dx2− 2

dy

dx− 3y = 2ex − 10 sin x

Answer. yc = c1e3x + c2e

−x

Step 1. S1 = ex, S2 = sin x, cosxStep 2. yc 6∈ S1 and yc 6∈ S2

Step 3. S = S1 ∪ S2 = ex, sin x, cosxStep 4. Let yp = Aex +B sin x+ C cosx.

Step 5. y′p = Aex +B cosx− C sin x, y′′p = Aex −B sin x− C cos x

[Aex − B sin x− C cos x]− 2[Aex +B cosx− C sin x] − 3[Aex +B sin x+ C cosx]

= 2ex − 10 sin x

−4Aex + (−4B + 2C) sinx+ (−4C − 2B) cosx = 2ex − 10 sin x

A = −1/2, B = 2, C = −1

yp = −12ex + 2 sin x− cosx

y = yc + yp = c1e3x + c2e

−x − 1

2ex + 2 sin x− cos x


7. Example

d2y

dx2− 3

dy

dx+ 2y = 2x2 + ex + 2xex + 4e3x

Answer. yc = c1ex + c2e

2x

Step 1. S1 = x2, x, 1, S2 = ex, S3 = xex, ex, S4 = e3xStep 2. S2 ⊂ S3 and yc ∈ S3 → S ′

3 = x2ex, xexStep 3. S = S1 ∪ S ′

3 ∪ S4 = x2, x, 1, x2ex, xex, e3xStep 4. yp = Ax2 +Bx+ C +De3x + Ex2ex + Fxex

Step 5.

2A −3B +2C = 0

−6A +2B = 0

2A = 2

2D = 4

−2E = 2

2E −F = 1

Thus we have A = 1, B = 3, C = 7/2, D = 2, E = −1, F = −3.

y = yc + yp = c1ex + c2e

2x + x2 + 3x+ 7/2 + 2e3x − x2ex − 3xex

4.4 Variation of Parameters

A. The method

Consider the general second order linear ODE

a2(x)d2y

dx2+ a1(x)

dy

dx+ a0(x)y = F (x). (4.4.1)

The corresponding homogeneous ODE is

a2(x)d2y

dx2+ a1(x)

dy

dx+ a0(x)y = 0. (4.4.2)

The complementary function of (4.4.2) is

c1y1(x) + c2y2(x).

The method is to determine

v1(x)y1(x) + v2(x)y2(x)

4.4 Variation of Parameters 91

to be a particular solution of (4.4.1). Hence we use the name, variation of parameters.

Assume yp = v1(x)y1(x) + v2(x)y2(x). Thus y′p = v1y

′1 + v2y

′2 + v′1y1 + v′2y2.

(1) We simplify y′p by demanding that

v′1y1 + v′2y2 = 0

Thus we have y′′p = v1y′′1 + v2y

′′2 + v′1y

′1 + v′2y

′2.

(2) Impose the basic condition yp be a solution of (4.4.1).

v′1y′1 + v′2y

′2 =

F (x)

a2(x)

Since W [y1(x), y2(x)] 6= 0 for y1 and y2 are linearly independent, we can solve (1) and (2) to obtain

v′1 = − F (x)y2(x)

a2(x)W [y1(x), y2(x)]

v′2 = +F (x)y1(x)

a2(x)W [y1(x), y2(x)]

Then we obtain a particular solution

yp(x) = v1(x)y1(x) + v2(x)y2(x)

B. Examples

1. Example

d2y

dx2+ y = tan x

Answer. yc(x) = c1 sin x+ c2 cosx. yp(x) = v1(x) sin x+ v2(x) cosx.

v′1 sin x +v′2 cos x = 0

v′1 cosx −v′2 sin x =tanx

1

Thus we have v′1 = sin x, v′2 = − sin x tanx = cosx− sec x.

∫

sec x dx =

∫

sec x · sec x+ tanx

sec x+ tanxdx

=

∫

sec2 x+ sec x tanx

sec x+ tan xdx

=

∫

1

sec x+ tan xd(sec x+ tanx)

= ln |sec x+ tanx|+ C


v1(x) =∫

sin x dx v2(x) =∫

cosx− sec x dx

= − cosx+ c3 = sin x− ln |sec x+ tanx|+ c4

yp(x) = (− cos x+ c3) sin x+ (sin x− ln | sec x+ tan x|+ c4) cosx

= c3 sin x+ c4 cos x− cosx ln | sec x+ tanx|y(x) = yc(x) + yp(x)

= c1 sin x+ c2 cos x− cosx ln | sec x+ tanx|

2. Higher Order Method

Consider the general nth order linear ODE

an(x)dny

dxn+ · · ·+ a1(x)

dy

dx+ a0(x)y = F (x). (4.4.3)

The corresponding homogeneous ODE is

an(x)dny

dxn+ · · ·+ a1(x)

dy

dx+ a0(x)y = 0. (4.4.4)

The complementary function of (4.4.4) is

c1y1(x) + c2y2(x) + · · ·+ cnyn(x).

The method is to determine

yp(x) = v1(x)y1(x) + v2(x)y2(x) + · · ·+ vn(x)yn(x)

to be a particular solution of (4.4.3). The method of variation of parameters is to solve the following

system:

v′1y1(x) +v′2y2(x) + · · · +v′nyn(x) = 0

v′1y′1(x) +v′2y

′2(x) + · · · +v′ny

′n(x) = 0

...

v′1y(n−1)1 (x) +v′2y

(n−1)2 (x) + · · · +v′ny

(n−1)n (x) =

F (x)

an(x)

3. Example

d3y

dx3− 6

d2y

dx2+ 11

dy

dx− 6y = ex

4.5 The Cauchy-Euler Equation 93

Answer. yc(x) = c1ex + c2e

2x + c3e3x. yp(x) = v1(x)e

x + v2(x)e2x + v3(x)e

3x.

v′1ex +v′2e

2x +v′3e3x = 0

v′1ex +2v′2e

2x +3v′3e3x = 0

v′1ex +4v′2e

2x +9v′3e3x = ex/1

Thus v′1 = 1/2, v′2 = −ex, v′3 = e−2x/2 and we have v1 = x/2, v2 = e−x, v3 = −e−2x/4.

yp(x) =1

2xex + e−xe2x + (−1

4)e−2xe3x

=1

2xex +

3

4ex

y(x) = yc(x) + yp(x)

= c1ex + c2e

2x + c3e3x +

1

2xex

4.5 The Cauchy-Euler Equation

A. The Equation and the Method of Solution

1. The so-called Cauchy-Euler equation is the equation that is in the form

anxn d

ny

dxn+ an−1x

n−1 dn−1y

dxn−1+ · · ·+ a1x

dy

dx+ a0y = F (x) (4.5.1)

where a0, a1, . . . , an are constants and an is not identically zero .

2. Theorem 4.5.1

The transformation x = et (i.e., t = ln x) reduces (4.5.1) to a linear ODE with constant coefficients.

Example

a2x2 d

2y

dx2+a1x

dy

dx+a0y = F (x)

−→ A2d2y

dt2+A1

dy

dt+A0y = G(x)

where A2 = a2, A1 = a1 − a2, A0 = a0 and G(t) = F (et).

dy

dx=

dy

dt

dt

dx=

1

x

dy

dtd2y

dx2=

d

dx(dy

dx) =

d

dx(1

x

dy

dt)

= − 1

x2

dy

dt+

1

x

d

dx(dy

dt)

= − 1

x2

dy

dt+

1

x

d2y

dt2dt

dx= − 1

x2

dy

dt+

1

x2

d2y

dt2


B. Some Examples

1. Example

x2 d2y

dx2− 2x

dy

dx+ 2y = x3

Answer. Let x = et, t = ln x. We have A2 = 1, A1 = −2 − 1 = −3, A0 = 2 and G(t) = F (et) = e3t.

Therefore

d2y

dt2− 3

dy

dt+ 2y = e3t

Consider the characteristic equation m2 − 3m + 2 = 0. We have yc = c1et + c2e

2t. Using the UC

method we set yp = Ae3t since yc 6∈ S = e3t.

9Ae3t − 9Ae3t + 2Ae3t = e3t, A = 1/2

y(t) = c1et + c2e

2t +1

2e3t

Since et = x, we find

y(x) = c1x+ c2x2 +

1

2x3.

2. Four-Steps Procedure

Four-Steps procedure to determine the coefficients of transformation from dny/dxn.

(1) Determine r(r − 1)(r − 2) · · · [r − (n− 1)].

(2) Expand the preceding as a polynomial.

(3) Replace rk by dky/dtk, k = 1, 2, . . . , n.

(4) Equate xn dny

dxnto the result in (3).

3. Example: n=3

(1) r(r − 1)(r − 2). (2) r3 − 3r2 + 2r. (3)d3y

dt3− 3

d2y

dt2+ 2

dy

dt.

(4) x3 d3y

dx3=

d3y

dt3− 3

d2y

dt2+ 2

dy

dt.

4. Example

x3 d3y

dx3− 4x2 d

2y

dx2+ 8x

dy

dx− 8y = 4 lnx

4.6 Linear Differential Equations and Difference Equations 95

Answer. Let x = et, t = ln x. We have

xdy

dx=

dy

dt

x2 d2y

dx2=

d2y

dt2− dy

dt

x3 d3y

dx3=

d3y

dt3− 3

d2y

dt2+ 2

dy

dt

Therefore the original ODE is transferred to

d3y

dt3− 7

d2y

dt2+ 14

dy

dt− 8y = 4t

Consider the characteristic equation m3 − 7m2 + 14m − 8 = 0. We have yc = c1et + c2e

2t + c3e4t.

Using the UC method we set yp = At +B since yc 6∈ S.

14A− 8At− 8B = 4t, A = −1/2, B = −7/8

y(t) = c1et + c2e

2t + c3e4t − 1

2t− 7

8

y(x) = c1x+ c2x2 + c3x

4 − 1

2ln x− 7

8.

4.6 Linear Differential Equations and Difference Equations

A. First Order Linear Differential Equations and Difference Equations

1. Homogeneous Differential Equations

Consider

a1dy

dx+ a0y = 0 (4.6.1)

where a1 and a0 are real constants. The polynomial equation in the unknown m:

a1 ·m1 + a0 = 0 (4.6.2)

is called the characteristic equation or the auxiliary equation of the equation (4.6.1). Solve (4.6.2)

and we have m = −a0/a1. Thus the general solution of (4.6.1) is

y(x) = c · emx = c · e−a0a1

·x

where c is an arbitrary constant.

Remark


The equation (4.6.1) is separable. Thus we can solve it by the method of separation variables.

1

ydy = −a0

a1dx

ln |y| = −a0a1

x+ c1

y(x) = c · exp(−a0a1· x)

2. Nonhomogeneous Differential Equations

Consider

a1dy

dx+ a0y = F (x) (4.6.3)

where a1 and a0 are real constants.


this by yc.




3. Exampledy

dx+ y = x

Answer. (1) Complementary function yc:

Consider the char. eqn. m+ 1 = 0. Thus m = −1.

yc(x) = c · e−x.

(2) Particular integral yp (UC method):

Let yp(x) = Ax+B. Thus A+ Ax+B = x. We have A = 1 and B = −1.

yp(x) = x− 1

(3) The general solution is

y(x) = yp(x) + yc(x) = x− 1 + c · e−x

Remark: This equation is linear and we can consider the integrating factor

µ(x) = exp

(∫

1 · dx)

= ex.

Thus we can mutiply the ODE by µ(x) and have

ex · dydx

+ ex · y = ex · xd

dx(ex · y) = ex · x

ex · y(x) = xex − ex + c

y(x) = x− 1 + c · e−x


4. Homogeneous Difference Equations

Consider

a1yn+1 + a0yn = 0 (4.6.4)

where a1 and a0 are real constants. The polynomial equation in the unknown m:

a1 ·m1 + a0 = 0 (4.6.5)

is called the characteristic equation or the auxiliary equation of the equation (4.6.4). Solve (4.6.5)

and we have m = −a0/a1. Thus the general solution of (4.6.4) is

yn = c ·mn = c ·(

−a0a1

)n

where c is an arbitrary constant.

Remark

a1yn+1 + a0yn = a1c

(

−a0a1

)n+1

+ a0c

(

−a0a1

)n

= c

(

−a0a1

)n(

−a1(a0a1

) + a0

)

= 0

If the initial value y0 is given then we can obtain c = y0.

We can also use the technique of recursive multiplication to solve it.

y1 =

(

−a0a1

)

y0

y2 =

(

−a0a1

)

y1

· · ·yn =

(

−a0a1

)

yn−1

Multiply these equations and we have yn =(

−a0a1

)n

· y0.5. Nonhomogeneous Difference Equations

Consider

a1yn+1 + a0yn = F (n) (4.6.6)

where a1 and a0 are real constants.

(1) The complementary solution of (4.6.6) is denoted by y(c)n .

(2) Any particular solution of (4.6.6) is denoted by y(p)n .

(3) The solution y(c)n + y

(p)n of (4.6.6) is called the general solution of (4.6.6).

6. Example

yn+1 − 2yn = 3n− 5


Answer. (1) Complementary solution y(c)n :

Consider the char. eqn. m− 2 = 0. Thus m = 2.

y(c)n = c · 2n.

(2) Particular solution y(p)n (UC method):

Let y(p)n = An +B. Thus A(n+ 1) +B − 2(An+B) = 3n− 5. We have A = −3 and B = 2.

y(p)n = −3n+ 2

(3) The general solution is

yn = y(c)n + y(p)n = −3n + 2 + c · 2n

Remark: if we have the initial value y0 then c = y0 − 2.

B. High-Order Differential and Difference Equations

1. Homogeneous and Nonhomogeneous Differential Equations

Consider the nth-order nonhomogeneous linear ODE (4.1.1) and the corresponding homogeneous

equation (4.1.2).


this by yc.




2. Homogeneous Difference Equations

Consider the nth-order homogeneous linear difference equation

anyn + an−1yn−1 + · · ·+ an−ryn−r = 0 (4.6.7)

where an, an−1, . . . , an−r are real constants. The polynomial equation in the unknown m:

anmr + an−1m

r−1 + · · ·+ an−r+1m+ an−r = 0 (4.6.8)

is called the characteristic equation or the auxiliary equation of the equation (4.6.7).

Case 1. Distinct Real Roots

Theorem 4.6.1

Consider the equation (4.6.7) with constant coefficients. If the characteristic equation (4.6.8) has the

r distinct real roots m1, m2, . . . , mr, then the general solution of (4.6.7) is

yn = c1mn1 + c2m

n2 + · · ·+ crm

nr


where c1, c2, . . . , cr are arbitrary constants.

Example

yn − 6yn−1 + 8yn−2 = 0, n ≥ 2

Answer. The char. eqn. is m2 − 6m+ 8 = 0. Thus m = 2 and 4.

yn = c14n + c22

n.

If the initial values are y0 = 0 and y1 = 2. Then we have c1 = 1 and c2 = −1. So yn = 4n − 2n.

Case 2. Repeated Real Roots

Theorem 4.6.2

Consider the equation (4.6.7) with constant coefficients. Assume that the characteristic equation

(4.6.8) has the s distinct real roots m1, m2, . . . , ms and s ≤ r. If the multiplicity of mi is ω(i),

i = 1, · · · , s then

for the root mi we have the solution

µi(n) =(

c(i)0 + c

(i)1 n+ c

(i)2 n2 + · · ·+ c

(i)

ω(i)−2nω(i)−2 + c

(i)

ω(i)−1nω(i)−1

)

mni

and the general solution of (4.6.7) is

yn = µ1(n) + µ2(n) + · · ·+ µs(n)

where c(i)0 , c

(i)1 , . . . , c

(i)

ω(i)−1are arbitrary constants.

Example

yn − 7yn−1 + 16yn−2 − 12yn−3 = 0, n ≥ 3

Answer. The char. eqn. is m3 − 7m2 + 16m− 12 = 0. Thus m = 2, 2 and 3.

yn = (c(1)0 + c

(1)1 n)2n + c

(2)0 3n.

Simply we may write yn = (c0 + c1n)2n + c23

n.

Case 3. Conjugate Complex Roots

Theorem 4.6.3

Consider the equation (4.6.7) with constant coefficients. Assume that the characteristic equation

(4.6.8) has the two conjugate complex roots m1 = reiθ and m2 = re−iθ (in polar form) If the

multiplicity of mi is 1 then

for the root m1 and m2 we have the solution

yn = C1rn cos(nθ) + C2r

n sin(nθ).


c1mn1 + c2m

n2 = c1r

neinθ + c2rne−inθ

= rn[c1(cosnθ + i sin nθ) + c2(cosnθ − i sinnθ)

= C1rn cos(nθ) + C2r

n sin(nθ),

where C1 = c1 + c2 and C2 = i(c1 − c2).

Example

yn+2 − 2yn+1 + 2yn = 0

Answer. The char. eqn. is m2− 2m+2 = 0. Thus m1,2 = 1± i. Hence r =√2 and θ = π/4, so that

the general solution is

yn = (√2)n(

C1 cosnπ

4+ C2 sin

nπ

4

)

.

3. Nonhomogeneous Difference Equations

Consider the nth-order homogeneous linear difference equation

anyn + an−1yn−1 + · · ·+ an−ryn−r = F (n) (4.6.9)

where an, an−1, . . . , an−r are real constants.

(1) The complementary solution of (4.6.9) is denoted by y(c)n .

(2) Any particular solution of (4.6.9) is denoted by y(p)n .

(3) The solution y(c)n + y

(p)n of (4.6.9) is called the general solution of (4.6.9).

Case 1. F (n) is a polynomial of n

Let F (n) = b0 + b1n+ · · ·+ bknk. We can use UC method. Let

y(p)n = A0 + A1n+ · · ·+ Ak+pnk+p

where

p = 0 if 1 is not a root of the char. eqn.

p = ω if 1 is a root of the char. eqn. and ω is the multiplicity of 1

Actually, we can use

y(p)n = Aωnω + Aω+1n

ω+1 + · · ·+ Ak+ωnk+ω.

Remark (Special Case, F (n) = b0)

In the special case when F (n) = b0 is a constant, a particular solution can easily be obtained by

setting y(p)n = A. Substitution of yn = A in the original difference equation leads to the determination

A =b0

an + an−1 + · · ·+ an−r


provided that the sum of the coefficients does not vanish.

Example

yn+2 − 2yn+1 + 2yn = 1

Answer. A particular solution can be found by y(p)n = 1/(1 − 2 + 2) = 1. The char. eqn. is

m2 − 2m+ 2 = 0. Thus m1,2 = 1± i. Hence r =√2 and θ = π/4, so that the general solution is

yn = y(p)n + y(c)n = 1 + (√2)n(

C1 cosnπ

4+ C2 sin

nπ

4

)

.

Example

yn − 7yn−1 + 10yn−2 = 4n− 5, n ≥ 2

Answer. (1) Complementary solution: the char. eqn. is m2 − 7m+ 10 = 0. Thus m = 2 and 5.

y(c)n = c12n + c25

n

(2) Particular solution: we can consider

y(p)n = A+Bn

because 1 is not a root of the char. eqn. So we have

(A+Bn)− 7(A+Bn− B) + 10(A+Bn− 2B) = 4n− 5.

Thus, A = 2 and B = 1 and y(p)n = 2 + n. The general solution is

yn = y(c)n + y(p)n = c12n + c25

n + n+ 2.

If the initial values are y0 = 2 and y1 = 4 then we have the solution yn = 5 · 2n − 5n + n− 2.

Example

yn − 2yn−1 + yn−2 = 2, n ≥ 2, y0 = 1, y1 = 1

Answer. (1) Complementary solution: the char. eqn. is m2 − 2m+ 1 = 0. Thus m = 1 and 1.

y(c)n = c1 + c2n

(2) Particular solution: Since the multiplicity of root 1 is that ω = 2, we can consider

y(p)n = An2


So we have

0− 2A+ 4A = 2.

Thus, A = 1 and y(p)n = n2. The general solution is

yn = y(c)n + y(p)n = c1 + c2n + n2.

Since the initial values are y0 = 1 and y1 = 1, we have c1 = 1 and c2 = −1. The solution is

yn = 1− n + n2.

Case 2. F (n) is an exponential function of n

Let F (n) = b · dn where b and d are constants. We can use UC method. Let

y(p)n = (A0 + A1n+ · · ·+ Apnp) dn

where

p = 0 if d is not a root of the char. eqn.

p = ω if d is a root of the char. eqn. and ω is the multiplicity of d

Example

yn + yn−1 − 2yn−2 = 3 · 2n, n ≥ 2

Answer. (1) Complementary solution: the char. eqn. is m2 +m− 2 = 0. Thus m = −2 and 1.

y(c)n = c1(−2)n + c2

(2) Particular solution: we can consider

y(p)n = A · 2n

because 2 is not a root of the char. eqn. So we have

A · 2n + A · 2n−1 − 2A · 2n−2 = 3 · 2n.

Thus, A = 3 and y(p)n = 3 · 2n. The general solution is

yn = y(c)n + y(p)n = 3 · 2n + c1(−2)n + c2.

Exercise

yn − 4yn−1 + 3yn−2 = 2 · 3n, n ≥ 2

Remark: Nonphysical Oscillations

Numerical difficulties with boundary layer problems.


• It may fail to convergence. (Typical iteration schemes)

• It may exhibit oscillations which are physical unrealistic.

• If an explicit time-stepping procedure is used, it may become unstable for unexpectedly small

time steps.

A simple example

−ǫ · u′′ + u′ = 0, on (0, 1), ǫ > 0,

u(0) = 0, u(1) = 1.

The characteristic equation is

−ǫλ2 + λ = 0

We have u(x) = c1e0 + c2e

x/ǫ. Initial conditions imply that c1 = −1/(e1/ǫ − 1) and c2 = 1/(e1/ǫ − 1).

u(x) =ex/ǫ − 1

e1/ǫ − 1

The standard finite element method (FEM) or central difference scheme is

−ǫ · uj+1 − 2uj + uj−1

h2+

uj+1 − uj−1

2h= 0, j = 1, 2, · · · , N − 1

u0 = 0, uN = 1.

Define the number, mesh Peclet number,

βdef=

h

ǫ> 0.

Rewrite the equation and we have

(1

2β − 1)uj+1 + 2uj + (−1

2β − 1)uj−1 = 0


(1

2β − 1)λ2 + 2λ+ (−1

2β − 1) = 0

We have uh(xj) = c1λj1 + c2λ

j2 where λ1 = 1 and the important char. root λ2

λ2 =1 + 0.5β

1− 0.5β.


0 0.2 0.4 0.6 0.8 1−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

X−Axis

Y−

Axi

s

9 mesh points17 mesh points65 mesh pointsExact solution

ε=0.01

Initial conditions imply that c1 = −1/(λN2 − 1) and c2 = 1/(λN

2 − 1) and we have

uh(xj) =λj2 − 1

λN2 − 1

.

It is clear that the approximate solution is oscillatory when β > 2, i.e., λ2 < 0.

References:

1. Elementary Numerical Analysis, Samuel D. Conte and Carl de Boor, 1987

2. Lecture Note, Prof. Ming-Luen Wu

3. Numerical Solution of Convection-Diffusion Problems, K. W. Morton, Publisher: Chapman &

Hall/CRC; 1 edition (May 15, 1996)

4. Numerical Solution of Partial Differential Equations by the Finite Element Method, C. Johnson,

1987

Chapter 5

Applications of Second-Order Linear

ODE

5.1 The Differential Equation of Vibrations of a Mass on a

Spring

A. Mechanical System – Spring Vibration

1. Hooke’s law

The magnitude of the force needed to produce a certain elongation, provided this elongation is not

too great. In mathematical form,

|F | = ks,

where F is the magnitude of the force, s is the amount of elongation, and k is a constant of propor-

tionality which we shall call the spring constant.

When a mass is hung upon a spring of spring constant k and thus produces an elongation of amount

s, the force F of the mass upon the spring therefore has magnitude ks. The spring at the same time

exerts a force upon the mass called the restoring force of the spring. This force is equal in magnitude

but opposite in sign to F and hence has magnitude −ks.

106 CHAPTER 5. APPLICATIONS OF SECOND-ORDER LINEAR ODE

L

Fig. (a)

L

l

m

Fig. (b)

L

l

O

x

m

Fig. (c)

(a) natural length L

(b) mass in equilibrium position; spring has stretched length L+ l.

(c) mass distance x below equilibrium position; spring has stretched length L+ l + x.

2. Forces Acting Upon the Mass

(a) F1, the force of gravity of magnitude mg.

F1 = mg ↓ positive (5.1.1)

(b) F2, the restoring force of the spring.

F2 = −k(x+ l) ↑ negative (5.1.2)

When x = 0, F1 + F2 = 0, thus we have mg = kl.

Replacing kl by mg in (5.1.2) we see that

F2 = −kx−mg. (5.1.3)

(c) F3, the resisting force of the medium, called the damping force. For small velocities it is ap-

proximately proportional to the magnitude of the velocity:

|F3| = a

∣

∣

∣

∣

dx

dt

∣

∣

∣

∣

(5.1.4)

where a > 0 is called the damping constant. When the mass is moving downward, F3 acts in

the upward direction and so F3 < 0.

F3 = −adx

dt, (a > 0) ↑ negative (5.1.5)

5.2 Free, Undamped Motion 107

(d) F4, any external impressed forces that act upon the mass.

F4 = F (t) (5.1.6)

We now apply Newton’s second law, F = ma, where F = F1 + F2 + F3 + F4. We find

md2x

dt2= mg − kx−mg − a

dx

dt+ F (t) (5.1.7)

or

md2x

dt2+ a

dx

dt+ kx = F (t) (5.1.8)

(e) It is a nonhomogeneous second-order linear differential equation with constant coefficients.

(f) a = 0: If a = 0, the motion is called undamped; otherwise (a 6= 0)it is called damped.

(g) F (t) = 0: If F (t) = 0, the motion is called free; otherwise it is called forced.

5.2 Free, Undamped Motion

1. Free, Undamped Motion

In this section we consider the free, undamped motion. a = 0 and F (x) = 0. The differential

equation (5.1.8) then reduces to

md2x

dt2+ kx = 0 (5.2.1)

where m(> 0) is the mass and k(> 0) is the spring constant. Let λ2 =k

m, we rewrite (5.2.1) in the

formd2x

dt2+ λ2x = 0 (5.2.2)

Hence the general solution of (5.2.2) can be written

x = c1 sinλt + c2 cosλt (5.2.3)

where c1 and c2 are arbitrary constants.

Initial conditions:

x(0) = x0, (initial displacement) (5.2.4)

x′(0) = v0, (initial velocity) (5.2.5)

Therefore

dx/dt|t=0 = x′(0) = v0 =⇒ [c1λ cosλt− c2λ sinλt]|t=0 = v0 (5.2.6)


We have

c2 = x0 and c1 =v0λ

The solution is

x(t) =v0λsin λt+ x0 cosλt (5.2.7)

An alternative form

x(t) = c[v0/λ

csin λt+

x0

ccos λt]

where c =√

(v0/λ)2 + x02. Let φ be defined by satisfying

v0/λ

c= − sin φ and

x0

c= cosφ. So

x(t) = c cos(λt+ φ) (5.2.8)

=√

(v0/λ)2 + x02 cos

(√

k

mt + φ

)

(5.2.9)

2. Simple Harmonic Motion

(1) The free, undamped motion of the mass is a simple harmonic motion.

(2) The constant c is called the amplitude of the motion and gives the maximum (positive) displace-

ment of the mass from its equilibrium position.

(3) The motion is a periodic motion.

x = c⇔√

k

mt+ φ = ±2nπ, n ∈ N ∪ 0

Therefore

t =

√

m

k(±2nπ − φ) > 0 (5.2.10)

The time interval between two successive maxima is called the period of the motion and it is given

by2π

λ=

2π√

k/m= 2π ·

√

m

k. (5.2.11)

(4) The reciprocal of the period, which gives the number of oscillations per second, is called frequency

( or natural frequency ) of the motion and it is given by

λ

2π=

√

k/m

2π=

1

2π·√

k

m. (5.2.12)

(5) The number of φ is called the phase constant (or phase angle). See Fig. 5.1.

3. Example

An 8-lb weight is placed upon the lower end of a coil spring suspended from the ceiling. The weight

comes to rest in its equilibrium position, thereby stretching the spring 6 in. The weight is then pulled

5.2 Free, Undamped Motion 109

0 0.5 1 1.5 2

−0.4

−0.3

−0.2

−0.1

0

0.1

0.2

0.3

0.4

π /8 2π/8 3π/8 4π/8

Period π /4

Amplitude 0.28

t = −φ/8

X

t

Figure 5.1: x(t) = 0.28 cos(8t + φ) where φ=-0.46

down 3 in. below its equilibrium position and released at t = 0 with an initial velocity of 1 ft/sec,

directed downward. Neglecting the resistance of the medium and assuming that no external forces

are present, determine the amplitude, period and frequency of the resulting motion.

(1) Formulation.

m = w/g = 8/32 (slugs) 6 in. = 1/2 ft.

F = ks ⇒ 8 = k/2 k = 16 (lb/ft)

8/32d2x/dt2 + 16x = 0 d2x/dt2 + 64x = 0

x(0) = x0 = 3/12 = 1/4 (ft) x′(0) = v0 = 1 (ft/sec)

The differential equation is

d2x

dt2+ 64x = 0

x(0) =1

4, x′(0) = 1


(2) Solution.

m2 + 64 = 0 m = 0± 8i

x(t) = c1 sin 8t+ c2 cos 8t x(0) = 1/4, x′(0) = 1⇒ c1 = 1/8, c2 = 1/4

x(t) =1

8sin 8t+

1

4cos 8t c =

√

(

1

8

)2

+

(

1

4

)2

=

√5

8

=

√5

8

(√5

5sin 8t+

2√5

5cos 8t

)

=

√5

8cos(8t+ φ) φ

.= −0.46 (radians)

The solution is

x(t).= 0.280 cos(8t− 0.46)

The amplitude of the motion is c =√58

.= 0.280 (ft). The period is 2π/8 = π/4 (sec). The frequency

is 4/π oscillations/sec. The graph is shown in Fig. 5.1.

5.3 Free, Damped Motion

We now consider the effect of the resistance of the medium upon the mass on the spring. Still

assuming that no external forces are present.

md2x

dt2+ a

dx

dt+ kx = 0 (5.3.1)

x(0) = x0, x′0(0) = v0 (5.3.2)

Let λ2 = k/m and 2b = a/m (for convenience). We have

d2x

dt2+ 2b

dx

dt+ λ2x = 0 (5.3.3)


m2 + 2bm+ λ2 = 0.

Thus m = −b ±√

b2 − λ2.

b2 − λ2 =a2

4m2− k

m=

a2 − 4mk

4m2

Case 1. Underdamping case, b2 − λ2 < 0 (i.e. a2 − 4mk < 0): Complex conjugate roots. Damped,

oscillatory motion.

Case 2. Critical damping case, b2 − λ2 = 0 (i.e. a2 − 4mk = 0): A real double root.

Case 3. Overdamping case, b2−λ2 > 0 (i.e. a2−4mk > 0): Distinct real roots. Overcritical damping

5.3 Free, Damped Motion 111

motion.

Let us discuss these three cases separately.

Case 1. Underdamping Motion

m = −b±√

b2 − λ2 = −b±√

λ2 − b2 i

x(t) = e−bt(c1 sin√

λ2 − b2 t + c2 cos√

λ2 − b2 t)

Let c =√

c21 + c22 > 0 and φ is determined by

c1√

c21 + c22= − sinφ

c2√

c21 + c22= +cosφ.

We have

x(t) = ce−bt cos(√

λ2 − b2 t + φ). (5.3.4)

Notice that

(1) The factor ce−bt is called the damping factor, or time-varying amplitude. c > 0 and b > 0 ⇒ce−bt → 0 as t→∞.

(2) cos(√

λ2 − b2 t+ φ) is of a period, oscillatory character.

(3) The oscillations are said to be ”damped out” because ce−bt → 0 as t → ∞. The motion is

described as damped, oscillatory motion.

(4) The time interval between two successive (positive) maximum displacement is still referred to

as the period.

(5) The period is2π

√

λ2 − b2

because of the oscillatory term, cos(√

λ2 − b2 t+ φ).

(6) The ratio of the amplitude at t = T and t = T −(

2π√

λ2 − b2

)

, one period before T , is the

constant

exp

(

− 2πb√

λ2 − b2

)

.

(7) The quantity

(

2πb√

λ2 − b2

)

is the decrease in the logarithm of the amplitude ce−bt over a time

interval of one period. It is called the logarithmic decrement.


0 1 2 3 4 5 6

−0.6

−0.4

−0.2

0

0.2

0.4

0.6 Period

+ce−bt

−ce−bt

1.0 2.0 3.0 4.0 5.0

Figure 5.2: Damped, oscillatory motion.

Now we consider the initial conditions (5.3.2).

x(t) = e−bt(c1 sin√

λ2 − b2 t+ c2 cos√

λ2 − b2 t) = ce−bt cos(√

λ2 − b2 t+ φ)

x(0) = x0 = c1 · 0 + c2 · 1 ⇒ c2 = x0

x′(0) = v0 = −b · c2 + c1√

λ2 − b2 ⇒ c1 =v0 + bx0√

λ2 − b2

Let c =√

c21 + c22 and we have

x(t) = ce−(a/2m)t cos

(√

k

m− a2

4m2t + φ

)

(5.3.5)

Notice that

(1) If damping were not present, a would be equal to 0.

x(t) = c cos

(√

k

m− 0 t + φ

)

where c1 = v0/λ =√mv0/

√k, c2 = x0 and c =

√

mv20/k + x20. Moreover,

cosφ =x0

cand sinφ = −

√

mv20/k

c.

The solution (5.3.5) of damped, oscillatory motion will reduce to the solution (5.2.9) of un-

damped, free motion.

5.3 Free, Damped Motion 113

(2) b2−λ2 < 0 ⇒ b < λ ⇒ a/2m <√

k/m ⇒ a <√2km. We can say that damped, oscillatory

motion occurs when

a <√2km. (5.3.6)

(3) The frequency of the oscillations cos

(√

k

m− a2

4m2t+ φ

)

is

1

2π

√

k

m− a2

4m2. (5.3.7)

(4) The frequency (5.3.7) of the oscillations is less than the natural frequency (5.2.12),1

2π

√

k

m, of

the corresponding undamped system.

Case 2. Critical Damping Motion

x(t) = (c1 + c2t)e−bt.

Notice that

(1) The motion is no longer oscillatory.

(2) Any slight decrease in the amount of damping will change the situation back to that of Case 1.

The motion in Case 2 is said to be critically damped.

(3) x(t)→ 0 as t→∞.

Case 3. Overdamping Motion

The characteristic equation is m2 + 2bm + λ2 = 0. Therefore m1 = −b +√

b2 − λ2 and m2 =

−b−√

b2 − λ2. So

x(t) = c1em1t + c2e

m2t.

Notice that

(1) No oscillations can occur.

(2) We can no longer say that every decrease in the amount of damping will result in oscillations,

as we could in Case 2. The motion in Case 3 is said to be overcritically damped ( or simply

overdamped).

(3) x(t)→ 0 as t→∞.


Example

An 32-lb weight is placed upon the lower end of a coil spring suspended from the ceiling. The weight

comes to rest in its equilibrium position, thereby stretching the spring 2 ft. The weight is then pulled

down 6 in. below its equilibrium position and released at t = 0 without any initial velocity. Assuming

that no external forces are present, determine the amplitude, period and frequency of the resulting

motion with the following resistance of the medium where dx/dt is the instantaneous velocity in feet

per second and a is the damping constant.

(a) 4 (dx/dt) (a= 4) (b) 8 (dx/dt) (a= 8) (c) 10 (dx/dt) (a= 10).

Formulation.

m = w/g = 32/32 = 1 (slug). 32 = k 2 and so k = 16 lb/ft.

Equations (a), (b) and (c) are

(a)d2x

dt2+ 4

dx

dt+ 16x = 0

(b)d2x

dt2+ 8

dx

dt+ 16x = 0

(c)d2x

dt2+ 10

dx

dt+ 16x = 0.

The initial conditions are

x(0) = 1/2, x′(0) = 0.

Solution.

(a) The auxiliary equation is m2+4m+16 = 0. Its roots are conjugate complex numbers −2±2√3i.

The general solution is

x(t) = e−2t(c1 sin 2√3t+ c2 cos 2

√3t).

ICs imply c1 =√3/6 and c2 = 1/2. The solution is

x(t) = e−2t

(√3

6sin 2√3t +

1

2cos 2√3t

)

=

√3

3e−2t cos

(

2√3t− π

6

)

Interpretation.

This is a damping oscillatory motion.

damping factor

√3

3e−2t

period 2π/2√3 =√3π/3

logarithmic decrement 2√3π/3

5.4 Forced Motion 115

(b) The auxiliary equation is m2 +8m+16 = 0. Its equal roots are −4, −4. The general solution is

x(t) = (c1 + c2t)e−4t.

ICs imply c1 = 1/2 and c2 = 2. The solution is

x(t) =

(

1

2+ 2t

)

e−4t

Interpretation.

This is a critically damped motion.

(c) The auxiliary equation is m2 + 10m+ 16 = 0. Its roots are −2, −8. The general solution is

x(t) = c1e−2t + c2e

−8t.

ICs imply c1 = 2/3 and c2 = −1/6. The solution is

x(t) =2

3e−2t − 1

6e−8t

Interpretation.

This is a overdamped motion.

5.4 Forced Motion

In this section we consider the model

md2x

dt2+ a

dx

dt+ kx = F (t) (5.4.1)

where F (t) is called the external impressed force, input force or driving force. A corresponding

solution is called an output or a response of the system to the driving force.

Of particular interest are periodic inputs. We shall consider a sinusoidal input, say,

F (t) = F1 cosωt

where F1 > 0 and ω are constants. Then (5.4.1) becomes

md2x

dt2+ a

dx

dt+ kx = F1 cosωt (5.4.2)

Dividing through by m and letting

a

m= 2b,

k

m= λ2,

F1

m= E1,


this takes the more convenient form

d2x

dt2+ 2b

dx

dt+ λ2x = E1 cosωt (5.4.3)

Of particular interest is that the damping is less than critical. We shall assume that

b < λ (i.e. a < 2√km).

Hence by (5.2.4) the complementary function of (5.4.3) can be written

xc(t) = ce−bt cos(√

λ2 − b2 t+ φ). (5.4.4)

We shall now find a particular integral by the method of undetermined coefficients. Let

xp(t) = A cosωt+B sinωt.

Then we have

dxp

dt= −ωA sinωt+ ωB cosωt,

d2xp

dt2= −ω2A cosωt− ω2B sinωt

−2bωA+ (λ2 − ω2)B = 0, (λ2 − ω2)A+ 2bωB = E1

A =E1(λ

2 − ω2)

(λ2 − ω2)2 + 4b2ω2, B =

2bωE1

(λ2 − ω2)2 + 4b2ω2.

The particular integral is

xp(t) =E1

(λ2 − ω2)2 + 4b2ω2

[

(λ2 − ω2) cosωt+ 2bω sinωt]

(5.4.5)

We now put this in the alternative ”phase angle” form. Let

cos θ =λ2 − ω2

√

(λ2 − ω2)2 + 4b2ω2

sin θ =2bω

√

(λ2 − ω2)2 + 4b2ω2

,(5.4.6)

and we have

xp(t) =E1

√

(λ2 − ω2)2 + 4b2ω2

cos(ωt− θ). (5.4.7)

The general solution of (5.4.2) is

x(t) = xc(t) + xp(t) = ce−bt cos(√

λ2 − b2 t+ φ) (5.4.8)

+E1

√

(λ2 − ω2)2 + 4b2ω2

cos(ωt− θ).

Observe that

5.5 Resonance Phenomena 117

(1) The first term, ce−bt cos(√

λ2 − b2 t + φ), represents the damped oscillation. If F (t) = 0, then

(5.4.8) reduce to (5.3.4).

(2) t→∞ ⇒ the first term → 0. The first term is thus called the transient term.

(3) t→∞ ⇒ x(t)→ the second term . The second term is thus called the steady-state term.

5.5 Resonance Phenomena

Define the function f of ω by

f(ω) =E1

√

(λ2 − ω2)2 + 4b2ω2

(5.5.1)

We know that

ω = 0 ⇒ f(0) =E1

λ2 and F (t) is the constant F1.

ω →∞ ⇒ f(ω)→ 0.

Let us consider 0 < ω <∞ and find the maximum of f(ω).

f ′(ω) = 0 ⇒ 4ω[2b2 − (λ2 − ω2)] = 0 ⇒ ω = 0 or ω =√

λ2 − 2b2.

(1) λ2 < 2b2:√

λ2 − 2b2 is a complex number. f ↓ 0. f has no extremum.

(2) λ2 = 2b2: f(ω) =E1

√

λ4 + ω4. f ↓ 0. f has no extremum.

(3) λ2 > 2b2: (i.e. a <√2km ) f(ω) has a relative maximum at ω1 =

√

λ2 − 2b2.

f(ω1) =E1

2b√

λ2 − b2.

Return to the original notation. We have

f(ω) =F1/m

√

(

k

m− ω2

)2

+( a

m

)2

ω2

(5.5.2)

f(ω) has a relative maximum at ω1 =√

λ2 − 2b2 =

√

k

m− a2

2m2.

f(ω1) =F1/m

a

m

√

k

m−( a

2m

)2.


If ω = ω1 then the forcing function is said to be in resonance with the system. The value ω1/2π is

called the resonance frequency of the system. For the special case m = k = E1 = 1,

f(ω) =1

√

ω4 + (−2 + a2)ω2 + 1.

See Fig. 5.3 for the relation between f(ω) and ω when a is approximated to 0. The graph of f(ω)

0 0.5 1 1.5 2 2.5 30

0.5

1

1.5

2

2.5

3

3.5

4

4.5

a=1.5

a=1.25

a=1.0

a=0.5

a=0.25

ω

f(ω

)

Figure 5.3: The relation between f(ω) and ω when a is approximated to 0.

is called the resonance curve of the system. For a given system with fixed m, k, and F1, there is a

resonance curve corresponding to each value of the damping coefficient a ≥ 0.

Notice that

(a) Resonance ←→ Max. amplitude.

(b) Resonance frequency

ω1

2π=

1

2π

√

k

m− a2

2m2. (5.5.3)


Frequency Formula and notation

Natural frequencyω0

2πdef=

1

2π

√

k

m

ω0 = λ =

√

k

m

Resonance frequencyω1

2πdef=

1

2π

√

k

m− a2

2m2

ω1 =√

λ2 − 2b2 =

√

k

m− a2

2m2

The corresponding free, damped frequencyωfd

2πdef=

1

2π

√

k

m− a2

4m2

ωfd =√

λ2 − b2 =

√

k

m− a2

4m2

The special case a = 0 will imply ω0 = ω1 = ωfd.

(c) Resonance can occur only if λ2 > 2b2. The underdamping case can occur only if λ2 > b2. Since

2b2 < λ2 ⇒ b2 < λ2, the damping must be less than critical in such a case.

(d) The resonance frequency is less than of the corresponding free, damped oscillation.

ω1

2π=

1

2π

√

λ2 − 2b2 <ωfd

2π=

1

2π

√

λ2 − b2

ω1

2π=

1

2π

√

k

m− a2

2m2<

ωfd

2π=

1

2π

√

k

m− a2

4m2

Now we will discuss forced motion with two case, undamped and damped cases.

Case 1. Undamped Forced Oscillation (a = 0)

md2x

dt2+ kx = F1 cosωt


(1) If ω 6= ω0, i.e. ω 6=√

k

m, then the UC set is cosωt, sinωt. From (5.4.5) we have

xp(t) =E1

(λ2 − ω2)2 + 0

[

(λ2 − ω2) cosωt+ 0]

(i.e. b = 0)

=E1

λ2 − ω2cosωt.

f(ω) =E1

λ2 − ω2=

F1/mkm− ω2

(amplitude of xp).

ω0

2π=

ω1

2π=

1

2π

√

k

m− 0 =

1

2π

√

k

mcycles/sec

= the ”natural frequency” of the undamped, free system.

xp(t) =F1/m

k/m− ω2cosωt

=F1

k· 1

1− (ω/ω0)2cosωt

ρ(ω) ≡ 1

1− (ω/ω0)2, ω → ω0 ⇒ ρ→∞

This phenomenon of excitation of large oscillations by matching input (ω) and natural frequency

(ω0) is known as resonance (ω = ω0).

Notice that

(a) The quantity ρ is called the resonance factor. See Fig. 5.4 for the special case ω0 = ω1 = 1.

(b)

ρ

k=

f(ω)

F1=

the amplitude of output (xp)

the amplitude of input

The ratio is the so-called amplification.

(2) If ω = ω0, i.e. ω =

√

k

m, then the UC set is t · cosωt, t · sinωt.

md2x

dt2+ kx = F1 cos

√

k

mt

d2x

dt2+

k

mx = E1 cos

√

k

mt

xc(t) = c1 sin

√

k

mt+ c2 cos

√

k

mt

xp(t) = At · sin√

k

mt +Bt · cos

√

k

mt


0 0.5 1 1.5 2 2.5 3−10

−8

−6

−4

−2

0

2

4

6

8

10

ω1

ω

ρ(ω

)

Figure 5.4: The relation between ρ and ω.

By the UC method we have A =E1

2

√

m

kand B = 0. The solution x(t) is that

x(t) = c1 sin

√

k

mt + c2 cos

√

k

mt +

E1

2

√

m

kt · sin

√

k

mt

= c · cos(√

k

mt+ φ

)

+E1

2

√

m

kt · sin

√

k

mt

Notice that

(a) The graph of the functionE1

2

√

m

kt · sin

√

k

mt is shown in Fig. 5.5 for the special case

m = k = E1 = 1.

(b) However, common sense intervenes and convinces us that before this exciting phenomenon

can occur the system will break down and the ODE will no longer apply!

(3) Another interesting and highly important type of oscillation is obtained when ω is closed to ω0.

ω.= ω0 but ω 6= ω0.


0 2 4 6 8 10 12 14

−6

−4

−2

0

2

4

6

t

1/2

⋅ t s

in(t

)

Figure 5.5: The particular integral xp(t) =12t sin(t).

x(t) = xc(t) + xp(t)

= c1 sinω0t + c2 cosω0t+F1/m

k

m− ω2

cosωt

= c1 sinω0t + c2 cosω0t+F1

m(ω20 − ω2)

cosωt

Initial conditions x(0) = 0 and x′(0) = 0 imply c1 = 0 and c2 = −F1

m(ω20 − ω2)

. Then we have

x(t) =F1

m(ω20 − ω2)

(cosωt− cosω0t)

=2F1

m(ω20 − ω2)

sinω0 + ω

2t · sin ω0 − ω

2t

Since ω is close to ω0, the difference ω0 − ω is small, so that the period of the last sine function is

large, and we obtain an oscillation of the type shown in Fig. 5.6 which is a solution of the following

example. It is so-called a beat.

Example (Undamped forced motion)


0 5 10 15 20 25 30 35 40

−1.5

−1

−0.5

0

0.5

1

1.5

← K sin((3−ω)t/2)

← −K sin((3−ω)t/2)

Figure 5.6: Forced undamped oscillation when the difference of the input and natural frequencies is

small (ω = 2.56).

2d2x

dt2+ 18x = 3cos(ωt), t > 0,

x(0) = 0, x′(0) = 0.

We consider two cases: (a) ω 6= 3 (b) ω = 3

(a) ω 6= 3

x(t) = xc(t) + xp(t) where

xp(t) =3

18− 2ω2cos(ωt)

xc(t) = c1 sin(3t) + c2 cos(3t)

c1 = 0, c2 =−3

18− 2ω2

x(t) = xp(t) + xc(t)

=−3

18− 2ω2cos(3t) +

3

18− 2ω2cos(ωt)

=6

18− 2ω2sin(

3 + ω

2t) sin(

3− ω

2t)

See Fig. 5.6 for the case ω = 2.65 (beat), Fig. 5.7 for ω = 5 and Fig. 5.8 for ω = 0.5.


0 5 10 15 20 25 30 35 40

−1

−0.5

0

0.5

1

ω = 5

Figure 5.7: Forced undamped oscillation, ω = 5.

0 5 10 15 20 25 30 35 40

−1

−0.5

0

0.5

1

ω = 0.5

Figure 5.8: Forced undamped oscillation, ω = 0.5.

(b) ω = 3

2d2x

dt2+ 18x = 3cos(ωt), t > 0,

x(0) = 0, x′(0) = 0.

ω = 3

We know that x(t) = xc(t) + xp(t) where

xp(t) =1

4t sin(3t)

xc(t) = c1 sin(3t) + c2 cos(3t)

c1 = 0, c2 = 0

Thus x(t) = 14t sin(3t). See Fig. 5.9 for the solution.

Case 2. Damped Forced Oscillation (a 6= 0)

md2x

dt2+ a

dx

dt+ kx = F1 cosωt

From (5.4.8) we have

x(t) = xc(t) + xp(t) = ce−bt cos(√

λ2 − b2 t+ φ) (5.5.4)

+E1

√

(λ2 − ω2)2 + 4b2ω2

cos(ωt− θ).

Notice that

(1) This is what happens in practice, because no physical system is completely undamped.

(2) The amplitude will always be finite.

(3) The amplitude f(ω) has a maximum at ω1 =

√

k

m− a2

2m2depending on a. This may be called


0 5 10 15 20 25 30 35 40−10

−8

−6

−4

−2

0

2

4

6

8

10

x(t)= t/4

x(t)= − t/4

Figure 5.9: ω = ω1

practical resonance.

(4) It is of great importance because it shows that some input may excite oscillations with such a

large amplitude that the system can be destroyed.

(5) Machines, cars, airplanes and bridges are vibrating mechanical systems, and it is sometimes

rather difficult to find constructions which are completely free of undesired resonance effects.

(6) Another expression of the amplitude of xp(t) (amplification)

From (5.5.1) and (5.5.2) we have

f(ω) =E1

√

(λ2 − ω2)2 + 4b2ω2

(5.5.5)

w20 = λ2 = k/m, ω1 =

√

k

m− a2

2m2, a2 = 2m2(ω2

0 − ω21) and

f(ω1) =F1/m

√

(ω20 − ω2

1)2 +

( a

m

)2

ω21

=F1/m

√

(

k

m− ω2

1

)2

+( a

m

)2

ω21

=F1/m

√

(ω20 − ω2

1)2 + 2(ω2

0 − ω21)ω

21

=F1/m

√

a4

4m4+

a2

m2ω21

=2mF1

a√

a2 + 4m2ω21

=2mF1

a√

2m2ω20 + 2m2ω2

1


Fig. 5.2 shows the amplification f(ω)/F1 as a function of ω for m = k = F1 = 1.

(7) Phase Angle ( or Phase Lag)

From (5.4.8) we have

xp(t) =E1

√

(λ2 − ω2)2 + 4b2ω2

cos(ωt− θ)

= f(ω) · cos(ωt− θ) where

cos θ =λ2 − ω2

√

(λ2 − ω2)2 + 4b2ω2

sin θ =2bω

√

(λ2 − ω2)2 + 4b2ω2

The angle θ is called the phase angle or phase lag because it measures the lag of the output with

respect to the input. Fig. 5.10 shows the phase lag as a function of ω for the case m = k = 1. If

ω < ω0, then θ < π/2; if ω = ω0, then θ = π/2, and if ω > ω0, then θ > π/2.

tan θ =2bω

λ2 − ω2=

a/m · ωω20 − ω2

.

If m = k = 1 then ω0 = 1 and

θ = tan−1 aω

1− ω2.

Example

An 64-lb weight is placed upon the lower end of a coil spring suspended from the ceiling. The spring

constant is 18 lb/ft. The weight is then pulled down 6 in. below its equilibrium position and released

at t = 0 without any initial velocity. The external force is F (t) = 3 cosωt.

(1) Assuming the damping force is 4(dx/dt) please find the resonance frequency.

(2) Assuming there is no damping force please find the value of ω that give rise to undamped

resonance.

m = w/g = 64/32 = 2 (slug). k = 18 and F (t) = 3 cosωt.

2d2x

dt2+ a

dx

dt+ 18x = 3 cosωt

x(0) = 1/2, x′(0) = 0.

(1) a = 4

2d2x

dt2+ 4

dx

dt+ 18x = 3 cosωt


0 0.5 1 1.5 2 2.5 30

pi/2

pi

a=1.5

a=1.25

a=1.0

a=0.5

a=0.25

a=0

ω

θ (ω

)

Figure 5.10: The phase lag function of ω.

The resonance frequency is

1

2π

√

k

m− a2

2m2=

1

2π

√

18

2− 42

2 · 22

=1

2π

√7

.= 0.42 (cycles/sec)

Thus resonance occurs when ω =√7

.= 2.65.

(2) a = 0

2d2x

dt2+ 18x = 3 cosωt

Undamped resonance occurs when the frequency ω/2π of the impressed force is equal to the natural

frequency.

xc(t) = c1 sin 3t+ c2 cos 3t.

The natural frequency is 3/2π. Thus ω = 3 gives rise to undamped resonance. The ODE in this case

becomesd2x

dt2+ 9x = 3/2 cos 3t.

The initial conditions, x(0) = 1/2, x′(0) = 0, imply the solution

x(t) =1

3cos 3t+

1

4· t · sin 3t.


Example (The Details)

2d2x

dt2+ a

dx

dt+ 18x = 3 cosωt

(1) a = 4, x(0) = 0, x′(0) = 0, (2) a = 0, x(0) = 1/2, x′(0) = 0.

(1) a = 4

2m2 + 4m+ 18 = 0 m = (−2 ±√4− 4 · 9)/2 = −1±

√8

xc(t) = e−t(c1 sin√8t+ c2 cos

√8t).

λ2 = k/m = 18/2 = 9, 2b = a/2 = 2, b = 1 < λ = 3.

xc(t) = e−bt(c1 sin√

λ2 − b2t + c2 cos√

λ2 − b2t)

= e−t(c1 sin√8t + c2 cos

√8t)

x′c(t) = −e−t(c1 sin

√8t+ c2 cos

√8t) + e−t(

√8c1 cos

√8t−

√8c2 sin

√8t)

For the particular integral we have

xp(t) =E1

(λ2 − ω2)2 + 4b2ω2

[


=3/2

(9− ω2)2 + 4ω2

[

(9− ω2) cosωt+ 2ω sinωt]

x′p(t) =

3/2

(9− ω2)2 + 4ω2

[

2ω2 cosωt− (9− ω2)ω sinωt]

Solve the initial conditions and we obtain the solution

x(0) = 0 ⇒ c2 +3/2

(9− ω2)2 + 4ω2(9− ω2) = 0, c2 = −

3/2

(9− ω2)2 + 4ω2(9− ω2)

x′(0) = 0 ⇒ (−c2 +√8c1) +

3/2

(9− ω2)2 + 4ω2· 2ω2 = 0

c1 = − 3/2√8 [(9− ω2)2 + 4ω2]

(9 + ω2)


= e−t

[

− 3/2 · (9 + ω2)√8 [(9− ω2)2 + 4ω2]

sin√8t− 3/2 · (9− ω2)

(9− ω2)2 + 4ω2cos√8t

]

+3/2

(9− ω2)2 + 4ω2

[

(9− ω2) cosωt+ 2ω sinωt]

(2) case 1. a = 0, ω 6= ω0 = λ =√

k/m = 3

2m2 + 0 ·m+ 18 = 0 m = ±3i

2d2x

dt2+ 18x = 3 cosωt

xc(t) = c1 sin 3t+ c2 cos 3t


λ2 = k/m = 18/2 = 9, 2b = a/2 = 0, b = 0 < λ = 3.

xc(t) = e−bt(c1 sin√

λ2 − b2t+ c2 cos√

λ2 − b2t)

= c1 sin 3t+ c2 cos 3t

x′c(t) = −3 · c1 sin 3t+ 3 · c2 cos 3t

For the particular integral we have

xp(t) =E1

(λ2 − ω2)2 + 4b2ω2

[


=E1

λ2 − ω2cosωt =

3/2

9− ω2· cosωt

x′p(t) = − 3/2

9 − ω2· ω sinωt

(a) Initial conditions: x(0) = 1/2, x′(0) = 0

x(0) = 1/2 ⇒ c2 +3/2

9− ω2= 1/2, c2 = 1/2− 3/2

9− ω2

x′(0) = 0 ⇒ 3c1 + 0 = 0, c1 = 0


=

(

1

2− 3/2

9− ω2

)

cos 3t +3/2

9− ω2cosωt

(b) Initial conditions: x(0) = 0, x′(0) = 0

x(0) = 0 ⇒ c2 +3/2

9− ω2= 0, c2 = −

3/2

9− ω2

x′(0) = 0 ⇒ 3c1 + 0 + 0 = 0, c1 = 0


= − 3/2

9− ω2cos 3t+

3/2

9− ω2cosωt

case 2. a = 0, ω = ω0 = λ =√

k/m = 3

2m2 + 0 ·m+ 18 = 0 m = ±3i

2d2x

dt2+ 18x = 3 cosωt

xc(t) = c1 sin 3t+ c2 cos 3t


UC set = t · sin 3t, t · cos 3t. For the particular integral we have

xp(t) =E1

(λ2 − ω2)2 + 4b2ω2

[


E1

λ2 − ω2cosωt

E1

2

√

m

k· t · sin

√

k

mt

=3/2

2

√

2

18· t · sin

√

18

2t =

1

4t sin 3t

x′p(t) =

1

4sin 3t+

3

4t cos 3t


= c1 sin 3t+ c2 cos 3t+1

4t sin 3t

(a) Initial conditions: x(0) = 1/2, x′(0) = 0

x(0) = 1/2 ⇒ c2 + 0 = 1/2, c2 = 1/2

x′(0) = 0 ⇒ 3c1 + 0 = 0, c1 = 0

x(t) =1

2cos 3t +

1

4t sin 3t

(b) Initial conditions: x(0) = 0, x′(0) = 0

x(0) = 0 ⇒ c2 + 0 = 0, c2 = 0

x′(0) = 0 ⇒ 3c1 + 0 + 0 = 0, c1 = 0

x(t) =1

4t sin 3t

5.6 Electric Circuit Problems

1. Unifying power of mathematics

The mass-spring system has a strictly quantitative analogy in electrical engineering, namely, the

LRC-circuit. This will be an impressive demonstration of the unifying power of mathematics, illus-

trating that entirely different physical systems may have the same mathematical model and can be

treated and solved by the same method.

2. Electric circuit

The simplest electric circuit is a series circuit in which we have a source of electric energy

(electromotive force, EMF) such as a generator or a battery, and a resistor, which uses energy, for

5.6 Electric Circuit Problems 131

example, an electric light bulb.

SwitchE R

Fig. 5.10. A electrical network

If we close the switch, a current will flow through the resistor, and this will cause a voltage drop,

that is, electric potentials at the two ends of the resistor will be different.

3. Three important laws

The following three laws concerning the voltage drops across these various elements (resistor, induc-

tor, capacitor) are known to hold:

(1)

ER = R · i (5.6.1)

where R is a constant of proportionality called the resistance and i is the current.

(2)

EL = L · didt

(5.6.2)

where L is a constant of proportionality called the inductance and i is the current.

(3)

EC =1

C· q (5.6.3)

where C is a constant of proportionality called the capacitance and q is the instantaneous charge

on the capacitor. For the relations between q, i and EC we have

i =dq

dt

q =

∫

i dt

EC =1

C

∫

i dt

4. Conventional symbols

E: Electromotive force (battery or generator)

R: Resistor

L: Inductor

C: Capacitor

The units in common use are listed in Table 5.6.1.

Table 5.6.1 The units used in circuits


Quantity and symbol Unit

EMF or voltage E volt (V)

current i ampere (A)

charge q coulomb

resistance R ohm (Ω)

inductance L henry (H)

capacitance C farad (F)

5. Kirchhoff’s Laws

(1) Kirchhoff’s Voltage Law (KVL, Form 1 )

The algebraic sum of the instantaneous voltage drops around a closed circuit in a specific

direction is zero.

(2) Kirchhoff’s Voltage Law (KVL, Form 2 )

The algebraic sum of the voltage drops across resistors, inductors, and capacitors is equal to

the total EMF in a closed circuit.

(3) Kirchhoff’s Current Law (KCL)

At any point of a circuit, the sum of the inflowing currents is equal to the sum of the outflowing

currents.

6. Example (RL-circuit)

E

R

L

RL-circuit: Ldi

dt+R · i = E(t)

Case A. Constant EMF (direct current, DC)


E(t) = E0 = Const. EL + ER = E(t)di

dt+

R

Li =

E0

Lµ(t) = exp

[∫

R

Ldt

]

= exp

(

R

Lt

)

µ · didt

+ µ · RLi = µ · E0

Ld(µ · i) = µ · E0

L

µ · i = E0

L

∫

µ dt+ c

i(t) =1

µ

[

E0

L

∫

µ dt+ c

]

= e−αt

[

E0

L

∫

eαt dt+ c

]

(α := R/L)

=E0

L· 1α+ e−αt · c =

E0

R+ c · e−Rt/L

The particular solution corresponding to the initial condition i(0) = 0 is

i(t) =E0

R(1− e−Rt/L)

=E0

R(1− e−t/τL)

where τL := L/R is called the inductive time constant of the circuit. See Fig. 5.11.

E0 / R

Time t

Cur

rent

i(t)

Figure 5.11: Current in an RL-circuit due to a constant EMF .

Case B. Periodic EMF (alternating current, AC)

Method I. Linear equation and integrating factor


We consider the integrating factor eαt where αdef= R/L.

E(t) = E0 sinωt

i(t) = e−αt

[

E0

L

∫

eαt sinωt dt+ c

]

(α = R/L)

= ce−Rt/L +E0

R2 + ω2L2(R sinωt− ωL cosωt)

= ce−Rt/L +E0√

R2 + ω2L2sin(ωt− δ)

where δ = tan−1(ωL/R).

Check

e−αt

[

E0

L

∫

eαt sinωt dt+ c

]

(α = R/L)

= ce−Rt/L +E0


Integration by parts. u = sinωt, dv = eαt dt, du = ω cosωt dt∫

eαt sinωt dt =1

αsinωt · eαt − ω

α

∫

eαt cosωt dt (u = cosωt, dv = eαt dt)

=1


α

[

1

αcosωt · eαt + ω

α

∫

eαt sinωt dt

]

Therefore(

1 +ω2

α2

)∫

eαt sinωt dt =1


α2cosωt · eαt

∫

eαt sinωt dt =α2

α2 + ω2

[

1


α2cosωt · eαt

]

=1

α2 + ω2

[

α sinωt · eαt − ω cosωt · eαt]

Thus, we have

e−αt

[

E0

L

∫

eαt sinωt dt+ c

]

= ce−αt + e−αt · E0

L· 1

α2 + ω2

[

α sinωt · eαt − ω cosωt · eαt]

= ce−Rt/L +E0

L· 1

(R/L)2 + ω2

[

R

Lsinωt− ω cosωt

]

= ce−Rt/L +E0


Method II. UC method


Let ip(t) = A cosωt+B sinωt. Then i′p(t) = Bω cosωt−Aω sinωt. Substituting them to the original

equation we have

LBω + RA = 0

−LAω + RB = E0

⇒

(R) ·A + (Lω) ·B = 0

(−Lω) · A + (R) · B = E0

By Cramer’s rule we have

=

∣

∣

∣

∣

∣

R Lω

−Lω R

∣

∣

∣

∣

∣

= R2 + ω2L2 A =

∣

∣

∣

∣

∣

0 Lω

E0 R

∣

∣

∣

∣

∣

B =

∣

∣

∣

∣

∣

R 0

−Lω E0

∣

∣

∣

∣

∣

Thus A =−ωLE0

R2 + ω2L2, B =

E0R

R2 + ω2L2and

ip(t) =E0


For the complementary function we know the characteristic equation is Lλ+R = 0. So λ = −R/L.

ic(t) = c · e−RL

Thus we have the solution

i(t) = ic(t) + ip(t) = ce−Rt/L +E0

R2 + ω2L2(R sinωt− ωL cosωt).

Interpretation. The exponential term will approach zero as t approaches to infinity. This means

that after sufficiently long time the current i(t) executes practically harmonic oscillations. The

oscillations of i(t) are in phase with those of EMF E(t) with a phase angle lag δ. See Fig. 5.12. The

phase angle δ as a function of ωL/R. If L = 0 then δ = 0. The oscillations of i(t) are in phase with

those of EMF E(t). See Fig. 5.13. Note that

(1) An electrical (or dynamical) system is said to be in the steady state when the variables describing

its behavior are periodic functions of time or constant, and it is said to be in the transient state

(or unsteady state) when it is not in the steady state. The corresponding variables are called

steady-state functions and transient functions, respectively.

(2) For the case A, E0/R is a steady-state function (or a steady-state solution). For the case B,

E0

R2 + ω2L2sin(ωt− δ)

is a steady-state function. Before the circuit reaches the steady state it is in the transient state.

(3) Such an interim or transient period occurs because inductors and capacitors store energy, and the

corresponding inductor currents and capacitor voltages can not be changed instantly. Practically,

this transient state will last only a short time.

7. Example

A circuit has in series an electromotive force given by E = 100 sin 40t V, a resistor of 10 Ω and an


0 \pi 2 \pi 3 \pi 4 \p 5 \pi−1

−0.5

0

0.5

1

1.5

2

Current i(t)

Exponential term

Time t

Cur

rent

i(t)

Figure 5.12: Current in an RL-circuit due to a periodic EMF (δ = π/4).

inductor of 0.5 H. If the initial current is 0, find the current at time t > 0.

0.5di

dt+ 10i = 100 sin 40t

di

dt+ 20i = 200 sin 40t

µ(t) = exp[

∫

20 dt] = e20t

i(t) = 2(sin 40t− 2 cos 40t) + c · e−20t

i(0) = 0 ⇒ c = 4

i(t) = 2(sin 40t− 2 cos 40t) + 4 · e−20t

Expressing the trigonometric terms in a ”phase-angle” form, we have

i(t) = 2√5

(

1√5sin 40t− 2√

5cos 40t

)

+ 4 · e−20t

i(t) = 2√5 sin(40t+ φ) + 4 · e−20t,

where φ is determined by

φ = cos−1(1/√5) = sin−1(−2/

√5)

.= −1.11 rad

i(t) = 2√5 sin(40t− 1.11) + 4 · e−20t.


0 1 2 3 4 5 6 7 8 9 100

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

π /2

ω L/ R

δ

δ = arctan( ω L/ R )

Figure 5.13: The phase angle function δ = tan−1(ωL/R).

Note that

(1) the phase angle φ.= −1.11 rad.

(2) the period π/20 is the same as that of E(t).

(3) the E(t) leads the steady-state current by approximately 1.11/40.

(sin(40t− 1.11) = sin[40(t− 1.11/40)])

8. LRC-circuit

We now consider the circuit shown in Fig. 5.14.

E

R

L

C Fig. 5.14 LRC-circuit

The ordinary differential equation is

Ldi

dt+Ri+

1

Cq = E (5.6.4)

Ld2q

dt2+R

dq

dt+

1

Cq = E. (i =

dq

dt) (5.6.5)


On the other hand, if we differentiate Eqn. (5.6.4) w.r.t. t, then we have

Ld2i

dt2+R

di

dt+

1

C

dq

dt=

dE

dt(5.6.6)

Ld2i

dt2+R

di

dt+

1

Ci =

dE

dt. (i =

dq

dt) (5.6.7)

Note that

(1) RL-circuit: if the circuit contains no capacitor, Eqn. (5.6.4) reduces to

Ldi

dt+Ri = E. (5.6.8)

(2) RC-circuit: if the circuit contains no inductor, Eqn(5.6.4) reduces to

Rdq

dt+

1

Cq = E. (5.6.9)

9. Unifying power of mathematics

The important comparison of the mechanical system and the electric system

An interesting and useful analogy for these two systems

Ld2q

dt2+R

dq

dt+

1

Cq = E(t) (5.6.10)

md2x

dt2+ a

dx

dt+ kx = F (t). (5.6.11)

See Table 5.6.2.

Table 5.6.2 The comparison of the mechanical system and the electric system

Mechanical system Electrical system

mass m inductance L

damping constant a resistance R

spring constant k reciprocal of capacitance 1/C

impressed force F (t) impressed voltage or EMF E(t)

displacement x charge q

velocitydx

dt= v current

dq

dt= i

10. Example

A circuit has in series an electromotive force given by E = 100 sin(60t) V, a resistor of 2Ω, an

inductor of 0.1 H, and a capacitor of 1/260 farads. (See Fig. 5.14) If the initial current and the

initial charge on the capacitor are both zero, find the charge on the capacitor at any time t > 0.


Answer. (1) Formulation L = 0.1, R = 2, C = 1/260, E = 100 sin(60t)

0.1d2q

dt2+ 2

dq

dt+ 260q = 100 sin(60t)

d2q

dt2+ 20

dq

dt+ 2600q = 1000 sin(60t)

q(0) = 0 , q′(0) = 0.

(2) Solution The auxiliary equation is

r2 + 20r + 2600 = 0, r = −10 ± 50i

The complementary function is

qc = e−10t(c1 sin 50t+ c2 cos 50t).

Employing the method of undetermined coefficients to find a particular integral, we write

qp = A sin 60t+B cos 60t.

Differentiating twice and substituting into the ODE we find that

A = −2561

and B = −3061


q(t) = qc(t) + qp(t)

= e−10t(c1 sin 50t+ c2 cos 50t) +

(

−2561

)

sin 60t+

(

−3061

)

cos 60t.

q(0) = 0 and q′(0) = 0 imply c1 =36

61and c2 =

30

61,

q(t) = e−10t

(

36

61sin 50t+

30

61cos 50t

)

+

(

−2561

)

sin 60t+

(

−3061

)

cos 60t

=6√61

61e−10t cos(50t− φ)− 5

√61

61cos(60t− θ)

.= 0.77e−10t cos(50t− 0.88)− 0.64 cos(60t− 0.69).

(3) Interpretation

The first term is the transient term. The second term is the steady-state term.

11. Reactance and Impedance

Consider the solution of the ordinary differential equation

Ldi

dt+Ri+

1

Cq = E. (5.6.12)


For a sinusoidal EMF E(t) = E0 sinωt we have dE/dt = E0ω cosωt.

ip(t) = a cosωt+ b sinωt. (5.6.13)

Substituting this into (5.6.12) we obtain

a =−E0S

R2 + S2, b =

E0R

R2 + S2(5.6.14)

where S is the so-called reactance given by

S = ωL− 1

ωC. (5.6.15)

check!

In any practical case, R 6= 0 so that the denominators of a and b are not zeros. Moreover, ip(t) =

i0 sin(ωt− θ) where

i0 =√a2 + b2 =

E0√R2 + S2

, tan θ = −ab=

S

R. (5.6.16)

The quantity√R2 + S2 is called the impedance which is somewhat analogous to the resistance

R = E/i (Ohm’s law) because√R2 + S2 =

E0

i0. (5.6.17)

The complementary function is

ic = c1eλ1t + c2e

λ2t

where λ1 and λ2 are the roos of the characteristic equation λ2 +R/Lλ+1/LC = 0. So λ1 = −α+ β

and λ2 = −α− β where

α =R

2L, β =

1

2L

√

R2 − 4L

C. (5.6.18)

If R > 0, ic(t)→ 0 as t→∞.

Hence, the transient current i = ic+ ip tends to steady-state current ip, and after some tie the output

will practically be a harmonic oscillation, which is given by ip and whose frequency is that of the

input.

12. Example

Find the current i(t) in an LRC-circuit with R = 100Ω, L = 0.1H , C = 10−3 Farads, E(t) =

155 sin 377t (hence 377/2π = 60Hz = 60 cycles/sec) and zero charge and current when time t = 0.

Find the reactance and the general solution.

Answer.

0.1d2i

dt2+ 100

di

dt+ 1000i = 155 · 377 · cos 377t.


The reactance S = 37.7− 1/0.377 = 35.

The steady-state current ip(t) = a cos 377t+ b sin 377t.

a =−155 · 351002 + 352

= −0.483, b =−155 · 1001002 + 352

= 1.381.

We have ip(t) = −0.483 cos 377t+ 1.381 sin 377t = 1.463 sin(377t− 0.34).

The characteristic equation is 0.1λ2 + 100λ+ 1000 = 0. Thus, λ1 = −10 and λ2 = −990.The general solution is

i(t) = c1e−10t + c2e

−990t + 1.463 sin(377t− 0.34).

The initial conditions q(0) = 0 and i(0) = 0 imply c1 = −0.043 and c2 = 0.526. Thus,

i(t) = −0.043e−10t + 0.526e−990t + 1.463 sin(377t− 0.34).

13. A Series Circuit and A Parallel Circuit

For a series LRC-circuit we know that the differential equation is

Ld2i

dt2+R

di

dt+

1

C

dq

dt=

dE

dt, (5.6.6)

Ld2i

dt2+R

di

dt+

1

Ci =

dE

dt. (i =

dq

dt) (5.6.7)

E

R

L

C Fig. 5.14 A series LRC-circuit

For a parallel LRC-circuit we know that applying Kirchhoff’s current law to the network in Figure

5.15 leads us to the equation

IS(t) =1

RV + C

dV

dt+

1

L

∫

V (s)ds. (5.6.19)

Upon differentiating and rearranging terms, the equation becomes

d2V

dt2+

1

RC

dV

dt+

1

LCV =

1

C

dIS(t)

dt. (5.6.20)

IS(t)Figure 5.15. A parallel

LRC-electrical network .


14. Example

Use the following consistent set of scaled units (referred to as the Scaled SI Unit System).

Table 5.6.3 The scaled units

Quantity Unit Symbol

Voltage volt V

Current milliampere (mA) I

Time millisecond (ms) t

Resistance kilohm (kΩ) R

Inductance henry (H) L

Capacitance microfarad (µF) C

Use the consistent set of scaled units as above. Consider the parallel LRC network shown in Fig.

5.16. Assume that at time t = 0, the voltage V (t) and its time rate of change are both zero. For the

given source voltage IS(t) = 1− e−t mA, determine the voltage V (t).

IS(t)C = 0.5µFR= 1 kΩL= 1 H

Figure 5.16. The parallel

LRC-electrical network .

IS(t) =1

RV + C

dV

dt+

1

L

∫

V (s)ds.

Upon differentiating and rearranging terms, the equation becomes

d2V

dt2+

1

RC

dV

dt+

1

LCV =

1

C

dIS(t)

dt,

V ′′ + 2V ′ + 2V = 2e−t.

Vc(t): λ2 + 2λ+ 2 = 0, λ = −1± i.

Vc(t) = c1e−t cos t + c2e

−t sin t.

Vp(t): Let Vp(t) = Ae−t. A− 2A+ 2A = 2, A = 2, Vp(t) = 2e−t.

V (t) = Vc(t) + Vp(t) = c1e−t cos t + c2e

−t sin t+ 2e−t.


Applying initial conditions we obtain

V (0) = 0 = c1 + 2

V ′(0) = 0 = −c1 + c2 − 2.

Thus c1 = −2, c2 = 0. The solution V (t) is then

V (t) = −2e−t cos t+ 2e−t,

= 2e−t(1− cos t). (Voltage)

Chapter 6

Systems of Linear Differential Equations

6.1 Basic Theory of Linear Systems

A. Introduction

1. Normal Form

The normal form in the case of two first-order ODEs in two unknown functions x and y is

dx

dt= a11(t)x+ a12(t)y + F1(t), (6.1.1)

dy

dt= a21(t)x+ a22(t)y + F2(t).

If F1(t) and F2(t) are zero for all t, then the system is called homogeneous; otherwise, the

system is said to be nonhomogeneous.

2. Example 6.1.1 (a)

dx

dt= 2x− y, (6.1.2)

dy

dt= 3x+ 6y,

is homogeneous.

Example 6.1.1 (b)

dx

dt= 2x− y − 5t, (6.1.3)

dy

dt= 3x+ 6y − 4,

is nonhomogeneous.

146 CHAPTER 6. SYSTEMS OF LINEAR DIFFERENTIAL EQUATIONS

3. Definition 6.1.1

By a solution of system (6.1.1) we shall mean an ordered pair (f, g) such that

df(t)

dt= a11(t)f(t) + a12(t)g(t) + F1(t),

dg(t)

dt= a21(t)f(t) + a22(t)g(t) + F2(t).

for all t such that a ≤ t ≤ b.

4. Example 6.1.2

The ordered pair (e5t,−3e5t) is a solution of the system (6.1.2). Note that (e3t,−e3t) is also a

solution of the system (6.1.2).

5. Theorem 6.1.1

Hypothesis. Let a11, a12, F1, a11, a11, F2 be continuous on a ≤ t ≤ b, t0 ∈ [a, b] and c1 and

c2 be two constants.

Conclusion. There exists a unique solution (x = f(t), y = g(t)) of system (6.1.1) such that

f(t0) = c1 and g(t0) = c2 and this solution is defined on the entire interval a ≤ t ≤ b.

B. Homogeneous Linear System

1. Consider the following homogeneous linear system

dx

dt= a11(t)x+ a12(t)y, (6.1.4)

dy

dt= a21(t)x+ a22(t)y.

Theorem 6.1.2

Hypothesis. Let (x = f1(t), y = g1(t)) and (x = f2(t), y = g2(t)) be two solutions, and c1 and

c2 be two constants.

Conclusion. Then

x = c1f1(t) + c2f2(t), (6.1.5)

y = c1g1(t) + c2g2(t),

is also a solution of the system (6.1.4).

2. Definition 6.1.2 (Linear Combination)

The solution (6.1.5) is called a linear combination of the solutions (f1(t), g1(t)) and (f2(t), g2(t)).

3. Theorem 6.1.3 (Restated)

Any linear combination of two solutions of the homogeneous linear system (6.1.4) is itself a

solution of the system (6.1.4).

6.1 Basic Theory of Linear Systems 147

4. Example 6.1.3

(e5t,−3e5t) and (e3t,−e3t) are two solutions of the system (6.1.2). Then

x = c1e5t + c2e

3t,

y = −3c1e5t − c2e3t,


5. Definition 6.1.3 (Linear Dependence)

Let (f1(t), g1(t)) and (f2(t), g2(t)) be two solution of the system (6.1.4). These two solutions

are linearly dependent on t ∈ [a, b] if ∃ c1, c2 not both zero, such that

c1f1(t) + c2f2(t) = 0,

c1g1(t) + c2g2(t) = 0,

for all a ≤ t ≤ b.

6. Definition 6.1.4 (Linear Independence)

Let (f1(t), g1(t)) and (f2(t), g2(t)) be two solutions of the system (6.1.4). These two solution

are linearly independent on t ∈ [a, b] if

c1f1(t) + c2f2(t) = 0,

c1g1(t) + c2g2(t) = 0,

implies c1 = c2 = 0.

7. Example 6.1.4 (a)

(e5t,−3e5t) and (2e5t,−6e3t) are linearly dependent.

Example 6.1.4 (b)

(e5t,−3e5t) and (e3t,−e3t) are linearly independent.

8. Theorem 6.1.4

There exists sets of two linearly independent solutions of the homogeneous linear system (6.1.4).

Every solution of the system (6.1.4) can be written as a linear combination of any two linearly

independent solution if (6.1.4).

9. Definition 6.1.5 (General Solution)

Let (f1(t), g1(t)) and (f2(t), g2(t)) be two linearly independent solutions of the system (6.1.4).

Then

x = c1f1(t) + c2f2(t),

y = c1g1(t) + c2g2(t),

is called the general solution of the system (6.1.4).


10. Theorem 6.1.5

Let (f1(t), g1(t)) and (f2(t), g2(t)) be two solutions of the system (6.1.4). Then a necessary and

sufficient condition that two solutions be linear independent on t ∈ [a, b] is that the determinant

(t) =

∣

∣

∣

∣

∣

f1(t) f2(t)

g1(t) g2(t)

∣

∣

∣

∣

∣

6= 0,

for all t ∈ [a, b].

11. Theorem 6.1.6

The determinant (t) of Theorem 6.1.5 either is identically zero or vanishes for no value of t

on the interval a ≤ t ≤ b.

12. Example 6.1.5 (a)

Consider the two solutions, (e5t,−3e5t) and (2e5t,−6e3t). (t) = 0 for all t ∈ [a, b].

Example 6.1.5 (b)

Consider the two solutions,(e5t,−3e5t) and (e3t,−e3t).

(t) =

∣

∣

∣

∣

∣

e5t e3t

−3e5t −e3t

∣

∣

∣

∣

∣

= 2e8t 6= 0,

for all t ∈ [a, b].

C. Nonhomogeneous Linear System

1. Consider the following nonhomogeneous linear system

dx

dt= a11(t)x+ a12(t)y + F1(t), (6.1.1)

dy

dt= a21(t)x+ a22(t)y + F2(t).

Theorem 6.1.7

Hypothesis. Let (x = f0(t), y = g0(t)) be any solution of the system (6.1.1) , and (x =

f(t), y = g(t)) be any solution of the corresponding homogeneous system (6.1.4).

Conclusion. Then

x = f1(t) + f0(t),

y = g1(t) + g0(t),


6.1 Basic Theory of Linear Systems 149

2. Definition 6.1.6 (General Solution)

Let (x = f0(t), y = g0(t)) be any solution of the nonhomogeneous system (6.1.1) , and

(x = f1(t), y = g1(t)) and (x = f2(t), y = g2(t)) be two linearly independent solutions of

the corresponding homogeneous system (6.1.4). Then

x = c1f1(t) + c2f2(t) + f0(t), (6.1.6)

y = c1g1(t) + c2g2(t) + g0(t),

is called the general solution of the system (6.1.1).

3. Example 6.1.6

dx

dt= 2x− y − 5t, (6.1.3)

dy

dt= 3x+ 6y − 4,

Sol. (2t+ 1,−t) is a solution of the nonhomogeneous system.

(e5t,−3e5t) and (e3t,−e3t) are two linearly independent solutions of the corresponding homo-

geneous system. Then the general solution is

x = c1e5t + c2e

3t + 2t+ 1

y = −3c1e5t − c2e3t − t

D. A Solution Method for Homogeneous Linear Systems with Constant Coefficients

1. Consider the following nonhomogeneous linear system

dx

dt= a1x+ b1y, (6.1.7)

dy

dt= a2x+ b2y.

Let us attempt to determine a solution of the form

x = Aeλt,

y = Beλt.


Then we have

dx

dt= Aλeλt = a1Ae

λt + b1Beλt,

dy

dt= Bλeλt = a2Ae

λt + b2Beλt.

We need to solve

(a1 − λ)A + b1B = 0,

a2A+ (b2 − λ)B = 0.

This system has a nontrivial solution if

(t) =

∣

∣

∣

∣

∣

a1 − λ b1

a2 b2 − λ

∣

∣

∣

∣

∣

= 0.

Then

pM(λ) = λ2 − (a1 + b2)λ+ (a1b2 − a2b1) = 0. (6.1.8)

This equation is called the characteristic equation associated with the system (6.1.4). In matrix

form,

d−xdt

def=

(

dxdtdydt

)

=

(

a1 b1

a2 b2

)(

x

y

)

def= M−x (6.1.4)

We want to determine a solution of the form

−x =

(

A

B

)

eλt.

Then we have

d−xdt

=

(

A

B

)

λ · eλt =(

a1 b1

a2 b2

)(

x

y

)

= M−x = M

(

A

B

)

eλt.

That is,

M

(

A

B

)

= λ ·(

A

B

)

.

We must find the eigenvalues, λ, and the corresponding eigenvectors, (A,B), of the matrix M

by the characteristic equation

pM(λ) = λ2 − (a1 + b2)λ+ (a1b2 − a2b1) = 0. (6.1.8)

2. Example 6.1.7

dx

dt= 2x− y, (6.1.2)

dy

dt= 3x+ 6y,

6.2 Matrices and Vectors 151

pM(λ) = λ2 − (2 + 6)λ+ (2 · 6− (−1) · 3) = 0, λ = 3, 5.

λ = 3:

A− 3I =

(

2− 3 −13 6− 3

)

=

(

−1 −13 3

)

.

The eigenvector is (1,−1).λ = 5:

A− 5I =

(

2− 5 −13 6− 5

)

=

(

−3 −13 1

)

.

The eigenvector is (1,−3).The general solution is

−x = c1

(

1

−1

)

e3t + c2

(

1

−3

)

e5t.

6.2 Matrices and Vectors

A. Matrix Multiplication and Inversion

1. Definition 6.2.1 (Inverse)

Let A be an n× n nonsingular matrix. The unique n× nmatrix B satisfies

AB = BA = In,

is called the inverse of A. We use the notation:

AA−1 = A−1A = In.

2. Definition 6.2.2 (Minor, Cofactor and the Matrix of Cofactor)

Let A ∈Mn(F) (i.e., an n× n matrix). A = (aij)n×n

(1) The minor of aij is defined to be the determinant of the (n− 1)× (n− 1) matrix obtained

from A by deleting the ith row and jth column of A. We denote it by mij .

(2) The cofactor of aij is defined to be the minor Mij of aij multiplied by the number (−1)i+j .

We denote it by cij .

cij = (−1)i+jmij . (6.2.1)

(3) The matrix of cofactor of the elements of A is defined to be the matrix obtained from A by

replacing each element aij of A by its cofactor cij. We denote it by cof A.


3. Example 6.2.1

A =

2 1 −14 3 −2−6 2 5

. Find cof A.

m23 =

∣

∣

∣

∣

∣

2 1

−6 2

∣

∣

∣

∣

∣

= 10. c23 = (−1)2+3m23 = −10.

Thus, the matrix of cofactor is

cof A =

∣

∣

∣

∣

∣

3 −22 5

∣

∣

∣

∣

∣

−∣

∣

∣

∣

∣

4 −2−6 5

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

4 3

−6 2

∣

∣

∣

∣

∣

−∣

∣

∣

∣

∣

1 −12 5

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

2 −1−6 5

∣

∣

∣

∣

∣

−∣

∣

∣

∣

∣

2 1

−6 2

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

1 −13 −2

∣

∣

∣

∣

∣

−∣

∣

∣

∣

∣

2 −14 −2

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

2 1

4 3

∣

∣

∣

∣

∣

=

19 −8 26

−7 4 −101 0 2

.

4. Definition 6.2.3 (Classical Adjoint)

Let A ∈ Mn(F). The classical adjoint of A is defined to be the transport of cof A. We denote

it by adj A (or A@).

5. Example 6.2.2

A =

2 1 −14 3 −2−6 2 5

. Find adj A.

adj A = (cof A)T =

19 −8 26

−7 4 −101 0 2

T

=

19 −7 1

−8 4 0

26 −10 2

.

6. Result A

Let A be an n× n nonsingular matrix. Then

A−1 =1

det(A)( adj A ),

where det(A) is the determinant of A and adj A is the classical adjoint matrix of A.


7. Example 6.2.3

A =

2 1 −14 3 −2−6 2 5

. Find A−1.

A−1 =1

det(A)( adj A ) =

1

4

19 −8 26

−7 4 −101 0 2

T

=1

4

19 −7 1

−8 4 0

26 −10 2

.

8. Definition 6.2.4 (Eigenvalues and Eigenvectors)

An eigenvalue (or characteristic value) of the matrix A is a number λ for which the equation

A−v = λ−v

has a nonzero solution −v .

An eigenvector (or characteristic vector) of the matrix A is a nonzero −v such that A−v = λ−vfor some number λ.

9. Example 6.2.4

A =

(

6 −32 1

)

, λ = 4, −v =

(

3

2

)

.

A−v =

(

6 −32 1

)(

3

2

)

=

(

12

8

)

= 4 ·(

3

2

)

= λ · −v

Thus λ = 4 is an eigenvalue and −v = (3, 2) 6= −0 is a corresponding eigenvector.

10. Result B

Solve the problem of determining the eigenvalues and eigenvectors.

A−v = λ−v has nonzero solutions −v if and only if (A− λI)−v = 0 has nonzero solutions, i.e.,

det(A− λI) = 0.

Definition 6.2.5

The characteristic polynomial of A is

pA(λ) = det(A− λI)

in the variable λ. The characteristic equation of A is pA(λ) = det(A− λI) in the unknown λ.

The eigenvalues of A are the roots.


11. Example 6.2.5

A =

(

6 −32 1

)

, Find all eigenvalues and eigenvectors.

det(A− λI) = 0, pA(λ) = λ2 − 7λ+ 12 = 0, λ = 4, 3.

λ1 = 4, (A− 4I)−v1 =

(

2 −32 −3

)(

x1

x2

)

=

(

0

0

)

We choose −v1 =

(

3

2

)

.

λ2 = 3, (A− 3I)−v2 =

(

3 −32 −2

)(

x1

x2

)

=

(

0

0

)

We choose −v2 =

(

1

1

)

.

Thus λ = 4 is an eigenvalue and −v = (3, 2) 6= −0 is a corresponding eigenvector and λ = 3 is

another eigenvalue and −v = (1, 1) 6= −0 is a corresponding eigenvector.

Let B1 = v1, v2 be a basis of R2,

P1 =

(

3 1

2 1

)

, P−11 =

(

1 −1−2 3

)

, and P−11 AP1 =

(

4 0

0 3

)

.

Let B2 = v2, v1 be a basis of R2,

P2 =

(

1 3

1 2

)

, P−12 =

1

(−1)

(

2 −3−1 1

)

, and P−12 AP2 =

(

3 0

0 4

)

.

The matrix A can be diagonalized.

12. Result C

Suppose each of the n eigenvalues λ1, . . . , λn of the matrix A is distinct (i.e., non-repeated);

and let v1, v2, . . . , vn be a set of respective corresponding eigenvectors of A. Then

v1, v2, . . . , vn is linearly independent.

13. Result D

Suppose that A ∈ Mn(F) has a eigenvalue of multiplicity m, where 1 < m ≤ n. Then

this repeated eigenvalue having multiplicity m has p linearly independent eigenvectors corre-

sponding to it, where

1 ≤ p ≤ m.

14. Result E

If A ∈Mn(F) has one or more repeated eigenvalues, then

there may exist less than n linearly independent eigenvectors of A.


15. Example 6.2.6

A =

1 1 2

0 1 3

0 0 2


det(A− λI) = 0, pA(λ) = (λ− 1)2(λ− 2) = 0, λ = 1, 1, 2.

λ1 = 1, (A− I)−v1 =

0 1 2

0 0 3

0 0 2

x1

x2

x3

=

0

0

0

We choose −v1 =

1

0

0

.

λ2 = 2, (A− 2I)−v2 =

−1 1 2

0 −1 3

0 0 0

x1

x2

x3

=

0

0

0

We choose −v2 =

5

3

1

.

There exists less than 3 linearly independent eigenvectors of A. Note that A can not be

diagonalized.

16. Result F

All of the eigenvalues of a real symmetric matrix are real numbers.

17. Result G

If A ∈Mn(F) is symmetric, then

there exist n linearly independent eigenvectors of A, whether the n eigenvalues of A are all

distinct or whether one or more of these eigenvalues is repeated.

18. Example 6.2.7

A =

7 −1 6

−10 4 −12−2 1 −1



det(A− λI) = 0, pA(λ) = (λ− 2)(λ− 3)(λ− 5) = 0, λ = 2, 3, 5.

λ1 = 2, (A− 2I)−v1 =

5 −1 6

−10 2 −12−2 1 −3

x1

x2

x3

=

0

0

0

We choose −v1 =

1

−1−1

.

λ2 = 3, (A− 3I)−v2 =

4 −1 6

−10 1 −12−2 1 −4

x1

x2

x3

=

0

0

0

We choose −v2 =

1

−2−1

.

λ3 = 5, (A− 5I)−v3 =

2 −1 6

−10 −1 −12−2 1 −6

x1

x2

x3

=

0

0

0

We choose −v3 =

3

−6−2

.

There exists 3 linearly independent eigenvectors of A. Note that A can be diagonalized.

19. Result H

Let Sk ∈Mk(F).

Sk =

0

1 0

1 0. . .

. . .

1 0

, STk =

0 1

0 1. . .

. . .

0 1

0

.

They are called the elementary nilpotent matrices. We note that

Sk

x1

x2

...

xk

=

0

x1

...

xk−1

, STk

x1

x2

...

xk

=

x2

...

xk

0

.

Let A ∈Mn(F). If B1 = v1, v2, . . . , vn satisfies that

(A− λI)n(vn) = 0, (A− λI)n−1(vn) 6= 0, and

vn−1 = (A− λI)vn

vn−2 = (A− λI)vn−1

· · ·v1 = (A− λI)v2

then (1) B1 is a (cyclic) basis for Fn.

(2) [A]B1 = STn + λIn (a Jordan fundamental matrix).

(3) Let B2 = vn, vn−1, . . . , v1. Then B2 is also a (cyclic) basis for Fn and [A]B2 = Sn+λIn.

6.3 Homogeneous Linear Systems with Constant Coefficients: Two Unknowns 157

20. Example 6.2.8

A =

0 1 2

−5 −3 −71 0 0

, Find the Jordan form.

det(A− λI) = 0, pA(λ) = −(λ+ 1)3 = 0, λ = −1, −1, −1.

(A + I) =

1 1 2

−5 −2 −71 0 1

, (A+ I)2 =

−2 −1 −3−2 −1 −32 1 3

, (A+ I)3 = 0.

We should choose v3 = (1, 0, 0) ∈ ker(A+ I)3 − ker(A+ I)2. Then v2 = (A+ I)v3 = (1,−5, 1)and v1 = (A+ I)v2 = (−2,−2, 2).B1 = v1, v2, v3,

P1 =

−2 1 1

−2 −5 0

2 1 0

, P−1

1 =

0 1/8 5/8

0 −1/4 −1/41 1/2 3/2

, and P−1

1 AP1 =

−1 1 0

0 −1 1

0 0 −1

.

B2 = v3, v2, v1,

P2 =

1 1 −20 −5 −20 1 2

, P−1

2 =

1 1/2 3/2

0 −1/4 −1/40 1/8 5/8

, and P−1

2 AP2 =

−1 0 0

1 −1 0

0 1 −1

.

6.3 Homogeneous Linear Systems with Constant Coefficients:

Two Unknowns

Let us consider the homogeneous linear system with constant coefficients:

dx

dt= a1x+ b1y, (6.3.1)

dy

dt= a2x+ b2y.

We attempt to obtain a solution of the form

−x =

(

x(t)

y(t)

)

=

(

A

B

)

eλt


It need to consider the characteristic equation (6.1.8)

pM(λ) = λ2 − (a1 + b2)λ+ (a1b2 − a2b1) = 0.

A. Case 1. The Roots of the Characteristic Equation are Real and Distinct

1. Theorem 6.3.1

Hypothesis. The roots λ1 and λ2 are real and distinct. −v1 = (A1, B1) and−v2 = (A2, B2) are

corresponding eigenvectors with respect to λ1 and λ2

Conclusion. The system (6.3.1) has two nontrivial linearly independent solutions. The general

solution is

−x =

(

x(t)

y(t)

)

= c1eλ1t−v1+c2e

λ2t−v2 = c1

(

A1

B1

)

eλ1t+c2

(

A2

B2

)

eλ2t =

(

c1A1eλ1t + c2A2e

λ2t

c1B1eλ1t + c2B2e

λ2t

)

.

2. Example 6.3.1

dx

dt= 6x− 3y, (6.3.2)

dy

dt= 2x+ y,

pA(λ) = λ2 − (6 + 1)λ+ (1 · 6− (−3) · 2) = 0, λ = 3, 4.

λ = 3:

A− 3I =

(

6− 3 −32 1− 3

)

=

(

3 −32 −2

)

.

The eigenvector is (1, 1).

λ = 4:

A− 4I =

(

6− 4 −32 1− 4

)

=

(

2 −32 −3

)

.

The eigenvector is (3, 2).


−x = c1

(

1

1

)

e3t + c2

(

3

2

)

e4t.

B. Case 2. The Roots of the Characteristic Equation are Real and Equal

6.3 Homogeneous Linear Systems with Constant Coefficients: Two Unknowns 159

1. Theorem 6.3.2

Hypothesis. The roots λ1 and λ2 are real and equal. Let λ denote their common value.

Further assume that the system (6.3.1) is not such that a1 = b2 6= 0 and a2 = b1 = 0.

Conclusion. The system (6.3.1) has two nontrivial linearly independent solutions of the form

−x1 =

(

x1(t)

y1(t)

)

=

(

A

B

)

eλt and −x2 =

(

x2(t)

y2(t)

)

=

(

A1t + A2

B1t +B2

)

eλt

where A1 and B1 are not both zero and B1/A1 = B/A.


−x =

(

x(t)

y(t)

)

= c1

(

A

B

)

eλt + c2

(

A1t + A2

B1t +B2

)

eλt.

Method II (A Cyclic Basis)

We can choose −v1 and −v2 such that

(A− λI)2 −v2 =−0 ,

(A− λI) −v2 = −v1 ,

and −v1 and −v2 are linearly independent. Then

the two solutions

−x1 = −v1 eλt,

−x2 = (−v1 t +−v2 )eλt,

are linearly independent.

Method III (Another Notations)

We can choose−ξ and −η such that

(A− λI)−ξ =

−0 ,

(A− λI) −η =−ξ ,

Then

the two solutions

−x1 =−ξ eλt,

−x2 = (−ξ t +−η )eλt,

are linearly independent.

Remark:−ξ ≡ −v1 and −η ≡ −v2 .


Proof. Let −x = c1−x1 + c2

−x2 = c1−ξ eλt + c2(

−ξ t +−η )eλt.

Then we check the first equation. (Note that A−ξ = λ

−ξ .)

d −x1

d t= (−ξ · λ) · eλt

= (A−ξ ) · eλt = A−x1.

Now we check the second equation. (Note that A−η = λ−η +−ξ .)

d −x2

d t= (−ξ ) · eλt + (

−ξ t+−η )λ · eλt

= (−ξ +−ξ t · λ+−η · λ)eλt

= (A−ξ · t + A−η ) · eλt = A−x2.

2. Example 6.3.2

dx

dt= 4x− y, (6.3.3)

dy

dt= x+ 2y,

pA(λ) = λ2 − 6λ+ 9 = 0, λ = 3, 3.

Then we want to find −v2 and −v1 .

A− 3I =

(

1 −11 −1

)

, (A− 3I)2 = 0

We can choose −v2 = (1, 0). Then −v1 = (A− 3I)−v2 = (1, 1). The general solution is

−x = c1

(

1

1

)

e3t + c2

((

1

1

)

t+

(

1

0

))

e3t.

Lecture V0305

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