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40

CHAPTER 2

SECOND ORDER LINEAR

DIFFERENTIAL EQUATIONS

Chapter Outline

2.0 Overview and Learning Outcomes

2.1 Introduction

2.2 Homogeneous Equations

2.3 Non Homogeneous Equations

2.3.1 Method of Undetermined Coefficients

2.3.2 Method of Variation of Parameters

2.4 Euler Equation

2.5 Mathematical Modelling using Second Order Linear Differential Equations

Practice 2 (Second Order Linear Differential Equation)

2.0 OVERVIEW AND LEARNING OUTCOMES

In this chapter, we will focus on solving second order linear differential

equations. We will start this topic with the definition of linear differential equation.

Students will be introduced to two type of second order linear differential equation

which is homogeneous equation and non homogeneous equation with constant

coefficients. Methods of solution second order linear differential equation also will be

discussed. Next, we will consider a non homogeneous equation with variable

coefficient named Euler’s equation and its solution and in the end of this topic

mathematical modelling involved will be discussed.

It is expected that at the end of this course students will be able to:

• Identify the form of the homogeneous linear equation and able to solve the second

order homogeneous linear equation.

• Solve the second order linear non homogeneous linear equation using the method

of undetermined coefficient and method of variation of parameter.

• Solve the Euler’s equation.

• Develop a mathematical model from science and engineering problems and find

for solutions.

2.1 INTRODUCTION

In previous chapter, we studied first order differential equation and its

methods of solution. Now, we proceed to linear differential equation of second order

that widely used in science and engineering and have a variety of applications. A

second order differential equation is the equation that involves the second derivative

of the independent variable.

41

A general second order linear differential equation can be written as

( )xfcydx

dyb

dx

yda =++

2

2

(2.2)

where cba and, are constants and ( )xf is a given function of x .

For ( ) ,0=xf equation (2.2) is called homogenous equation

02

2

=++ cydx

dyb

dx

yda

and if ( ) 0≠xf , equation (2.2) is known as non homogenous equation

( ).2

2

xfcydx

dyb

dx

yda =++

Example 2.1

State whether or not each of the following equation is linear. If linear, determine

whether or not equation is homogeneous.

a) 03542

2

=−+ ydx

dy

dx

yd b) 0

2

23 =−+ y

dx

dyx

dx

ydx

c) xeydx

dy

dx

yd=+− 35

2

2

d) 71052

2

=−+ ydx

dy

dx

yd

e) 02

2

2

=− ydx

yd f) 2

2

2

24 xydx

dy

dx

ydy =++

g) xydx

dy

dx

ydsinln4

2

2

=++ h) xxedx

dy

dx

yd y cos52

2

=++

Solution

The answer for these examples is given as follows.

a) A linear homogeneous equation, constant coefficients.

b) A linear homogeneous equation, variable coefficients.

c) A linear non homogeneous equation, variable coefficients.

d) A linear non homogeneous equation, constant coefficients.

d) A nonlinear equation because of the term 2y .

e) A nonlinear equation because of the term 2

2

dx

ydy .

A differential equation of nth order is said to be linear if it can be written in the

form

( )xfyadx

dya

dx

yda

dx

yda

n

n

nn

n

n =++++ −

−

− 011

1

1 .. (2.1)

where variable y and its derivatives are first degree and coefficients

( )xfaaaa nn and,,..,, 011− are functions in term of x . If all the coefficients

011 ,,..,, aaaa nn − constant, equation (2.1) is called linear differential equation with

constant coefficients. If one not satisfies, it calls linear differential equation with

variable coefficients.

Definition 2.1 (Linear Differential Equation)

42

f) A nonlinear equation because of the term yln .

g) A nonlinear equation because of the term ye .

2.2 HOMOGENEOUS EQUATIONS

In this subtopic, we shall discuss the solution for homogeneous equation with constant

coefficients. In general, the equation for second order homogeneous equation is

written as

02

2

=++ cydx

dyb

dx

yda (2.3)

where cba and, are constants.

If mxey = is the solution for equation (2.3), hence .and 2

2

2mxmx em

dx

ydme

dx

dy==

Substitute y and its derivatives into (2.3) , hence

( ) ( ) ( ) 02 =++ mxmxmx ecmebema

( ) 02 =++ mxecbmam

Since 0≠mxe for any values of x , then

02 =++ cbmam . (2.4) Equation (2.4) is known as characteristic equation or auxiliary equation.

Alternatively, a simple way to find characteristic equation by replacing

1yand,2

2

2

⇒⇒⇒ mdx

dym

dx

yd.

Equation (2.4) has three forms of roots.

(i) Real and distinct roots, if 042 >− acb .

(ii) Real and equal roots, if 042 =− acb .

(iii) Two complex roots, if 042 <− acb .

Let 21 andmm are the characteristic roots of equation (2.4).

a) If 042 >− acb . Hence 21 mm ≠ . Then xmey 1

1 = and xmey 2

2 = are the solutions of

the homogeneous equation. Then, the general solution written as:

constants,,21 BABeAeyxmxm +=

b) If 042 =− acb . Hence 21 mmm == . The characteristic equation has only one

root .m The general solution written as:

( ) constants,, BAeBxAy mx+=

Hence, mxmx xeyey == 21 and .

43

c) If 042 <− acb , the characteristic equation has two complex roots,

., 21 imim βαβα −=+= .

Then the general solution written as : ( ) ( ) constants,, DCDeCey xixi βαβα −+ +=

By using Euler formula:

θsinθcosandθsinθcos θθ ieie ii −=+= − , then

( ) ( )

{ }( ) ( ){ }

( ) ( ){ }{ } ( ) ( )iDCBDCAxBxAe

xiDCxDCe

xixDxixCe

DeCee

DeCey

xixi

xixi

−=+=+=

−++=

−++=

+=

+=−

−+

and,sincos

sincos

sincossincos

ββ

ββ

ββββ

α

α

α

ββα

ββα

Conclusion:

If characteristic equation has two complex roots, im βα ±= , the general solution

could be written as:

( )xBxAey ββα sincos +=

Hence, xeyxey xx ββ αα sin,cos 21 == .

Example 2.2 (Real and distinct roots)

Determine the general solution of each of the following differential equations.

a) 024102

2

=+− ydx

dy

dx

yd b) 09

2

2

=− ydx

yd

Solution

a) The characteristic equation is 024102 =+− mm

The roots are real and distinct 6,4=m .

The general solution is .64 xx BeAey +=

b) The characteristic equation is 092 =−m

The roots are real and distinct 3,3−=m .

The general solution is .33 xx BeAey += −


Solve the initial value problem.

a) 062

2

=−+ ydx

dy

dx

yd; ( ) 40 =y , ( ) 30 =′y

b) 032

2

=−dx

dy

dx

yd; ( ) 30 =y , ( ) 60 =′y

Solution

a) The characteristic equation is .062 =−+ mm

Since the roots are real and distinct ,2,3−=m

then, the general solution is .32 xx BeAey −+=

Differentiating this solution, we get .32 32 xx BeAey −−=′

44

To satisfy the initial conditions, we required that

( ) 40 =+= BAy ................................ (i)

( ) .3320 =−=′ BAy .............................. (ii)

By solving simultaneous equation (i) and (ii), we find 3=A and 1=B . Thus, the

required solution of the initial value problem is given by

xx eey 323 −+= .

b) The characteristic equation is .032 =− mm

The roots are real and distinct .3,0=m

Then the general solution is .3xBeAy +=

Differentiating this solution, we get .3 3xBey =′


( ) 30 =+= BAy ............................... (i)

( ) .630 ==′ By ............................. (ii)

From (ii), we have 2=B and so (i) gives 1=A .

Thus, the required solution of the initial value problem is given by xey 321+= .

Exercises (Real and distinct roots)

Solve the following differential equation.

a) 022

2

=−dx

dy

dx

yd b) 0823

2

2

=−+ ydx

dy

dx

yd; ( ) 10 =y , ( ) 10 −=′y

Answer

a) xBeAy 2+= b) 4

2 37 3

10 10

xxy e e−= +

Example 2.4 (Real and equal roots)


a) 0962

2

=+− ydx

dy

dx

yd b) 09124

2

2

=+− ydx

dy

dx

yd

Solution

a) The characteristic equation is 0962 =+− mm

can be factored ( ) .032 =−m

So, the only root is real and equal .3=m

The general solution is ( ) .3xeBxAy +=

b) The characteristic equation is 09124 2 =+− mm

can be factored as ( ) .0322 =−m

So, the only root is real and equal .2

3=m

The general solution is ( ) .2

3x

eBxAy +=

45



a) 0442

2

=+− ydx

dy

dx

yd; ( ) 40 =y , ( ) 20 −=′y

b) 0253092

2

=+− ydx

dy

dx

yd; ( ) 30 =y , ( ) 40 =′y

Solution

a) The characteristic equation is .0442 =+− mm

Since the roots are real and equal ,2=m

then the general solution is ( ) xeBxAy 2+=

and its derivative are .22 222 xxx BeBxeAey ++=′


( ) 40 == Ay ............................. (i)

( ) .220 −=+=′ BAy ........................... (ii)

Substituting (i) into (ii) gives 10−=B .

The solution for IVP is ( ) xexy 2104 −= .

b) The characteristic equation is .025309 2 =+− mm

Since the roots are real and equal ,3

5=m

then the general solution is ( )x

eBxAy 3

5

+=

and its derivative are .3

5

3

53

5

3

5

3

5xxx

BeBxeAey ++=′


( ) 30 == Ay .............................. (i)

( ) .43

50 =+=′ BAy ........................... (ii)

Substituting (i) into (ii) gives 1−=B .

The solution for IVP is ( )x

exy 3

5

3−= .

Exercises (Real and equal roots)


a) 010252

2

=+− ydx

dy

dx

yd b) 044

2

2

=++ ydx

dy

dx

yd; ( ) 30 =y , ( ) 10 =′y

Answer

a) ( )x

eBxAy 5

1

+= b) ( ) xexy

23

−+=

46

Example 2.6 (Complex roots)


a) 0842

2

=++ ydx

dy

dx

yd b) 09

2

2

=+ ydx

yd

Solution

a) The characteristic equation is .0842 =++ mm

By the quadratic formula, the roots are ( )( )

( ) 2

164

12

81444 2 −±−=

−±−=m

.22 im ±−=

So 2−=α and 2=β . The general solution of the differential equation is

( )xBxAeyx

2sin2cos2 += − .

b) The characteristic equation is .092 =+m

By the quadratic formula, the roots are 22 99 im =−=

.39 2 iim ±==

So 0=α and 3=β . The general solution of the differential equation is

xBxAy 3sin3cos += .



a) 01062

2

=+− ydx

dy

dx

yd; ( ) 10 −=y , ( ) 20 =′y

b) 0742

2

=++ ydx

dy

dx

yd; ( ) 60 =y , ( ) 80 =′y

Solution

a) The characteristic equation is .01062 =+− mm

By the quadratic formula, the roots are

( ) ( ) ( )( )

( ) 2

46

12

1014662 −±

=−−±−−

=m

.3 im ±= So 3=α and 1=β . The general solution is

( )xBxAey x sincos3 += ............... (i)

and so ( ) ( ).cossinsincos3 33 xBxAexBxAey xx +−++=′ ........... (ii)

Substituting ( ) 10 −=y into (i) gives ( ) .10 −== Ay

Substituting ( ) 20 =′y and 1−=A into (ii) gives ( ) .230 =+=′ BAy

So, 5=B . The solution for IVP is ( )xxey x sin5cos3 +−= .

b) The characteristic equation is .0742 =++ mm

By the quadratic formula, the roots are ( ) ( )( )

( ) 2

124

12

714442 −±−

=−±−

=m

.32 im ±−= So 2−=α and 3=β .

47

The general solution is ( )xBxAey x 3sin3cos2 += − ....... (i)

and so

( ) ( ).3cos33sin33sin3cos2 22 xBxAexBxAey xx +−++−=′ −− ....... (ii)

Substituting ( ) 60 =y into (i) gives ( ) .60 == Ay

Substituting ( ) 20 =′y and 1−=A into (ii) gives ( ) .8320 =+−=′ BAy

So, 3

16=B .

The solution for IVP is

+= −

xxeyx

3sin3

163cos6

2 .

Exercises (Complex roots)


a) 0322

2

=+− ydx

dy

dx

yd b) 094

2

2

=+− ydx

dy

dx

yd; ( ) 00 =y , ( ) 30 −=′y

Answer

a) ( )xBxAey x 2sin2cos += b)

−= xey

x5sin

5

32

2.3 NON HOMOGENEOUS EQUATIONS

Now, we will discuss the solution for non homogeneous equation with

constant coefficients. For this purpose, we will discuss briefly two methods in this

section, Method of Undetermined Coefficients and Method of Variation of

Parameters. In general, the equation for second order non homogeneous equation is

written as

( )xfcydx

dyb

dx

yda =++

2

2

(2.5)

where cba and, are constants and ( )xf is function of x .

The solution of equation (2.5) is pc yyy += wherecy , the complementary solution or

homogeneous solution is and py is the particular solution.

The substitution the homogeneous solution following to three cases,

,21 xmxm

c BeAey += ( ) mx

c eBxAy += and ( )xBxAeyc ββα sincos += will give value

( )xf in equation (2.5) equal to 0. So, we need to find another term of solution such

that satisfies equation (2.5). Previously in Section 2.2, we used to find cy in order to

solve homogeneous equation. Now, in the next part, we will find out how to find py .

48

2.3.1 METHOD OF UNDETERMINED COEFFICIENTS

Consider ( ) .0,2

2

≠=++ axfcydx

dyb

dx

yda

The basic idea behind this approach is as follows.

(I) ( )xf a polynomial of degree n.

(II) ( )xf an exponential form kxCe , (C, k constants).

(III) ( ) kxCkxCxf sinorcos= , (C, k constants).

(IV) ( )xf is a linear combination involving Case (I), Case (II) and Case (III).

(V) ( )xf is a product function between Case (I), Case (II) and Case (III).

Method of Solution

In general, there are 3 steps to find .py

1. Identify type of )(xf the value of r, non negative integer.

2. Differentiate py for pp yy ′′′ and then substituting into given differential

equation.

3. Equate the coefficients of corresponding powers of x and solve the resulting

equations for undetermined coefficients, then we get: 011 and,..,, BBBB nn − .

CASE I: ( )xf a polynomial of degree n.

Suppose nxf degreeofpolynomialais)( , then the particular solution for this case

given by 1

1 1 0( )r n n

p n ny x B x B x B x B−−= + + + +L

1 1 0where , ,.., andn nB B B B− constants and

,..2,1,0=r is the non negative integers such that no term in py exist similarly in .cy

For example

1. )(5)( Cxyxf r

p =→=

2. )(96)( DCxxyxxf r

p +=→−=

3. )(12)( 22 EDxCxxyxxf r

p ++=→+=

4. )(76)( 2323 FExDxCxxyxxxf r

p +++=→++−=

5. ( ) )()( 0

1

10

1

1 BxBxBxyBxBxBxf n

n

n

n

r

p

n

n

n

n +++=→+++= −−

−− LL

Example 2.8

Solve the following differential equation .5103 2

2

2

xydx

dy

dx

yd=−+

Solution

Find cy : Auxiliary Equation 01032 =−+ mm . Then

49

xx

c BeAey

mm

mm

25

21 2,5

0)2)(5(

+=

=−=

=−+

−

Find py

Cy

DCxy

EDxCxy

yyEDxCxy

r

EDxCxxyxxf

p

p

p

Cpp

r

p

2

2

,Thus

)intermanytocorrespondintermNo(

0

)(5)(

2

2

22

=′′

+=′

++=

++=

=

++=→=

[ ] [ ] [ ][ ] [ ] [ ] 22

22

2

5103210610

510232

implies5103equationgiveninand,Substitute

xEDCDCxCx

xEDxCxDCxC

xyyyyyy ppp

=−++−+−

=++−++

=−′+′′′′′

Equating coefficients power x,

100

19

10

32gives01032]Constant[

10

3

2

1

5

3

10

6gives0106][

2

1gives510][ 2

−=+

==−+

−=

−===−

−==−

DCEEDC

CDDCx

CCx

.100

19

10

3

2

1

100

19

10

3

2

1

225

22

−−−+=+=

−−−=++=

− xxBeAeyyy

xxEDxCxy

xx

pc

p

Example 2.9

Solve the following differential equation .63 xyy =′+′′

Solution

Find cy : Auxiliary Equation 032 =+ mm . Then

BAeBeAey

mm

mm

xxx

c +=+=

−==

=+

−− 303

21 3,0

0)3(

Find py :

Cy

DCxy

yyDxCxyr

yByDDCxy

r

DCxxyxxf

p

p

Cpp

Cpp

r

p

2

2

)intermanytocorrespondintermNo(,1choosing,Next

)intermtocorrespondinTerm(

0

)(6)(

2

=′′

+=′

+==

+=

=

+=→=

50

[ ] [ ][ ] [ ] xDCCx

xDCxC

xyyyyy ppp

6326

6232


=++

=++

=′+′′′′′

Equating coefficients power x

( )3

21

3

2

3

2gives032]Constant[

1gives66][

−=−=−==+

==

CEDC

CCx

.3

2

3

2

23

22

xxBAeyyy

xxDxCxy

x

pc

p

−++=+=

−=+=

−

Exercises


a) 2792

2

=+ ydx

yd b) 4844 2

2

2

+=+− xydx

dy

dx

yd

Answer

a) cos3 sin 3 3y A x B x= + + b) ( ) 2 22 4 4xy A Bx e x x= + + + +

CASE II: ( )xf an exponential form kxCe , (C, k constants)

Suppose , e form lexponentiaan )(xk

Cxf then the particular solution for this case

given by )( kxr

p Cexy = where ,..2,1,0=r is the nonnegative integers such that no

term in py corresponds to .cy

For example

1. )(5)( 33 xr

p

x Cexyexf =→=

2. )(9)(77 xr

p

xCexyexf

−− =→−=

Example 2.10

Solve the following differential equation .44 xeyyy −=+′+′′

Solution

Find cy : Auxiliary Equation 0442 =++ mm . Then

( )

.

2

02

22

21

2

xx

c BxeAey

mm

m

−− +=

−==

=+

Find py :

51

x

p

x

p

Cp

x

p

xr

p

x

Cey

Cey

yyCey

r

Cexyexf

−

−

−

−−

=′′

−=′

=

=

=→=


0

)()(

[ ] [ ] [ ].1gives

44


==

=+−+

=+′+′′′′′

−−

−−−−

−

CeCe

eCeCeCe

eyyyyyy

xx

xxxx

x

ppp

.22 xxx

pc

xx

p

eBxeAeyyy

eCey

−−−

−−

++=+=

==

Example 2.11

Solve the following differential equation .106 3

2

2xey

dx

dy

dx

yd=−−

Solution

Find cy : Auxiliary Equation 062 =−−mm . Then

( )

xx

c BeAey

mm

mm

32

21 3,2

0)3(2

+=

=−=

=−+

−

Find py :

3 3

3 3 3

( ) 10 ( )

0

(Term in correspond to term in )

x r x

p

x x x

p p C

f x e y x Ce

r

y Ce Ce y Be y

= → =

=

=

xxxxx

p

xx

p

Cp

x

p

CeCxeCeCeCxey

CeCxey

yyCxeyr

33333

33

3

69339

3

)intermanytocorrespondintermNo(,1choosing,Next

+=++=′′

+=′

==

[ ] [ ] [ ][ ] [ ] xxx

xxxxxx

x

ppp

eCCeCCCxe

eCxeCeCxeCeCxe

eyyyyyy

333

333333

3

106639

106369


=−+−−

=−+−+

=−′−′′′′′

Equating coefficients power 3xe ,

.2

2

2gives105][

332

33

3

xxx

pc

xx

p

x

xeBeAeyyy

xeCxey

CCe

++=+=

==

==

−

52

Exercises


a) xeyyy 222 =+′−′′ b) xeyy 52=+′′ ; ( ) 10 =y , ( ) 20 −=′y

Answer

a) 22x x xy Ae Bxe e= + + b) xexxy 5

13

1sin

13

31cos

13

12+−=

CASE III: ( ) kxCkxCxf sinorcos= , (C, k constants)

Suppose ( ) kxCkxCxf sinorcos= , then the particular solution for this case given by

)(kxr

p Cexy = where ,..2,1,0=r is the nonnegative integer such that no term in

py corresponds to .cy

For example

1. ( ) 5sin 4 ( cos 4 sin 4 )r

pf x x y x C x D x= → = +

2. ( ) 6sin 2 ( cos 2 sin 2 )r

pf x x y x C x D x= − → = +

3. ( )xDxCxyxxxf r

p 6sin6cos6sin86cos7)( +=→−=

Example 2.12

Find the general solution for the given equation .cos2

2

xydx

yd=+

Solution

Find cy : Auxiliary Equation 012 =+m . Then

xBxAy

im

m

c sincos

12

+=

±

−=

Find py :

( ) cos ( cos sin )

0

cos sin (Terms cos and sin in correspond to

terms cos and sin in )

r

p

p p

C

f x x y x C x D x

r

y C x D x C x D x y

A x B x y

= → = +

=

= +

[ ] [ ]

[ ] [ ] xDCxxDxC

xDxDxDxxCxCxCxy

xDxCxDCxxDxDxxCxCxy

yyxDxxCxyr

p

p

Cpp

cos2sin2

coscossinsinsincos

cossinsincoscossin

)intermanytocorrespondintermNo(sincos,1choosing,Next

+−+−−=

++−−−−=′′

+++−=+++−=′

+==

[ ] [ ] [ ][ ] [ ][ ] [ ] xxDxC

xxCxDCxxDxDxC

xxDxxCxxDCxxDxC

xyyyyy ppp

coscos2sin2

coscos2sin2

cossincoscos2sin2

impliescosequationgiveninand,Substitute

=+−

=++−++−−

=+++−+−−

=+′′′′′

53

Equating coefficients power sin x and cos x,

.0gives02][sin

2

1gives12][cos

==−

==

CCx

DDx

.sin2

1sincos

sin2

1sincos

xxxBxAyyy

xxxDxxCxy

pc

p

++=+=

=+=

Example 2.13

Find the general solution for the given equation .sin4222

2

xydx

dy

dx

yd=++

Solution

Find cy : Auxiliary Equation 0222 =++ mm . Then ( )

( )( ) xxx

c BxeAeeBxAy

m

m

−−− +=+=

−=

=+

rootRepeated1

012

Find py :

)intermanytocorrespondintermNo(sincos

0

)sincos(sin4)(

Cpp

r

p

yyxDxCy

r

xDxCxyxxf

+=

=

+=→=

xDxCy

xDxCy

p

p

sincos

cossin

−−=′′

+−=′

[ ] [ ] [ ][ ] [ ][ ] [ ] xxDxC

xxCDCxDCD

xxDxCxDxCxDxC

xyyyyyy ppp

sin4cos2sin2

sin4cos2sin2

sin4sincoscossin2sincos

impliessin42equationgiveninand,Substitute

=+−

=++−++−−

=+++−+−−

=+′+′′′′′

Equating coefficients cos x and sin x,

( ) .cos2

cos2sincos

.2gives42][sin

0gives02][cos

xeBxAyyy

xxDxCy

CCx

DDx

x

pc

p

−+=+=

−=+=

−==−

==

−

Exercises

a) Consider the differential equation .2sin32cos292

2

xxydx

yd+=+

Find the complementary equation ( cy ), the particular integral ( py ) and complete

solution from the given equation.

b) Find the general solution for the given equation .cos422

2

xydx

dy

dx

yd=−+

54

Answer

a)

)2sin32cos2(5

13sin3cos xxxBxAy +++=

b) xxBeAey xx sin1850

1cos

1850

4367 +−+= −

CASE IV: ( ) ( ) ( ) ( )xfxfxfxf n±±±= ..21 is a linear combination involving CASE

(I, II and III).

For this case, we need to find piy for each if for .,..,2,1 ni =

For example

1. ( ) ( )xGxFxEDxCxxyxxxf rr

p 4sin4cos4sin2)( 22 ++++=→+=

2. ( ) ( )xExDxCexyxexfrxr

p

x2sin2cos2sin63)(

55 ++=→−= −−

3. ( ) ( )xrr

p

x EexDCxxyexxf 334)( ++=→+=

Example 2.14

Find the general solution for the given equation .10442

2xexy

dx

yd −−=+

Solution

Find cy : Auxiliary Equation 042 =+m . Then

xBxAy

im

m

c 2sin2cos

2

42

+=

±

−=

Find py :

( ) ( ) xx exfxxfexxf −− −==→+= 10and4104)( 21

( )

0


0

)(4(a)

1

1

1

11

=′′

=′

+=

=

+=→=

p

p

Cpp

r

p

y

Cy

yyDCxy

r

DCxxyxxf

[ ] [ ][ ] [ ] xDxC

xDCx

xyyyyy ppp

444

440

implies44equationgiveninand,Substitute 111

=+

=++

=+′′′′′


xy

DD

CCx

p =

==

==

1

.0gives04]constant[

1gives44][

55

( )

x

p

x

p

Cp

x

p

xr

p

x

Cey

Cey

yyCey

r

Cexyexf

−

−

−

−−

=′′

−=′

=

=

=→−=

2

2

2

22


0

)(10(b)

[ ] [ ][ ] xx

xx

x

ppp

eeC

xCeCe

eyyyyy

−−

−−

−

−=

=+

−=+′′′′′

105

44

implies104equationgiveninand,Substitute 222

Equating coefficients power xe−

x

p

x

ey

CCe

−

−

−=

−=−=

2

2gives105][

2

.22sin2cos21

x

ppcpc exxBxAyyyyyy −−++=++=+=

Example 2.15

Consider the differential equation .sin5622 2

2

2

xxydx

dy

dx

yd+=++ Find the

complementary equation ( cy ), the particular integral ( py ) and complete solution from

the given equation.

Solution

Find cy : Auxiliary Equation 0222 =++ mm . Then

ii

m ±−=±−

=−±−

=−±−

= 12

22

2

42

2

842

( ) xBexAeexBxAy xxx

c sincossincos −−− +=+=

Find py :

( ) ( )

( )

Cy

DCxy

yyEDxCxy

r

EDxCxxyxxf

xxfxxfxxxf

p

p

Cpp

r

p

2

2


0

)(6(a)

sin5and6sin56)(

1

1

1

2

1

2

1

2

1

2

2

1

2

=′′

+=′

++=

=

++=→=

==→+=

[ ] [ ] [ ][ ] [ ] [ ] 22

22

2

111

6222222

6222


xEDCxDCxC

xEDxCxDCxC

xyyyyyy ppp

=+++++

=+++++

=+′+′′′′′

56


[ ]

( )1333

033gives0222]constant[

3gives022][

3gives62

2

1

2

−=−=

=+−=−−==++

−=−==+

==

xxxxy

DCEEDC

CDDCx

CCx

p

( )

xDxCy

xDxCy

yyxDxCy

r

xDxCxyxxf

p

p

Cpp

r

p

sincos

cossin

)intermanytocorrespondintermNo(sincos

0

)sincos(sin5(b)

2

2

2

22

−−=′′

+−=′

+=

=

+=→=

[ ] [ ] [ ][ ] [ ][ ] [ ] xxDCxDC

xxDCDxCDC

xxDxCxDxCxDxC

xyyyyyy ppp

sin5sin2cos2

sin5sin22cos22

sin5sincos2cossin2sincos

impliessin522equationgiveninand,Substitute 222

=+−++

=+−−+++−

=+++−+−−

=+′++′′′′′

Equating coefficients of ,sinandcos xx

[ ]

xxxDxCy

CDD

DDDCx

DCDCx

p sincos2sincos

2,155

522gives52][sin

2implies02][cos

2 +−=+=

−==∴=

=+−−=+−

−==+

( ) ( ) xxxxexBxAyyyyyy x

ppcpc sincos213sincos21 +−−++=++=+= −

Exercises

a) Find the general solution for the given equation xeydx

dy

dx

yd 5

2

2

612 −+=−+ .

b) Find the particular solution for the given equation

2)0(',1)0(,1262

2

−==+−=−− yyxeydx

dy

dx

yd x

Answer

a)

4 3 51 1

2 8

x x xy Ae Be e− −= + − + b) 2 311 7 1 12

6 6 6 3

x x xy e e e x−= + + − +

57

CASE V: ( )xf is a product involving Case (I), Case (II) and Case (III).

1. xexxfx

2sin2)(2=

2. xexf x 6cos)( 5−=

3. xxexf 3)( =

For ,..2,1,0=r is the non negative integer, py can be summarized in table below

following cases considered.

Example 2.16

Find the general solution for the given equation .22 xxeyyy =−′−′′

Solution

Find cy : Auxiliary Equation 022 =−−mm . Then

( )( )

xx

c BeAey

mm

mm

2

21 2,1

012

+=

=−=

=+−

−

Find py :

xr

p

x eDCxxyxexf )(2)( +=→=

)intermanytocorrespondintermNo()(

0

Cp

xxx

p yyDeCxeeDCxy

r

+=+=

=

xxxxxxx

p

xxx

p

DeCeCxeDeCeCeCxey

DeCeCxey

++=+++=′′

++=′

2

[ ] [ ] [ ][ ] [ ][ ] [ ] xxx

xxx

xxxxxxxxx

x

ppp

xeDCeCxe

xeDDCDCeCCCxe

xeDeCxeDeCeCxeDeCeCxe

xeyyyyyy

222

2222

222


=−+−

=−−−++−−

=+−++−++

=−′−′′′′′

No )(xf py

1 ( ) kx

n exP ⋅ kxn

n

n

n

r

p eBxBxBxBxy )( 01

1

1 ++++= −− L

2 ( )( ) xxP

orxxP

n

n

β

β

cos

sin

⋅

⋅

xCxCxCxCx

xBxBxBxBxy

n

n

n

n

r

n

n

n

n

r

p

β

β

sin)(

cos)(

01

1

1

01

1

1

+++++

++++=−

−

−−

L

L

3

xCe

orxCe

kx

kx

β

β

cos

sin

⋅

⋅

( )xDxCexy kxr

p ββ sincos +⋅=

4 ( ) xexP kx

n βsin⋅⋅ xeBxBxBxBxy kxn

n

n

n

r

p βsin)( 01

1

1 ⋅⋅++++= −− L

5 ( ) xexP kx

n βcos⋅⋅ xeBxBxBxBxy kxn

n

n

n

r

p βcos)( 01

1

1 ⋅⋅++++= −− L

58

Equating coefficients of xx xee and ,

xxx

pc

xx

p

x

x

exBeAeyyy

exeDCxy

CDDCe

CCxe

−−++=+=

−−=+=

−===−

−==−

− 12

1

12

1)(

.2

1

2

1gives02][

1gives22][

2

Exercises

a) Find the general solution for the given equation xeyyy x 2sin22 =+′−′′ .

b) Find the general solution for the given equation .cos222

2

xxydx

dy

dx

yd=+−

Answer

a)

( ) xeexBxAy xx 2sin3

1sincos −+=

b)

( ) ( ) xxxeBxAy x sin1cos +−−+=

2.3.2 METHOD OF VARIATION OF PARAMETERS

This method can be used in solving non homogeneous differential equation

( )xfcydx

dyb

dx

yda =++

2

2

where cba and, are constants, and ( )xf function of x in the form of polynomial, an

exponential functions, a trigonometric function, linear combination and product of

(polynomial , exponential functions, a trigonometric function).

In this method, the general solution is in the form

21 vyuyy +=

where ( )xuu = , ( )xvv = and 21, yy are the solution corresponds to homogeneous

equation respectively where u and v given by

2

1

( )

( ).

y f xu dx A

aW

y f xv dx B

aW

= − +

= +

∫

∫

59

The Method of Solution

Consider: ( )xfcydx

dyb

dx

yda =++

2

2

.

1. Determine a and ( )xf .

2. Determine 21, yy the independent solution for homogeneous equation.

3. Find Wronskian, .21

21

yy

yyW

′′=

4. Find u and v given by .)(

,)( 12 ∫∫ +=+−= Bdx

aW

xfyvAdx

aW

xfyu

5. The general solution is given by .21 vyuyy +=


Solve the following differential equation using method of variation of parameter.

.6 3xeyyy =−′−′′

Solution

From given differential equation 06 =−′−′′ yyy , the values of 1=a and ( ) xexf 3= .

Find the homogenous solution 21 ByAyyc +=

The characteristic equation .062 =−−mm

with real and distinct roots, .3,2−=m

The homogeneous solution is .32 xx

c BeAey += −

So we have xey 2

1

−= and xey 3

2 = xey 2

1 2 −−=′ and .3 3

2

xey =′

Calculate the Wronskian, ( ) ( ) .52332

2332

32

32

xxxxx

xx

xx

eeeeeee

eeW =−−=

−= −−

−

−

Next find ( )

Aedxedxe

eeu xx

x

xx

+−=−=−= ∫∫ 5533

25

1

5

1

5

and ( )

.5

1

5

1

5

32

Bxdxdxe

eev

x

xx

+=== ∫∫−

The general solution is 21 vyuyy +=

xxx eBxeAe 325

5

1

25

1

++

+−= −

.5

1

25

1 3233 xxxx BeAexee +++−= −


Find the solution of the differential equation ( ) xexyyy 41168 +=+′−′′ if given conditions ( ) 10 =y and ( ) 10 −=′y .

60

Solution

From given differential equation ( ) xexyyy 41168 +=+′−′′ , the values of 1=a and

( ) ( ) xexxf 41+= .


The characteristic equation 01682 =+− mm with real and equal root, .4=m

The homogeneous solution is .44 xx

c BxeAey +=

So we have xey 4

1 = and xxey 4

2 = xey 4

1 4=′ and .4 44

2

xx exey +=′

Calculate the Wronskian,

( ) ( ) .4444

844444

444

44

xxxxxx

xxx

xx

eexeexeeexee

xeeW =−+=

+=

Next find ( )( )

( ) Axx

dxxxdxe

exxeu

x

xx

+−−=+−=+

−= ∫∫ 231

1

1 23

8

44

and ( )( )

( ) .2

11

12

8

44

Bxx

dxxdxe

exev

x

xx

++=+=+

= ∫∫


xx xeBxx

eAxx 4

24

23

223

+++

+−−=

xxxx ex

ex

BxeAe 42

43

44

26+++= .............(i)

and

2

46

42

442

443

442

444 xexe

xee

xBeBxeAey xxxxxxx ++++++=′ ......(ii)

Substituting ( ) 10 =y into (i) yields 1=A .

Substituting ( ) 10 −=′y and the value of A into (ii) we get 14 −=+ BA gives 5−=B .

The general solution is given by xxxx ex

ex

xeey 42

43

44

265 ++−= .


Find the general solution of the differential equation . cosec xyy =+′′

Solution

From given differential equation xyy cosec=+′′ , the values of 1=a and

( ) xxf cosec= .


The characteristic equation is 012 =+m with complex roots, im ±= .

The homogeneous solution is xBxAyc sincos += so we have

xy cos1 = and xy sin2 =

xy sin1 −=′ and .cos2 xy =′

Calculate the Wronskian, ( ) .1sincoscossin

sincos22 =−−=

−= xx

xx

xxW

61

Next find ( ) Axdxdxxxu +−=−=−= ∫∫ cosecsin

and ( ) .sinlnsin

cos coseccos Bxdx

x

xdxxxv +=== ∫∫


( ) ( ) xBxxAx sinsinlncos +++−=

.sinlnsincossincos xxxxxBxA +−+=

Example 2.20 (IVP Real and distinct roots)

Find the solution of the differential equation xyyy 286 =+′−′′ which satisfies the

conditions ( ) 00 =y and ( ) 10 =′y .

Solution

From given differential equation xyyy 286 =+′−′′ , the values of 1=a and

( ) xxf 2= .


The characteristic equation 0862 =+− mm with real and distinct roots, 4,2=m .

The homogeneous solution is xx

c BeAey 42 +=

So we have xey 2

1 = and xey 4

2 = x

ey2

1 2=′ and .44

2

xey =′

Calculate the Wronskian,

( ) ( ) .22442

62442

42

42

xxxxx

xx

xx

eeeeeee

eeW =−==

Next find

( )

( )( ) Aeex

dxxedxe

xeu xxx

x

x

++=−=−= −−−∫∫ 222

6

4

4

1

221

2

and ( )

( )( ) .16

1

421

2 444

6

2

Beex

dxxedxe

xev

xxx

x

x

+−−=== −−−∫∫


xxxxxxeBee

xeAee

x 444222

16

1

44

1

2

+−−+

++= −−−−

16

3

4

42 +++=x

BeAe xx .......................(i)

and so 4

142 42 ++=′ xx BeAey .......................(ii)

Substituting ( ) 00 =y into (i) gives16

3−=+ BA .

Substituting ( ) 10 =′y gives .4

342 =+ BA

By solving simultaneous equation above, we get 4

3−=A and

16

9=B .

Hence, the general solution is given by 16

3

416

9

4

3 42 +++−=x

eey xx .

62

Exercises


a) xxeydx

dy

dx

yd 2

2

2

234 =+− b) xeyyy

−=+′−′′ 44 ; ( ) 80

9y = − , ( ) 1

09

y′ = − .

Answer

a) xxx xeBeAey 23 2−+= ; Aexeu xx ++−= , Bexev xx +−−= −−

b) 2 2 12

9

x x xy e xe e−= − + + ; Aexe

uxx

++=−−

93

33

, Be

vx

+−=−

3

2.4 EULER EQUATION

In previous section, we already study the solution of homogeneous equation and non

homogeneous equation with constants coefficients. Now we consider second order

linear differential equation with the independent variable coefficient.

A differential equation in the form of

( )xfyadx

dyxa

dx

ydxa

dx

ydxa

n

nn

nn

nn

n =++++ −

−−

− 011

11

1 .. (2.6)

where naaa ,..,, 10 are constants, is known as Euler’s equation of nth

order.

A second order Euler’s equation can be written as:

( )xfcydx

dybx

dx

ydax =++

2

22 (2.7)

where cba and, are constants.

The Method of Solution

Substitute tex = or equivalent xt ln= andxdx

dt 1= . Then

xdt

dy

dx

dt

dt

dy

dx

dy 1⋅=⋅= or

dt

dy

dx

dyx = (2.8)

Next,

=

dt

dy

dx

d

dx

dyx

dx

d

xdt

yd

dx

dt

dt

dy

dt

d

dx

dy

dx

ydx

12

2

2

2

⋅=

=+ since

=xdx

dt 1.

2

2

2

22

dt

yd

dx

dyx

dx

ydx =+ or

dx

dyx

dt

yd

dx

ydx −=

2

2

2

22 .

From equation (2.8),

dt

dy

dt

yd

dx

ydx −=

2

2

2

22 (2.9)

Substitute (2.8) and (2.9) into (2.7) and then

( )tefcydt

dyb

dt

dy

dt

yda =+

+

−

2

2

or

63

( ) ( ).2

2tefcy

dt

dyab

dt

yda =+−+ (2.10)

Equation (2.10) is the Euler’s equation with constant coefficients.

Example 2.21

Solve the following differential equations.

a) 092

22 =−+ y

dx

dyx

dx

ydx b) 0984

2

22 =+− y

dx

dyx

dx

ydx

c) 0852

22 =++ y

dx

dyx

dx

ydx

Solution

a) Given 092

22 =−+ y

dx

dyx

dx

ydx .

The differential equation is an Euler’s equation where 9,1,1 −=== cba and

( ) .0=xf By using substitutions dt

dy

dx

dyxex t == , and

dt

dy

dt

yd

dx

ydx −=

2

2

2

22 , we get

( ) ( ).2

2tefcy

dt

dyab

dt

yda =+−+

Substitute 1,1 == ba and ,9−=c then

092

2

=− ydt

yd

which is the second order homogeneous differential equation with constants

coefficients we have study previous section.

The general solution is given by

( ) tt BeAety 33 += − (Refer example 2.2b)

( ) ( )33 tt eBeA +=−

where A and B are two arbitrary constants.

Substituting tex = or xt ln= ,

( ) ( ) ( )33xBxAxy += −

.33 BxAx += −

b) Given .09842

22 =+− y

dx

dyx

dx

ydx

The differential equation is an Euler’s equation where 9,8,4 =−== cba and

( ) .0=xf By using substitutions tex = , we get

( ) ( )2

2.td y dy

a b a cy f edt dt

+ − + =

Substitute 8,1 −== ba and ,9=c then

.091242

2

=+− ydt

dy

dt

yd


( ) ( )t

eBtAty 2

3

+= (Refer example 2.4b)

( )( )2

3teBtA +=

64



( ) ( )( )2

3

ln xxBAxy +=

( ) 2

3

ln xxBA+=

c) Given 0852

22 =++ y

dx

dyx

dx

ydx .

The differential equation is an Euler equation where 8,5,1 === cba and ( ) .0=xf By

using substitutions tex = , we get

( ) ( ).2

2tefcy

dt

dyab

dt

yda =+−+

Substitute 5,1 == ba and ,8=c then

.0842

2

=++ ydt

dy

dt

yd


( ) ( )tBtAety t 2sin2cos2 += − (Refer example 2.6a)

( ) ( )tBtAet 2sin2cos2

+=−



( ) ( ) ( )xBxAxxy ln2sinln2cos2 += −

( ).ln2sinln2cos2 xBxAx += −

Example 2.22

Solve the following initial value problems.

a) 02492

22 =+− y

dx

dyx

dx

ydx ; ( ) ,11 =y ( ) 21 =′y .

b) 0952

22 =+− y

dx

dyx

dx

ydx ; ( ) ,11 =y ( ) 11 −=′y .

Solution

a) Given 02492

22 =+− y

dx

dyx

dx

ydx .

The differential equation is an Euler equation where 24,9,1 =−== cba and

( ) .0=xf By using substitutions tex = , we get

( ) ( )2

2.td y dy

a b a cy f edt dt

+ − + =


.024102

2

=+− ydt

dy

dt

yd


( ) ttBeAety

64 += (Refer example 2.2a)

( ) ( )64 tt eBeA +=



65

( ) ( ) ( )64xBxAxy +=

.64 BxAx += ...............................(i)

Differentiate (i) with respect to x ,

( ) .64 53 BxAxxy +=′ .............................(ii)

Given ( ) 11 =y , from (i), we get

.1 BA += .............................(iii)

Given ( ) 21 =′y , from (ii), we get .642 BA += .............................(iv)

By solving the simultaneous equation (iii) and (iv), we obtained 1−=B and .2=A

The particular solution is

( ) .2 64 xxxy −=

b) Given 0952

22 =+− y

dx

dyx

dx

ydx .

The differential equation is an Euler equation where 9,5,1 =−== cba and ( ) .0=xf

By using substitutions tex = , we get

( ) ( )2

2.td y dy

a b a cy f edt dt

+ − + =


.0962

2

=+− ydt

dy

dt

yd


( ) ( ) teBtAty 3+= (Refer example 2.4a)

( )( )3teBtA+=



( ) ( )( )3ln xxBAxy +=

( ) .ln3

xxBA += ..........................(i)

Differentiate (i) with respect to x ,

( ) .ln33 222 xBxBxAxxy ++=′ …...................(ii)

Given ( ) 11 =y , from (i), we get

.1=A ........................(iii)

Given ( ) 11 −=′y , from (ii), we get .31 BA +=− .........................(iv)

By substituting (iii) into (iv), we obtained .4−=B

The particular solution is

( ) ( ) .ln413

xxxy −= Example 2.23


a) 3

2

22 6 xydx

ydx =− b) ( ) 4

2

22 1ln167 xxy

dx

dyx

dx

ydx +=+−

Solution

a) Given 3

2

22 6 xydx

ydx =− .

66

The differential equation is an Euler equation where 6,0,1 −=== cba and

( ) .3xxf = By using substitutions tex = , we get

( ) ( )2

2.td y dy

a b a cy f edt dt

+ − + =

Substitute 0,1 == ba and ,6−=c then

.6 3

2

2tey

dt

dy

dt

yd=−−


( ) tttt teeBeAety 3332

5

1

25

1+−+= − (Refer example 2.17)

( ) ( ) ( ) ( )3332

5

1

25

1 tttt eteeBeA +−+=−



( ) ( ) ( ) ( ) ( ) xxxxBxAxy ln5

1

25

1 3332 +−+= −

.ln5

1

25

1 3332 xxxBxAx +−+= −

b) Given ( ) 4

2

22 1ln167 xxy

dx

dyx

dx

ydx +=+− .

The differential equation is an Euler equation where 16,7,1 =−== cba and

( ) ( ) .1ln 4xxxf += By using substitutions tex = , we get

( ) ( )2

2.td y dy

a b a cy f edt dt

+ − + =


( ) .1168 4

2

2tety

dt

dy

dt

yd+=+−


( ) tttt et

et

BteAety 42

43

44

26+++= (Refer example 2.18)

( ) ( ) ( ) ( )42

43

44

26

tttt et

et

eBteA +++=



( ) ( ) ( ) ( ) ( ) ( ) ( )42

43

44

2

ln

6

lnln x

xx

xxxBxAxy +++=

( ) ( ) 4

2

4

3

44

2

ln

6

lnln x

xx

xxBxAx +++=

( ) ( )

.2

ln

6

lnln

23

4

+++=

xxxBAx

Exercises


67

a) 0622

22 =−+ y

dx

dyx

dx

ydx b) 025219

2

22 =+− y

dx

dyx

dx

ydx c)

01052

22 =+− y

dx

dyx

dx

ydx

Solution

a) ( ) 32 −+= BxAxxy b) ( ) ( )5

3lny x A B x x= + c) ( ) ( )xBxAxxy lnlncos3 +=

2.5 MATHEMATICAL MODELLING USING SECOND ORDER LINEAR

DIFFERENTIAL EQUATIONS

Spring/Mass Systems: Free Undamped Motion

Hooke's Law stated that the restoring force F of a spring opposite to the direction of

elongation and proportional to its total elongation. The equation given by F = ks

where F, the restoring force, s, amount of elongation and k, spring constant. For

example, if a mass weighing 14 pounds stretches a spring ½ foot, then 14 = k(1/2) and

k = 28 lbs/ft.

Before proceed to Newton’s Second Law, we define the weight, W = mg where mass

is measured in slugs, grams, or kilograms. For example, g = 32 ft. /s2 or 9.8 m / s

2 or

980 cm/ s2.

The condition for equilibrium is mg = ks or mg - ks = 0. If the mass is

displaced by an amount x from its equilibrium position, the restoring force of the

spring is then k(x + s). Assume that there are no retarding forces acting on the system

and assuming that the mass vibrates free of other forces (free motion), we can write

Newton’s Second Law with resultant force and the weight given by

2

2( )

d xm k s x mg

dt= − + −

= kx mg ks kx− + − = − (2.11)

The negative sign in (2.11) show that the restoring force acts opposite to the direction

of motion. Next,

2

2

d xm kx

dt= −

Divide by m both side equations

2

2

d x kx

dt m

−=

Let 2 k

mω =

Then

2

2

2

d xx

dtω= − and

22

20

d xx

dtω+ = (2.12)

The last equation is called Simple Harmonic Motion or Free Undamped Motion.

Simple harmonic motion (SHM) is the motion of a simple harmonic oscillator, a

motion that is neither driven nor damped. In order to solve equation (2.12), we need

68

two initial condition, 0(0)x x= , the amount of initial displacement, 1'(0)x x= , the

initial velocity of the mass.

Method of Solution

To solve2

2

20

d xx

dtω+ = , we need to find the auxiliary equation associated to second

order homogenous equation.

So, we have 2 2 0m ϖ+ = so m iϖ= ± .

The solution given by,

( ) ( ) ( )tBtAtx ωω sincos += (2.13)

The period of free vibrations for equation (2.13) is given by 2

Tπω

= and the

frequency, πω2

1==

TF .

Equation (2.13) is called Equation of Motion.

By using two initial conditions, we should able to determine A and B.

Example 2.24 (Free Undamped Motion)

A mass weighing 4 pounds stretches a spring 8 inches. At t = 0, the mass is released

from a point 10 inches below the equilibrium position with an upward velocity of 3/4

ft/sec. Determine the equation of free motion. Determine the period of free vibrations

and its frequency.

Solution

8 inches = 8/12 ft. = 2/3 ft. = s.

Weight, W = mg implies 2

4 1

32 / sec 8

W lbsm slug

g ft= = =

From Hooke's Law, F = ks, we have

.3

24

= k

So k = 6 lb/ft.

Since 2

2

d xm kx

dt= − we have x

dt

xd6

8

12

2

−= and .0482

2

=+ xdt

xd

x(0) = 10 inches = 10 5

12 6ft= and

3'(0)

4x

−= (the mass has an initial velocity in the

negative or upward direction)

From 0482

2

=+ xdt

xd, 482 =ω so .48=ω

The general solution to the DE is

( ) ( ) ( ).48sin48cos tBtAtx +=

Note that

( ) ( ) ( ).48cos4848sin48 tBtAtx +−=′

69

Since 5

(0)6

x = , then

( ) ( )0sin0cos6

5BA += implies .

6

5=A

Since 3

'(0)4

x−

= ,

( ) ( ).0cos480sin486

5

4

3B+−=−

So .484

3B=−

Then .16

3

1634

3

484

3−=−=−=B

Thus, the equation of motion is

( ) ( ) ( )tttx 48sin16

348cos

6

5−= .

The period of free vibrations, .3248

22 ππωπ

===T

The frequency is .32

32

11

ππ=

==

TF

70

Example 2.25

A mass with 10 kg is attached to a spring and it stretches a spring 0.2 m. Initially, the

mass is released from a point 1 m below the equilibrium position with an upward

velocity of 5 m/s with a given ./8.9 2smg = Determine the equation of free motion,

the period of free vibrations and its frequency.

Solution

m=10kg, s=0.2m, ./980 2scmg =

F=mg gives NmgF 98)8.9(10 ===

F = ks, ( ) ⇒= 2.098 k k = 490

2

2

d xm kx

dt= − gives

⇒−= xdt

xd49010

2

2

0492

2

=+ xdt

xd

where 7492 =⇒= ωω

049equationsticCharacteri 2 =+m

iim 749areRoots ±=±=

General Solution

( ) ( ) ( )tBtAtx 7sin7cos +=

x(0)=1 and ( ) 50 −=′x (the mass has

an initial velocity in the negative or

upward direction)

1st Condition (x(0)=1m)

( ) ( ) ( )0sin0cos10 BAx +==

gives 1=A

2nd

Condition ( ( ) 50 −=′x )

( ) ( ) ( )tBtAtx 7cos77sin7' +=

( ) ( ) ( )0cos70sin175 B+=−

gives 7

5−=B

a) Equation of the motion

( ) ( ) ( )tttx 7sin7

57cos −=

b) The period of free vibrations

.7

22 πωπ

==T

c) The frequency is

.2

7

7

211

ππ=

==

TF

71

PRACTICE 2

1. State whether or not each of the following equation is linear. If linear, determine

whether the coefficient of the equation constant or variable.

a) ( ) 012

2

=++− ydx

dyx

dx

ydx b) 032

2

2

=++ ydx

dy

dx

yd

c) 44232

2

=++ ydx

dy

dx

yd d) ( )xey

dx

dy

dx

yd x 2cos372

2

=++

e) xydx

dy

dx

ydy =−+ 23

2

22 f) 02

2

2

=+ ydx

yd

g) 32552

2

=+− yxedx

dy

dx

yd h) xxy

dx

dy

dx

ydsinln4

2

2

=−+

2. Find the general solution for each differential equation.

a) 0192

2

=− ydx

yd b) 0149

2

2

=++ ydx

dy

dx

yd

c) 0722

2

=−+ ydx

dy

dx

yd d) 01892

2

2

=−+ ydx

dy

dx

yd

e) 0922

2

=−− ydx

dy

dx

yd f) 03612

2

2

=+− ydx

dy

dx

yd

g) 02540162

2

=+− ydx

dy

dx

yd h) 04129

2

2

=++ ydx

dy

dx

yd

i) 025102

2

=++ ydx

dy

dx

yd j) 042025

2

2

=+− ydx

dy

dx

yd

k) 084 =+′+′′ yyy l) 0452

2

=++ ydx

dy

dx

yd

m) 04112

2

=+ ydx

yd n) 0408

2

2

=++ ydx

dy

dx

yd

o) 0=+′+′′ yyy

3. Find the particular solution for each of the following differential equation.

a) 03762

2

=−+ ydx

dy

dx

yd; ( ) 10 =y , ( ) 20 =′y

b) 09162

2

=− ydx

yd; ( ) 10 −=y , ( ) 10 =′y

c) 0442

2

=++ ydx

dy

dx

yd; ( ) 10 =y , ( ) 20 =′y

d) 049142

2

=++ ydx

dy

dx

yd; ( ) 13 =−y , ( ) 23 =−′y

e) 0162

2

=+ ydx

yd; ( ) 60 −=y , ( ) 70 =′y

72

4. Find the general solutions of the following differential by using the method of

undetermined coefficients.

a) 1168 2 −=+′−′′ xyyy b) 1cos29 2 −=+′′ xyy

c) xyyy 3cos10209 =+′+′′ d) 1223 +=+′+′′ xyyy

e) xeyyy

7944

−=+′−′′ f) xyyy 4sin96 =+′+′′

g) xeyyy 4472 =−′−′′ h) xeyyy x 2cos2 2 +=+′+′′

i) xxyyy 2cos2 2 +=+′+′′ j) xeyyy +=+′−′′ 202

k) xxeyyy =−′+′′ 43

5. Solve the following initial value problems using the method of undetermined coefficients.

a) ( ) ( )5 50 , 0 1, 0 3y y x y y′′ ′ ′− = = = −

b) ( ) ( ) 10,20,8152 3 −=′==−′+′′ yyeyy x

c) ( ) ( )4 sin 2 , 0 2, 0 3y y x y y′′ ′+ = = − =

d) ( ) ( ) 10,00,sin23 −=′=+=+′−′′ yyxeyyy x

e) ( ) ( )65

10,00,sin52 =′==+′+′′ yyxeyyy x

6. By using the method of variation parameter, determine the general solution of the

following differential equations.

a) x

x

e

eyyy

+=+′−′′

165

2

b) xeyyy x ln2 =+′−′′

c) xyy 2sec24 =+′′ d) xyy 2cos44 =−′′

e) 21

22

x

eyyy

x

+=+′−′′ f) xyy =+′′

g) 2 3 xy y y e−′′ ′− − = h) 2444

x

eyyy =+′−′′

i) xyy sec2=+′′ j) xeyyy −=−′+′′ 62

k)3

21

4

x

y y y e′′ ′− + = l) xyy 3csc9 =+′′

7. Solve the following initial value problems.

a) xeyy x sin22 =′−′′ ; ( ) 20 −=y , ( ) 10 =′y

b) xeydx

dy

dx

yd 2

2

2

444 =+− ; ( ) 10 =y , ( ) 20 −=′y

c) xyy sin=+′′ ; ( ) 00 =y , ( )4

30 =′y

d) ( )xeyyy +=+′−′′ 122 ; ( ) 10 =y , ( ) 20 =′y

e) x

eyyy 2

5

89124 =+′−′′ ; ( ) 40 =y , ( ) 60 −=′y

8. Solve the following Euler equations.

a) 01

2

2

=−dx

dy

xdx

yd b) 01525

2

22 =++ y

dx

dyx

dx

ydx

c) 0752

22 =++ y

dx

dyx

dx

ydx d) 0853

2

22 =−+ y

dx

dyx

dx

ydx

73

e) 0452

22 =++ y

dx

dyx

dx

ydx f) 03

2

22 =+− y

dx

dyx

dx

ydx

g) 0932

22 =+− y

dx

dyx

dx

ydx

9. Solve the following initial value problems.

a) 02

2

2

=−dx

dy

xdx

yd; ( ) 11 =−y , ( ) 21 =−′y .

b) 04

32

2

=+− yxdx

dy

dx

ydx ; ( ) 11 −=y , ( ) 01 =′y .

10. Solve the following differential equations.

a) xydx

dyx

dx

ydx ln285

2

22 =+− b) xxy

dx

dyx

dx

ydx ln233 2

2

22 =+−

c) x

xy

dx

dyx

dx

ydx

+=+−

164

2

2

22 d) ( )1ln43 2

2

22 +=+− xxy

dx

dyx

dx

ydx

e) 22

22 1

87x

ydx

dyx

dx

ydx =++ f)

x

xy

dx

dyx

dx

ydx

ln43

2

2

22 =+−

11. Assuming that there is no external force acting on the spring system, the Newton’s

second law with the net or resultant force and weight, W, is given 2

2

d xm kx

dt= −

and Hooke’s law state that a restoring force F opposite to the direction of elongation

and proportional to the amount of elongation, s, simply stated as ksF = where k is the

spring constant and m is a mass attached to the spring. By defining, mgW = , find the

equation of the motion if a mass weighing 2 pounds stretches a spring 6 inches.

At t = 0, the mass is released from a point 8 inches below the equilibrium position

with an upward velocity of 3/4 ft/sec given ./32 2sftg = Then, determine the period

of free vibrations and its frequency.

12. A mass weighing 10 N stretches a spring 2 cm. At t = 0, the mass is released from

a point 10 cm below the equilibrium position with an upward velocity of 7 cm/s with

a given ./980 2scmg = Find the equation of free motion. Next, determine the period

of free vibrations and its frequency.

13. A mass weighing 14 N stretches a spring 0.2 m. Initially, the mass is released

from a point 1 m below the equilibrium position with an upward velocity of 2/3 m/s

with a given ./8.9 2smg = Determine the equation of free motion, the period of free

vibrations and its frequency. Assume that there is no external force acting on the

spring system. The Newton’s second law with the net or resultant force and weight,

W, is given 2

2

d xm kx

dt= −

where k is the spring constant and m is a mass attached to the spring.

74

14. If a mass weighing 20 N stretches a spring 0.2 m, find the equation of the motion.

At t = 0, the mass is released from a point 0.8 m below the equilibrium position with

an upward velocity of 56 cm/s given ./980 2scmg = Then, determine the period of

free vibrations and its frequency. According to the Newton’s second law with the net

or resultant force and weight, W and define mgW = , is given 2

2

d xm kx

dt= −


15. If a mass with 1 kg stretches a spring 49 cm, find the equation of the motion. At

t = 0, the mass is released from a point 5cm below the equilibrium position with an

upward velocity of 10 cm/s given ./980 2scmg = Then, determine the period of free

vibrations and its frequency. According to the Newton’s second law with the net or

resultant force and weight, W and define mgW = , is given 2

2

d xm kx

dt= −


Answer

1. a) Linear homogeneous, variable coefficients

b) Linear homogeneous, constant coefficients

c) Linear non homogeneous, constant coefficients

d) Linear non homogeneous, variable coefficients

e) Nonlinear equation, 2

22

dx

ydy f) Nonlinear equation, 2y

g) Nonlinear equation, ye h) Nonlinear equation, yln

2. a) xx BeAey 1919 += − b) xx BeAey 27 −− +=

c) xx BeAey 89 += − d) x

x BeAey 2

3

6 += −

e) ( ) ( )xx BeAey 101101 +− += f) ( ) xeBxAy 6+=

g) ( )x

eBxAy 4

5

+= h) ( )x

eBxAy 3

2−

+=

i) ( ) xeBxAy 5−+= j) ( )x

eBxAy 5

2

+=

k) ( )xBxAey x 2sin2cos2 += l)

+=−

xBxAeyx

5

1sin

5

1cos5

2

m) xBxAy11

2sin

11

2cos +=

n) ( )xBxAey x 62sin62cos4 += −

o)

+=

−xBxAey

x

2

3sin

2

3cos2

1

75

3. a) xx

eey 2

3

3

1

11

10

11

21−

−= b) xx

eey 4

3

4

3

6

1

6

7+−=

−

c) x

exy 2

1

2

51

−

+= d) ( ) 217928 −−+= xexy

e) xxy 4sin4

74cos6 +−=

4. a) ( )256

10

16

1

16

1 24 −+++= xxeBxAy x

b) xxBxAy 2cos5

13sin3cos ++=

c) xxBeAey xx 3sin85

273cos

85

1145 +++= −−

d) 12 −++= −− xBeAey xx

e) ( ) xx eeBxAy 72

9

1 −++=

f) ( ) xxeBxAy x 4sin625

74cos

625

243 −−+= −

g) xxx

xeBeAey 442

1

9

1++=

−

h) ( ) xxeeBxAy xx 2cos25

32sin

25

4

9

1 2 −+++= −

i) ( ) xxxxeBxAy x 2cos25

32sin

25

4642 −++−++= −

j) ( ) xx exeBxAy 2

2

120 +++= k) xxxx xeexBeAey

25

1

10

1 22 −++= −

5. a) ( ) xxey x 2565

1 25 −−−=

b) xxxxeeey

335 ++= −

c) xxxxy 2cos4

12sin

8

132cos2 ++−=

d) xxxeeey xxx sin10

1cos

10

3

5

1

2

1 2 ++−+−=

e) ( ) ( ) xx exxexxy sin7cos465

12sin2cos4

65

1+−++= −

6. a) ( ) ( ) ( )xxxxx eeeBeeAy −++++−= 1ln11 232 ; ( ) Aeu x ++= −1ln ,

( ) Beev xx +++−= −− 1ln

b) xxxx exxex

BxeAey 22

4

3ln

2−++= ; A

xx

xu ++−=

4ln

2

22

,

Bxxxv +−= ln

c) cos 2 sin 2 cos 2 ln cos 2 sin 2y A x B x x x x x= + + + ; Axu ++= 2cosln ,

Bxv += 2

d) xBeAey xx 2cos2

122 −+= − ; Axe

xe

uxx

+−−= 2sin4

2cos4

22

,

76

Bxe

xe

vxx

++−=−−

2sin4

2cos4

22

e) ( ) xxexeBxeAey xxxx 12 tan21ln −++−+= ; ( ) Axu ++−= 21ln ,

Bxv += −1tan2

f) xxBxAy ++= sincos ; Axxxu +−= sincos , Bxxxv ++= cossin

g) xxxx exeBeAey −−− −−+=3

12 ; Axu +−= , Bev x +−= −3

3

1

h) 22 2 21

2

x x x

y Ae Bxe x e= + − ; 22u x A= − + , Bxv +=

i) xxxxxBxAy sin2coslncos2sincos +++= ; Axu += cosln2 ,

Bxv += 2

j) xxx eBeAey −− −+= 32 ; Aeu x +−= 2 , Bev x +−= −2

k) x

xx

eBxeAey 2

3

22 ++= ; Aexeu xx ++−= , Bev x +=

l) xxBxxAy 3sin3sinln9

13cos

3

1

++

−= ; Axu +−=3

1,

Bxv += 3sinln9

1

7. a) xeey xx sin3 2 −+−= ; Axexe

uxx

+−=2

sin

2

cos,

Bxexe

vxx

+−−=−−

2

sin

2

cos

b) xxx exxeey 2222 24 +−= ; Axu +−= 22 , Bxv += 4

c) 1 1

sin sin 2 cos sin cos 2 cos4 4 2

xy x x x x x x= + − − ;

Axxu +−=2

12sin

4

1,

1cos 2

4v x B= − +

d) 23 2x x xy e xe x e= − + + + ; 22 2x xu xe e x A− −= + − + , 22 2xv e x B−= − + +

e) xxx

exeey 2

5

2

3

2

3

2142 +−= ; Aexeu xx ++−= 22 , Bev x += 2

8. a) ( ) 2BxAxy +=

b) ( ) ( ) 5

1

ln xxBAxy +=

c) ( ) ( )xBxAxxy ln3sinln3cos2 += −

d) ( ) 3

4

2 BxAxxy += −

e) ( ) ( ) 2ln −+= xxBAxy

f) ( ) ( )xBxAxxy ln2sinln2cos +=

g) ( ) ( )xBxAxxy ln5sinln5cos2 +=

9. a) ( ) 35 1

4 4y x x= + b) ( ) ( ) 2ln21 xxxy +−=

77

10. a) ( )16

3

4

ln42 +++=x

BxAxxy

b) ( ) xxBxAxxy ln2 23 −+=

c) ( ) ( ) ( ) ( )1232 1ln11 −++++−= xxxBxxAxy

d) ( ) ( ) ( )3 2 21ln ln 3 ln

6y x A B x x x x

= + + +

e) ( ) xx

xBAxxy ln2

1

4

12

24 +

−+= −−

f) ( ) ( ) ( )xxxxxBAxy lnlnlnln22 ++=

11. ( ) ( ) ( )tttx 8sin32

38cos

3

2−= ,

48

22 ππωπ

===T , .41

π==

TF

12. ( ) ( ) ( )tttx 107sin10

1107cos10 −= ,

107

22 πωπ

==T .2

1071

π==

TF

13. ( ) ( ) ( )tttx 7sin21

27cos −= ,

7

22 πωπ

==T , .2

71

π==

TF

14. ( ) ( ) ( )tttx 7sin87cos80 −= , 7

22 πωπ

==T , .2

71

π==

TF

15. ( ) ( ) ( )tttx 52sin552cos5 −= , 552

22 ππωπ

===T , .51

π==

TF

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