Chap 3 Differential Equations

7/23/2019 Chap 3 Differential Equations

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WLB 10102 / MATHEMATICS 2 Chapter 3 Differential Equations

Page 1 of 29

3.1 Definition of A Differential Equation

A differential equation is an equation involving derivatives or differentials. The following are some

examples of differential equations.

Example 1:

( ) ( )32'23'' y x y =+ where

dx

dy y =' ,

2

2

''dx

yd y =

Example 2:

2

y x

y

dx

dy

=+

Example 3:

t Qdt

dQ

dt

Qd 2sin423

2

2

=+−

Example 4:

y x

y x

dx

dy

−

+= or ( ) ( ) 0=−++ dy x ydx y x

Example 5:

02

2

2

2

=∂

∂+

∂

y

V

dx

V

3.2 First Order Differential Equations (solution of differential equation)

The order of a differential equation is given by the highest derivative involved in the equation.

equation.aldifferentiorder- thirdcalledis 04x

edx

dyy

3dx

y3

d

equation.aldifferentiorder-second calledis 0sinx2

y2

dx

y2

d xy

equation.aldifferentiorder-first calledis 0 2

=+−

=−

=− ydx

dy x

To solve a differential equation, we have to find the function for which the equation is true. This means

that we have to manipulate the equation so as eliminate all the differential coefficients and leave a

relationship between y and x. The rest of this part is devoted to two methods of solving first order

differential equations.

CHAPTER 3 DIFFERENTIAL EQUATIONS

Ordinary Differential Equations : involving

only one independent

variable

Partial Differential Equations: involvin two or more inde endent




Page 2 of 29

3.2.1 Solution of First Order Differential Equations by Separation of Variables

( ) x f = term of x

( ) yg = term of y

To solve this type of equation, all the functions x are gathered with dx and all the functions y are

gathered with dy. Then, solve the equation by direct integration.

( ) ( ) yg x f dx

dy=

( ) ( )∫∫ = dx x f dy

yg

1

( ) ( ) 21 c xF c yF +=+

( ) ( )

tsconstanare1c,2cc,

and1

c2

cc where −=+=∴ c xF yF

Therefore ( ) ( ) c xF yF += is the general solution for this differential equation.

Caution:

The general solution, ( ) ( ) c xF yF += must only has an arbitrary constant, c even though initially

there are two arbitrary constants, 21,cc .

I. The solution of equations of the form )( x f dx

dy=

A diferential equation of the form )( x f dx

dy= is solved by direct integration, i.e.

∫= dx x f y )(

General Form:

( ) ( ) yg x f dx

dy=

Other Form:

( )

( )

( ) ( )

( ) ( ) 0

0

=+

=+

=

=

ygdx

dy x f

dx

dy yg x f

ygdx

dy

x f dx

dy




Page 3 of 29

Example 7:

Determine the general solution of342 x

dx

dy x −=

Solution:

Rearranging342 x

dx

dy x −= gives:

233

424242

x x x

x

x x

x

dx

dy−=−=

−=

Integrating both sides gives:

∫

−= dx x

x y

242

i.e. ,3

4ln2 3

c x x y +−= which is the general solution.

Example 8:

Find the particular solution of the differential equation ,325 =+ xdx

dy

given the boundary conditions5

21= y .2when = x

Solution:

Since 325 =+ xdx

dy then

5

2x-

5

3

5

23=

−=

x

dx

dy

Hence5

2

5

3∫

−= dx

x y

i.e. ,

55

3 2

c x x

y +−= which is the general solution.

Substituting the boundary conditions 2and 5

21 == x y to evaluate c gives:

c,5

4-

5

6

5

21 += from which, c = 1.

Hence the particular solution is .155

3 2

+−= x x

y




Page 4 of 29

Example 9:

Solve the equation .1 when2θgiven,5θ

2 ===

− t

dt

d t t

Solution:

Rearranging gives:

2

5θ and

2

5θ

t t

dt

d

t dt

d t −==−

Integrating gives:

∫

−= dt

t t

2

5θ

i.e. ,ln2

5

2θ

2

ct t

+−= which is the general solution.

When ct +−=== 1ln2

5

2

12 thus,1 ,2θ from which, .

2

3=c

Hence the particular solution is ( ).3ln52

1θ 2

+−= t t

Example 10:

The bending moment M of the beam is given by xw xlwdx

dM and where),( −−= are constants.

Determine M in terms of x given:2

2

1wl M = when .0= x

Solution:

wxwl xlw

dx

dM +−=−−= )(

Integrating with respect to x gives:

,2

2

cwx

wlx M ++−= which is the general solution.

When .0 ,2

1 2== xwl M

Thus cw

wlwl ++−=2

)0()0(

2

1 22

from which, .2

1 2wlc =

Hence the particular solution is:




Page 5 of 29

( )

( )2

22

22

21or

22

1 i.e.

2

1

2

xlw M

xlxlw M

wlwx

wlx M

−=

+−=

++−=

Exercise 1 Further problems on equations of the form )( x f dx

dy=

1) In Problems (a) to (b), solve the differential equations.

a) x xdx

dy24cos −=

b) 332 xdxdy x −=

c) .1 when2given,3 ===+ x y xdx

dy

d) .3

πθ when

3

2given,0θsin

θ3 ===+ y

d

dy

e) .0 when1given,321

==−=+ x ydx

dy x

e x

2) The gradient of a curve is given by:

x x

dx

dy3

2

2

=+

Find the equation of the curve if it passes through the point .3

1 ,1

II. The solution of equations of the form )( y f dx

dy=

A differential equation of the form )( y f dx

dy= is initially rearranged to give

)( y f

dydx = and then the solution is obtained by direct integration, i.e.

∫ ∫=)( y f

dydx




Page 6 of 29

Example 11: Find the general solution of .23 ydx

dy+=

Solution:

Rearranging ydx

dy23+= gives:

y

dydx

23+=


∫ ∫ +

= y

dydx

23

Thus, by using the substitution ),23( yu += we have

c y x ++= )23ln(2

1

Example 12: Determine the particular solution of ydx

dy y 3)1( 2

=− given that .6

12 when1 == x y

Solution:

Rearranging gives:

dy y

ydy

y

ydx

−=

−=

3

1

33

12

Integrating gives: ∫ ∫

−= dy

y

ydx

3

1

3

i.e. ,ln3

1

6

2

c y y

x +−= which is the general solution.

When .2which,from ,1ln31

61

612 thus,

612 ,1 =+−=== cc x y


2ln3

1

6

2

+−= y y

x

Example 13:

At time t minutes, the rate of change of temperature of a cooling body is proportional to the temperature

C T o of that body at that time. Initially, C T o 72= . Show that kt eT −= 72 where k is a constant.




Page 7 of 29

Given also that C T o 32= when 10=t , find how much longer it will take the body to cool to C

o 72

under these conditions.

Solution:

We can form the differential equation

kT dt

dT −=

Separating the variables and solving give

dt k T

dT t T

⋅−= ∫∫ 072

∴ [ ] [ ]

t T

kt T 072ln −=

∴ kt T −=− 72lnln

∴ kt T

−=

72ln

∴ kt

eT −

=72

∴ kt

eT −

= 72

as required.

When C T t o32,10 == . Substituting intokt

eT −

= 72 gives the value of the constant .k

k e

107232 −

=

∴ 9

4

72

3210==

− k e

∴

=−

9

4ln10k

∴

−=

9

4ln

10

1k

Now if t is the time it takes the body to cool to C o27 , we have

kt e−

= 7227

∴ 8

3

72

27==

−kt e

Hence




Page 8 of 29

=−

8

3lnkt

∴

=

8

3ln9

4ln10

1t

∴ 1.12

9

4ln

8

3ln

10 =

=t

So it will take the body 12.1 – 10 = 2.1 minutes longer to cool to C o27 .

Exercise 2 Further problems on equations of the form )( y f dxdy =

1) In Problems (a) to (c), solve the differential equations.

a) ydx

dy32 +=

b) ydx

dy 2cos2=

c) ( ) .

2

1 when1given,522

===+ x y y

dx

dy y

2) The velocity of a chemical reaction is given by

,in timensferedamount tratheis where),( t x xak dt

dx−=

andconstantais ak ionconcentrattheis at time .0 when0 == xt Solve the equation and

determine .of in terms t x

III. The solution of equations of the form )()( y f x f dx

dy⋅=

A differential equation of the form )()( y f x f dx

dy⋅= where )( x f is a function of x

only and )( y f is a function of y only, may be rearranged as

,)()(

dx x f y f

dy= and then the solution is obtained by direct integration, i.e.

∫∫ = dx x f y f

dy)(

)(




Page 9 of 29

Example 13:

Solve the equation 14 2−= y

dx

dy xy

Solution:

Seperating the variable gives:

dx x

dy y

y 1

1

42

=

−


∫ ∫

=

−dx

x

dy

y

y 1

1

42

Using the substitution ,12−= yu the general solution is:

( ) c x y +=− ln1ln2 2

Example 14: Determine the particular solution of ,2θ θ23 −=

t e

dt

d given that .0θ when0 ==t

Solution:

),)((22θ θ23θ23 −−

== eeedt

d t t

by the laws of indices.

Seperating the variable gives:

,2θ 3

θ2 dt e

e

d t =

−

i.e. dt ed e t 3θ22θ =


∫ ∫= dt ed e t 3θ2 2θ

Thus the general solution is:

cee t +=

3θ2

3

2

2

1

When : thus,0θ ,0 ==t

cee +=00

3

2

2

1, from which, .

6

1

3

2

2

1−=−=c




Page 10 of 29


6

1

3

2

2

1 3θ2−=

t ee or 143

3θ2−=

t ee .

Example 15:

Find the curve which satisfies the equation ( )dx

dy x xy

21+= and passes through the point (0, 1).

Solution:

Seperating the variables gives:

( ) y

dydx

x

x=

+21


( ) c y x +=+ ln1ln2

1 2

When x = 0, y = 1 thus ,1ln1ln2

1c+= from which, .0=c

Hence the particular solution is ( ) .ln1ln2

1 2 y x =+

i.e. ( ) ,ln1ln 2

1

2 y x =+ from which, ( ) .1 2

1

2 y x =+

Hence the equation of the curve is ( )21 x y += .

Example 16:

For an adiabatic expansion of a gas ,0=+V

dV C

P

dPC pv where pC and vC are constants. Given

,v

p

C C n = show that constants.=

n pV

Solution:

Seperating the variables gives:

V

dV C

P

dPC pv −=




Page 11 of 29


∫ ∫−==V

dV C

P

dPC pv

i.e. k V C pC pv +−= lnln .

Dividing throughout by constant vC gives:

vv

p

C

k V

C

C p +−= lnln

Since . where,lnln then,vv

p

C

k K K V n pn

C

C ==+=

i.e. ,lnorlnln K pV K V p nn==+ by the laws of logarithms.

Hence constant.i.e. ,eK

==

nn

pV pV

Exercise 3 Further problems on equations of the form )()( y f x f dx

dy⋅=

1) In Problems (a) to (d), solve the differential equations.

a) x ydx

dycos2=

b) )13()12( 2+=− x

dx

dy y , given 1= x when .2= y

c) y xe

dx

dy −=

2, given 0= x when .0= y

d) 0)1()1(2 =++−dx

dy y x x y , given 1= x when .1= y

3.2.2

The Linear First Order Differential Equations

( ) ( ) xQ xP , are the function of x

1st step:

To solve this type of equation, Firstly we have to find the integral factor, ( ) xV .

General Form:

( ) ( ) xQ y xPdx

dy=+




Page 12 of 29

( ) ( )∫

= dx xP

e xV

For example: a) If ( ) xee xV x

xP xdx x ==

∫==

ln1

,1

)(

( ) x xV =∴

2nd step:

Multiply both side of differential equation with integral factor, ( ) xV . Then integrate the equation.

( ) ( ) ( ) ( ) xQ xV y xPdx

dy xV =

+

( ) ( ) ( ) ( ) ( ) xQ xV y xP xV dx

dy xV =+

⇒ ( ) ( )[ ] ( ) ( ) xQ xV y xV dx

d

dx

dy xV =+

⇒ ( )[ ] ( ) ( ) xQ xV y xV dx

d =

( )[ ] ( ) ( )∫∫ ⋅=⋅ dxxQxVdxyxVdx

d Integrate

Example 17:

Solve2

3

2

1

3 x y

x

x

dx

dy=

++ .Express y in terms of x .

Solution:

Given,2

3

2

1

3 x y

x

x

dx

dy=

++ . (linear differential equation)

Find the integral factor:

xee

xee

xe

x x

x x

x

11

2

lnln

2lnln2

ln

==

==

=

−−

( ) ( ) ( ) solution.generalaisdxxQxVyxV ∫ ⋅=∴




Page 13 of 29

( ) ( )

( ) ∫=∴

=

+=

⋅

+

dx x

x

e xV

x xQ x

x xP

13

23

2

3

2

,1

3

( )

( )13

1ln 3

+=

= +

x

e x

To find the general solution, use the formula

( ) ( ) ( ) dx xQ xV y xV ∫ ⋅=

( ) ( ) dx x x y x .11 233

∫ +=+∴

( )dx x x .132

∫ +=

( ) C x ++= 13

1 3

( )( ) ( )11

1

3

133

3

++

+

+=

x

C

x

x y

∴ ( )13

13+

+= x

C y

Example 18:

Solve23 x y

dx

dy x =+ if given

5

2= y when 1−= x .

Solution:

equationaldifferentilinear.................. 3

3 2

x y xdx

dy

x y

dx

dy x

=+

=+

Find the integral factor:

( ) ( )

( ) ∫=∴

==

⋅dx xe xV

x xQ x

xP

3

,3




Page 14 of 29

3

ln

ln3

3

x

e

e

x

dx x

=

=

∫= ⋅


( ) ( ) ( ) dx xQ xV y xV ∫ ⋅=

xdx x y x ⋅=∴ ∫33

dx x∫=4

c x

+= 5

5

5

2,1 =−= y x when

( ) ( )

c+−

=−∴5

51

5

231

5

1−=∴ c

5

1

5

53

−=∴ x

y x

Example 19:

Solve the following differential equation

1

12

−=+

x y

xdx

dy

Solution

1

12

−=+

x y

xdx

dy …… linear differential equation

Find the integral factor

( ) ( )

( ) ∫=∴

−==

⋅dx xe xV

x xQ

x xP

2

1

1,

2




Page 15 of 29

xe ln2

=

2ln x

e=

2 x=


( ) ( ) ( ) dx xQ xV y xV ∫ ⋅=

∴ dx x

x y x ∫

−=

1

22

dx x-

x∫

++=

1

11

∴ ( ) c x x x y x +−++= 1ln2

22

Example 20:

a) Find the general solution of the equation

1)1(

)1(3)2( =

+

−+− y

x

x

dx

dy x

b) Given the boundary conditions that 5= y when 1−= x , find the particular solution of the equation

given in (a).

Solution

a)

i) Rearranging gives:)2(

1

)2)(1(

)1(3

−=

−+

−+

x y

x x

x

dx

dy which is of the form ,QPy

dx

dy=+ where

.)2(

1

and )2)(1(

)1(3

−=

−+

−=

xQ x x

x

P

ii) ∫ ∫ −+

−= ,

)2)(1(

)1(3dx

x x

xPdx which is integrated using partial fractions.

Let

)2)(1(

)1()2(

)2()1()2)(1(

)1(3

−+

−+−

≡

−+

+≡

−+

−

x x

x B x A

x

B

x

A

x x

x




Page 16 of 29

from which, ).1()2(33 ++−=− x B x A x

When ,1−= x

,36 A−=− from which, .2= A

When ,2= x

,33 b= from which, .1=b

Hence

[ ])2()1(ln

)2ln()1ln(2

2

1

1

2

)2)(1(

)1(3

2−+=

−++=

−+

+=

−+

−

∫∫

x x

x x

dx x x

dx x x

x

iii) Integrating factor

[ ] )2()1()( 2)2()1(ln 2

−+==∫

= −+

x xee xV x xPdx

iv) Substituting in equation ( ) ( ) ( ) dx xQ xV y xV ∫ ⋅= gives:

dx xdx x

x x y x x ∫∫ +=−

−+=−+222 )1(

)2(

1)2()1()2()1(

v) Hence the general solution is .)1(31)2()1( 32

c x y x x ++=−+

b) When .0which,from ,0)3)(0(5 thus5 ,1 =+=−=−= cc y x

Hence32 )1(

3

1)2()1( +=−+ x y x x

i.e)2()1(3

)1(2

3

−+

+=

x x

x y

and hence the particular solution is

)2(3

)1(

−

+=

x

x y

Example 21 ( Mixing Problem)

Consider a large tank holding 1000L of water into which a brine solution of salt begins to flow at a

constant rate of 6L/min. The solution inside the tank is kept well stirred and is flowing out of the tank at a

rate of 6L/min. If the concentration of salt in the brine entering the tank is 1kg/L, determine when theconcentration of salt in the tank will reach 0.5kg/L.




Page 17 of 29

Solution :

Step 1: Set up the mathematical model.

Let x(t) be the quantity of salt in the tank at time t.

The rate of change of salt in the tank is .rateoutput rateinput dt

dx−=

Input rate = quantity of the salt × volume of the water

= (1 kg/l)× (6 l/min) = 6 kg/min.

Output rate =

1000

xkg/l × (6 l/min) =

1000

6 xkg/min.

Then, the rate of change is

500

36

1000

66

x x

dt

dx−=−= .

Step 2 : Solve the differential equation.

Initial value, when t = 0, x = 0.

( )500

10003

500

33000 x x

dt

dx −=

−=

∫∫ =−

dt dx x 500

3

1000

1

( )

( ) ct x

ct x

+−=−

+=−−

500

31000ln

500

31000ln

( )

( )

( )

−=

=−

=

⋅=−

==

⋅==−

=

−

−

−+−

+−−

t

t

c

c

t c

ct

ct x

e x

e x

e

ee

t x

eee x

ee

500

3

500

3

0

500

3

500

3

500

3

1000ln

11000

10001000

.1000

01000

:0,0

)1000(

Step 3 : Solve the particular problem in the question.

Find the time t when the concentration of the salt reach x = 0.5kg/L.

Equation 1




Page 18 of 29

.min52.1153

5006931.0

6931.0500

3

)5.0ln(ln

5.05.01

11000

500

500

3

500

3

500

3

=

−−=

−=−

=

=−=

−=

−

−

−

t

t

e

e

e

t

t

t

Example 22 : (Radioactive decay)

Radioactive material decays at a rate proportional to the amount present For a certain radioactivesubstance, approximately 10% of the original quantity decomposes in 25 years. Find the half-life of this

radioactive material, that is find the time that elapses for the quantity of material to decay to one-half of

its original quantity.

Solution:

Step 1: Set up the mathematical model.

Let Q0 represent the original quantity present and Q is the amount present at any time t (in years). The

observed rate of decay process is

dt

dQ∝ Q

kQdt

dQ= .

Step 2 : Solve the differential equation.

kdt dQQ

kQdt

dQ

=

=

1

Then, we have

C kt Q

kdt dQQ

+=

=∫ ∫ln

1

When t = 0, Q = Q0,

0

0

ln

)0(ln

QC

C k Q

=

+=




Page 19 of 29

Then, the equation is,

.

ln

lnln

lnln

0

0

0

0

0

kt

kt

eQQ

eQ

Q

kt Q

Q

kt QQ

Qkt Q

=

=

=

=−

+=

Step 3 : Solve the particular problem in the question.

First, we have to find the value of the constant k. Given that after 25 years, 10% of the material has

decayed.

That is, for t = 25, 00 9.0)10.01( QQQ =−= .

Put inside Equation 1,

.00421.025

9.0ln

9.0ln25

ln9.0ln

9.0

9.0

25

25

)25(

00

−==

=

=

=

=

k

k

e

e

eQQ

k

k

k

Now to determine the half-life of the radioactive represents which is represent the time for the material to

decay to one-half of its original quantity, that is .2

0QQ =

Then,2

0QQ = put inside Equation 2,

.1652ln

00421.0

2ln

2

1

20

0

yearst

kt

e

eQQ

kt

kt

=−

−=

=−

=

=

Equation 2




Page 20 of 29

Example 23:

The current I in a series circuit with constant inductance L, constant resistance R and a constant voltage

V applied is described by the differential equation

V RI dt

dI

L =+

.

Find the equation for the current I as a function of time t .

Answer :

−=

− t

L

R

e R

V I 1

Exercise 4 Further problems on linear first order differential equations

1) Solve the following differential equations.

a) ydx

dy x −= 3

b) )21( y xdx

dy−=

c) yt dt

dyt −=− 5

d) 11

=+ ydx

dy

x

e) y xdx

dy2=+

2) y xdx

dy x 21 3

−=

+ , given 1= x when .3= y

3) Solve 241

=+ ydx

dy

x given the boundary conditions 0= x when .4= y

4) Show that the solution of the equation x

y

dx

dy

−=+1 is given by ,2

3 2

x

x y

−

= given 1= x when

.1= y

5) Determine the particular solution of 0=+− y xdx

dy, given that 0= x when .2= y

6) Given the equation y xdx

dy x −

+=

2

2 show that the particular solution is ),2ln(

2+= x

x y given the

boundary conditions that 1−= x when .0= y




Page 21 of 29

3.3 Second Order Differential Equations

3.3.1 Second Order Differential Equations of the form 02

2

=++ cydx

dyb

dx

yd a

(homogeneous differential equations)

Procedure to solve differential equations of the form 02

2

=++ cydx

dyb

dx

yd a

a) Rewrite the differential equation

02

2

=++ cydx

dyb

dx

yd a

as 0)( 2=++ ycbDaD

b) Substitute m for D and solve the auxiliary equation .for02

mcbmam =++

c) If the roots of the auxiliary equation are:

i) real and different, say α=m and β=m , then the general solution is

x x Be Ae y βα+=

ii) real and equal, say α=m twice, then the general solution is

xe B Ax y α)( +=

iii) complex, say ,βα jm ±= then the general solution is

( ) x B x Ae y x βsinβcosα+=

d) Given boundary conditions, constants A and B, may be determined and the particular solution of

the differential equation obtained.

Example 24:

Determine the general solution of .03522

2

=−+ ydx

dy

dx

yd Find also the particular solution given that

when and 4 ,0 == y x .9=dx

dy

Solution:

a) 0352 2

2

=−+ ydx

dy

dx

yd in D- operator form is ,0)352(

2

=−+ y D D where .dx

d D ≡




Page 22 of 29

b) Substituting m for D gives the auxiliary equation

.0352 2=−+ mm

Factorizing gives: ,0)3)(12( =+− mm from which, .3or2

1−== mm

c) Since the roots are real and different the general equation is .32

1

x x

Be Ae y −+=

d) When x = 0, y = 4,

Hence B A +=4 ……………….. equation (1)

Since x

x Be Ae y 32

1

−+= then

x x

Be Aedx

dy 32

1

32

1 −−=

When 9 ,0 ==dx

dy

x thus B A 32

1

9 −=

……………….. equation (2)

Solving the simultaneous equations (1) and (2) gives A = 6 and B = -2.

Hence the particular solution is x

xee y

32

1

26 −−= ■

Example 25:

Find the general solution of 0162492

2

=+− ydt

dy

dt

yd and also the particular solution given the

boundary conditions that when .3 ,0 ===dt

dy yt

Solution:

a) 0162492

2

=+− ydt

dy

dt

yd in D- operator form is ,0)16249( 2

=+− y D D where .dt

d D ≡


.016249 2=+− mm

Factorizing gives: ,0)43)(43( =−− mm i.e.3

4=m twice.

c) Since the roots are real and equal, the general equation is .)( 3

4t

e B At y +=

d) When t = 0, y = 3, then

,)0(3 0e B+= i.e. B = 3.




Page 23 of 29

Sincet

e B At y 3

4

)( += then ,3

4)( 3

4

3

4t t

Aee B At dt

dy+

+= by the product rule.

When 3 ,0 == dt

dy

t thus00

3

4

)0(3 Aee B ++=

i.e. A B +=3

43 from which, A = -1, since B = 3.

Hence the particular solution ist

et y 3

4

)3( +−= ort

et y 3

4

)3( −= ■

Example 26 :

Solve the differential equation 01362

2

=++ ydxdy

dx yd , given that when .7and ,3y ,0 ===

dxdy x

Solution:

a) 01362

2

=++ ydx

dy

dx

yd in D-operator form is ,0)136( 2

=++ y D D where .dx

d D ≡


.01362

=++ mm

Using the quadratic formula:

)1(2

)13)(1(466 2−±−

=m

ii

232

46

2

)16(6±−=

±−=

−±−=

c) Since the roots are complex, the general equation is ).2sin2cos(3 x B x Ae y x+=

−

d) When x = 0, y = 3, hence ),0sin0cos(3 0 B Ae += i.e. A = 3.

Since )2sin2cos(3 x B x Ae y x+=

− then

( ) ( ),2sin2cos32cos22sin233

x B x Ae x B x Aedt

dy x x+−+−=

−−

by the product rule,

( ) ( )[ ] x B A x A Bedt

dy x 2sin322cos323+−−=

−

When ,7 ,0 ==

dx

dy x hence ( ) ( )[ ]0sin320cos327 0

B A A Be +−−=




Page 24 of 29

i.e. ,327 A B −= from which, B = 8, since A = 3.

Hence the particular solution is )2sin82cos3(3 x xe y x

+= −

■

Exercise 5 Further problems on differential equations of the form 02

2

=++ cydx

dybdx

yd a

1) In Problems (a) to (b), determine the general solution of the given differential equations.

a) 0262

2

=−− ydt

dy

dt

yd

+=

− t t

Be Ae y 2

1

3

2

b) 0θθ4θ42

2

=++dt d

dt d

+=

− t

e B At 2

1

)(θ

c) 0522

2

=++ ydx

dy

dx

yd [ )2sin2cos(3

x B x Ae y x

+= −

]

2) In problems (a) to (d), find the particular solution of the given differential equations for the stated

boundary conditions.

a) ;06562

2=−+ y

dx

dy

dx

yd when and 5 ,0 == y x .1−=

dx

dy

+=

− x x

ee y 23

32

23

b) ;0542

2

=+− ydt

dy

dt

yd when and 1 ,0 == yt .2−=

dt

dy

−=

t t

ee y 34 4

1

c) ,0)25309( 2=++ y D D when and 0 ,0 == y x .2=

dx

dy

=

− x

xe y 3

5

2

d) ;0962

2

=+− xdt

dx

dt

xd when and 2 x,0 ==t .0=

dt

dx

t et x3)31(2 −=

e) ;01362

2

=++ ydx

dy

dx

yd when and 4 ,0 == y x .0=

dx

dy )2sin32cos2(2 3

x xe y x

+= −

f) ,0θ)125204( 2 =++ D D when and 3θ ,0 ==t .5.2θ =dt d )5sin25cos3(θ 5.2 t t e t += −




Page 25 of 29

3.3.2 Second Order Differential Equations of the form )(2

2

xQcydx

dyb

dx

yd a =++

(non-homogeneous differential equations)

Procedure to solve differential equations of the form )(2

2

xQcydx

dyb

dx

yd a =++

a) Rewrite the differential equation

)(2

2

xQcydx

dyb

dx

yd a =++

as )()( 2 xQ ycbDaD =++

b) Substitute m for D and solve the auxiliary equation .for02

mcbmam =++

c) Obtain the complementary solution, c y , which is achieved using the same procedure as in

Section 3.3.1(c) , page 19.

d) To determine the particular solution, p y , firstly assume a particular integral which is suggested

by )( xQ , but which contains undetermined coefficients. Table 3.1 provides some guidance on

choosing the general form of the particular solution to a nonhomogeneous differential equation.

e) Substitute the suggested P.I. into the differential equation )()( 2 xQ ycbDaD =++ and equate

relevant coefficients to find the constants introduced.

f) The general solution is given by pc y y y +=

g) Given boundary conditions, arbitrary constants in the complementary solution, may be

determined and the particular solution of the differential equation obtained.




Page 26 of 29

Example 27 :

Solve the differential equation .422

2

=−+ ydx

dy

dx

yd

Solution:

a) 422

2

=−+ ydx

dy

dx

yd in D-operator form is ,4)2( 2

=−+ y D D where .dx

d D ≡


.022

=−+ mm

Factorizing gives:

0)2)(1( =+− mm , from which .2or1 −== mm

c) Since the roots are real and different, the complementary solution, .2 x x

c

Be Ae y −+=

d) Since the term on the right hand side of the given equation is a constant, i.e. ,4)( = xQ let the

particular solution also be a constant, say .k y p =

e) Substituting k y p = into ,4)2( 2=−+ p y D D gives .4)2( 2

=−+ k D D Since

0)( and 0)( 2== k Dk D then ,42 =− k from which .2−=k Hence the the particular solution

.2−= p y

f) The general solution is given by pc

y y y += i.e. 22−+=

− x x Be Ae y ■

Example 28:

Determine the particular solution of the equation ,932

2

=−dx

dy

dx

yd given the boundary conditions that

when .0 and 0 ,0 ===dx

dy y x

Solution:

a) 932

2

=−dx

dy

dx

yd in D-operator form is .9)3( 2

=− y D D where .dx

d D ≡


.032=− mm

Factorizing gives:

0)3( =−mm , from which .3or0 == mm

c) Since the roots are real and different, the complementary solution, .3 x

c Be A y +=




Page 27 of 29

d) Since the complementary solution contains a constant (i.e. A ) then let the particular solution,

.kx y p =

e) Substituting kx y p = into ,9)3( 2=− p y D D gives .9)3( 2

=− kx D D Since

0)( and )( 2 == kx Dk kx D then ,930 =− k from which .3−=k Hence the particular solution

.3 x y p −=

f) The general solution is given by pc y y y += i.e. .33 x Be A y x−+= ■

g) When ,0 ,0 == y x thus ,00 0−+= Be A

i.e. B A +=0 ……………….. equation (1)

;33 3−=

x Be

dx

dy ,0 when0 == x

dx

dy thus 330

0−= Be from which, .1= B From equation

(1), .1−= A

Hence the particular solution is

,311 3 xe y x

−+−=

i.e. 133−−= xe y x

■

Example 29:

Solve the differential equation .2312112 2

2

−=+− x ydx

dy

dx

yd

Solution:

a) 23121122

2

−=+− x ydx

dy

dx

yd in D-operator form is .23)12112( 2

−=+− x y D D where

.dx

d D ≡


.0121122

=+− mm

Factorizing gives:

,0)4)(32( =−− mm from which, .4or2

3== mm

c) Since the roots are real and different, the complementary solution, .42

3

x x

c Be Ae y +=

d) Since 23)( −= x xQ is a polynomial, let the particular solution , .bax y p +=

e) Substituting bax y p += into 23)12112( 2 −=+− x y D D p gives:




Page 28 of 29

23))(12112( 2−=++− xbax D D ,

i.e. ,23)(12)(11)(2 2−=+++−+ xbaxbax Dbax D

i.e. 231212110 −=++− xbaxa

Equating the coefficient of x gives: ,312 =a from which, .41=a

Equating the constant terms gives:

21211 −=+− ba

i.e. 2124

111 −=+

− b , from which,

4

3

4

11212 =+−=b .

16

1=∴b


1

4

1+=+= xbax y p

f) The general solution is given by pc y y y += i.e.16

1

4

142

3

+++= x Be Ae y x x

■

Example 30:

Solve the equation xe y

dx

dy

dx

yd 4

2

2

32 =+− given the boundary conditions that when

.3

14 and

3

2 ,0 =−==

dx

dy y x

Solution:

a) xe y

dx

dy

dx

yd 4

2

2

32 =+− in D-operator form is xe y D D 42 3)12( =+− where .

dx

d D ≡


.0122

=+− mm

Factorizing gives:

,0)1)(1( =−− mm , from which, 1=m twice.

c) Since the roots are real and equal, the complementary solution, ( ) . x

c e B Ax y +=

d) Since xe xQ 43)( = is an exponential function, let the particular solution , .4 x

p ke y =

e) Substituting x

p ke y4

= into xe y D D 42 3)12( =+− gives:

x x eke D D 442 3)12( =+− ,

i.e. x x x x ekeke Dke D 44442 3)(1)(2)( =+−

i.e. x x x x ekekeke 4444 3816 =+−




Hence ,39 44 x x eke = from which, .3

1=k


1 44 x x

p eke y ==

f) The general solution is given by pc y y y += i.e. ( ) x x ee B Ax y4

3

1++= ■

g) When ,3

2 ,0 −== y x thus ,

3

1)0(

3

2 00ee B ++=− from which, .1−= B

.3

4)()( 4 x x x

e Aee B Axdx

dy+++=

0When = x , ,3

14=

dx

dy thus

3

4

3

13++= A B from which, ,4= A since .1−= B Hence the

particular solution is

x xee x y

4

3

1)14( +−= ■

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Chap 3 Differential Equations