y p = ( r , d x Differential Equations - 2 Differential Equations–2 47 Example 1: Solve the following differential equations. 1) Solve 2 2 7 44 0 d y dy y dx dx Solution: The given

Differential Equations - 2

2.1 LINEAR DIFFERENTIAL EQUATIONS WITH CONSTANTCOEFFICIENTS

A linear differential equation is that in which the dependent variable and its

derivatives occur only in the first degree and are not multiplied together.

Thus

1 2

1 2 11 2.........

nn ny

n nn n n

dd y dy dyp p p p y X

dxdx dx dx

where p1, p2, p3 ......... pn and X are functions of x only.

A differential equation of the form

1 2

1 21 2............

n nny y

nn n n

d dd yk k k y X

dx dx dx

...(1)

where k1, k2, k3 .......... kn are constants, is called a linear differential

equation with constant coefficients.

Let 2 3

2 32 3

, , ...........n

nn

d d d dD D D D

dx dx dx dx

where D is a differential operator.

Then equation (1) becomes

1 2 31 2 3 ............n n n n

nD k D k D k D k y X

or f(D) Y = X ....(2)

where f(D) = Dn + k1Dn–1 + k2 Dn–2 + .......... + kn

2V a =

3rdx

dtx

y

O

pr,=( )

DIFFERENTIAL EQUATIONS–2

47Differential Equations–2

Example 1: Solve the following differential equations.

1) Solve 2

27 44 0

d y dyy

dxdx

Solution: The given equation in symbolic form is

(D2 – 7D – 44) y = 0

Its AE is D2 – 7D – 44 = 0

(D + 4) (D – 11) = 0

D = – 4, D = 11

Its solution is

y = c1e–4x + c2 e11x

2) Solve (D4 – 5D2 + 4) y = 0

Solution: The given equation is (D4 – 5D2 + 4) y = 0

Its, AE is D4 – 5D2 + 4 = 0

D4 – 4D2 – D2 + 4 = 0

D2 (D2 – 4) – 1 (D2 – 4) = 0

(D2 – 4) (D2 – 1) = 0

(D – 2) (D + 2) (D – 1) (D + 1) = 0

D = 2, –2, 1, –1 are the roots.

Hence its solution is

y = c1e2x + c2e–2x + c3ex + c4e–x.

3) Solve 3 2

3 22 4 8 0

d y d y dyy

dxdx dx

Solution: The given equation is

(D3 – 2D2 – 4D + 8) y = 0

Its A.E. is D3 – 2D2 – 4D + 8 = 0

D2 (D – 2) – 4(D – 2) = 0

(D – 2) (D – 2) (D + 2) = 0

D = 2, 2, –2 are the roots

Its solution is

y = (c1x + c2) e2x + c3 e–2x

48 Engineering Mathematics–II

4) Solve 4

44

0d y

m ydx


(D4 – m4) y = 0

Its AE is D4 – m4 = 0

(D2 – m2) (D2 + m2) = 0

(D – m) (D + m) (D – mi) (D + mi) = 0

D = m, –m, mi and –mi

Its solution is

y = c1emx + c2e–mx + (c3 cos x + c4 sin x)

5) Solve (D4 + m4) y = 0


(D4 + m4) y = 0

Its AE is D4 + m4 = 0

D4 + 2m2 D2 + m4 – 2m2D2 = 0

(D2 + m2)2 – 2m2 D2 = 0

(D2 + m2 + 2 mD) (D2 + m2 – 2 mD) = 0

2 2 2 22 0, 2 0D mD m D mD m

D = – 2 22 2 4

;2

m m m D =

2 22 2 4

2

m m m

= 2 2

m mi ; =

2 2

m mi

Its solution is

y = 2 21 2 3 4cos sin cos sin

2 2 2 2

mx mxmx mx mx mx

e c c e c c


EXERCISE

Solve the following differential equations.

1. (D2 + 1)2 (D2 + D + 1)y = 0

Ans.: y = (c1x + c2) cos x + (c3x + c4) sin x

+

1

25 6

3 3cos sin

2 2

xe c x c x

2.3 2

3 26 11 6 0

d y d y dyy

dxdx dx

Ans.: y = c1e–x + c2 e–2x + c3 e–3x

3.3 2

3 23 3 0

d y d y dyy

dxdx dx

Ans.: y = (c1x2 + c2 x + c3) ex

4.2

24 0

d y dyy

dxdx

Ans.: y = c1(2 3) (2 3)

2x xe c e

5.4 3 2

4 3 24 5 36 36 0

d y d y d y dyy

dxdx dx dx

Ans.: y = (c1x + c2) e–2x + c3 e3x + c4 e–3x

2.3 INVERSE OPERATOR 1

( )f D

Definition: 1

( )f D X is that function of x, free from arbitrary constants which

when operated upon by f(D) gives X.

Thus f(D) 1

( )X X

f D

.


f(D) and 1

( )f D are inverse operators.

Note the following important results:

1.1

( )X

f D is the particular integral of f(D) y = X.

2.1

X X dxD

.

3.1 ax axX e X e dx

D a

.

2.4 RULES FOR FINDING THE PARTICULAR INTEGRAL

Consider the differential equation

f (D) y = X

PI = 1

( )X

f D

Type I(A): When X is of the form eax, where a is any constant.

PI = 1 1 1

( ) ( ) ( )ax axX e e

f D f D f a . Provided f(a) 0.

Rule: In f(D), put D = a and PI will be calculated, provided f(a) 0.

Type I(B): When X is of the form eax but f(D) = 0 has got ‘a’ as its root

(Failure case of type 1).

PI =1 1

( ) 0( ) 0

ax axe e f af D

Our method fails.

Now, since ‘a’ is a root of f(D), (D – a) must be a factor of f(D) which

therefore can be written as (D – a) (D)

PI =1

( ) ( )axe

D a D


= 1

1( ) ( )

axe

a D a a

= 1

1( )

axe

a D

= .( )

axex

a

1

D

Similarly, if f(D) = (D – a)p (D) then

PI = 1

. 1( )

ax

p

e

a D

= .( ) !

ax pe x

a p

Rule: Put D = a in those factors of f(D) which do not vanish for D = a and

then make the question as PI of a product of eax and 1 which is calculated.

Note: In case of sinh ax and cosh ax we take

i) sinh ax = 2

ax axe eii) cosh ax =

2

ax axe e

Example 1: Solve (D2 + 3D + 5) y = e2x

Solution: Given equation is

(D2 + 3D + 5) y = e2x

AE is D2 + 3D + 5 = 0

D =3 9 20 3 11

2 2 2i

CF =

3

21 2

11 11cos sin

2 2

xe c x c x

And, PI =2

2

1

3 5

xeD D

Put D = 2 in f (D)


=2 21 1

4 6 5 15x xe e

G.S. = CF + PI

y =

322

1 211 11 1

cos sin2 2 15

xxe c x c x e

Example 2: Solve 2

24 4 2sinh 2 .

d y dyy x

dxdx


(D2 + 4D + 4) y = 2 sinh 2x

= 2 2

2 22( )

2

x xx xe e

e e

AE is D2 + 4D + 4 = 0

(D + 2)2 = 0 D = –2, –2 are the roots.

CF = (c1x + c2) e–2x

And PI for e2x =

22 2

2 2

1 1

16( 2) (2 2)

xx x e

e eD

...(1)

Also PI for e–2x = 2 2

2

1 1

0( 2)

x xe eD

Our method fails.

Then to find PI put D – 2 for D in f(D) and take out e–2x

PI = 22

11

( 2 2)

xeD

= 2

2 22

11

2!x x xe eD

....(2)

GS = CF + PI1 – PI2

y = (c1x + c2) e–2x + 2 2

2

16 2!

xxe x

e .


Example 3: Solve (D3 + 3D2 + 3D + 1) y = e–x.


(D3 + 3D2 + 3D + 1) y = e–x

AE is D3 + 3D2 + 3D + 1 = 0

(D + 1)3 = 0 D = –1, –1, –1 are the roots.

CF = (c1x2 + c2x + c3) e–x

And PI = 3

1

( 1)

xeD

= 3

1(1)

( 1 1)

xeD

= e–x 3

3

11

3!x x

eD

G.S. = CF + PI

y =3

21 2 3( )

3!x x x

c x c x c e e

Example 4: Solve (D2 + 4D + 3)y = e–3x.


(D2 + 4D + 3) y = e–3x

AE is D2 + 4D + 3 = 0

(D + 1) (D + 3) = 0 D = –1, D = –3

CF = c1e–x + c2e–3x

PI =31

( 1) ( 3)xe

D D

put D = –3 in f(D) then

PI = 31

0xe method fails

PI = 3 3 31

1 1( 3 1) ( 3 3) ( 2) 2

x x xe e ex

D D

G.S. = CF + PI


y =3

31 2

2

xx x e

c e c e x

Type II(A): When X is of the form sin ax or cos ax.

PI =2 2

1 1sin or cos

( ) ( )ax ax

f D f D

=2 2

1 1sin or sin

( ) ( )ax ax

f a f a

In this case put the quantity – a2 in the place of D2, we can not put anything for D as is imaginary. D3 on substitution shall become D2 . D = –a2D,D4 will be D2 . D2 = (–a2) (–a2).

In otherwords f(D) shall be reduced to a linear factor of the form lD – mor D1 + m. Then multiply both numerator and denominator by a conjugatefactor lD + m or lD – m respectively, the denominator shall be reduced tol2 D2 – m2 in which again put D2 = –a2 and it will become a constant i.e.,l2 (–a2) – m2 = –a2 l2 – m2.

PI = 2 2 2

cos sinla ax m ax

a l m

Type II B: (Failure case)

When X is of the form sin ax or cos ax but f(D) becomes zero when weput D2 = –a2.

PI = 2 2

1 1sin or cos

( ) ( )ax ax

f D f D

Put D2 = –a2

PI =1 1

sin or cos0 0

ax ax

Our method fails

Then PI = 2 2

1sin sin

2

xax ax dx

D a

and PI = 2 2

1cos cos

2

xax ax dx

D a


Example 1: Solve (D2 + D + 1) y = sin 2x.


(D2 + D + 1) y = sin 2x

AE is D2 + D + 1 = 0

D =1 1 4 1 3

2 2 2i

CF =

1

21 2

3 3cos sin

2 2

xe c x c x

And PI = 2

1sin 2

1x

D D

Put D2 = – z2

=1 1

sin 2 sin 24 1 3

x xD D

= 2

( 3)sin 2

9

Dx

D

Put again D2 = –22

=(sin 2 ) (3sin 2 )

4 9

D x x

=2cos2 3sin 2

13

x x

G.S. = CF + PI

y =

1

21 2

3 3 1cos sin (2cos 2 3sin 2 )

2 2 13

xe c x c x x x

Example 2: Solve (D2 – 5D + 6) y = cos 3x


(D2 – 5D + 6) y = cos 3x

AE is D2 – 5D + 6 = 0

(D – 2) (D – 3) = 0 D = 2, 3 are the roots.

CF = c1e2x + c2 e3x


And PI = 2

1cos3

5 6x

D D

Put D2 = –32

=1

cos39 5 6

xD

=1

cos35 3

xD

= 2

(5 3)cos3

25 9

Dx

D

Put D2 = –32 again

= 5 (cos3 ) 3(cos3 )

25 ( 9) 9

D x x

= 15sin3 3cos3

234

x x

=(15sin 3 3cos3 )

234

x x

G.S. = C.F. + P.I.

y =2 3

1 21

(15sin 3 3cos )234

x xc e c e x x

Example 3: Solve (D2 + 4)y = cos 2x.


(D2 + 4) y = cos 2x

AE is D2 + 4 = 0 D = ± 2i

C.F. = c1 cos 2x + c2 sin 2x

And

PI = 2

1cos2

4x

D

Put D2 = –22

=1

cos20

x Our method fails.


P.I. = 2

1cos 2 cos2

24

xx x dx

D

= sin 2

sin 22 2 4

x x xx

G.S. = C.F. + P.I.

y = c1 cos 2x + c2 sin 2x + sin 24

xx .

Example 4: Solve (D2 + 4) (D2 + 1) y = cos 2x + sin x.


(D2 + 4) (D2 + 1) y = cos 2x + sin x

AE is (D2 + 4) (D2 + 1) = 0

D = ± 2i and D = ± i

CF = (c1 cos 2x + c2 sin 2x) + (c3 cos x + c4 sin x)

Now, PI for cos 2x =2 2

1cos 2

( 4) ( 1)x

D D

=1 1

cos2 cos2( 4 4) ( 4 1) 0

x x

PI = 2

1 cos2

( 4 1) 4

x

D

= 1 sin 2

cos2 sin 23 2 6 2 12

x x x xx dx x

...(1)

2 2( put 2D in those factors which do not vanish)

And PI for sin x =2 2

1sin

( 4) ( 1)x

D D

put D2 = –12 in those factors which do not vanish.

=1 1

sin ( cos )( 1 4) 2 3 2

x xx dx x

= cos6

xx ....(2)


G.S. = CF + PI1 + PI2y = c1 cos 2x + c2 sin 2x + c3 cos x

+ c4 sin x – sin 2 cos12 6

x xx x

Type III: When X is of the form xm.

PI =1

( )mx

f D

To find the PI of this type one should remember the following

expansions.

i) (1 – D)–1 = 1 + D + D2 + D3 + ..........

ii) (1 + D)–1 = 1 – D + D2 – D3 + ..........

iii) (1 – D)–2 = 1 + 2D + 3D2 + 4D3 + .........

iv) (1 + D)–2 = 1 – 2D + 3D2 – 4D3 + ...........

v) (1 – D)–3 = 1 + 3D + 6D2 + 10D3 + .........

vi) (1 + D)–3 = 1 – 3D + 6D2 – 10D3 + .........

Example 1: Solve 2

22 5 2 5 2d y dy

y xdxdx

.


(2D2 + 5D + 2) y = 5 + 2x

AE is 2D2 + 5D + 2 = 0

(2D + 1) (D + 2) = 0 D = – 1

,2

D = –2

C.F. = c1

122

2

xxe c e

P.I. = 2

1(5 2 )

2 5 2x

D D

=

121 (2 5 )

1 (5 2 )2 2

D Dx

=

21 (2 5 )1 ........ (5 2 )

2 2

D Dx


=1 5

5 2 . 22 2

x

= x

G.S. = CF + PI

y =

122

1 2

xxc e c e x

Example 2: Solve (D2 + 2D + 1) y = x2 + ex – sin x + 2x.


(D2 + 2D + 1) y = x2 + ex – sin x + 2x

AE is D2 + 2D + 1 = 0

(D + 1)2 = 0 D = –1, –1

C.F. = (c1x + c2) e–x

1) PI for x2 =2

2

1

(1 )x

D

= (1 + D)–2 (x2)

= (1 – 2D + 3D2 – 4D3 ...... ) x2

= x2 – 4x + 6

ii) PI for ex =2

1

2 1D D ex

Put D = 11 1

1 2 1 4x xe e

iii) PI for sin x =2

1sin

2 1x

D D

Put D2 = –12, =1 1

sin1 2 1 2

xD D

=1 1

sin cos2 2

x dx x 1

D

iv) PI for 2x = 2

1

2 1D D 2x

=log 2

2

1

2 1

xeD D


y = 41 2 3 4 4 4 4 4

1 24 1cos sin sinax axc e c e c ax c ax x bx

a a b a

Type IV: When X is of the form eaxV, where V is any function of x.

PI =1

( )axe V

f D

=1

( )axe V

f D a

Rule: It means that take out eax and in f(D) write D + a for every D so that

f(D) becomes f(D + a) and then operate 1

( )f D a with V alone by previous

method.

Example 1: Solve (D2 – 4D + 3) y = e2x sin 3x.


(D2 – 4D + 3) y = e2x sin 3x

A.E. is D2 – 4D + 3 = 0

(D – 1) (D – 3) = 0 D = 1, 3

C.F. = c1ex + c2 e3x

And PI =2

2

1sin3

4 3

xe xD D

=2

2

1sin 3

( 2) 4( 2) 3

xe xD D

= e2x 2

1sin3

4 4 4 8 3x

D D D

=2

2

1sin 3

1

xe xD

put D2 = –32

=2 1

sin 39 1

xe x


=21

sin 310

xe x

G.S. = CF + PI

y = c1ex + c2 e3x – 21

sin 310

xe x

Example 2: Solve (D2 – 5D + 6) y = x (x + ex)

Solution:

Given equation is

(D2 – 5D + 6) y = x (x + ex)

= x2 + xex

AE is D2 – 5D + 6 = 0

(D – 2) (D – 3) = 0 D = 2, 3

CF = c1e2x + c2e

3x

And

i) PI for x2 =2

2

1

5 6x

D D

=

12

21 ( 5 )1 ( )

6 6

D Dx

=

2 2 221 ( 5 ) ( 5 )

1 ........... ( )6 6 6

D D D Dx

=

22 21 5 25

1 .......... ( )6 6 6 36

D DD x

=21 2 10 50

6 6 6 36

xx

=21 10 19

6 6 18

xx


= 2

1sin

( 1 1)

xe x xD

= 2

1( sin )xe x x

D

=1

sinxe x x dxD

= 1

( cos ) ( sin )xe x x xD

= ( cos sin )xe x x x dx = e–x [–x sin x – (–1) (–cos x) – cos x]

= e–x [–x sin x – 2 cos x]

= – e–x (x sin x + 2 cos x)

G.S = CF + PI

y = (c1x + c2) e–x – e–x (x sin x + 2 cos x)

Example 4: Solve 2

22 cosxd y dy

y xe xdxdx


(D2 _ 2D + 1) y = xex cos x

AE is D2 – 2D + 1 = 0

(D – 1)2 = 0 D = 1, 1

CF = (c1x + c2) ex

And,

PI =2

1cos

( 1)

xxe xD

= 2 2

1 1cos cos

( 1 1)

x xe x x e x xD D

=1

. cosxe x x dxD


= 1

sin 1 ( cos )xe x x xD

= ( sin cos )xe x x x dx= ex [x (– cos x) + sin x + sin x]

= ex (–x cos x + 2 sin x)

GS = CF + PI

y = (c1x + c2) ex + ex (–x cos x + 2 sin x)

Example 5: Solve 2

22

2x x xd y dye e e y x

dxdx


22

22 xd y dy

y x edxdx

AE is D2 + 2D + 1 = 0

(D + 1)2 = 0 D = –1, 1

CF = (c1x + c2) e–x

And, PI =2

2

1

( 1)

xx eD

=2

2

1xe xD

=3

21 1

3x x x

e x dx eD D

=4

3 .3 3 4

x xe e xx dx

=4

12

xe x

GS = CF + PI

y = (c1x + c2) e–x + 41

12xe x


G.S. = CF + PI

y = c1 cos 2x + c2 sin 2x + 1

9 (3x sin x – 2 cos x)

Example 2: Solve 2

22

sind y

y x xdx


(D2 + 1) y = x2 sin x

AE is D2 + 1 = 0 D = ± i

CF = c1 cos x + c2 sin x

PI =2

2

1sin

1x x

D

= 2

1

1D imaginary part of x2 eix

= Imaginary part of 22

1

( ) 1

ixe xD i


2

1

2 1 1

ixe xD Di


211 ( )

2 2ix De x

Di i


22

11 ......... ( )

2 2 4

ix

i

D De x

D i i

= Imaginary part of 2 2 2

2 2 4

ixe xx

Di i

= Imaginary part of 2 1

2 2

ixex ix

Di

= Imaginary part of

3 2

2 3 2 2

ixe x ix x

i


= Imaginary part of 3 2(2 3 )12

ixex x ix

i


12i (cos x + i sin x) (2x3 – 3x + i 3x2)


12i[(cos x) (2x3 – 3x) – (sin x) 3x2

+ i (2x3 – 3x) sin x + 3x2 cos x]

=1

12 (2x3 – 3x) cos x – 3x2 sin x

Taking imaginary part only.

GS = CF + PI

y = c1 cos x + c2 sin x – 1

12 (2x3 – 3x) cos x – 3x2 sin x.

Example 3: Solve 2

22

secd y

a y axdx


(D2 + a2) y = sec ax

AE is D2 + a2 = 0 D = ± ai

CF = c1 cos ax + c2 sin ax

And, PI= 2 2

1secax

D a

=1

sec( ) ( )

axD ai D ai

Resolve into partial fractions

=1 1 1

sec2

axai D ai D ai

=1 1 1

sec sec2

ax axai D ai D ai

Now,


PI =2 2

1 1tan tan

( ) ( )ax ax

D ai D aiD a

=1 1 1

tan ,2

axai D ai D ai

by partial fractions

=1 1 1

tan tan2

ax axai D ai D ai

Now 1

tan taniax iaxax e ax e dxD ai

= tan (cos sin )iaxe ax ax i ax dx

=

2sinsin

cosiax axe ax i dx

ax

= sin (sec cos )iaxe ax i ax ax dx

=cos sin

log(sec tan )iax ax i i axe ax ax

a a a

= 1

(cos sin ) log(sec tan )iaxe ax i ax i ax axa

=1

log(sec tan )iax iaxe e i ax axa

=1

1 log(sec tan )iaxie ax axa

Changing i to –i we get

1tanax

D ai=

11 log(sec tan )iaxie ax ax

a

PI = 1 1 11 log(sec tan 1 log(sec tan )

2iax iaxie x x e ax ax

ai a a

= 2

1

a log (sec ax + tan ax) . cos ax


GS = CF + PI

y = c1 cos ax + c2 sin ax – 2

1

a cos ax log (sec ax + tan ax)

2.5 INITIAL VALUE PROBLEMS

Example 1: Solve y + 4y + 4y = 0 given y (0) = 3, y(0) = 1


(D2 + 4D + 4) y = 0 with x = 0, y = 3 and x = 0, y = 1

AE is D2 + 4D + 4 = 0

(D + 2)2 = 0 D = –2, –2

The solution is

y = (c1x + c2) e–2x ....(1)

and y = (c1x + c2) e–2x (–2) + c1e–2x ....(2)

Given y = 3 when x = 0 and y1 = 1 when x = 0

c2 = 3 and c1 = 7

The complete solution is

y = (7x + 3) e–2x

Example 2: Solve the initial value problem 2

24 5 2cosh 0

d y dyy x

dxdx

given y = 0, 1dy

dx at x = 0.


(D2 + 4D + 5) y = –2 cosh x

=2( )

( )2

x xx xe e

e e

AE is D2 + 4D + 5 = 0

D =4 16 20

22

i

CF = e–2x (c1 cos x + c2 sin x)


= cos sin sin . cos2 2

x xa x dx a x dx

=cos sin

cos . sin2 2

x x x xa a

GS = CF + PI

y = c1 cos x + c2 sin x – 1 1

cos cos sin sin2 2

a x x a x x

and y = 1 21 1

sin cos cos cos cos . sin2 2

c x c x a x a x x

1 1sin sin sin . cos

2 2a x a x x

Given y = 0, y = 0 when x = 0

0 = c1 c1 = 0

and 0 = 21

cos2

c a 21

cos2

c a .


y =1 1 1

cos sin cos cos sin sin2 2 2

a x ax x a x x

=1 1

cos (sin cos ) sin sin2 2

a x x x a x x

Example 4: Solve the initial value problem 2

25 6 0

d y dyy

dxdx given

y (0) = 0, (0) 15dy

dx .


(D2 + 5D + 6) y = 0

AE is D2 + 5D + 6 = 0

(D + 2) (D + 3) = 0 D = –2, –3

CF is y = c1e–2x + c2 e–3x

anddy

dx= – 2c1e–2x – 3c2 e–3x


At x = 0, y = 0 and 15dy

dx

0 = c1 + c2

15 = –2c1 – 3c2 Solving c2 = –15 and c1 = 15


y = 15e–2x – 15e–3x

= 15 (e–2x – e–3x)

2.6 SIMULTANEOUS DIFFERENTIAL EQUATIONS OF FIRSTORDER

The differential equation in which there is one independent variable and two

or more than two dependent variables are called simultaneous linear

differential equations. Here we consider simultaneous linear equation with

constant coefficients only.

Example 1: Solve sin ; cosdx dy

y t x tdt dt

Solution: The simultaneous equation are

Dx + y = sin t ....(1)

Dy + x = cos t ....(2) where d

Ddt

Multiply (1) by D

D2x + Dy = cos t ....(3)

Dy + x = cos t ....(4)

Subtracting (4) from (3) we get

D2x – x = 0

AE is (D2 – 1) = 0 D = 1, –1

x = c1 et + c2 e–t ....(1)

And Dx + y = sin t

y = sin t – 1 2t td

c e c edt

y = sin t – c1et – c2 e–t


2. Solve (D – 2)2 y = 8 (e2x + sin 2x + x2)

Ans.: y = (c1x + c2) e2x + 4x2 e2x + cos 2x + 2x2 + 4x + 3.

3. Solve 4

4cos cosh

d yy x x

dx

Ans.: y = c1ex + c2 c–x + c3 cos x + c4 sin x –

1cos cosh

5x x .

4. Solve 2

2cosec

d yy x

dx

Ans.: y = c1 cos x + c2 sin x + sin x log sin x – x cos x.

5. Solve (D2 – 4D + 3) y = sin 3x cos 2x

= 1

sin 5 sin2

x x

Ans.: y = c1ex + c2 e3x +

1 1(10cos5 11sin5 ) (sin 2cos )

884 20x x x x

6. Solve (D2 – 3D + 2) y = 6 e–3x + sin 2x.

Ans.: y = c1 ex + c2 e2x + 33 1

(3cos 2 sin 2 )10 20

xe x x

7. Solve (D2 – 4) y = (1 + ex)2

Ans.: y = c1 e2x + c2e–2x

21 2 1

4 3 4x xe xe

8. Solve 2

32

3 2 sin 2xd y dyy xe x

dxdx

Ans.: y = c1ex + c2 e2x +

31 1(2 3) (3cos2 sin 2 )

4 20xe x x x

9. Solve 2

24 4 tan 2

d yy x

dx

Ans.: y = c1 cos 2x + c2 sin 2x – cos 2x log (sec 2x + tan 2x)


Case VII: If X = eax (a0 + a1x + a2x2 + .......... + an xn) sin bx or cos bx then

the particular integral is of the form.

y = eax {(A0 + A1x + ........ + Anxn) sin bx + (B0 + B1x + .... + Bnxn)

cos bx}

Note: However, when X = tan x or sec x, this method fails, since the

number of terms obtained by differentiating X = tan x or sec x is

infinite.

Example 1: Solve by the method of undetermined coefficients,

y – 3y + 2y = x2 + x + 1

Solution: Given differential equation is

y – 3y + 2y = x2 + x + 1 ....(1)

AE is D2 – 3D + 2 = 0

(D – 1) (D – 2) = 0 D = 1, 2

CF = c1ex + c2 e2x

PI is of the form

y = A0 + A1x + A2x2

y = 0 + A1 + 2A2x

y = 0 + 2A2

Substituting in (1)

2A2 – 3 (A1 + 2A2x) + 2 (A0 + A1x + A2x2) = x2 + x + 1

2A2x2 + (2A1 – 6A2) x + 2A0 – 3A1 + 2A2 = x2 + x + 1

Comparing the coefficients

2A2 = 1, 2A1 – 6A2 = 1, 2A0 – 3A1 + 2A2 = 1

Solving, A2 =1

2, A1 = 2, A0 = 3

PI = 3 + 2x + 1

2x2

GS = CF + PI

y = c1ex + c2e2x + 21

3 22

x x


Example 2: Solving by the method of undetermined coefficients the

equation y – 2y + 5y = e2x.


y – 2y + 5y = e2x ...(1)

AE is D2 – 2D + 5 = 0

D = 1 ± 2i

CF = ex (c1 cos 2x + c2 sin 2x)

PI is of the form

y = Ae2x

y = 2Ae2x and y = 4Ae2x

Substituting in (1)

4Ae2x – 2 + 2Ae2x + 5Ae2x = e2x

5A e2x = e2x

5A = 1 A = 1

5

PI =1

5 e2x

G.S. = CF + PI

y = ex (c1 cos 2x + c2 sin 2x) + 1

5 e2x

Example 3: Solve by the method of undetermined coefficients, the

equation y – 5y + 6y = sin 2x.


y – 5y + 6 y = sin 2x ....(1)

AE is D2 – 5D + 6 = 0 D = 2, 3

CF = c1e2x + c2 e3x

PI is of the form

y = A cos 2x + B sin 2x

y = –2A sin 2x + 2B cos 2x

y = –4A cos 2x – 4B sin 2x

Substituting in (1)


PI =1 1 3 1

sin cos4 2 10 10

x x x

G.S. = CF + PI

y = c1ex + c2 e–2x –

1 1 3 1sin cos

4 2 10 10x x x

Example 5: Solve by the method of undetermined coefficients the equation

y + 4y = x2 + e–x.

Solution: The given differential equation is

y + 4y = x2 + e–x ....(1)

AE is D2 + 4 = 0 D = ± 2i

CF = c1 cos 2x + c2 sin 2x

Particular integral is of the form

y = A0 + A1x + A2x2 + Be–x

y = A1 + 2A2x – Be–x

y = 2A2 + Be–x

Substituting in (1)

(2A2 + Be–x) + 4(A0 + A1x + A2x2 + Be–x) = x2 + e–x

(4A0 + 2A2) + 4A1x + 4A2x2 + 5Be–x = x2 + e–x

Equating the coefficients, we get

4A0 + 2A2 = 0 Solving

4A1 = 0 A0 = – 1

8, A1 = 0

4A2 = 1 A2 = 1

4,

1

5B

PI = 21 1 1

8 4 5xx e

G.S. = CF + PI

y = c1 cos 2x + c2 sin 2x – 1 1 1

8 4 5xx e .


Example 6: Solve by the method of undetermined coefficients

y – y – 4y = x + cos 2x .

Solution:

The given differential equation is

y – y – 4y = x + cos 2x ....(1)

AE is D2 – D – 4 = 0

D =1 1 16 1 17

2 2

CF =

(1 17 ) (1 17 )

2 21 2

x xc e c e

Particular integral is of the form

y = A0 + A1x + B0 cos 2x + B1 sin 2x

y = A1 – 2B0 sin 2x + 2B1 cos 2x

y = –4B0 cos 2x – 4B1 sin 2x

Substituting in (1)

(–4 B0 cos 2x – 4B1 sin 2x) – (A1 – 2B0 sin 2x + 2B1 cos 2x)

– 4 (A0 + A1x + B0 cos 2x + B1 sin 2x) = x + cos 2x.

Equating the coefficients

–4 A0 – A1 = 0 Solving

–4 A1 = 1 0 11 1

,16 4

A A

–8B0 – 2B1 = 1 0 12 1

,17 34

B B

2B0 – 8B1 = 0

PI =1 1 2 1

cos2 sin 216 4 17 34

x x x

G.S. = CF + PI

y =

(1 17 ) (1 17 )

2 21 2

1 1 2 1cos 2 sin 2

16 4 17 34

x x

c e c e x x x


Example 7: Solve by the method of undetermined coefficients

2

2sin

d yy x

dx .


y + y = sin x ...(1)

AE is D2 + 1 = 0 D + ± i


Note that ‘sin x’ is common in CF and RHS of the equation andtherefore particular integral is of the form

y = x (A cos x + B sin x)

y = x (–A sin x + B cos x) + (A cos x + B sin x)

y = x (–A cos x – B sin x) + (–A sin x + B cos x)

+ (– A sin x + B cos x)

Substituting in (1)

–2A sin x + 2B cos x = sin x


–2A = 2 and 2B = 0

A = 1

2 and B = 0

PI =1

cos2x x

GS = CF + PI

y = c1 cos x + c2 sin x – 1

cos2x x

Example 8: Solve by the method of undetermined coefficients the equation

(D2 + 1) y = 4x – 2 sin x.

Solution: The given differential equation is

y + y = 4x – 2 sin x ....(1)

AE is D2 + 1 = 0 D = ± i




2A2 – 3 A1 + 2A0 = 0 A0 = 17 3

,4 2

A

2A2 = 1 21

2A B = – 1

PI =27 3 1

4 2 2xx x xe

G.S. = CF + PI

y =2 2

1 27 3 1

4 2 2x x xc e c e x x xe

Example 10: Solve by the method of undetermined coefficients the

following equations.

i) (D2 + 1) y = 4x cos x – 2 sin x ....(1)

Hint: CF = c1 cos x + c2 sin x

Since ‘cos x and sin x’ are common in CF and RHS of (1) particular

integral is of the form.

y1 = x [(A0 + A1x) cos x + (A2 + A3x) sin x]

and y2 = PI for 2 sin 2x

= x {B0 cos x + B1 sin x}

y = y1 + y2

ii) (D2 – 1) y = 10 sin2x

Hint: = 10(1 cos2 )

5 5cos22

xx

PI is of the form

y = A0 + {B0 cos x + B1 sin x}

iii) (D2 – 1) y = e–x (2 sin x + 4 cos x)

Hint: PI is of the form

y = e–x (A cos x + B sin x)

iv) (D3 – D) y = 3ex + sin x ....(1)

Hint: D(D2 – 1) = 0 D = 0, D = 1 D = –1

CF = c1 + c2ex + c3 e–x


ex is common in CF and RHS of (1)

PI is of the form

y = A0 xex + (B0 sin x + B1 cos x)

v) (D2 – 5D + 6) y = e2x + sin x

Hint:

PI is of the form

y = Axe2x + (B0 sin x + B1 cos x)

2.8 APPLICATIONS TO LINEAR DIFFERENTIAL EQUATIONS

The applications of linear differential equations to various physical problems

play a dominant role in uniflying seemingly different theories of mechanical

and electrical systems just by renaming the variables. This analogy has an

important practical application. Since electrical circuits are easier to

assembe, less expensive and accurate measurements can be made of

electrical quantities.

A. Simple harmonic mothion (S.H.M.)

A particle is said to execute simple harmonic motion if it moves in a straight

line such that its acceleration is always directed towards a fixed point in the

line is proportional to the distance of the particle from the fixed point.

A AO Px

2x

a

Let O be the fixed point in the line AA. Let P be the position of theparticle at any time t where

OP = x

Since the acceleration is always directed towards O i.e., the accelerationis in the direction opposite to that in which x increases, the equation ofmotion of the particle is

2

2

d x

dt = – 2x or (D2 + 2)x = 0 where

dD

dt ...(1)

which is the linear differential equation with constant coefficient.


The solution of equation (1) is

x = c1 cos t + c2 sin t ....(2)

Velocity of the particle at P is

dx

dt= – c1 sin t + c2 cos t....(3)

If the particle starts from rest at A, where OA = a then from (2) at t = 0,x = 0. c1 = a.

and from (3), at t = 0, 20 0dx

cdt

x = –a sin t ...(4)

anddx

dt=

22

21 cos 1

xa t a

a ...(5)

dx

dt= 2 2a x ...(6)

Hence equation (4) gives the displacement of the particle from the fixedpoint O at any time t.

Equation (6) gives the velocity of the particle at any time t. Equation (6)also shows that the velocity is directed towards O and decreases as xincreases.

Example 1: In the case of a stretched elastic horizontal string which has

one end fixed and a particle of mass m attached to the other, find the

equation of motion of the particle given that l is natural length of the

string and e is its elongation due to a weight mg. Also find the

displacement of the particle when initially s = s0, v = 0.

Solution: Let OA = l be the elastic horizontal string with the end O fixedand a particle of mass m attached at A.

O PA T

l

Let P be the position of the particle at any time t

Let OP = s so that the elongation AP = s – l.

Now, for the elongation e tension = mg


For the elongation (s – l), tension = ( )mg s l

e

Since tension is the only horizontal force acting on the particle, itsequation of motion is

2

2

d smdt

= – T

2

2

d smdt

=( )mg s l

e

or2

2

d s

dt=

g gls

e e

2

2

d s gs

edt =

2 or gl g gl

D se e e

....(1) when

dD

dt

Which is the linear differential equation

Its AE is 2 0g g

D D ie e

CF = c1 2cos sin

g gt c t

e e.

And, PI =0

2 2

1 1 1.tgl gl gl

e lg g ge e eD D ee e

The complete solution of (1) is

s = 1 2cos sin

g gc t c t l

e e ...(2)

When t = 0, s = s0 from (2) we get

s0 = c1 + 0 + l or c1 = s0 – l

Also,


b) Damped Oscillations

If the motion of the mass m be subject to anadditional force of resistance, theoscillations are said to be damped. Thedamping force may be constant orproporational to velocity. The latter type ofdamping is important and is usually calledviscous damping.

Now, if the damping force be

proportional to velocity saydx

rdt

then

the equation of motion becomes

2

2

d xmdt

= ( )dx

mg k e x rdt

= dx

kx rdt

Takingr

m= 2 2 and

k

m .

We get

22

22 0

d x dxx

dtdt ....(3)

Which is the linear differential equationwith constant coefficient.

c) Forced Oscillations(Without dumping)

If the point of the support of the spring is alsovibrating with some external periodic force,then the resulting motion is called the forcedoscillatory motion.

Taking the external periodic force to bemp cos nt the equation of the motion is

2

2( ) cos

d xm mg k e x mp ntdt

= – kx + mp cos nt

A

B

P

mg

x

e

k e x ( + )

O

rdxdt

damper

A

B

P

mg

x

e

k e x ( + )

O

mp nt cos


Taking 2k

m the equation becomes

22

2

d xx

dt = p cos nt ....(4)

Which is linear differential equation.

The solution of equation (4) is

x = c1 cos t + c2 sin t + 2 2

cosp nt

n

d) Forced Oscillations (With damping)

If in addition, there is a damping force proportional to velocity say: dx

rdt

then the above equation becomes.

A

B

P

mg

x

e

k e x ( + )

O

rdxdt

damper

mp nt cos

2

2

d xm

dt= mg – k (e + x) + mp cos nt – r

dx

dt

= – k x + mp cos nt – dx

rdt


Taking 22 and

r k

m m then the equation becomes

22

22 cos

d x dxx p nt

dtdt

Which is a linear differential equation and its solution is

x = 2 2 2 22 2

1 2 2 2 2 2

( )cos 2 sin

( ) 4

tst p n nt n te c e c e

n n

With the increase of time, the free oscillations die away while the

forced oscillations continue giving the steady state motion.

C. Oscillatory Electrical Circuits

a) L – C Circuit

Consider an electrical circuit containing an inductance L and capacitance C.

C V

L

i

Let i be the current and q be the charge in the condenser plate at anytime t then the voltage drop across

L =2

2

di d qL L

dt dt

and voltage drop across q

CC

As there is no applied e.m.f. in the circuit, therefore by Kirchooff’s firstlaw, we have

2

20

d q qL

Cdt


Dividing by L and taking 21

LC we get

22

20

d qq

dt ...(1)

Which is linear differential equation and it reprsents free electrical

oscillations of the current having period 2

2 Lc

.

b) L – C – R Circuit

Consider the discharge of a condenser through an inductance L and the

resistance R. Since the voltage drop across L, C and R are respectively.

C V

LR

i

2

2, and

d q q dqL R

C dtdt

By Kirchhoff’s law, we have

2

2+

d q dq qL R

dt Cdt = 0

Taking 21

2 and R

L LC we get

22

22 0

d q dqq

dtdt ....(2)

Which is the linear differential equation.


By Kirchhoff’s law the equation is

2

2cos

d q dq qL R p nt

dt Cdt

Dividing by L

2

2cos

d q R dq q pnt

L dt LC Ldt

Taking 21

2 and R

L LC , we have

22

22 cos

d q dq pq nt

dt Ldt

Which is the linear differential equation.

Note: The L – C – R circuit with a source of alternating e.m.f. is anelectrical equivalent of the mechanical phenomena of forced oscillations withresistance.

Example 1: In an L – C – R circuit, the charge q on a plate of a condenser

is given by

2

2sin

d q dq qL R E pt

dt Cdt

The circuit is tunned to resource so that 2 1p

LC . If initially the

current i and the charge q be zero, show that for small values of R

L. The

current in the circuit at time t is given by sin2

Etpt

L.


2 1sinLD RD q E pt

C

....(1)

AE is 2 1

0LD RDC


D =

22

2

41

2 2 4

LR R

R RC

L L LCL

As R

L is small, we have

D =1

2 2

R Ri ip

L LLC

CF = 21 2( cos sin )

Rt

Le c pt c pt

= 1 21 ( cos sin )2

Rtc pt c pt

L

2

rejecting the terms in etc.R

L

And PI =2

1sin

1E pt

LD RDC

=2

sinsin

1

E pt Ept dt

RLp RDC

2 1 pLC

= cosE

ptRp

Thus the complete solution is

q = 1 21 cos sin cosRt E

c pt c pt ptL Rp

...(ii)

i = dq

dt= 1 21 ( sin cos )

Rtc pt p c pt p

L

1 2( cos sin ) sin

R Ec pt c pt pt

L R....(iii)


When t = 0, q = 0, i = 0

From (ii) 0 = 1 1

E Ec c

Rp Rp

and from (iii) 0 = 12 1 2 22 2 2

RcR Ec p c c

L Lp Lp

i = 21 sin cosRt E

pt c pt pL Rp

2cos sin sin

2 2

R E E Ept pt pt

L Rp RLp

= sin is small2

Et Rpt

L L

D. Deflection of Beams

Example 1: The deflection of a strut of length l with one end (x = 0) built

in and the other supported and subjected to end trust p satisfies the

equation.

2 22

2( )

d y a Ra y l x

pdx

Prove that the deflection curve is sin

cosR ax

y l ax l xp a

and

al = tan al.


(D2 + a2)y =

2

( )a R

l xp

....(1)

Its AE is D2 + a2 = 0 D = ± ai

CF = c1 cos ax + c2 sin ax

And PI =

2

2 2

1( )

a Rl x

pD a


=

122

21 ( )

2

w Dx lx

EIa a

=2

2

2

2

wx lx

p a

where p = EIa2

Thus the complete solution of (1) is

y = c1 cosh ax + c2 sinh ax – 2

2

2

2

wx lx

p a

...(ii)

At the end O, t = 0 when x = 0

(ii) gives O = 1 12 2

w wc c

pa pa

At the end A, y = 0 when x = l

(ii) gives, O = c1 cosh al + c2 sinh al – 2

w

pa

c2 sinh al = 12 2(1 cosh )

w wal c

pa pa

c1 = 2tanh

2

w al

pa

Substituting in (ii)

y = 2

2 2

2cosh tanh sinh

2 2

w al wax ax x lx

ppa a

Which is the deflection of the beam at N.

Thus the central deflection = at 2

ly x

y =

2

2cosh tanh sinh 1

2 2 2 8

w al al al wl

ppa

=

2

2sec 1

2 8

w al wlh

ppa


Also the bending moment is maximum at the point of maximum

deflection 2

lx

.

The maximum bending moment

2

2at

2

d y lEI xdx

=

2( ) at 2 2

w lpy x lx x

= sech 12

w al

a

EXERCISES

1. A particle is executing simple harmonic motion with amplitude 20 cm

and time 4 seconds. Find the time required by the particle in passing

between points which are at distance 15 cm and 5 cm from the centre offorce and are on the same side of it.

Ans.: 0.38 sec.

2. An elastic string of natural length a is fixed at one end and a particle ofmass m hangs freely from the other end. The modulus of elasticity ismg. The particle is pulled down a further distance l below its

equilibrium position and released from rest. Show that the motion of the

particle is simple harmonic and find the periodicity.

3. The differential equation of a simple pendulum is

2202

d x

w xdt

= F0 sin nt, where w0 and F0 are constants. If initially x = 0,

0dx

dt determine the motion when w0 = n.

Ans.: x = 02

(sin cos )2

Fnt n nt

n

4. An e.m.f. E sin pt is applied at t = 0 to a circuit containing a capacitance

C and inductance L. The current i satisfies the equation 1di

L i dtdt C


2.9 QUESTION BANK

A. Choose the correct answer from the given alternatives.

1. A general solution of a linear differential equation of nth order contains.

(a) one constant (b) n–constants

(c) zero constants (d) two constants

2. The solution of the differential equation 2

23 0

d y dy

dxdx is

(a) y = c1x + c2e3x (b) y = c1 + c2e3x

(c) y = (c1 + c2x) e3x (d) c1x = c2

3. The solution of the differential equation (D2 + a2) y = 0 is

(a) y = c1eax + c2e–ax (b) y = c1 cos ax + c2 sin ax

(c) y = (c1x + c2) cos ax (d) y = ex (c1 cos x + c2 sin x)

4. The solution of the differential equation 2

22 0

d y dyy

dxdx is

(a) y = c1e–x + c2ex (b) y = (c1x + c2) ex

(c) y = (c1x – c2) e–x (d) y = ex (c1 cos x + c2 sin x)

5. The complementary function of y+ 2y + y = xe–x sin x is

(a) c1ex + c2e–x (b) (c1x + c2) e–x

(c) (c1x + c2) ex (d) c1 + e2 ex

6. Particular integral of the differential equation (D2 + 5D + 6) y = ex is

(a) ex (b)12

xe(c)

30

xe(d)

6

xe

7. Particular integral of the differential equation (D2 + 4) y = sin 2x is

(a) sin 22

xx (b) cos2

4

xx (c) cos 2

2

xx (d) cos 2

4

xx

8. When X = eaxV where V is any function of x then particular integral is


18. Solve 3 2

3 23 3 0

d y d y dyy

dxdx dx

19. Solve 3

30

d yy

dx

20. Solve 2

22 10 0, (0) 4, (0) 1

d y dyy y y

dxdx

C. Questions carrying six marks

21. Solve (D3 + 1) y = cos (2x – 1)

22. Solve 2

22

2 4d y dy

x xdxdx

23. Solve (D – 2)2y = 8(e2x + sin 2x + x2)

24. Solve (D2 – 4D + 3) y = sin 3x cos 2x.

25. Solve (D2 – 2D + 1) y = xex sin x.

26. Solve 2

22

secd y

a y axdx

27. Solve 2

2 22

4 4 8 sin 2xd y dyy x e x

dxdx

28. Solve the following simultaneous equations

7 , 2 5dx dy

x y x ydt dt .

29. Solve sin ; cosdx dy

y t x tdt dt given x = 2, y = 0 when t = 0.

30. Solve 5 2 5, 2 0dx dy

x y x ydt dx given x = 0, y = 0, when

t = 0.


(26) y = c1 cos ax + c2 sin ax + 2

1 1sin cos log cosx ax ax ax

a a

(27) y = (c1 + c2x) e2x – e2x [4x cos 2x + (2x2 – 3) sin 2x]

(28) x = e6t [c1 cos t + c2 sin t]

y = e6t [(c1 – c2) cos t + (c1 + c2) sin t]

(29) x = et + e–k, y = e–t – et + sin t

(30)31 1

(1 6 ) (1 3 )27 27

tx t e t

y = 32 2

(2 3 ) (2 3 )27 27

tt e t

_________

Documents

y p = ( r , d x Differential Equations - 2 Differential Equations–2 47 Example 1: Solve the following differential equations. 1) Solve 2 2 7 44 0 d y dy y dx dx Solution: The given