Assignment 1

Maximum Marks 30

Due Date 24, October 2004 Assignment Weight age 2%

Question 1 Solve the differential equation

( )22( 1) 2 1dy

x y xdx

− + = +

Solution

2 2( 1) 2 ( 1)dy

x y xdx

− + = +

2 2

2 2

( 1) 2 ( 1)

( 1) [( 1) 2 ] 0

x dy ydx x dx

x dy x y dx

⇒ − + = +

⇒ − − + − =

2 2[( 1) 2 ] ( 1) 0x y dx x dy⇒ + − − − = -----------(1) 2 2( 1) 2 1

2 2

here M x y N x

M Nx

y x

= + − = −∂ ∂= − = −∂ ∂

i.e., given differential equation is inexact, to make it exact we have to find an I.F.

Consider 2

2 2 2( )

11

M N

xy xP x

N xx

∂ ∂−− + −∂ ∂ = = =

+−

So, I.F. is 2

2 ln( 1)12

1

( 1)

dxxxu e e

x

−− ++∫= = =

+

Multiplying the (1) by I.F., we get

2

2 ( 1)[1 ] 0

( 1)( 1)

y xdx dy

xx

−− − =++

2

2 11

1( 1)

y xhere M N

xx

−= − = −++

2 2

2 2

( 1) ( 1)

M N

y xx x

∂ − ∂ −= =∂ ∂+ +

i.e., now the given differential equation is exact.

sinF F

ce M Nx y

∂ ∂= =∂ ∂

2

2 11

1( 1)

F y F xand

x y xx

∂ ∂ +⇒ = − = −

∂ ∂ −+

Integrating first w.r.t. “x “. 2

( )( 1)

yF x g y

x= + +

+

Now, taking partial derivative w.r.t.”y” 2

'( )1

Fg y

y x

∂ = +∂ +

Putting the value of F

y

∂∂

in the above equation

'( ) 1

( )

g y

g y y

⇒ = −⇒ = −

Thus, 2

( , )( 1)

(1 )( , )

( 1)

yF x y x y c

x

y xF x y x c

x

= + − =+−= + =+

Question 2 (a)Find the value of k, so that the differential equation is exact.

2 2(2 ) (2 1) 0x xxy ye dx x y ke dy+ + + − =

Solution We have to find the value of k for which the given differential equation is exact.

2 22 2 1x xHere M xy ye N x y ke= + = + −

4 4x xM Nxy e xy ke

y x

∂ ∂= + = +∂ ∂

By the given condition M

y

∂∂

should equal to N

x

∂∂

4 4x xxy e xy ke⇒ + = + 1k⇒ =

(b)Solve the differential equation by separating the variables. 2

1ln

dy yy x

dx x

+ =

Solution By separating the variables, we have

2 2( 1) ln

y dy dx

y x x=

+

Solving the L.H.S. by partial fraction

2

11 1

[ ]( 1) ln( 1)

dxxdyy x xy

− =+ +

Now, integrating on both side

2

ln ln1 ln lnln(1 )

1

x xy dx C

y x x+ + = + +

+ ∫

Question 3

Solve the differential equation 2 9 20

6 2 10

dy x y

dx x y

+ −=+ −

Solution Since the given differential equation is non homogenous,so putting

x=X+h y=Y+k

2 2 9 9 20 2 9 2 9 20

6 6 2 2 10 6 2 6 2 10

dY X h Y k X Y h k

dX X h Y k X Y h k

+ + + − + + + −= =+ + + − + + + −

Put ( )( )

2 9 20 0 1

6 2 10 0 2

h k

h k

+ − = − − − − −

+ − = − − − − −

Multiplying (1) by 3 and subtract (2) from it, we get, 25k -50=0

k=2 Putting in (1), to get the value of “h”

h=1 Now the given differential equation is homogenous for h=1 and k=2

i.e. 2 9

6 2

dY X Y

dX X Y

+=+

Putting now, Y VX

dY dVV X

dX dX

=

= +

2

2 9

6 2

2 3 2

6 2

dV VV X

dX V

dV V VX

dX V

+⇒ + =

++ −

⇒ =+

Now separating the variables

(6 2 )

(2 1)( 2)

V dV dX

V V X

+ = −+ −

After partial fraction of L.H.S., we get 2 2

2 1 2

dXdV

V V X

− + = − + −

Now integrating on both sides ln(2 1) 2( 2) ln lnV V X C− + + − = − +

Using the properties of natural logarithm, we get 2( 2)

2 1

V C

V X

− =+

SinceY

VX

= , so solution becomes

2( 2 )

2

Y XC

Y X

− =+

Since 1 2x X y Y= + = + 1 2X x Y y⇒ = − = − Now the solution becomes

( )22

(2 5)

y xC

y x

−=

+ −

Assignment 2

Maximum Marks 30

Due Date 04, November 2004 Question 1 (a)Solve the differential equation

2

2 2 xdyxy e

dx−+ = , y (0) =1

Solution

2

2

2 2 2 2

2

2

2

2 2

.

2 2

( )2

( ) 2

x

xdx x

x x x x

x

x

dyxy e

dx

I F e e

dye xye e e

dx

d ye

dx

d ye dx

−

−

+ =

∫= =

+ =

=

=

Integrate both sides

2

2

2

2 2

( ) 2

2

(0) 1

1

2 1

2

x

x

x

x x

d ye dx

ye x c

applying y

c

ye x

y xe e− −

=

= +=

=

= +

= +

(b)Solve the differential equation.

( )32 1dy

y y xdx

+ = −

Solution

( )3

3 2

2

3

3

2 1

2 1

2

2

1

1

.

(1 )

( )(1 )

( )

dx x

x x x

xx

x x x

dyy y x

dxdy

y y xdx

put v y

dv dyy

dx dxdv dy

ydx dx

dvv x

dxdv

v xdx

I F e e

dve ve x e

dx

d vex e

dx

d ve e dx xe dx

− −

−

−

−

− −

− − −

−−

− − −

+ = −

+ = −

=

= −

− =

− + = −

− = −

∫= =

− = −

= −

= −

Integrating both sides

2

x x x x

x x

ve e xe e c

ve xe c

v x c

y x c

− − − −

− −

−

= − + + += +

= += +

Question 2 (a)Find the orthogonal trajectories of the family.

2 3ay x= Where “a” is the variable parameter

Solution

2 3ay x= This is the family of cubical parabolas. First we eliminate the value of the parameter

32 3

2

2

32

2

2 3

2 3

2 3

3

2

xay x a

y

dyay x

dx

x dyy x

y dx

dyx y

dxdy y

dx x

= → =

=

=

=

=

Next we write the DE or the orthogonal family 1322

3

3 2

dyydxx

dy x

dx y

ydy xdx

= −

= −

= −

Now integrate both sides

2 2

2 2

3

23 2

y x c

y x c

= − +

+ =

A family of similarly situated ellipses

Question 3

Show that the half-life of radioactive substance is, in general,

( )2 1

1

2

ln 2

ln

t tt

AA

−=

Where 1 1 2 2 1 2( ) ( ), .A A t and A A t t t= = <

Solution

Let A (t) be the amount at time t and 0A the initial amount of the radioactive substance i.e.

A (0) = 0A Now we solve the problem

1 1 2 2, ( ) ( )dA

kA A A t and A A tdx

= = =

dA

kdtA

=

Integrate both sides

1ln A kt c= +

Taking exponent on both sides 1

1

1

ln

0

0

0

sin (0)

kt cA

ckt

ckt

kt

e e

A e e

A Ce whereC e

ceA A

C A

A A e

+=

=

= ==

=

=

If “t” the half-life then the constant is given k by

002

1

2ln 2

ln 2(3)

kt

kt

AA e

e

kt

tk

=

=

− =−= − − −

Now applying 1 1 2 2( ) ( )A A t and A A t= = we get

2

1 0

2 0

(1)

(2)

kt

kt

A A e

A A e

= − − −

= − − −

Divide equation (1). By equation (2) we get

1 2( )1

2

11 2

2

1

2

1 2

ln ( )

ln

( )

k t tAe

A

Ak t t

A

AA

kt t

−=

= −

=−

Substitute that value in equation (3)

1

2

1 2

2 1

1

2

ln 2

ln

( )

( ) ln 2

ln

tA

A

t t

t tt

AA

−=

−−=

Assignment 3

Maximum Marks 30

Due Date 17, November 2004 Question 1 (a) Determine whether the functions in problems are linearly independent or dependent on ( )∞∞− , .

( ) ( ) ( )1 2 3cos 2 , sin 2 , 1f x x f x x f x= = =

Solution

( ) ( ) ( )

( )

1 2 3

2 2

cos 2 , sin 2 , 1

cos 2 sin 2 1

2sin 2 2cos 2 0

4cos 2 4sin 2 0

cos2 0 sin 2 (0) 1(8sin 2 8cos )

8 0

f x x f x x f x

x x

x x

x x

x x x x

= = =

−− −

= − + += ≠

Which shows set of functions is linearly independent.

(b) Verify that the given functions form a fundamental set of solutions of the differential equation on the indicated interval. Form the general solution.

'' 4 0; cosh 2 ,sinh 2 ,( , )y y x x− = −∞ ∞

Solution To verify that set forms fundamental set of solutions we need to see whether both are linearly independent or not so we see wronskian is non-zero or not

( )

( )2 2

cosh 2 ,sinh 2

cosh 2 sinh 2

2sinh 2 2cosh 2

2 cosh 2 sinh 2

2

W x x

x x

x x

x x

=

= −

=

Is non-zero So the above set forms fundamental set of solutions

Question 2

(a)Verify that the given two-parameter family of functions is the general solution of the non-homogeneous differential equation.

1 2'' 8 5 ; cos 2 2 sin 2 2y y x y c x c x+ = = +

Solution

( )( )

1 2

1 2

1 2

cos 2 2 sin 2 2

' 2 2 sin 2 2 cos 2 2

'' 8 cos2 2 sin 2 2

y c x c x

y c x c x

y c x c x

= +

= − +

= − +

Now we substitute above in the associated homogeneous differential equation, if equation satisfies then two parameter family forms general solution of the associated homogeneous differential equation Consider the associated homogeneous equation

( )1 2 1 2

'' 8 0

8( cos 2 2 sin 2 2 ) 8 cos 2 2 sin 2 2 0

y y

c x c x c x c x

+ =

+ − + =

It shows that two parameter family forms general solution of the associated homogeneous equation.

(b) Find the 2nd solution of each of Differential equations by reducing order or by using the formula.

2 2 3'' 4 ' 6 0 ;x y xy y y x x− + = = +

Solution

2 2 3'' 4 ' 6 0 ;x y xy y y x x− + = = + It can also be written as

2

4 6'' ' 0y y y

x x− + =

Then 2nd solution is given by

( )( )

( )( )

2 1 21

4

2 32 22 3

4

2 32 22 3

pdx

dxx

dxx

ey y dx

y

ey x x dx

x x

ey x x dx

x x

−

− −

∫=

∫= +

+

∫= +

+

∫

∫

∫

( )( )

( )( )

4ln2 3

2 22 3

42 3

2 24 1

xey x x dx

x x

xy x x dx

x x

= ++

= ++

∫

∫

( )( )

2 32 2

1

1y x x dx

x= +

+∫

( ) ( )

( ) ( )

2 1

2 32

22

22

1

2 11

11

xy x x

y x xx

y x

− ++= +

− +

= − ++

= −

Question 3

Solve the given differential equations subject to the indicated initial conditions. ''' 5 '' 3 ' 9 0, (0) 1, '(0) 2, ''(0) 3y y y y y y y− + + = = = =

Solution Given differential equation

''' 5 '' 3 ' 9 0y y y y− + + = Put

2

3

'

''

'''

mx

mx

mx

mx

y e

y me

y m e

y m e

====

Substituting in the given differential equation, we have

( )3 25 3 9 0mxe m m m− + + =

Since mxe non zero∀ x, the auxiliary equation

( )3 25 3 9 0m m m− + + =

By using synthetic division we find the roots of the cubic equation -1 satisfied the equation so

1)1 5 3 9

1 6 9

1 6 9 0

− −

− −

−

( )( )( )

( )( )

3 2

2

2

5 3 9 0

1 6 9 0

1 3 0

1,3,3

m m m

m m m

m m

m

− + + =

→ + − + =

+ − == −

( ) 31 2 3

x xy c e c c x e−= + +

Now applying initial conditions

( ) 31 2 3

1 2

(0) 1

1 (1)

'(0) 2

x x

y

y c e c c x e

c c

y

−

== + +

= + − − −=

( )

( )

3 31 2 3 3

1 2 3

3 3 31 2 3 3 3

1 2 3

' 3

2 3 (2)

'' 9 3 3

''(0) 3

3 9 9 (3)

x x x

x x x x

y c e c c x e c e

c c c

y c e c c x e c e c e

y

c c c

−

−

= − + + += − + + − − −

= + + + +=

= + + − − −

Now by solving simultaneously above three equation

1 2

1 2 3

1 2 3

1 (1)

2 3 (2)

3 9 9 (3)

c c

c c c

c c c

= + − − −= − + + − − −= + + − − −

1 2 3

3

3 25 4, ,

28 28 73 25 4

28 28 7x x

c c c

y e x e−

= = = −

= + −

Is required solution

Assignment 4

Maximum Marks 30

Question 1

Solve the given differential equations subject to the indicated initial conditions by using method of undetermined coefficient (superposition approach)

( ) 2'' 4 ' 4 3 , (0) 2, '(0) 5xy y y x e y y−− + = + = =

Solution Complimentary function To find

cy , we first solve the associated homogeneous equation

'' 4 ' 4 0y y y− + =

We put mxey = , mxemymxmey 2 , =′′=′ Then the associated homogeneous equation gives

2( 4 4) 0mxm m e− + = Therefore, the auxiliary equation is

2 4 4 0 as 0, mxm m e x− + = ≠ ∀ Using the quadratic formula, roots of the auxiliary equation are 2,2m = Thus we have real and distinct roots of the auxiliary equation 1 22 and 2m m= =

Hence the complementary function is

1 22( ) xy c c x e

c= +

Next we find a particular solution of the non-homogeneous differential equation.

Particular Integral Since the input function

( ) 2( ) 3 xg x x e−= +

Is a quadratic polynomial Therefore, we assume that

2( ) xpy Ax B e−= +

Then / 2 // 2( 2 2 ) and (4 4 4 )x xp py Ax B A e y Ax B A e− −= − − + = + −

Therefore // / 24 4 (16 16 8 ) xp p py y y Ax B A e−− + = + −

Substituting in the given equation, we have

2 2

2 2 2 2

(16 16 8 ) (3 )

16 (16 8 ) 3

x x

x x x x

Ax B A e x e

Axe B A e xe e

− −

− − − −

+ − = ++ − = +

Or Equating the coefficients of 2 2,x xxe e− − , we have

16A 1= , -8A+16B 3=

Solving this system of equations leads to the values

1 7,

16 32A B= =

Thus a particular solution of the given equation is

2 7

16 2

x

p

ey x

− = +

.

Hence, the general solution of the given non-homogeneous differential equation is given by

pyc

yy +=

Or 2

1 2

72( )16 2

xexy c c x e x−

= + + +

Question 2 Find the annihilator operator of the following functions.

(1) 212 8 sin 4x x x+ −

Solution

Suppose that

212 8 sin 41 2( ) , y ( )x x xy x x+= = Then

( )

( )( )

3 3 21

2 2 2

2 2

D D 12 8 0,

Since the differential operator

2

is the annihilator operator of the following functions.

Then if we take 0, 4

Then the differential operator

2

sinx

y x x

D D

D D

e xα

α α β

α β

α α

β

= + =

− + +

= =

− + ( )( )( ) ( )

2

2 22 22

16 16 sin 4 0.D y D x

β+

+ = + =

Therefore, the product of two operators

( )23 2D 16D +

Annihilates the given function 2( ) 12 8 sin 4f x x x x= + − (2) 2 5sin cosx xe x e x−

Solution

Suppose that

2 5sin cos1 2( ) , y ( )x xe x e xy x x= =

Since the differential operator

( )( )2 2 22D Dα α β− + +

is the annihilator operator of the following functions.

cosxe xα β And sinxe xα β Then if we take

2, 1α β= = Then the differential operator

( )( )2 2 22D Dα α β− + +

( ) ( )

( )( )( ) ( )

2 2 21

2 2 2

2 2 52

4 5 4 5 sin 0,

And if we take

5, 1

Then the differential operator

2

10 26 10 26 cos 0.

x

x

D D y D D e x

D D

D D y D D e x

α β

α α β

− + = − + =

= =

− + +

− + = − + =

Therefore, the product of two operators

( )( )2 210 26 4 5D D D D− + − +

Annihilates the given function 2 5( ) sin cosx xf x e x e x= −

Question 3

Solve the given differential equations using method of variation of parameter.

'' ' tany y x+ =

Solution

'' ' tany y x+ = This equation is already in standard form y’’+ P(x)y’ +Q(x)y = f(x) Therefore, we identify the function f(x) as f(x) = tanx To find

cy , we first solve the associated homogeneous equation

'' ' 0y y+ =

We put mxey = , mxemymxmey 2 , =′′=′ Then the associated homogeneous equation gives

2( ) 0mxm m e+ = Therefore, the auxiliary equation is

2 0 as 0, mxm m e x+ = ≠ ∀ Using the quadratic formula, roots of the auxiliary equation are

( 1) 0

0, 1

m m

m

+ == −

Thus we have real and distinct roots of the auxiliary equation 1 20 and 1m m= = −

Hence the complementary function is

1 2xy c c e

c−= +

We construct the determinants Since 1 1y = , 2

xy e−= so

1 2

1

2

1( , )

0

0tan

tan

1 0tan

0 tan

xx

x

xx

x

eW y y e

e

eW e x

x e

W xx

−−

−

−−

−

= = −−

= = −−

= =

We determine the derivatives of the function 1u and 2u

11 1

22 2

1 2

tan' tan

tan' tan

x

x

xx

xp

W e xu u xdx

W eW x

u u e xdxW e

y u u e

−

−

−

−

−= = → =−

= = → = −−

= +

∫

∫ V

Is required particular solution

Assignment 5

Maximum Marks 30

Question 1

Solve the given differential equations.

24 '' 0, ( 1) 2, '( 1) 4x y y y y+ = − = − =

Solution Suppose that mxy = , then

1−= mmxdx

dy , .)1( 2

2

2−−= mxmm

dx

yd

Substituting in the differential equation, we get: 2 24 '' 0 (4 ( 1) 1) (4 4 1) 0m mx y y x m m x m m+ = = − + = − + =

If 24 4 1 0m m− + = or 2(2 1) 0m − = .

Since 1

1

2m = , the general solution is

1 1

2 21 2

1 2

1 2

-1

-1

ln

ln

2 1 1 ln( 1)

ln (z) = ln z +i arg(z) where arg(z)= tan y/x , z=x+iy

since -1 is also complex number as -1+0i

so

ln( -1)=ln -1 +itan 0/-1 =ln1+i0=0

y c x c x x

y c x c x x

c c

= +

= +

= − + − −

1 2

1 1

( 1) 2

2= (0)

2= 2i=

Applying y

c i c i

c i c

− =+→ −

1 2

1 2 2

1 2 2

ln

' ln2 2

' ln2 2

Now y c x c x x

c c c xy x

xx xc c c

y xx x x

= +

= + +

= + +

1 2 2

1 2

2

2

2

'( 1) 4

4 ln( 1)2 2

422i

42

4 1

5

2 5 ln

Applying y

c c c

i i ic c

i ic

i ic

ic i

y i x i x x

− =

= + − +

= +

−= +

+ =

=

= − +

Question 2 Write the solution of the given initial value problem in form ( ) ( )sinx t A tω φ= +

.1 '' 10 0

(0) 1, '(0) 1

x x

x x

+ == =

Solution

2

2

2

2

2

2

.1 10 0

110 0

10

100 0

d xx

dt

d xx

dt

d xx

dt

+ =

+ =

+ =

Put mtmt emdt

xdex 2

2

2

, ==

Then, the auxiliary equation is

2 100 0m + =

⇒ 0 10m i= ±

Therefore, the general solution is:

( ) 1 2cos10 sin10x t c t c t= +

Now we apply the initial conditions.

( ) 1 20 1 .1 .0 1x c c= ⇒ + =

Thus 1 1c =

So that ( ) 2(1)cos10 sin10x t t c t= +

210sin10 10 cos10dx

t c tdt

= − +

Therefore ( )0 1x′ = ( ) 210 0 10 .1 1c⇒ − + =

Thus 2

1

10c =

Hence, the solution of the initial value problem is

( ) 1cos10 sin10

10x t t t= +

Thus amplitude of motion is given by

( )2

2 1 1011 1.000498

10 10A

= + = =

And the phase angle is defined by

sin1 10

0101 /10 101

φ = = >

cos1/10 1

0101 /10 101

φ >= =

Therefore

tan 10φ =

or φ = ( )1tan 10 84.2894 radians− =

Hence the required form of the solution is

( ) ( )( ) ( )

sin 84.2894

1.000498sin 10 84.2894

x t A t

x t t

ω φ= +

= +

Question 3

Solve the given differential equations by alternative method of Cauchy –Euler equation.

23

5'' 9 ' 20x y xy y

x+ − =

Solution First consider the associated homogeneous differential equation.

23

5'' 9 ' 20x y xy y

x+ − =

With the notation 22

2

, Ddx

dD

dx

d == , the differential equation becomes:

2 2( 9 20) 0x D xD y+ − =

With the substitution tx e= or lnt x= , we have: ∆=xD , )1(22 −∆∆=Dx So the homogeneous differential equation becomes: [ ( 1) 9 20] 0y∆ ∆ − + ∆ − =

2( 9 20) 0y∆ + ∆ − =

or 2

29 20 0

d y dyy

dt dt+ − =

Put mty e= then the AE is:

2 9 20 0m m+ − = or 10,2m = −

10 21 2

t tcy c e c e−∴ = + , as tx e=

10 21 2cy c x c x−= +

For py we write the differential equation as:

2 5

9 20 5'' 'y y y

x x x+ − =

we make the identification 5

5( )f x

x= . Now with 10

1y x−= , 22 xy = , and

10 21 2py u x u x−= + , where 1u and 2u are functions given by

11

Wu

W=′ , 2

2W

uW

=′ ,

with 10 2

1 9 2 11 9911

122 10 12

10 2

x xW x x x

xx x

−− − −

−= = + = =

− ,

35

2 5,1

0

5 2 x

xW

x x−= = − and

10

1511 5

52

0

10 5 x

xW

x x

−

− −= =

−

So that 3 6

9

55

1 12 12x x

x

u = = −′ and 15

69

55

2 12 12x

xx

u = =′

7 75 51 12 84

x xu dx∴ = − = −∫

and 6 5

5 12

12 12x xu dx= = −∫ .

Therefore

10 7 3 3 35

5 1 5 1 1284 84 12 712

xpx

y x x x x x− − − − − − − − = −

= + =

Hence the general solution is: c py y y= +

10 31

72

1 2y xc x c x− −−= +

Assignment 6

Maximum Marks 30

Question 1

(a)The first four terms of a power series in x for the given function.

22 4 6

1 ...2 3 4

x x x − + − +

Solution

22 4 6

2 4 6 2 4 6

2 4 6 2 2 4 6

4 2 4 6 6 2 4 6

2

1 ...2 3 4

1 ... 1 ...2 3 4 2 3 4

1 1 ... 1 ...2 3 4 2 2 3 4

1 ... 1 ... ...3 2 3 4 4 2 3 4

1

x x x

x x x x x x

x x x x x x x

x x x x x x x x

x

− + − +

= − + − + − + − +

= − + − + − − + − + +

− + − + − − + − + +

= −4 6 2 4 6 8

4 6 8 10 6 8 10 12

0 2 4 6

2 4

... ...2 3 4 2 4 6 8

... ... ...3 6 9 12 4 8 12 16

, , ,

1 1 1 1 1 1 1 1 11

2 2 3 4 3 4 6 6 4

x x x x x x

x x x x x x x x

now we just collect the terms of x x x x

x x

+ − + − − + +

+ − + − + − − + − + +

= + − − + + + + − − − −

6

4 62 11 5

112 6

x

x xx

+

= − + − +

⋯

⋯

Question 1

(b) Find the interval of convergence of the given power series.

( ) ( )1

15

10

kk

kk

x∞

=

−−∑

Solution Firstly we need to find radius of convergent

0 0

( )kk k

k k

a c x a∞ ∞

= == −∑ ∑ =

( ) ( )1

15

10

kk

kk

x∞

=

−−∑ , a=5,

( )1

10

k

k kc

−= =

Then R=

( )

( )( )( )

1

11

1

11 1010lim lim lim 10

1 1 10

10

k

k kkn

k k kn n nn

k

c

c

+

+→∞ →∞ →∞++

−−

= = =− −

( )x a− <R

( 5)x − <10

-10<x-5<10

-10+5<x<15

-5<x<15

This means that the power series converges ifx belongs to the interval

( 5,15)−

The series diverges outside this interval i.e. when 5x > − or 15x < . The convergence of the power series at the numbers 5− and 15 must be investigated separately by substituting into the power series.

a) When we substitute 5x = − , we obtain

( ) ( )1

15

10

kk

kk

x∞

=

−−∑ =

( ) ( )

( ) ( ) ( )

( )

1

1

2

1

15 5

10

1 110

10

1

kk

kk

k kk

kk

k

k

∞

=

∞

=∞

=

−− −

− −=

= −

∑

∑

∑

which is a divergent by divergent test.

b) When we substitute 15x = , we obtain

( ) ( )1

15

10

kk

kk

x∞

=

−−∑ =

( ) ( )

( ) ( )

( )

1

1

1

115 5

10

110

10

1

kk

kk

kk

kk

k

k

∞

=

∞

=∞

=

−−

−=

= −

∑

∑

∑

which is divergent, by divergent test. Hence, the interval of convergence of the power series is

( )5,15− . This means that the series is convergent for those vales of x which satisfy

5 15x− < <

Question 2

Show that the indicial roots do not differ by an integer. Use the method of frobenius to obtain two linearly independent series solutions about the regular singular point 00 =x Form the general solution

on (0,∞ )

22 '' ( 1) ' 0x y x x y y− − − =

Solution As 0=x is regular singular points of the differential equation

22 '' ( 1) ' 0x y x x y y− − − =

We try a solution of the form =y ∑∞

=

+

0

.n

rnn xc

Therefore ∑∞

=

−++=′0

1.)(n

rnn xcrny

And ∑∞

=

−+−++=′′0

2.)1)((n

rnn xcrnrny

22 '' ( 1) 'x y x x y y− − − = 2 2

0

2 ( )( 1) n rn

n

x n r n r c x∞

+ −

=

+ + −∑ ( )2 1

0

( ) n rn

n

x x n r c x∞

+ −

=− − +∑ - ∑

∞

=

+

0

.n

rnn xc

= 2 2

0

2( )( 1) n rn

n

n r n r c x∞

+ + −

=+ + −∑

1 2 1 1

0 0

( ) ( )n r n rn n

n n

n r c x n r c x∞ ∞

+ − + + − +

= =− + + +∑ ∑ ∑

∞

=

+−0

.n

rnn xc

=0

2( )( 1) n rn

n

n r n r c x∞

+

=+ + −∑ 1

0 0

( ) ( )n r n rn n

n n

n r c x n r c x∞ ∞

+ + +

= =− + + +∑ ∑ ∑

∞

=

+−0

.n

rnn xc

=0

2( )( 1) n rn

n

n r n r c x∞

+

=+ + −∑ 1

0 0

( ) ( 1)n r n rn n

n n

n r c x c x n r∞ ∞

+ + +

= =− + + + −∑ ∑

=0

2( )( 1) n rn

n

n r n r c x∞

+

=+ + −∑ 1

0 0

( 1) ( )n r n rn n

n n

c x n r n r c x∞ ∞

+ + +

= =+ + − − +∑ ∑

=0

2( )( 1) n rn

n

n r n r c x∞

+

=+ + −∑ 1

0 0

( 1) ( )n r n rn n

n n

c x n r n r c x∞ ∞

+ + +

= =+ + − − +∑ ∑

= [ ] 1

0 0

( 1) 2 2 1 ( )n r n rn n

n n

c x n r n r n r c x∞ ∞

+ + +

= =+ − + + − +∑ ∑

= ( ) 00

1

1 (2 1) ( 1)(2 2 1)r nn

n

x r r c x n r n r c x∞

=

− + + + − + +

∑ 1

0

( ) nn

n

n r c x∞

+

=

− +

∑

1−= nk nk =

Then ( ) ( ) 10 1

0

1 (2 1) [( 1 1)(2 2 2 1) ] 0r kk k

k

x r r c k r k r c c k r x∞

++

=

− + + + + − + + + − + =

∑

( ) 10 1

0

1 (2 1) [( )(2 2 3) ( )] 0r kk k

k

x r r c k r k r c c k r x∞

++

=

− + + + + + − + =

∑

( ) 10 1

0

1 (2 1) ( ) [(2 2 3) ] 0r kk k

k

x r r c k r x k r c c∞

++

=

− + + + + + − =

∑

This implies ( ) 01 (2 1) 0r r c− + =

1( )[(2 2 3) ] 0k kk r k r c c++ + + − = , 0,1,2,...k =

Since nothing is gained by taking 00 =c , we must then have

( )1 (2 1) 0r r− + = [Called the indicial equation and its roots 1

1,2

r = − are called

indicial roots or exponents of the singularity.]

And 1 (2 2 3)k

k

cc

k r+ =+ +

, 0,1,2,...k = Substitute 1

1

2r = −

and 2 1r = in the above equation and these values will give two different recurrence relations:

For 1

1

2r = − , 1 (2 2 3)

kk

cc

k r+ =+ +

0,1,2,...k = (1)

For 2 1r = , 1 (2 2 3)k

k

cc

k r+ =+ +

0,1,2,...k = (2)

Iteration of (1) gives 0 0 01 2( 1/ 2) 3 1 3 2

c c cc = = =

− + − +

0 01 12 22 2( 1/ 2) 3 5 1 2.4 2!2

c cc cc = = = =

+ − + −

0 02 23 34 2( 1/ 2) 3 7 1 2.4.6 3!2

c c ccc = = = =

+ − + −

3 3 0 04 46 2( 1/ 2) 3 9 1 2.4.6.8 4!2

c cc cc = = = =

+ − + −

In general 0 , 0,1,2,...!2n n

cc n

n= =

Iteration of (2) gives

0 01 0 2 3 5

c cc = =

+ +

01 12 2 2 3 7 5.7

cc cc = = =

+ +

02 13 4 2 3 9 5.7.9

cccc = = =

+ +

3 014 6 2 3 11 5.7.9.11

cccc = = =

+ +

In general 0

5.7.9...(2 3)n

cc

n=

+, 1,2,...n =

Thus we obtain two series solutions

1

021 0

1

1!2

nn

n

cy c x x

n

∞−

=

= +

∑ (3)

12 0

1

11

5.7.9...(2 3)n

n

y c x xn

∞

=

= + +

∑ . (4)

By the ratio test it can be demonstrated that both (3) and (4) converge for all finite values of x. Also it should be clear from the form of (3) and (4) that neither series is a constant multiple of the other and therefore, )(1 xy and )(2 xy are linearly independent on the x-axis. Hence by the superposition principle

)()( 2211 xyCxyCy += =2 2

3 31

1

1

!5.8.11.14...(3 2)

n

n

C x xn n

∞ +

=

+ + ∑

21

11

!1.4.7...(3 2)n

n

C xn n

∞

=

+ + −

∑ , ∞<x

is an other solution of the differential equation. On any interval not containing the origin, this combination represents the general solution of the differential equation

Question 3

(a) Find the general solution of the Bessel differential equation

2 2 10

361x y xy x y

′′ ′+ + − =

On ( )∞ ,0

Solution

The Bessel differential equation is

( ) 0222 =−+′+′′ yvxyxyx (1)

2 2 10

361x y xy x y

′′ ′+ + − =

(2)

Comparing (1) and (2), we get 2 1

361v = , therefore

1

19v = ±

So general solution of (1) is ( ) ( )1 1/19 2 1/19J Jy C x C x−= + Question 3

(b) Express the given Bessel function in terms of sin x andcosx , and power ofx .

( ) ( )3/ 2 3/ 2J x and J x−

If

( ) ( ) ( )1 1

1/ 2 1/ 2

2

2 2( ) sin , ( ) cos

v v vv

J x J x J xx

J x x J x xx xπ π

− +

−

+ =

= =

Solution To find ( )3/ 2J x

Consider

( ) ( ) ( )1 12

v v vv

J x J x J xx− ++ =

For 12

v =

( ) ( ) ( ) ( )1 1 11 12 2 2

12 2J x J x J xx− ++ =

( ) ( ) ( )

( ) ( ) ( )

1 3 12 22

3 1 12 22

1

1

J x J x J xx

J x J x J xx

−

−

+ =

= −

As given (we know)1/ 2 1/ 22 2

( ) sin , ( ) cosJ x x J x xx xπ π−= =

( )

( )

32

32

1 2 2sin cos

2 sincos

J x x xx x x

xJ x x

x x

π π

π

= −

= −

To find ( )3/ 2J x−

Consider

( ) ( ) ( )1 12

v v vv

J x J x J xx− ++ =

For 12

v = −

( ) ( ) ( ) ( )1 1 11 12 2 2

12 2J x J x J xx− − − + −

−+ =

( ) ( ) ( )

( ) ( ) ( )

3 1 12 22

3 1 12 22

1

1

J x J x J xx

J x J x J xx

−−

−−

+ = −

= − −

As given (we know)1/ 2 1/ 22 2

( ) sin , ( ) cosJ x x J x xx xπ π−= =

( )

( )

32

32

1 2 2cos sin

2 cossin

J x x xx x x

xJ x x

x x

π π

π

−

= − −

− = − −

Assignment 6

Maximum Marks 30

Question 1

(a)The first four terms of a power series in x for the given function.

22 4 6

1 ...2 3 4

x x x − + − +

Solution

22 4 6

2 4 6 2 4 6

2 4 6 2 2 4 6

4 2 4 6 6 2 4 6

2

1 ...2 3 4

1 ... 1 ...2 3 4 2 3 4

1 1 ... 1 ...2 3 4 2 2 3 4

1 ... 1 ... ...3 2 3 4 4 2 3 4

1

x x x

x x x x x x

x x x x x x x

x x x x x x x x

x

− + − +

= − + − + − + − +

= − + − + − − + − + +

− + − + − − + − + +

= −4 6 2 4 6 8

4 6 8 10 6 8 10 12

0 2 4 6

2 4

... ...2 3 4 2 4 6 8

... ... ...3 6 9 12 4 8 12 16

, , ,

1 1 1 1 1 1 1 1 11

2 2 3 4 3 4 6 6 4

x x x x x x

x x x x x x x x

now we just collect the terms of x x x x

x x

+ − + − − + +

+ − + − + − − + − + +

= + − − + + + + − − − −

6

4 62 11 5

112 6

x

x xx

+

= − + − +

⋯

⋯

Question 1

(b) Find the interval of convergence of the given power series.

( ) ( )1

15

10

kk

kk

x∞

=

−−∑

Solution Firstly we need to find radius of convergent

0 0

( )kk k

k k

a c x a∞ ∞

= == −∑ ∑ =

( ) ( )1

15

10

kk

kk

x∞

=

−−∑ , a=5,

( )1

10

k

k kc

−= =

Then R=

( )

( )( )( )

1

11

1

11 1010lim lim lim 10

1 1 10

10

k

k kkn

k k kn n nn

k

c

c

+

+→∞ →∞ →∞++

−−

= = =− −

( )x a− <R

( 5)x − <10

-10<x-5<10

-10+5<x<15

-5<x<15

This means that the power series converges ifx belongs to the interval

( 5,15)−

The series diverges outside this interval i.e. when 5x > − or 15x < . The convergence of the power series at the numbers 5− and 15 must be investigated separately by substituting into the power series.

a) When we substitute 5x = − , we obtain

( ) ( )1

15

10

kk

kk

x∞

=

−−∑ =

( ) ( )

( ) ( ) ( )

( )

1

1

2

1

15 5

10

1 110

10

1

kk

kk

k kk

kk

k

k

∞

=

∞

=∞

=

−− −

− −=

= −

∑

∑

∑

which is a divergent by divergent test.

b) When we substitute 15x = , we obtain

( ) ( )1

15

10

kk

kk

x∞

=

−−∑ =

( ) ( )

( ) ( )

( )

1

1

1

115 5

10

110

10

1

kk

kk

kk

kk

k

k

∞

=

∞

=∞

=

−−

−=

= −

∑

∑

∑

which is divergent, by divergent test. Hence, the interval of convergence of the power series is

( )5,15− . This means that the series is convergent for those vales of x which satisfy

5 15x− < <

Question 2

Show that the indicial roots do not differ by an integer. Use the method of frobenius to obtain two linearly independent series solutions about the regular singular point 00 =x Form the general solution

on (0,∞ )

22 '' ( 1) ' 0x y x x y y− − − =

Solution As 0=x is regular singular points of the differential equation

22 '' ( 1) ' 0x y x x y y− − − =

We try a solution of the form =y ∑∞

=

+

0

.n

rnn xc

Therefore ∑∞

=

−++=′0

1.)(n

rnn xcrny

And ∑∞

=

−+−++=′′0

2.)1)((n

rnn xcrnrny

22 '' ( 1) 'x y x x y y− − − = 2 2

0

2 ( )( 1) n rn

n

x n r n r c x∞

+ −

=

+ + −∑ ( )2 1

0

( ) n rn

n

x x n r c x∞

+ −

=− − +∑ - ∑

∞

=

+

0

.n

rnn xc

= 2 2

0

2( )( 1) n rn

n

n r n r c x∞

+ + −

=+ + −∑

1 2 1 1

0 0

( ) ( )n r n rn n

n n

n r c x n r c x∞ ∞

+ − + + − +

= =− + + +∑ ∑ ∑

∞

=

+−0

.n

rnn xc

=0

2( )( 1) n rn

n

n r n r c x∞

+

=+ + −∑ 1

0 0

( ) ( )n r n rn n

n n

n r c x n r c x∞ ∞

+ + +

= =− + + +∑ ∑ ∑

∞

=

+−0

.n

rnn xc

=0

2( )( 1) n rn

n

n r n r c x∞

+

=+ + −∑ 1

0 0

( ) ( 1)n r n rn n

n n

n r c x c x n r∞ ∞

+ + +

= =− + + + −∑ ∑

=0

2( )( 1) n rn

n

n r n r c x∞

+

=+ + −∑ 1

0 0

( 1) ( )n r n rn n

n n

c x n r n r c x∞ ∞

+ + +

= =+ + − − +∑ ∑

=0

2( )( 1) n rn

n

n r n r c x∞

+

=+ + −∑ 1

0 0

( 1) ( )n r n rn n

n n

c x n r n r c x∞ ∞

+ + +

= =+ + − − +∑ ∑

= [ ] 1

0 0

( 1) 2 2 1 ( )n r n rn n

n n

c x n r n r n r c x∞ ∞

+ + +

= =+ − + + − +∑ ∑

= ( ) 00

1

1 (2 1) ( 1)(2 2 1)r nn

n

x r r c x n r n r c x∞

=

− + + + − + +

∑ 1

0

( ) nn

n

n r c x∞

+

=

− +

∑

1−= nk nk =

Then ( ) ( ) 10 1

0

1 (2 1) [( 1 1)(2 2 2 1) ] 0r kk k

k

x r r c k r k r c c k r x∞

++

=

− + + + + − + + + − + =

∑

( ) 10 1

0

1 (2 1) [( )(2 2 3) ( )] 0r kk k

k

x r r c k r k r c c k r x∞

++

=

− + + + + + − + =

∑

( ) 10 1

0

1 (2 1) ( ) [(2 2 3) ] 0r kk k

k

x r r c k r x k r c c∞

++

=

− + + + + + − =

∑

This implies ( ) 01 (2 1) 0r r c− + =

1( )[(2 2 3) ] 0k kk r k r c c++ + + − = , 0,1,2,...k =

Since nothing is gained by taking 00 =c , we must then have

( )1 (2 1) 0r r− + = [Called the indicial equation and its roots 1

1,2

r = − are called

indicial roots or exponents of the singularity.]

And 1 (2 2 3)k

k

cc

k r+ =+ +

, 0,1,2,...k = Substitute 1

1

2r = −

and 2 1r = in the above equation and these values will give two different recurrence relations:

For 1

1

2r = − , 1 (2 2 3)

kk

cc

k r+ =+ +

0,1,2,...k = (1)

For 2 1r = , 1 (2 2 3)k

k

cc

k r+ =+ +

0,1,2,...k = (2)

Iteration of (1) gives 0 0 01 2( 1/ 2) 3 1 3 2

c c cc = = =

− + − +

0 01 12 22 2( 1/ 2) 3 5 1 2.4 2!2

c cc cc = = = =

+ − + −

0 02 23 34 2( 1/ 2) 3 7 1 2.4.6 3!2

c c ccc = = = =

+ − + −

3 3 0 04 46 2( 1/ 2) 3 9 1 2.4.6.8 4!2

c cc cc = = = =

+ − + −

In general 0 , 0,1,2,...!2n n

cc n

n= =

Iteration of (2) gives

0 01 0 2 3 5

c cc = =

+ +

01 12 2 2 3 7 5.7

cc cc = = =

+ +

02 13 4 2 3 9 5.7.9

cccc = = =

+ +

3 014 6 2 3 11 5.7.9.11

cccc = = =

+ +

In general 0

5.7.9...(2 3)n

cc

n=

+, 1,2,...n =

Thus we obtain two series solutions

1

021 0

1

1!2

nn

n

cy c x x

n

∞−

=

= +

∑ (3)

12 0

1

11

5.7.9...(2 3)n

n

y c x xn

∞

=

= + +

∑ . (4)

By the ratio test it can be demonstrated that both (3) and (4) converge for all finite values of x. Also it should be clear from the form of (3) and (4) that neither series is a constant multiple of the other and therefore, )(1 xy and )(2 xy are linearly independent on the x-axis. Hence by the superposition principle

)()( 2211 xyCxyCy += =2 2

3 31

1

1

!5.8.11.14...(3 2)

n

n

C x xn n

∞ +

=

+ + ∑

21

11

!1.4.7...(3 2)n

n

C xn n

∞

=

+ + −

∑ , ∞<x

is an other solution of the differential equation. On any interval not containing the origin, this combination represents the general solution of the differential equation

Question 3

(a) Find the general solution of the Bessel differential equation

2 2 10

361x y xy x y

′′ ′+ + − =

On ( )∞ ,0

Solution

The Bessel differential equation is

( ) 0222 =−+′+′′ yvxyxyx (1)

2 2 10

361x y xy x y

′′ ′+ + − =

(2)

Comparing (1) and (2), we get 2 1

361v = , therefore

1

19v = ±

So general solution of (1) is ( ) ( )1 1/19 2 1/19J Jy C x C x−= + Question 3

(b) Express the given Bessel function in terms of sin x andcosx , and power ofx .

( ) ( )3/ 2 3/ 2J x and J x−

If

( ) ( ) ( )1 1

1/ 2 1/ 2

2

2 2( ) sin , ( ) cos

v v vv

J x J x J xx

J x x J x xx xπ π

− +

−

+ =

= =

Solution To find ( )3/ 2J x

Consider

( ) ( ) ( )1 12

v v vv

J x J x J xx− ++ =

For 12

v =

( ) ( ) ( ) ( )1 1 11 12 2 2

12 2J x J x J xx− ++ =

( ) ( ) ( )

( ) ( ) ( )

1 3 12 22

3 1 12 22

1

1

J x J x J xx

J x J x J xx

−

−

+ =

= −

As given (we know)1/ 2 1/ 22 2

( ) sin , ( ) cosJ x x J x xx xπ π−= =

( )

( )

32

32

1 2 2sin cos

2 sincos

J x x xx x x

xJ x x

x x

π π

π

= −

= −

To find ( )3/ 2J x−

Consider

( ) ( ) ( )1 12

v v vv

J x J x J xx− ++ =

For 12

v = −

( ) ( ) ( ) ( )1 1 11 12 2 2

12 2J x J x J xx− − − + −

−+ =

( ) ( ) ( )

( ) ( ) ( )

3 1 12 22

3 1 12 22

1

1

J x J x J xx

J x J x J xx

−−

−−

+ = −

= − −

As given (we know)1/ 2 1/ 22 2

( ) sin , ( ) cosJ x x J x xx xπ π−= =

( )

( )

32

32

1 2 2cos sin

2 cossin

J x x xx x x

xJ x x

x x

π π

π

−

= − −

− = − −