Differential Equations - Solved Assignments - Semester Fall 2006

8/9/2019 Differential Equations - Solved Assignments - Semester Fall 2006

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Assignment 1 of MTH401 (Fall 2006)

Q.No:1 2( 2 2) ( 2 ) 0 xy x y dx x x dy+ + + + + = --------------(A)

Solution:

[ ]( 2) 1( 2) ( 2) 0

( 1)( 2) ( 2) 0

( 2) ( 1)( 2)

( 1)( 2)

( 2)

1.( 2) (1)

( 2)

( ) . ( )

( 1)( )

( 2)

( ) 2

( ) 0

2

x y y dx x x dy

x y dx x x dy

x x dy x y dx

dy x y

dx x x

dy x y

dx x x

dyh x x g y

dx

suchthat

xh x

x x

and

g y y

Solving g y

y

+ + + + + =

+ + + + =

+ = − + +

+ += −

+

+= − + − − − − − − − − − −

+

=

+= −

+

= +

=

+ 0

2 ( )

i.e. constant solution to the given (1).

y a

=

= − − − − − −− −

For nonconstant solution, separating variables in Eq. (1)

1

2 ( 2)

dy xdx

y x x

+= −

+ +

Integrating,

2

2

1

2 2

( 2) 1 (2 2)( 2)

( 2) 2 2

dy xdx

y x x

d y xdx dy d y

y x x

+= −

+ +

+ += − = +

+ +

∫ ∫

∫ ∫ ∵



2

2

2 2

1

12 2

2

1 ( 2 )ln 2

2 2

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

2ln 2 ln( 2 ) ln

d x x y

x x

y x x c d x x x dx

y x x c−

+⇒ + = −

+

⇒ + = − + + + = +

+ = + +

∫

∵

As ln lnm y ym=

And 1 2ln ( )c c say=

12 2

2ln 2 ln ( 2 ) y x x c− + = +

Taking antilog to base ‘e’ on both sides1

2 222 ( 2 ) y x x c

−⇒ + = +

So, exact explicit form is

22

2 ( ) ( )2

c y b where c c say

x x= − − − − − − − =

+

Finally, combining both (a) & (b); so that solution (constant and

nonconstant) are

22

2

2

c y

x x

y

= −

+ = −

Q.No:22 2 2( ) 0 x xy y dx x dy+ + − =

Solution:2 2 2( ) 0 x xy y dx x dy+ + − =

2 2 2

2 2

2

( )

( , ) (1)

x xy y dx x dy

dy x xy y f x y

dx x

⇒ + + =

+ +⇒ = = − − − − − − −

2 2 2 2

2 2

2 2 2

2 2

0

Replace (1)

( )( )

( , )

tx x and ty y in

dy t x tx ty t y

dx t x

t x xy y

t x

t f x y

→ →

+ +⇒ =

+ +=

=

⇒ Given diff.eq is homogeneous of degree zero.



Put solution put

( )

2 2 2

2

2

( )(1)

1

(1, )

yv

x

dy d vx

dx dxdv

v xdx

dv x x vx v xv x

dx x

v v

f v

=

⇒ =

= +

+ +∴ ⇒ + =

= + +

=

( )

2

2

2

1

1 (2)

11

( ). ( )

dvv x v v

dx

dv x v

dx

dvv

dx x

h x g v

+ = + +

⇒ = + − − − − − − −

= +

=

Where,

21( ) ( ) 1h x and g v v

x= = +

For constant solution solving

2

2

For constant solution,put ( ) 0

1 0

1

g v

v

v

v i

yi

x

y ix

=

⇒ + =

⇒ = −

⇒ = ±

= ±

= ±

Which are imaginary.

Now separating variables in (2) to find nonconstant solutions;

2

2

(2)1

;

1

dv dx

v x

Onintegrating

dv dx

v x

∴ ⇒ =+

=+

∫ ∫



[ ]

[ ][ ]

1tan ln

tan ln

tan ln

tan ln

v x c

v x c

y

x c x

y x x c

−= +

= +

= +

= +

i-e is the required real solution

Q.No:32 3 2(2 cos 3 ) ( sin ) 0 (1) x y x y dx x x y y dy+ + − − = − − − − −

Solution:

2

2

2 cos 3

2 sin 3

Here

M x y x y

M x y x

y

= +

∂⇒ = − +

∂

3 2

2

sin

3 2 sin

(1) .

N x x y y

N x x y

x

M N

y x

is exact

= − −

∂= −

∂

∂ ∂=

∂ ∂

⇒

∵

L.H.S. of (1) is an exact differential i.e. there exist a function f(x,y) s.t.

2

3 2

2 cos 3 (2)

sin (3)

f x y x y M

x

f x x y y N

y

∂= + = − − − − − −

∂

∂= − − = − − − − − −

∂

Integrating (2) w.r.t. ‘x’,we get2( , ) 2 cos 3 ( ) f x y x y dx x y dx h y= + +∫ ∫

Where h(y) is constant of integration w.r.t.’x’2

2 3

( , ) (2cos ). (3 ) ( )

cos ( ) (4)

f x y y xdx y x dx h y

x y x y h y

= + +

= + + − − − − − −

∫ ∫

Now diff. w.r.t.’y’,we get



2 3

3 2 3 2

2

22 3

sin ( ) (3)

sin ( ) sin (Using value of N)

( )

( )2

(2)

( , ) cos2

f x y x h y N from

y

x x y h y x x y y

h y y

yh y

So required solution

y f x y x y x y c

∂′= − + + =

∂

′= − + = − −

′⇒ = −

−⇒ =

∴ = + − =

Alternate Method:

2

2

2 cos 3

2 sin 3

Here

M x y x y

M x y x

y

= +

∂⇒ = − +

∂

3 2

2

sin

3 2 sin

(1) .

N x x y y

N x x y

x

M N

y x

is exact

= − −

∂= −

∂

∂ ∂=

∂ ∂

⇒

∵

Therefore diff.eq: has solution given by

y-Constant

2

Students remember above formula

for the solution of exact Differential equati

( )

(2 cos 3 ) (

on

{

}

)

Mdx terms of N not containing x dy c

x y x y dx y dy c

+ =

+ + − =

∫ ∫

∫ ∫

22

22 3

(2cos ) 3

2

cos2

y y x dx y x dx c

y x y yx c

+ − =

+ − =

∫ ∫





42 2( ) ; ( )

3 p x q x x

x= − =

So, integrating factor for (2) is

( )

2

( )

2

2ln

ln

2

2

2

432 2

2

3

2

332

. .

1.(2); .

( . .) ( ) .

1 2 1. ( . )

3

2

3

2.

3 3

1 2.

9

p x dx

dx x

x

x

I F e

e

e

e

x

Multiplying it witheq suchthatit becomes exact x

Thereforeits solution is givenby

v I F q x I F dx c

y x dx c x x

x dx c

xc

y x c

x

−

−

−

−

∫=

∫=

=

=

=

=

= +

= +

= +

= +

= +

⇒

∫

∫

∫

2

5 23

35 2 2

2

9

2( )9

.

y x cx

y x cx

Hencetheresult

= +

= +



Question 2marks 10

Find the orthogonal trajectories of the family of the curves 2 4 y cx= .

Solution:2 4 y cx= ---------- (A)

Diff. w.t.r. ‘x’; we get



2

2

2

2 4

2(1)

(1) .

1

2

2

2

. . .

1

2

1

2

1ln

2

.

( )

,

x D

c

x

Dc

x

c

dy y c

dx

dy c

dx yeq for theorthogonal family

dy

cdx

y

y

c

dy y

dx c

i e Seperablelinear eq So

dydx

y c

Integrating

dydx

y c

y x Dc

y e

e e

y Ae B

Hence familyof curv

− +

−

−

=

⇒ − − − − − −

⇒

= −

= −

= −

= −

= −

= − +

=

=

⇒ = − − − − −

∫ ∫

( ) ( ).

.

e A are orthogonalto family B

Hence the result



Question 3 marks 10

The population of a certain country is known to increase at a rate proportional to the number of

people presently living in the country. If after two years the population has doubled, and afterthree years the population is 20,000, estimate the number of people initially living in the

country.

Solution:

Suppose that 0P is initial population of country and P(t), the population at any time ‘t’ then

population growth is governed by the diff.eq.

0

0

0 0

2

(0.3457)

0

(1)

'( ) (2)

(2) 2

(2) 2

2

2 ln(2)

ln 2

2

0.693147

2

0.3457

(2) ( ) (3)

2 ;

(3) 20,000

kt

kt

k

t

dPkP

dt

It s solution isP t P e

According to givencondition

P P

P P e

e

k

k

k

P t P e

According to nd condition

P Pu

= − − − − −

= − − − − −

=

∴ ⇒ =

⇒ =

⇒ =

⇒ =

=

=

∴ ⇒ = − − − −

=

(0.3457)(3)

0

1.03972

0

0 1.03972

(3)

(3)

20,000

20,000

7071.07

7071

t in

P P e

P e

Pe

P

=

=

=

=

≈

i.e. initial population of community.

Hence the result.




Maximum Marks 30

Due Date 11

November 2006

Assignment Weightage 2%

Question 1 marks

10

Find the Wronskian of the functions;

(2 ln ) & (2ln ). x Cos x x Sin xα α

Also find the values of ‘x’, for which above functions are linearly dependant and independent.

Solution:

Let

[ ] [ ]

1 2

' 1 1 ' 1 1

1 2

1 1

cos(2 ln ) sin(2ln )

cos(2ln ) 2 sin(2ln ) ; sin(2 ln ) 2 cos(2ln )

cos(2ln ) 2sin(2ln ) ; sin(2ln ) 2cos(2 ln )

f x x and f x x

f x x x x f x x x x

x x x x x x

α α

α α α α

α α

α α

α α

− − − −

− −

= =

= − = +

= − = +

[ ] [ ]

1 2

1 2 ' '

1 2

1 1

2 1 2 2 1 2

( , )

cos(2ln ) sin(2ln )

cos(2ln ) 2sin(2ln ) sin(2ln ) 2cos(2ln )

sin(2ln )cos(2 ln ) 2cos (2ln ) sin(2ln )cos(2ln ) 2sin (2 ln )

f f Wronskian of both functions W f f

f f

x x x x

x x x x x x

x x x x x x x x

α α

α α

α α

α α

α α

− −

− −

= =

=− +

= + − −

=2 12 x

α −

Linear InDependence:



{ }

1 2

1 2

1 2

2 1

2 1

( , )

( , ) 0

.

. . ( , ) 0

2 0

0

0

0

As both f and f are derivableon

Sotheir W f f

Their linearindependence

i e W f f

x

x

x

Functions are linear independent x R

α

α

−

−

−∞ ∞

≠

⇒

≠

⇒ ≠

≠

≠

∴ ∀ ∈ −

Linear dependence:

1 2

21

1

1 2

2

1

1 21

1

1 2

1

cos(2ln ) sin(2ln ) 0 (1)

cos(2ln ) sin(2ln ); 0

.

(1) (2 ln )

2ln ( ) 0

1

ln ( )2

Taking

C x x C x x

C x x x x C

C

Both functions arelinearly dependent if at least oneC or C is not zero

C Now Cot x

C

C x Cot C

C

C

x Cot C

x

α α

α α

−

−

+ = − − − − −

= − ≠

⇒ − =

= − =

= −

∵

1 2

1

1( )

2

. . .

.

C Cot

C e

i e Exponential function and defined for all real numbers

Functions arelinear dependent x R

−−

=

⇒

⇒ ∀ ∈

Question 2

marks 10

Solve the IVP,

7 19 13 0; (0) 0, (0) 2 & (0) 12 y y y y y y y′′′ ′′ ′ ′ ′′+ + + = = = = −

Solution:Corresponding Auxiliary equation of differential equation is

3 27 19 13 0m m m+ + + = By inspection, we can see that m = -1 be its one root.

So using the Synthetic Division Method,

11 7 19 13

0 1 6 13

1 6 13 0

−

− − −



Now, depressed equation is

[ ]

[ ]

2

3

1 2 3

1 2

2 3

1 2 3 2

6 13 0

6 36 52

2

6 16

2

6 4

2

3 2 , 3 2

cos 2 sin 2 (1)

1 (0) 0 (1)

(1) 0 (2)

. .(2);

3 cos 2 sin 2 2 sin 2 2

x x

x x x

m m

m

i

m i i

Solution is

y C e e C x C x

Put st condition y in

C C

Diff eq

y C e e C x C x e C x

− −

− − −

+ + =

− ± −=

− ± −=

− ±=

= − + − −

∴

= + + − − − − −

=

∴ ⇒ + = − − − − − −

′ = − − + + − +[ ]

[ ] [ ]

[ ] [ ]

[ ]

3

/

1 2 3

1 2 3

3 3

1 2 3 2 3

3 3

2 3 2

cos 2 (3)

2 (0) 12;

2 3 0 0 2

3 2 2 (4)

(3) . . ' '

9 cos 2 sin 2 3 2 sin 2 2 cos 2

4 cos 2 4 sin 2 3 2 sin 2 2

x x x

x x

C x

Put nd condition y

it C C C

C C C

Differentiating w r t x

y C e e C x C x e C x C x

e C x C x e C x

− − −

− −

− − − − −

=

⇒ = − − + + +

− − + = − − − − −

′′ = + + − − +

+ − − − − +[ ]

[ ] [ ] [ ] [ ]

3

//

1 2 3 2 3

1 2 3

3

2 3

1 2 3

1 2 3

cos 2 (5)

3 (0) 12;

12 9 3 2 4 3 2

9 6 4 6

12 5 12 (6)

:(2),(4)&(6)

;

0; 0; 1

.(1) sin2 x

C x

Put rd conditon y we have

C C C C C

C C C C C

C C C

Solving eqs simultaneously

wehave

C C C

Eq i e the req y e x ui−

− − − −

= −

− = + − + − −

= + − − −

− = + − − − − − − − −

= =

−=

=

⇒ .red solution



Question 3 marks 10

If the rate of spread of dengue virus is proportional to infected & non infected persons. Then

what will be number of infected persons after 2 weeks provided that a man carrying this virus,come back to his community of 10000.Further it is provided that after one week, carriers

increase to 98.

Solution:

( )

(0) 1 { 0}

Assuming 10,000.

,

10,000 ; (0) 1

logistic

a man carries dengue virus initially enter community

x i e no of carrier at t

no one leaves from community of

So initial value problem is

dykt t x

dt

i e equation when comparing with

dP

∴ = − =

= − =

−

∵

( )

00

0 0

10,000 10,000

( ) 10000 ,

'

( ) ; (0) 1

,

10,000 10,000( ) (1)

9999 1 9999,

at

kt kt

P a bP such that a k b k dt

It s solution is given by

aPP t Here P x

bP a bP eSo

k x t

k e e

it is given that after one week

−

− −

= − = =

= = =

+ −

= = −−−−−+ +

∵



10,000(7 )

70,000

70,000

98

(7) 9810000

(1) 981 9999

98 98(9999 ) 10,000

10000 9898 9999

log,

10000 9870,000 ln( ) ln

98 99991 10000 98

ln70,000 98 9999

k

k

k

number of carriers i e

x

e

e

e

Taking anti we have

k e

k

−

−

−

= −

=

∴ ⇒ =+

+ =

−=

×

−

−− =

×

−= −

×

10,000(0.0000656388)

0.0000656388

10000(1) ( ) (2)

1 9999, ;

14 (2)

(14) 4947.9 (14) 4948 8

t

k

x t e

So for after two weeks no of carriers

Put t in

x x

−

=

∴ ⇒ = −−−−+

=

≈≅ ⇒




Maximum Marks 30

Due Date 11

November 2006

Assignment Weight

age 2%

Question 1 marks

10

Using the method of undetermined coefficient, find the appropriate form of particular

solution of differential equation;

( ) ( )5 1 1 y y x Cosx x Sinx+ = + + −′.

Solution:

( ) ( )

( ) ( )

( ) ( )

1 2

1

1 0 1 0

2

1 0 1 0

( ) ( ).

( ) ( 1)sin

sin cos (2)

( ) ( 1)cos

sin cos

5 1 1 (1)

( ) ( 1)sin ( 1)cos

( ) x g x

g x x x is given as

C x C x D x D x

and for g x x x is given as

E x E x F x F x

y y x Cosx x Sinx

g x x x x x

Let g x g

Here assumed solution for

+

= −

+ + + − − − −

= +

+ + + − − −

+ = + + − −−−−′

= − + +

=

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( )

1 0 1 0 1 0 1 0

1 1 0 0 1 1 0 0

1 0 1 0

(3)

(2) & (3) . . . (1);

sin cos sin cos

sin cos

sin cos

;

p

p

p

i i i

Combining for R H S of we have particular solution as

y C x C x D x D x E x E x F x F x

y C E x C E x D F x D F x

y A x A x B x B x

Where

A C E i

−

∴

= + + + + + + +

= + + + + + + +

= + + +

= + = 0,1

; 0,1 j j j

B D F j= + =

Hence the result.

Question 2

marks 10



Solve3 y y Tan x′′ + = .

By using method of variation of parameters.

Solution:Associated auxiliary equation is

( )

2

1 2

1

1 2

2 2

1 0

cos sin

cos ; sin

cos sin,

sin coscos sin

1

c

c

m

m i

y C x C x

Here

y x y x

x xW y y

x x x x

+ =

⇒ = ±

⇒ = +

= =

∴ =−

= +

=

Now



4

1 33

3

2 23

4 3' '1 21 23 2

4 3

1 23 2

2

2

2

0 sin sin

costan cos

cos 0 sin

cossin tan

sin sin

cos cos

sin sin;

cos cos

tan .sin

(sec 1)sin

sec .sin sin

sec tan

x xW

x x x

and

x xW

x x x

So

W W x xu u

W x W x

x xu dx u dx

x x

x x dx

x x dx

x x dx x dx

x xd

= = −

= =−

= = − = = −

⇒ = − ⇒ =

=

= −

= −

=

∫ ∫

∫

∫∫ ∫

( )

1 2 1 2

4

2 3

1

1 2

1

cos

sec cos

Re

cos sin ( ) ( ) ( ) ( )

sincos sin cos sin sec co

o

.

s

( )

c s

c p

x x

x x

quired solution is

y y y

C x C x y x u x y x u x

xC x C x x dx x x

Here u x hasvery elongated approach for its solution Soit is not need to solve

x x

+

= +

∴

= +

= + + +

= + + − + +

∫

∫

.

Hence the result.

However1

u is calculated as



( )( )

( )( )

( )

.i

4

1 3

2 2

2

2

3

3

3

3

1

2

nt

1 1.

1

2

2ln (1)

,

.

II I by egration b

Sin xu dx

Cos x

Cos x Cos x dxCos x Cosx

Sec x Secx Cosx dx

Sec x Secx Secx Cosx dx

Sec xdx Secxdx Cosxdx

Sec xdx Secx Tanx Sinx

Nowlet I Sec xdx

Sec x Secx

= −

− −= −

= − − −

= − − − +

= − + −

= − + + −

=

=

∫

∫

∫

∫

∫ ∫ ∫

∫

∫

⋯⋯⋯

( )

( )

2

2

3

1

1

1

. . .

. .

. 1

ln

2 ln

1ln

2

y patrs

dx

Secx Tanx Tanx Secx Tanx dx

Secx Tanx Secx Tan xdx

SecxTanx Secx Sec x dxSecxTanx Sec xdx Secxdx

SecxTanx I Secx Tanx

I SecxTanx Secx Tanx

I SecxTanx Secx Tanx Pu

= −

= −

= − −

= − +

= − + +

= + +

= + + − − − −

∫

∫

∫

∫∫ ∫

1

(1)

1 ln 2ln2

1 3ln

2 2

t in

u SecxTanx Secx Tanx Secx Tanx Sinx C

SecxTanx Secx Tanx Sinx

∴ = − + + + + − +

= − + + −



Question 3 marks 10

Find the expression for the motion of the mass, provided that a mass of 2kg is suspended from

a spring with a known spring constant of 10N/m & allowed to come to rest under no air

resistance. It is then set in motion by giving an initial velocity of 150 cm/sec.

Solution:

Equation of motion is



2

2

2

0 & .

5 0

5 0

5

( ) cos 5 sin 5 (1)

0 ; (0) 0

(1) 0

0

(1) ( ) sin 5 (2). . .

c

k x x as motion goes under undamping no air resistance

m

d x x

dt Its auxilary equation is

m

m i

x t A t B t

At t x

A

A

x t B t Differentiating w r t

+ =

∴ + =

⇒

+ =

⇒ = ±

∴ = + − − − − − − − −

= =

⇒ =

⇒ =

⇒= − − − − − − − −

ɺɺ ∵

0

' ';

5 cos 5 (3)

0 ; (0) 150 / sec

(3) 150 5

1500.6708

5

150(2) ( ) sin 5

5

( ) (0.6708)sin 5

. . e .

t we obtained

dxv B t

dt

At t v v cm

B

B

x t t

x t t

i e xpression for the motion

= = − − − − − − − −

= = =

∴ ⇒ =

⇒ = =

∴ ⇒ =

=

Hence the result.




Maximum Marks 30

Due Date 30 December

22, 2006

Assignment Weight

age 2%

Question 1 marks

10

For the motion described by the harmonic oscillator having displacement at time‘t’ given by

( ) ,7

( ) 6 1.8166

x t Sin t = +

find the first value of time for which the mass passes throughthe equilibrium position heading downward.

Hint: Find the sign of velocity at different values of relevant t’s and then decide the

appropriate one.

Solution:

For spring mass system

1 2

1 2

( ) 0

( ) 07

sin(6 1.816) 06

6 1.816 ;

6 1.816 1 ,6 1.816 2 ,...... .

0.2209, 0.7445,.........

x t downword direction of motion

and

x t equilibrium position of particle

t

t n n Z

t t etc

t t

π

π π

+

> ⇒

> ⇒

⇒ + =

⇒ + = ∈

⇒ + = + =

= =



[ ]

[ ]

1

2

7( ) sin(6 1.816)

6

7 cos(6 1.816)

7 cos 6(0.2209) 1.816 0

7 cos 6(0.7445) 1.816 0

, 0 0

at t

at t

Now

x t t

dxt

dt

dx

dt

dx

dt

dxFor the first time through equilibrium position x from positive side and at this pos

dt

= +

= +

∴ = + <

= + >

→ >

2

.

0 0.7445

0.7445

ition

dxat t

dt

required value of time

⇒ > =

∴ =

Hence the result.

Question 2

marks 10

For the harmonic oscillator under damped motion having position at time’t’given as

2

( ) 2 3 3 (1)3t

x t e Cos t Sin t −

= − − − −− −

find the time when the graph of the solution (1) touches the graph of exponential functiont e−± .

Solution:The graph of the solution



( )

( ) ( )

22

2 22 2

2( ) 2cos3 sin 3

3

' '

22cos3 sin 3 1 ...........(1)3

. . .

2

2 2 2 32cos3 sin 3 2 cos3 sin 3

3 3 2 22 2

3 3

t

t

x t e t t

touches the graph of e at those values of t for which

t t

Taking L H S

t t t t

−

−

= − −

±

− − = ±

−

− − − = − + − +

− + − − + −

[ ]

[ ]

1

2 10 2 2 1cos3 sin 33 32 10 2 10

3 3

2 10 3 1cos3 sin 3

3 10 10

3 1sin cos

10 10

3tan 3

1

tan (3) 1.25

2 2 102cos3 sin 3 sin .cos3 cos .sin 3

3 3

2 10.sin 3

3

2 10

3

t t

t t

Put and

t t t t

t

φ φ

φ

φ

φ φ

φ

−

− = + −

− = −

−= = −

⇒ = =

= =

∴ − − = +

= +

= [ ]

[ ]

[ ]

1

sin 3 1.25

2 10(1) sin 3 1.25 1

3

3sin 3 1.25

2 10

33 1.25 sin 2

2 10

0.49 2

2 0.755;

3

t

t

t

t n

n

nt n Z

π

π

π

−

+

∴ ⇒ + = ±

+ = ±

+ = +

= +

−= ∈

. Hence the result.



Question 3 marks 10

Find the general solution of non-homogeneous Euler –Cauchy equation

( )2 4 ln x y xy y Sin x+ + =′′ ′ on the interval ( )0,∞ .Also discuss why solution of

above equation does not exist on the compliment of the interval ( )0,∞ .

Solution:



( )

( )2 2

2 2

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

4 sin(ln ) .................(2) ;

;

ln ; ( 0)

, ( 1)

4 ln

t

We rewrite the equation in the form

d

x D xD y x where D dx

Making transformation

x e

x t x

d xDy D y x D y D D y D

dt

Then the given di

x y xy y Sin x

+ + = =

=

⇒ = >

′ ′ ′ ′∴ = = − =

+ + =′′ ′

∵

( )

( )'2

1 2

1 4 sin

4 sin ..............(3)

.

(3)

cos 2 sin 2

;

c p

c

p

fferential equation is transformed into the equation

D D y D y y t

D y t

which is a differential equation coefficients

General solution of is given by

y y y

where y c t c t

For y

′ ′ ′− + + =

+ =

∴

= +

= +

( ) ( )

'

''

1

cos sin

sin cos ...............(4)

(4);

(3) 4 sin

cos sin 4 cos sin sin

3cos 3 sin sin

3 0 , 3

cos(

1

2

10

(2

ln

3

)

p

p

p p

c p

y A t B t

y A t B t

From

y y t

A t B t A t B t t

t B t t

where from we obtain

A B

A B

General solution of is y y y

y c

= +

⇒ = − +

∴ ⇒ + =

⇒ − − + + =

⇒ + =

= =

⇒ = ⇒ =

∴=

=

+

( ) ](

( )

2

0,

1) sin(2ln ) sin(ln

,0

0

)3

ln

,

the complement of Set of all non positive real numbers

and the solution contain Set of all non positive real numbers

solution

x c x x

x wh

does not exists in the complement

ich is not defined for

of

∞ = −∞ = −

−

∴ ∞

+ +

∵




Maximum Marks 30

Due Date 05

January 2007

Assignment Weight

age 2%

Question 1

mark 05+05

(a) Discuss the validity of the recursive relation;

2

3 1

1

11

n

n

n

c if n Evens

c if n Oddsc if n and n Odds

−

+ ∈

= ∈> ∈

, to become the coefficients of

the power series solution;n

nn o

c x∞

=∑ of an ordinary differential equation (independent

variable’x’) by developing its corresponding sequence.

(b) Determine the singular points of the differential equation;

( ) ( ) ( )

2 23 2

2 3 3 1 0 x x x y x x y x y− − + − − + =′′ ′

Hence classify these points under regularity and irregularity.

Solution: (a)

2

3 1

1

1

1

n

n

n

For given recursive relation

c if n Evens

c if n Odds

c if n and n Odds−

+ ∈

= ∈

> ∈



1

2 2 1

2

3 3(3) 1 8 8 4

2

4 2

2

4 4 2

2

5 3(5) 1 14 14 7

2

3(7) 1 20 2

1, 2,3,4,5,........ :

1 1

2 1 1 1 1 2

3 1 1

( 1) 1 2 2 2 4

4 1 1 2 1 3

5 1 1

1 1 (

Put n to develop the progression

n c

n c c c

n c c c c c

c c

n c c c

n c c c c c

c c c

−

−

−

=

= ⇒ =

= ⇒ = + = + = + =

= ⇒ = = = + = +

= + + = + = + =

= ⇒ = + = + = + =

= ⇒ = = = + = +

= + = + = 0 10

2

5 10 5

2

5 5

1) 1 2

( 1) 2 3

3

0 3 . .

.

c

c c c

c c

i e false

recursive relation is invalid

+ + = +

= + + = +

⇒ = +

⇒ =

⇒

(b): For given differential equation

3 2 2 2

2

3 2 2 3 2 2

2

2 2 2 2 2 2

2 2 2

( 2 3 ) ( 3) ( 1) 0

( 3) ( 1)0

( 2 3 ) ( 2 3 )

( 3) ( 1)0

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

1 10 .............(1)

( 1) ( 3) ( 1)

(1)

x x x y x x y x y

x x x y y y

x x x x x x

x x x y y y

x x x x x x

y y y x x x x x

Computing with stand

′′ ′− − + − − + =

− +′′ ′⇒ + − =

− − − −

− +′′ ′+ − =

− + − +

′′ ′+ − =+ − +

2 2 2

. .

( ) ( ) 0

1 1( ) ; ( )

( 1) ( 3) ( 1)

s

1 ; 0 ; 3

ard form i e

y P x y Q x y

P x Q x x x x x x

Here ingularities are

x x x

′′ ′+ + =

∴ = = −+ − +

= − = =



2

2 2

, :

1

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

( 1)( 1)

( 1) ( ) ...............( )( 3)

( ) 1 ( ) .

1 s

For regularities and irregularities we proceed as follows

For x

Taking x P x a

x x x

x Q x b x x

Here a is analytic at x but b is not analytic

x is irregular ingul

= −

+ =

+− +

+ =−

= −

⇒ = −

2

2

2

2

2

2

.

0

1( ) . . 0

( 1)

1

( ) . . 0( 3) ( 1)

0 s .

3

3( 3) ( ) . . 3

( 1)

1( 3) ( ) . . 3

( 1)

3

arity

For x

xP x i e analytic at x x

x Q x i e also analytic at x x x

x is regular ingularity

For x

x x P x i e analytic at x

x x

x Q x i e also analytic at x x x

x is re

=

= =+

−= =

− +

⇒ =

=

−− = =

+

−− = =

+

⇒ = s .gular ingularity

Hence tha result.

Question 2

marks 10Using the power series method to find the general solution of the 2

ndorder homogeneous

differential equation

0 y y+ =′′ .

Solution:

Given differential equation is

0

0 1

1 1

1

1 2

, ;

0 ..........(1)

n n

n n

n n

n n

n n

n n

y c x c c x

So term by term differentiation of the proposed series solution yields

y nc x c nc x

y y Assuimng that solution of given differential equation exists in the form

∞ ∞

= =

∞ ∞

− −

= =

= = +

′ = = +

′

+ =′′

∑ ∑

∑ ∑

2

2

( 1) n

n

n

y n n c x∞

−

=

′ = −∑



[ ]

2

0

2 1

2

0 0

2

0

. . (1)

( 1)

2; 0,1, 2,....... 1 2 ;

( 1)( 2)

( 1)( 2) (1);

(1) ( 1)( 2)

n n

n n

n n

k k

k k

k k

k

k k

k

k

L HS of

n n c x c x c

Put k n k n such that k in st and nd terms respectively

k k c x c x

k k c c x Put in

k k c

y y

y y

∞ ∞

−

= =

∞ ∞

+

= =

∞

+

=

+

⇒

− + +

= − = =

+ + +

= + + +

∴ ⇒ + +

+ =′′

+ =′′

∑ ∑

∑ ∑

∑

[ ]

{ }

2

0

0 1 1

0 1 1

2

2

0

, , , ......, , ,.... .

, , ,......, , ,.... .

; ( 1)( 2) 0

; 0,1,2,3, ..........( 1)( 2)

k

k

k

k k

k k

k

k k

k k

c x

Set x x x x x is linearly independent

coefficient of x x x x x are all zero

for x k k c c

cc k

k k

From iteration of t

∞

=

+

+

+

+

+ =

⇒

⇒ + + + =

⇒ = − =+ +

∑

∵

0 02

1 13

2

0 024

2

3 1 15

2 3

04 16

01.2 2!

1 2.3 3!

( 1)12

3.4 2! 3.4 4!

( 1)13

4.5 3! 4.5 5!

( 1) ( 1)14

5.6 4! 5.6 6!

5

his recurrence relation it follows that

c ck c

c ck c

c cck c

c c ck c

cc ck c

k

= ⇒ = − = −

= ⇒ = − = −

− = ⇒ = − = − − =

− = ⇒ = − = − − =

− −= ⇒ = − = − =

= ⇒2 3

5 1 1

7

0 1

2 3 4 5 6 7

0 1 2 3 4 5 6 7

2 22 3 40 0 01

0 1

( 1) ( 1)1

6.7 5! 6.7 7!

;

........

( 1) ( 1) ( 1)( 1)

2! 3! 4!

c c cc

Above iteration leaves c and c arbitrary

required solution is given by

y c c x c x c x c x c x c x c x

c c cc y c c x x x x

− −= − = − =

∴

= + + + + + + + +

− − −−= + + + + +

3 32 6 70

1 1

2 4 6 3 5 7

0 1

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

5! 6! 7!

1 ...... ......2! 4! 6! 3! 5! 7!

cc x x c x

x x x x x xc c x

− −+ + +

= − + − + + − + − +



2 4 6

3 5 7

0 1

'

cos 1 ......2! 4! 6!

sin ......3! 5! 7!

cos sin

By Maclaurin s series

x x x x

x x x x x

y c x c x

∴

= − + − +

= − + − +

∴ = +

Hence the result.

Question 3 marks 10

In question ‘2’find the two linearly independent solutions by the Euler’s method

(The Auxiliary equation method).Also(a) Compare this general solution with power series solution and discuss which method is

more general and why?

(b) Explain the linear independence of the two solutions.

Solution:

For differential equation

[ ] [ ]

( ) ( )

2

2

2

0 .........(1)

(1) ( 1) 0

1 0 ( 0 )

cos sin cos sin

cos sin

cos sin ( )

mx

mx

mx

mx

mx

ix ix

y y

Put y e

y me

y m e

m e

m e x R

m i

required solution is given as

y Ae Be

A x i x B x i x

A B x Ai Bi x

y C x D x where C A B and D i A B

H

−

′′ + =

=

′⇒ =

′′⇒ =

∴⇒ + =

⇒ + = ≠ ∀ ∈

⇒ = ±

∴

= +

= + + −

= + + −

= + = + = −

∵

{ }

.

'

, , ,

ere both solution of power series and by auxilary equations method are identical

Power series method is more general than Euler s method because

it is applicable even when coefficients of y y y are transcendantal functions

but Eul

′′ ′

' .er s method holds for constant and polynomials coefficients



(b) Linear Independence:Here two solutions are

2 4 6

1

3 5 7

2

1 ................ cos2! 4! 6!

................ sin3! 5! 7!

;

cos sin 0.................( )

sin cos 0 .........( )

( ) & ( )

x x x y x

x x x y x x

For linear independance

Put A x B x a

Differentiating A x B x b

solving a b A B

= − + − + =

= − + − + =

+ =

⇒ − + =

⇒ = 0.both solutions are linearly independent

=⇒

Hence the result.




Maximum Marks 30

Question 1

mark 05+05

Show that differential equation

( ) ( ) ( ) 0r x y q x p x yλ

′+ + =′ is equivalent to Legendre’s and Bessel’s differential

equation for suitable replacements of ( ), ( ), ( ) and r x p x q x λ .Hint: Differentiate first term of given differential equation and compare

( ), ( ), ( )r x p x q x and λ for Legendre’s and Bessel’s equations.

Solution:

Given differential equation is

[ ]

( )2

2

( ) ( ) 0...........(1)

'

1 2 ( 1) 0..............(2)

(1) & (2) , ,

( ) 1

( ) ( ) ( ) 0

( ) ( ) q x p x y

Legendre s equation is given as

x y xy n n y

Comparing for coefficients of y y and y we have

r x x

r x y q x p x y

r x y r x y λ

λ

+ =

′′ ′− − + + =

′′ ′

∴ = −

′+ + =′

⇒ + +′′ ′ ′

∵

( ) 2

( ) 0 ; ( 1) ; ( ) 1

(2) ' (1)

r x x

q x n n p x

Legendre s equation is special case of

λ

′⇒ = −

= = + =

∴ ⇒

2 2 2

2

, '

( ) 0

10 ..........(3)

Now Bessel s equation is

x y xy x v y

Dividing by x xy y x v y x

′′ ′+ + − =

′′ ′∴ ⇒ + + + − =

2

(1) & (3) , ,

1( ) ( ) 1 ; ( ) ; ( ) ;

' (1).

Comparing for coefficients of y y and y we have

r x x r x q x x p x v

x Hence Bessel s equation is also another special case of

λ

′′ ′

′= ⇒ = = = − =

Question 2

marks 10



(a) For set of Legendre’s polynomials { }( ); 0,1,2,3nP x n = ,verify that

1

1

( ) ( ) 0,n mP x P x dx

−

=∫ provided that m n≠

(b) Generate Legendre’s polynomials 54( )( ) and P xP x using Rodrigue’s formula for

4,5n =

Solution:

(a) As Legendre’s polynomials for 0,1,2,3,4,r = ……are given as

( )

( )

( )( )

( )

0

1

2

2

3

3

1 1

2 3

1 1

1

2 3

1

1

5 3 3

1

1

6 4 4 2

1

( ) 1

( )

1( ) 3 1

21

( ) 5 32

2,3 2 3

( ) ( ) ( ) ( )

13 1 5 3

4

1

15 9 5 34

1 15 9 5 3

4 6 4 4 2

1(0)

4

0

,

n m

P x

P x x

P x x

P x x x

Consider the case m n for as

P x P x dx P x P x dx

x x x dx

x x x x dx

x x x x

Similarly fo

− −

−

−

−

=

=

= −

= −

≠ ≠

∴ =

= − −

= − − +

= − − +

=

=

∫ ∫

∫

∫

0 1,1 2,3 1 .,r all other cases like etc we have≠ ≠ ≠

1

1

( ) ( ) 0n mP x P x dx−

=

∫

Hence verified.



(b)





( )

( )

( ) ( )

( )

2

44

2

4 4 4

4 34 4

2 2

4 3

33

2

3

36 2 2 2

3

'

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

2 !

4

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

2 4!

1 1

4 1 .2

8 3( ) (1) 3

nn

n n n

Rodrigues formula for Legendre s polynomials is

d P x x

n dx

Put n

d P x x

dx

d d d Here x x

dxdx dx

d x x

dx

d x x x x

dx

= −

=

= −

− = −

= −

= − +

∵

( )2 3

37 5 3

3

4 2

4 2

4

4 2

4

7 5 33 3

(1) (1)

8 3 3

8 210 180 18 .............(3)

(2)

1( ) .8 210 180 18

16.24

1( ) 35 30 3

8

x x x x It is obtained by taking three times the derivatives of

d x x x x

dx

x x

Put in

P x x x

P x x x

P

− + −

−

= − + −

= − +

∴ ⇒

= − +

= − +

( )

( )

( )

( )

55

2

5 5 5

2

52 1

5 5 4 3 2 2 3 4 5

0 8 6 4 2

5 552 10 8 6 4 2

5 5

5

5 10 10

5 (1);

1( ) . 1 ...........(4)

2 5!

& 1

1 5 10 10 5 1

1 5 10 10 5 1

30240 3360

5

ut n in

d P x x

dx

Here

By Binomial Theorm

a

a x b

x x x x x x

d d x

b a a b

x x x x xdx d

a b a b ab

x

x

b

=

= −

= =

− = − + − + −

∴ − = − + −

− = − + − +

=

−

+ −

−3

5 3

5

5 3

5

5 35

10 8 6 4 25 10 10 5 1

0 7200

(4)

1( ) . 30240 33600 7200

32.120

1( ) .480 63 70 15

32.120

1( ) 63 70 158

x x x x x It is obtained by taking five times the derivatives of x x

Put in

P x x x x

P x x x x

P x x x x

− + − + −

+

∴ ⇒

= − +

= − +

= − +



Hence the result.

Question 3 marks 10

Solve, if possible, the following differential equation by either systematic elimination or by use

of determinants:

( ) ( )

( )

1 1 2

3 2 1

D x D y

x D y

+ + − =

+ + = −

Solution:



( ) ( )

( )

( )

( ) ( )

( ) ( )( ) ( )( )

( )2

1 .( ) ;

3 6

3 1

________________________________________________

3 3 3 2

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

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

3 .( )&1 3 1

1 1 2 1

D with eq b then subtracting

y

y

D D D

D x D y a

x D y b

Multplying with eq a D x D

D x D D D

+

=

= −

− − −

− − + +

+ + − =

+ + = −

+ + −

+ + + + +

( )

( )

2

5

2

2

2

2

2

2

2

6 1

3 3 3 2 7

5 7

5 7..................(1)

'

5 0............( )

( ) 5 0

5 0 0

5

(1

mt

mt

mt

mt

mt

y

D D D y

D y

d y y

dt

It s associated homogenous equation

d y y c has solution

dt

y e

y me

y m e

c m e

m e

m i

For

= +

− − − − =

⇒ + = −

⇒ + = −

+ =

=

′ =

′′ =

∴ ⇒ + =

⇒ + = ≠

⇒ = ±

∴

∵

1 2

' ''

1 2

) cos 5 sin 5

0

7

(1) 5 7 5

7( ) cos 5 sin 5 ..................( )

5

c

p p p

c p

y c t c t

Now assuimng

y A y y

A A

y y y

y t c t c t i

⇒ = +

= ⇒ = =

∴ ⇒ = − ⇒ = −

= +

⇒ = + −



( ) ( )

( )( ) ( )( ) ( )

( ) ( )( ) ( ) ( )

( )

2

2

2

2

2 1 .( ) ;

2

3 1

________________________________________________

3 2 3 3 4 1

3 3

3 3........

( ); .( )&

2 1 2 1 2

1 1 2 1

D D with eq b then subtracting

y

y

D D D x

D x

d x x

dt

For x t Multplying with eq a

D D x D D D

D x D D D

+ −

=

= −

− − −

+ + − + = −

⇒ + =

⇒ + =

+ + + + − +

− + − + −

2

3 4

' ''

3 4

..........(2)

'

3 0

3

cos 3 sin 3

0

(2) 3 3

1

( )

( ) cos 3 sin 3 1..................( )

, (

c

p p p

c p

It s auxilary equation associated to homogenous equation is

m

m i

x c t c t

Assuming x B x x

B

B

x t x x

x t c t c t ii

Hence system

+ =

⇒ = ±

= +

= ⇒ = =

∴ ⇒ =

⇒ =

∴ = +

⇒ = + +

) & ( ) ( ) & ( ).i ii give required solution of a b




Maximum Marks 30

Due Date

27January 2007

Assignment Weight

age 2%

Question 1

mark 05+05

Verify that the vector ‘ X ’ is the solution of the given system

2dx

x ydt

dy x

dt

= +

= −

where1 4

3 4t t X e te

= +−

.

Solution:

Given system is

2

2 1

1 0

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

dx x y

dt

dy x

dt

reducing the given system in matrix form

dx

xdt

dy y

dt

x xd A

y ydt

X AX where assumig

= +

= −

⇒=

−

⇒ =

′⇒ =

2 1 ( ) ( ) ( ); ;

1 0 ( ) ( ) ( )

dx

x t x x t x t xd dt A X X

y t y y t dy y t ydt

dt

′ ′

′= = = = = = = ′ ′−

( )

( )

1 4( )

3 4

1 44

3 4 3 4

t t

t t t

t t t

Here given vector is

X t X e te

e t e te X

e te e t

= = +

−

+ += =

− −



( )

( )

( ){ }

( ){ }

( )

( )

. . . ' '

1 41 4

3 4 3 4

1 4 4

3 4 4

4 5..................(2)

4 1

t t

t

t

t

t

t

Differentiating w r t t

d e t e t d d dt

X

d dt dt e t e t dt

e t X

e t

t e

t

⇒

+ +

= =

− −

+ +′ =

− −

+ =

− −

( )

( )

. . . (1)

1 42 1

1 0 3 4

2 1 1 4

1 0 3 4

2 8 3 4

1 4

5 4...............(3)

4 1

(2) & (3), .(1) . .

t

t

t

t

t

Now taking R H S of

e t

AX e t

t e

t

t t e

t

t e

t

From we conclude that eq i e X AX shows its validity f

⇒

+ = − −

+ =

− −

+ + − =

− −

+ =

− −

′ =1 4

3 4

.

t t or given vector X e te

X is solution of given system

= +

− ⇒

Hence the result.

Question 2

marks 10

If the vectors ‘1

1

1

t X e

= −

’ and ‘2

2 8

6 8

t t X e te

= + −

’ are the solutions of the system

X AX =′

then determine whether these vectors form a fundamental set t R∀ ∈

byinvestigating their linear independence.

Solution:

Two vectors solutions 1 2& X X form a fundamental set { }1 2, X X if 1 2& X X are

linearly independent if Wronskian of ( )1 2 1 2& , 0 X X W X X = ≠ .

Thus we have to show that

( )1 2, 0W X X ≠ for given vector solutions



( )

( )

1

2

1

1

2 8 2 8

6 8 6 8

1 42

3 4

t

t

t

t t

t t

t t

t

t

e X e

e

e te

X e te e te

e t

e t

−

= =

−

+ = = = − −

+=

−

Wronskian of

( )( )

( )

( )

1 2 1 2

2

1 2

1 2

2 1 4& ,

2 3 4

1 1 4.2 2 3 4 1 4 8 0 ;

1 3 4& .

& .

t t

t t

t t t t

e e t X X W X X

e e t

t e e e t t e t R

t X X are linearly independant

X X form a fundamental system

+= =

− −

+= = − + + = ≠ ∀ ∈

− −

⇒

⇒

Hence the result.Question 3 marks 10

Solve the following system of differential equations

4 5

2 6

dx x y

dt dy

x ydt

= +

= − +

Solution:

Given system is

4 5

2 6

dx x y

dt

dy x y

dt

= +

= − +

it’s matrix form is



( )

4 5

2 6

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

( ) 4 5,

( ) 2 6

det 0

4 5

2 6

dx

xdt

dy y

dt x xd

A y ydt

X AX

x t where X A coefficient of matrix

y t

Characteristic equation of coefficient matrix

A I λ

λ

= −

⇒ =

′⇒ =

= = =

−

− =

−

− −

( ) ( )

( ) ( )2

2

0

4 6 10 0

6 4 10 0

10 24 10 0

10 34 0

10 100 136 10 65 3 .

2 2

ii which are Eigenvalues

λ

λ λ

λ λ

λ λ

λ λ

λ

=

− − + =

⇒ − − + =

− + + =

− + =

± − ±= = = ±



( )( )

( )

( )

( )

1

1

1

2

1

2

1 2

1 2

5 3

4 5 3 50 0

2 6 5 3

1 3 5 0

2 1 3 0

1 3 5 0 ..............(1)

2 1 3 0 ...................(2)

(

Eigen vector for i is given by

i k A I k

k i

k i

i k

i k k

k i k

It has trivial solution

λ

λ

= +

− + − = ⇒ =

− − +

− − ⇒ =

− − −

⇒ − − + =

− + − =

( )

( )( )

( )

1

2

1

2

1

2

2 2

1

2

0,0)

,

3 1(1)

5

, 5

3 1

5. . 1 .

1 3

5 3

4 5 3 50 0

2 6 5 3

1 3 5

02 1 3

For nontrivial solution

i k k

By inspection taking k

k i

k i e st Eigen vector i

Eigen vector for i given as

i A I k k

i

k i

i k

λ

λ

+⇒ =

=

⇒ = +

⇒ =

+

= −

− −− = ⇒ =

− − −

− +

⇒ = − +

( )

( )

( )

( )

1 2

1 2

1 2

2

1

2 1

2

1 3 5 0 ................(3)

2 1 3 0.................(4)

(3) & (4) (0,0).

(4) 2 1 3 0

1 3

2

2 1 3

2

i k k

k i k

Again has trivial solution as For nontrivial solution

k i k

i k k

By inspection taking k k i

nd Eigen vector k

⇒ − + + =

− + + =

⇒ − + + =

+=

= ⇒ = +

∴ =

1

2

5 3

1 1

(5 3 )

2 2

1 3

2

5

1 3

1 3

2

t i

t i t

i

Consequently two solutions are

X k e ei

i X k e e

λ

λ

+

−

+ =

= =

+

+ = =



By superposition principle general solution is

( )

( )

( ){ }

( ){ }

1 1 2 2

(5 3 ) (5 3 )

1 2

(5 3 ) (5 3 )

1 2

(5 3 ) (5 3 )

1 2

5 3 3

1 2

5 3 3

1 2

5 1 3

1 3 2

5 1 3( )

( ) 1 3 2

( ) 5 1 3

( ) 1 3 2

5

i t i t

i t i t

i t i t

t i t i t

t i t i t

X c X c X

i X c e c e

i

c e i c e x t

y t c i e c e

x t e c e i c e

and y t e c i e c e

Consider c

+ −

+ −

+ −

−

−

= +

+ = +

+

+ + =

+ +

⇒ = + +

⇒ = + +

( ) ( )

{ } { }

{ } { }

[ ]

[ ] ( ) ( )

1 2 1 2

5

5

5

1 2

5

3 4 1 2 3 4

, 1 3 , 1 3 , 2

( ) cos3 sin 3 cos3 sin 3

( ) cos3 sin 3 cos3 sin 3

( ) cos3 sin 3

( ) cos3 sin 3 , , ,

t

t

t

t

A i c B i c C c D

x t e A t i t B t i t

and y t e C t i t D t i t

x t e c t c t

and y t e c t c t where c A B c i A B c C D c i C D

= + = + = =

⇒ = + + −

⇒ = + + −

∴ = +

= + = + = − = + = −

Hence the result.

Documents

Differential Equations - Solved Assignments - Semester Fall 2006