differential equations - solved assignments - semester spring 2006

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

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Assignment 1 - Spring2006

Maximum Marks 30

Due Date: May 04, 2006

Question 1

Solve the (separable) differential equation

Solution

First we simplify the right hand side until the equation is clearly separable.2( 4) ( 2)( 2)

( 2)2 ( 2)

dy y x y x xy x

dx x x

The equation is separable & has constant solutions

y=0

For non-constant solutions separate variables, integrate both sides, and simplify:

Question 2


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Solve the differential equation

( )

( )

dy x y

dx x y

Solution:

It is easy to check that the function

( , )x y

f x yx y

is a homogeneous function.

Tosolve the differential equation

( ) ( ) 0 x y dx y x dy

Subsitute y vx

dy vdx xdv

2

2

2

2

( )

( )

( )( ) ( )

( )(1 ) (1 )

( )(1 ) (1 )

( 1 ) (1 ) 0

( 1 ) (1 ) 0

(1 ) (1 )

(1 ) 1

(1 )

vdx xdv x vx

dx x vx

vdx xdv x vx x vx dx

x vdx xdv v x v dx

vdx xdv v v dx

v v v dx xdv v

v dx xdv v

xdv v v dx

vdv dx

v x

This is separable equation.

Solving by separation of variables all solutions are implicitly given by

1 2ln | | tan ( ) 1/ 2 ln(1 ) x v v c

Going back to the function y through the substitution vxy , we get


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1 2ln | | tan ( / ) 1/2 ln(1 ( / ) ) x y x y x c

Question 3


3 4 3(2 ) ( 2) 0 xy y dx xy dy (By integrating factor technique)

Solution:

We rewrite the equation to get

3 4 3

3 4 3

2 3

(2 ) ( 2) 0

( , ) 2

6 4

xy y dx xy dy

M x y xy y and N xy

Now

M N xy y and y

y x

This clearly implies that the equation is not exact.

Let us find an integrating factor. We have

.3

N M

x y

M y

Therefore, an integrating factor u(y) exists and is given by

3

3ln 3( )dy

yyu y e e y

The new equation is

3 3 4 3 3

3

(2 ) ( 2) 0

(2 ) ( 2 ) 0

y xy y dx y xy dy

x y dx x y dy


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which is exact.

Let us find F(x,y). Consider the system:

3

2

2

Fx y

x

Fx y

y

Let us integrate the first equation. We get2( , ) ( )F x y x xy y

We differentiate F w. r. t. yand use the second equation of the system

3( ) 2F

x y x yy

Differentiate with respect toy and use the second equation of the system to get

3( ) 2y y

2( )y y

Thus the solution is 2 2C x xy y


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Maximum Marks 25


Question 1 marks 05


42 2

3

dyy x

dx x

Sol:The equation is 1st order linear differential equation, here

p(x) = -2/x and q(x) = 42

3x

Hence the integrating factor is given by

( ) 2/ 2ln 2

2 4 2 3

5 2

2 2 2

( )

2 2 2.

23 3 9

9

p x dx xdx xu x e e e x

Therefore the general solution is

x x dx c x dx c x c

y x cx x x x

Question 2 marks 10


3dy x y xydx


6/31

Sol:

The given differential can be written as

31dy y ydx x

whichis a Bernoulli equation with

1( ) , ( ) 1, 3 p x q x n

x

Dividing with3

y we get

23 1

dy yy

dx x

----------------- 1

Thereforewe substitute

1 3 2v y y

Differentiating w.r.t. x we have

3 1

2

dy dvy

dx dx

So that the equation number 1 reduces to

1( ) 1

2

2( ) 2

dv v

dx x

dvv

dx x

This is a linear equation. To solve this we find the integrating factor)(xu

2( ) , ( ) 2 p x q x

x

2

2ln 2( )dx

xxu x e e x


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The solution of the linear equation is given by

2

2

( ) ( ) 2

( )

u x q x dx c x dx cv

u x x

12

2

2x Cv x Cx

x

To go back to y we substitute2v y

Therefore the general solution of the given DE is

2 2 y x Cx

2

2

1

(2 )y

x Cx

2 1 /2

1

(2 )y

x Cx

Question 3 marks 10

A certain radioactive material is known to decay at the rate proportional to the amount present. If

initially there are 100 milligrams of the material present and if after two years it is observed that 5

percent of the original mass has decayed, find

a) An expression for the mass at any time tb) The time necessary for 10 percent of the original mass to have decayed.

Sol:

Let N denote the amount of material present at time t, then decay of material is governed by thedifferential equation

0dN

kNdt

The differential equation is separable and linear; its solution is

kt N ce

Now at t=0, we are given that N=100. Put the values in above equation


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0

100

kce

c

Thus

ktN e ---------------- 1

At t=2, 5 percent of the original mass of 100mg, has decayed. Hence at t=2,

N= 100-5 = 95, substituting these values in above equation we get

295 100

1 19ln

2 20

0.026

ke

k

k

Substituting these values in equation number 1 we obtain the amount of mass present at any time t as

0.026tN e -------------------2

where t is measured in years.

b)Now we require the time for 10 percent of the original mass to have decayed.

The mass after decaying of 10 percent of the original mass ( i.e. 100 mg) is

100-10 = 90

Now put the value in equation number 2 to get the time

0.026

4.05

te

t

So we require 4.05 years to decay 10 percent of the original mass.


9/31


Maximum Marks 25


Question 1

Show by computing the Wronskian that the given functions are linearly independent or dependent on

( , )

2 3

1 2 3( ) , ( ) , ( )f x x f x x f x x

Sol:

2 3

2 3 2

2 2 3

3 3

3

2 3

, , 1 2 3

0 2 6

[ 2 6 3 2 ] [ 6 2 ] 0

6 4

2

, , 0

( , ) .

x x x

W x x x x x

x

x x x x x x x

x x

x

W x x x

hence given functions are linearlyindependent on for all x R

Question 2


9 20 0 y y y

Sol:


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2 4

4 2

4 2

, ,

9 20 0

( 9 20) 0

0

mx

mx mx mx

mx mx mx

mx

mx

Put y e

y me y m e y m e

Putting these values in the given equation

m e m e e

m m e

Since e

So the characteristic equation is4 2

1 2 3 4

2 2 5 5

1 2 3 4

9 20 0

int

( 2)( 2)( 5)( 5) 0

2 , 2, 5, 5

x x x x

m m

Which canbe factor o

m m m m

The roots are

m m m m

Hence the soluiton is

y c e c e c e c e

Question 3

Solve4 4 y y x

Sol:

For the complementary solution consider the associated homogeneous equation

4 4 0

mx

y y

To solvethis equation we put

y e

Then the auxiliary equation is

21 0

1

m

m


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Roots of the auxiliary equation are real and distinct. Therefore the complementary function is

1 2

x x

c y c e c e

From the complementary solution we find

1

2

x

x

y e

y e

( , ) 2x xW e e

y +p(x) y +Q(x)y=f(x)

xy -y=

4

xHere f(x) =

4

1

2

0

4/ 4

0

4/ 4

x x

x

x x

x

e xeW

x e

e xeW

e x

1

2

1

2

1

8

2

8

1 1( 1)

8 8

1 1( 1)

8 8

x

x

x x

x x

w xeu

w

w xeu

w

u xe dx x e

u xe dx x e

A particular solution if the non-homogenous equation is

1 1( 1) ( 1)

8 8

4

x x x x

p y x e e x e e

x

hence the general solution is


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1 24

x x

c p c

x y y y y c e c e


13/31


Maximum Marks 30

Due Date: June 29, 2006

Question 1 Marks 5

A 32-lb weight, attached to spring, stretches it 6 inches. Find the equation of the motion if the

weight is released from rest from a point 8 inches above the equilibrium position.

Sol:

weight of the body is

W= 32 lbsW= mg

m= W/g=32/32=1

Since stretch s= 6 inches = 1/2 foot

By Hooks Law

32 = k(1/2)

k=64 lbs/ft

Hence the equation of simple harmonic motion

2

2

2

2

2

2

64

64 0

d xm kx

dt

becomes

d xx

dt

d xx

dt

Since the initial displacement is 8 inches=

2

3 ftInitial velocity is zero as the object is released from rest, so the initial conditions are

2(0) , (0) 0

3x x

To solve the initial value problem2

264 0

d xx

dt


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15/31

2

22 3

(0) 1, (0) 0

d x dx x Sin t

dt dt

x x

Sol:

First consider the associated homogeneous differential equation

2

20

d x dxx

dt dt

Put2

2

2, ,mt mt mt

dx d x x e me m e

dt dt

Then the auxiliary equation is2 2 1 0m m

m= -1, -1

hence

1 2

t t

c x c e c te

To find the particular integral of non- homogeneous differential equation we use the underminedcoefficients, we assume that

cos 3 sin 3p

y A t B t

cos 3 sin 33 sin3 3 cos3

9 cos3 9 sin3

p

p

p

y A t B t

y A t B t

y A t B t

so that

9 cos 3 9 sin 3 2( 3 sin 3 3 cos 3 ) cos 3 sin 3 2sin 3

( 9 6 )cos3 ( 9 6 )sin3 2sin3

A t B t A t B t A t B t t

A B A t B A B t t


16/31

( 9 6 )cos3 ( 9 6 )sin 3 2sin 3

in ,

8 6 0

8 6 2

A B A t B A B t t

Equat g coefficients

A B

and

B A

B=-4/25

A=-3/25

3 4cos3 sin 3

25 25p

y t t

Hence the general solution of the differential equation

1 2

3 4cos3 sin 3

25 25

t t x c e c te t t

Question 3 Marks 10

Solve using tx e

22

25 5 0

5

d y dy yx x

dx dx

Sol:

22

20

25

The differential can be written as

d y dy yx x

dx dx


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22

2

2 2

2 2

2

2

2

025

1( ) 0

25

,

, ( 1)

1[ ( 1) ] 0

25

1[ ] 0

25

1

025

t

The differential can be written as

d y dy yx x

dx dx

x D xD y

With the subsitution x e we have

xD x D

so the equation becomes

y

y

or

d y

ydt

Put 2

,

10

25

1

5

mt y e we get the auxiliary equation as

m

or m i

1 2

1 2

1 1cos sin

5 5ln ln

cos sin5 5

y c t c t

x xc c


18/31


Maximum Marks 30

Due Date: July , 2006

Question 1 Marks 10

Find solution of the differential equation

0y y

In the form of power series of x.

Sol:If we assume that a solution of the given equation exists in the form

0

0 1

n nn n

n n y c x c c x

Now differentiation of the proposed series solution gives

1 11

1 2

n nn n

n n

y nc x c nc x

Now we have

2

2

1 nnn

y n n c x

Substituting the expression for y and y , we obtain

2

2 0

1 n nn nn n

y y n n c x c x

Using, 2n k and n k , we get

20 0

2 1 k kk kk k

y y k k c x c x

or 20 2 1k

k kk

y y k k c c x

Substituting in the given differential equation, we obtain

20

2 1 0kk kk

k k c c x

From this last identity we conclude that


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22 1 0k kk k c c

or 2

, 0,1,2,2 1

kk

cc k

k k

From iteration of this recurrence relation it follows that

0 0

2

1 1

3

02

4

3 1

5

04

6

5 1

7

2.1 2!

3 2 3!

4.3 4!

5.4 5!

6.5 6!

7.6 7!

c c

c

c cc

ccc

c cc

ccc

c cc

This iteration leaves both0

c and1

c arbitrary. From the original assumption we have

2 3 4 5 6 7

0 1 2 3 4 5 6 7

2 3 4 5 6 70 0 01 1 10 1

2! 3! 4! 5! 6! 7!

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

c c cc c cc c x x x x x x x

or

2 4 6 3 5 70 1

1 1 1 1 1 11

2! 4! 6! 3! 5! 7! y c x x x c x x x x

is a general solution. When the series are written in summation notation,

2

1 0

0

1

2 !

n

n

n

y x c xn

and

2 1

2 1

0

1

2 1 !

n

n

n

y x c xn

Question 2 Marks 10

Determine the singular points of the differential equation and classify them as regular or irregular.

22 ( 2) 3 ( 2) 0 x x y xy x y


20/31

Sol:

Dividing the equation by 22 ( 2)x x we have

23 1 0

2( 2) 2 ( 2)xyy y

x x x

The singular points are x=0 and x=2.

Now a singular point x0 is regular singular point if (x-x0)P(x) ad (x-x0)^2 Q(x) are analytic at x0.

X=0 is a regular singular point.

X=2 is an irregular singular point.

Question 3 Marks 10

Solve

22 7 ( 1) 3 0x y x x y y (Use frobenius method)

Just find the values of r. (No need to find the general solution)

Sol:

The assumption rn

n

nxcy

0

leads to

22 7 ( 1) 3 x y x x y y

2 2 2 1

0 0 0

2 ( )( 1) (7 7 ) ( ) 3n r n r n r n n n

n n n

x n r n r c x x x n r c x c x


21/31

2 2 2 1

0 0 0

1

0 0 0 0

1

0 0 0

0

2 ( )( 1) (7 7 ) ( ) 3

2 ( )( 1) 7 ( ) 7 ( ) 3

2 ( 1) 7 3 7 ( ) 2 ( )(

n r n r n r

n n n

n n n

n r n r n r n r

n n n n

n n n n

r r r n r

n

n

x n r n r c x x x n r c x c x

n r n r c x n r c x n r c x c x

r r c x rc x c x n r c x n r n

1

1 1

2 1

0

0 1

2

0

0

1)

7 ( ) 3

(2 5 3) 7 ( ) [( )(2 2 2) 7( ) 3]

(2 5 3) 7 ( )

n r

n

n

n r n r

n n

n n

r n r n r

n n

n n

r

n

n

r c x

n r c x c x

r r c x n r c x c x n r n r n r

r r c x n r c

1 2 2

1

2 1 2 2

0

0 1

2 [2 2 5 5 3]

[(2 5 3) 7 ( ) 2 [2 2 5 5 3]]

n r n r

n

n

r n n

n n

n n

x c x n r n r

x r r c n r c x c x n r n r

Thus 22 5 3 0r r so that 1 21/ 2, 3r r


22/31


Maximum Marks 25

Due Date: July 20 , 2006

Question 1 Marks 5

Find the general solution of the equation

2 2 16 6 6 06

x y xy x y

on ,0

Sol:

The Bessel differential equation is

0222 yvxyxyx

2 2 1 036

Given differential equation can be written as

x y xy x y

Comparing above two, we get 21

36v , therefore

1

6v

So general solution is 1 1/6 2 1/6J J y C x C x


23/31

Question 2 Marks 10

Solve, if possible, the given system of differential equation by systematic elimination

4 3

3

dxx y

dt

dyx

dt

Sol:

The given system of linear differential equations can be written in the differential

operator form as

4 33

( 4) 3 0

3 0

Dx x y Dy x

D x y

Dy x

First eliminate x by multiplying 3 to first equation and operating (D-4) on second

equation and then by adding the two equation we get2

[ 4 9] 0 D D y

The auxiliary equation for y is given by2m 4m 9 0

The roots of the auxiliary equation are

2 13m Therefore

(2 13) (2 13)

1 2

t t y c e c e

Now eliminate y from equation by apply D on first equation and multiply second

equation by 3 and then by adding we get


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2

2

1

2

( 4 9) 0

4 9 0

2 13

2 13

D D x

D D

roots of the auxiliary equation are

m

m

Therefore

(2 3)(2 3)

3 4

ttx c e c e

2 13 2 13

1 2

2 13 2 13

3 4

t t

t t

Nowwe havethe two solution

y t C e C e

x t C e C e

Put these value of x and y in the second equation of given differential equation

2 13 2 13 2 13 2 13

1 2 3 4

2 13 2 13

1 3 2 4

1 3

2 4

3 1 4 2

3 0

2 13 3 2 13 3 0

3 02 13

2 13 3 0

2 13 2 13

3 3

t t t t

t t

D C e C e C e C e

C C e C C e

C C

C C

hence

C C and C C

Now x and y can be written as

2 13 2 13

1 2

2 13 2 13

1 2

2 13 2 13

3 3

t t

t t

y t C e C e

x t C e C e


25/31

Question 3 Marks 10

Solve, if possible, the given system of differential equation by use of determinants

4 3 , 3 4dx dy

x y x ydt dt

Solution:First we write the differential equations of the system in terms of the differential operator D

4 3 0

3 ( 4) 0

D x y

x D y

We form the determinant

4 3 0 3 4 0, ,

3 4 0 4 3 0

D D

D D

Since the 1

st

determinant is non-zero

24 3 7 03 4

DD

D

Therefore, we write the decoupled equations

4 3 0 3

3 4 0 4

Dx

D D

4 3 4 03 4 3 0

D DyD

After expanding we find that

2 7 0D x

2 7 0D y The auxiliary equation for both of the differential equations is:

2 7 0 7m m

The auxiliary equation has real and distinct roots

7 71 2

7 73 4

t t

t t

x c e c e

y c e c e


26/31

Hence, the general solution of the two decoupled equations

7 71 2

7 73 4

( )

( )

t t

t t

x t c e c e

y t c e c e

Substituting these solutions for x and y into the first equation of the given system, we obtain

1 2 43 1c3, .4 7 4 7

c c c

Hence, the general solution of the given system of differential equations is

7 74

7 73 4

3 1c3 .

4 7 7 4

t t

t t

x e c e

y c e c e


27/31

Assignment 7 Spring 2006

Maximum Marks 15

Due Date: July 20 , 2006

Question 1 Marks 10

Find the eigenvalues and eigenvectors of

1 0 2

1 0 1

2 0 4

A

SOL:

Eigenvalues

The characteristic equation of the matrix A is

1 0 2

det 1 0 1 0

2 0 4

A I

Expanding the determinant, we obtain

2

( 5) 0 Hence the eigenvalues of the matrix are

1 2 30 0 5 , ,

Eigenvectors

For 1 0 we have

1 0 2 0

0 | 0 1 0 1 0

2 0 4 0

A

R2-R1


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R3 -2R1

1 0 2 0

0 0 1 0

0 0 0 0

R1 + 2R2

1 0 0 0

0 0 1 0

0 0 0 0

-R32

1 0 0 0

0 0 0 0

0 0 1 0


29/31

Thus we have the following equations in 1 2,k k and 3k . The number 2k can be chosen arbitrarily

1 0k , 2 31, 0k k

1

0

1

0

K

For 3 5

4 0 2 0

5 | 0 1 5 1 0

2 0 1 0

A

R1+ 3R2

R3-2R2

1 15 5 0

1 5 1 0

0 10 3 0

-R1

1 15 5 0

1 5 1 0

0 10 3 0

R2- R1

1 15 5 0

0 20 6 0

0 10 3 0

R2 +2R2

1 15 5 0

0 0 0 0

0 10 3 0

R23


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1 15 5 0

0 10 3 0

0 0 0 0

1/10 R2

1 15 50

30 1 0

100

0 0 0

R1-15R2

11 0

2 03

0 1 0100

0 0 0

Hence we obtain the following two equations involving 1 2,k k and 3k .

2 3

3

10k k

,1 3

1

2K k

Choosing3

103

k we have 1 25 , 13

k k .

Hence we have an eigenvector corresponding to the eigenvalues2

5

2

5

3

1

10

3

K

Question 2 Marks 5

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


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53 4 1;4 2

t X X X e

Sol:

Since51

2

tX e

And

55

5

1 5

2 10

tt

t

eX e

e

Futher

AX =3 4

4

51

2

te

=

5

5

5

10

t

t

e

e

= X

Thus the vectors X satisfy the homogeneous linear system.

differential equations - solved assignments - semester spring 2006

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