Transcript
Page 1: Differential Equations - Solved Assignments - Semester Spring 2005

Assignment 1 (Spring 2005)

Maximum Marks 60

Due Date 07, April 2005

Assignment Weight age 2%

Question 1

(a) Define initial value and boundary value problem elaborate with the help of examples (at least one).

05

Solution

INITIAL VALUE PROBLEM

Differential equation of first order or greater in which the dependent variable y or its derivative are

specified at one points such as

0 0

,dy

f x ydx

y x y

If equation of the second order

2

2 1 02

'

0 0 0 1, '

d y dya x a x a x y g x

dx dx

y x y y x y

Where 0y and 1'y are arbitrary constants

Is called the initial value problem and the 0 0y x y '

0 1, 'y x y is called the initial conditions

BOUNDARY VALUE PROBLEM

Differential equation of order two or greater in which the dependent variable y or its derivative are

specified at different points such as

2

2 1 02

0 0 1 1,

d y dya x a x a x y g x

dx dx

y x y y x y

Where 0y and 1'y are arbitrary constants

Is called the boundary value problem and the 0 0 1 1,y x y y x y is called the boundary conditions

NOT THE PART OF THE SOLUTION (MOTIVATION OF THE CONDITIONS)

Page 2: Differential Equations - Solved Assignments - Semester Spring 2005

Basically when we solve the differential equations then we find some constant in the solution so we

need to elaborate these constants to solve this problem we formulate two kinds of conditions(given

above) which basically helps us to elaborate the constants finding in the solution of the differential

equation.

Still the question is how many constants are present in the differential equation. Is there any hard and

fast rule? yes! It depends that what is the order of the differential equation. If we have the first order

DE then its solution contains must one constant similarly if second order DE then its solution contains

must two constants……and so on and definitely to elaborate the constant we always need same

number of condition.

Question 1

(b) Solve the following differential equation.

2 20; (0) 1

xdy ydxxdx ydy y

x y

10

Solution

Page 3: Differential Equations - Solved Assignments - Semester Spring 2005

2 2

2 2 2 2

2 2

2 2

2 2 2 2

3 2 2 3

2 2 2 2

2 2 3

( ) 0 , 0 1

0

( , )

( , )=

,

2 1 2

xdy ydxxdx ydy y

x y

y xx dx y dy

x y x y

f yM x y x

x x y

f xN x y y

y x y

y xM x N y

x y x y

x xy y x y y xM N

x y x y

xy x y y xM

y

2

22 2

3 3 2 2 3 3 2

22 2

3 3 2 2 3 3 2

22 2

2 2

22 2

2 2 2 3

22 2

3 3 2 2 3 3 2

22 2

3 3 2

2 2 2 2 2

2 2 2 2 2

2 1 2

2 2 2 2 2

2 2

xy y

x y

x y xy x y x y xy yM

y x y

M x y xy x y x y xy y

y x y

M y xnow

y x y

xy x y x x y y xN

x x y

x y xy x y x y xy xN

x x y

N x y xy x

x

2 3 3 2

22 2

2 2

22 2

2 2 2y x y xy x

x y

N y x M

x yx y

2 2 ( , )

soour givenequationis exact

nowas we know

f yM x y x

x x y

Integrating both sides with respect to “x”

Page 4: Differential Equations - Solved Assignments - Semester Spring 2005

2 2 2 2

21

1

2 2

21

'

2 2

2

( , )

1( , ) ( )

1( , ) tan ( )

2

1sin tan

( , ) tan ( ) (1)2

1( )

1

f x y Mdx

yf x y x dx xdx y dx y

x y x y

x xf x y y y

y y

dx xce

x a a a

x xf x y y

y

f xy

xy y

y

f

2'

2 2 2

'

2 2

2 2

'

2 2 2 2

2

21

21

( )

( )

( , )

( )

'( )

( )2

1

2( , ) tan

2 2

tan2

y xy

y y x y

f xy

y y x

as we knowthat

f xN x y y

y x y

we will get

x xy y

x y y x

y y

yy

now putting this valuein equation no

so

x y yf x y

x

x yc

x

21

2

2

1

2 2

1 2tan

2 2 2 2

y

applying initial condition we get

c

x y y

x

Page 5: Differential Equations - Solved Assignments - Semester Spring 2005

Question 2

(a) Define the order and degree of DE also write down the order and degree of following DE.

2 " ' 2

22 "' ' 2

1 ( 1) 0

2 19 4 (4 25) 0

x y xy x y

x y xy x

04

Solution

Order of Differential Equation:

Order of the differential equation is the highest order derivative in a differential equation e.g.3y''+(y') =0 has order two.

Degree of the Differential Equation:

Degree of the differential equation is the power of the highest order derivative in the differential e.g.

y''+y=2 has degree one. Although, (y'') 3+y'+y=0 has degree three.

Order

(1) 2

(2) 3

Degree

(1) 1

(2) 2

Question 2

(b) Find I.F from 3rd

and 4th

method of exactness i.e. (xM+yN) & (xM-yN) respectively if possible. If

not then explain the reason.

3 3 5 44 6 0x y xy dx x x y dy 05

Solution

Integrating factor for 3rd

case can not be found due to that 0xM yN is not homogenous

So first case is not applicable as M= 3 34x y xy , N= 5 46x x y

3 3 5 4

4 2 3 5 4 2

4 +y 6

4 + 6

xM yN x x y xy x x y

xM yN x y x y x y x y

To check it is homogeneous

Page 6: Differential Equations - Solved Assignments - Semester Spring 2005

Say

4 2 3 5 4 2

5 4 5 2 3 6 5 6 4 2

5 4 2 3 5 4 2

5

4 + 6 ,

, , ,

, 4 + 6

, 4 +t 6

, ,

n

xM yN x y x y x y x y f x y

f tx ty t f x y for some real number n

but

f tx ty t x y t x y t x y t x y

f tx ty t x y x y x y tx y

f tx ty t f x y

For 4th case it is not possible because condition ( ) ( ) 0yf xy dx xg xy dy is not being fulfilled here.

As

3 3 5 4

3 2 4 3

4 6 0

4 6 0

x y xy dx x x y dy

y x xy dx x x x y dy

But 3 24x xy and 4 36x x y are not the function of xy

NOTE (RMEMBER)

f xy Means each term should have the product of xy e.g.

2 2f xy xy x y And 2 2 4 4f xy xy x y x y

Both are functions of xy

Question 2

(c) Define ordinary and partial differential equation as well as check whether the following are

ordinary differential equation or partial differential equation.

2

21 4 2

d x dy dz

du du du Where x, y and z are dependent variable and u is independent variable

2

22 3 4

d x dx

du du Where x is dependent variable and u is independent variable

2 2

2 23 3

x x

u v

Where x is dependent variable, u and v are independent variable

2 2

2 24 4

u v

x y

Where u, v are dependent variable, x and y are independent variable 06

Solution

ORDINARY DIFFERENTIAL EQUATION (ODE)

Page 7: Differential Equations - Solved Assignments - Semester Spring 2005

If an equation contains only ordinary derivatives of one or more dependent variables depending on

only one independent variable is known as ODE.

PARTIAL DIFFERENTIAL EQUATION (PDE)

If an equation contains partial derivatives of one or more dependent variables depending on two or

more independent variables is known as PDE

(1) ODE

(2) ODE

(3) PDE

(4) PDE

Question 3

(a) Define explicit and implicit solution. Check whether 1 1 1y x c i.e. solution of the DE

01 1

dx dy

x y

can be defined explicitly or not? If it can be define then write down the explicit form?

05

Solution

Explicit solution:

A solution of the differential equation of the form y= f (x) is called the explicit solution of the

differential equation or roughly speaking if we can separate the dependent an independent variable in

the solution then we say solution is explicit. For example dy / dt = y2-1/x has the solution y = 3+x

2/3-

x2.

It is explicit solution because dependent and independent variable are separate here of it is form of

y = f(x).

Implicit solution:

A solution of the differential equation of the form G (x ,y)=0 is called the implicit solution of the

differential equation or roughly speaking if you cannot separate the dependent an independent variable

then that solution is said to be implicit.

For example, dy/dt = 1+1/y2

has the solution y – tan-1

(y) = t+c. it is implicit solution because you

cannot separate the dependent and independent variable or it is of the form of G (x, y) =0.

Solution can be defined explicitly as follows.

Page 8: Differential Equations - Solved Assignments - Semester Spring 2005

1 ( 1)( 1)

11

( 1)

11

( 1)

1 1

1

y x c

yx c

yx c

x cy

x c

Question 3

(b) Whether following differential equation is separable or not if it is separable then solve it?

2 2

2 2 3

1 ( 5 5) 20 20 0

2 y

r r dr r r d

dyt t e

dt

10

Solution

2 2

2 2

2

2

2

1 ( 5 5) 20 20 0

( 5 5) 20 20

5 1 5 20 1 20

1 5 1 20

1 20

51

5 5 20

1 5

5 5 20

1 5 5

25

1 5

r r dr r r d

r r dr r r d

r dr r d

r dr r d

rdr d

r

drd

r

drd d

r

drd d

r

Integrate both sides

25

1 5

ln 1 25ln 5

drd d

r

r c

Page 9: Differential Equations - Solved Assignments - Semester Spring 2005

3

3

3

2 2 3

2 2 3

2 3

2

3

32

3 3

32

3

32

3

3 3

3 3

3 3

3 3

3

1 1

2

0

1

1

1

3

3 1

1 3

3 1

ln(1 )

3 3

ln(1 ) 3

(1 )

(1 ) .

(1 ) ,

y

y

y

y

y

y y

y

y

y

y

y

y

y t c

y t c

y t

dyt t e

dt

dy t t e dt

dy t e dt

dyt dt

e

e dyt dt

e e

e dyt dt

e

edy t dt

e

e tc

e t c

e e

e e e

e c e say c e

3c

Question 4

(a) Define a homogenous function. Check whether 2

3 3,

x yf x y

xy x y

is homogeneous or not explain.

Also check whether the DE 2

3 3( )

dy x y

dx xy x y

is solvable by method of homogenous equation or not if

not then give reason? 05

Solution

Homogeneous Function:

A function ,f x y is said to be homogeneous if , , ,nf tx ty t f x y for somereal number n

e.g. 2

2

2x, =

3xy+yf x y is homogeneous function

Page 10: Differential Equations - Solved Assignments - Semester Spring 2005

2

3 3

3 2

4 3 3

2

3 3

1

,

,

,

, ,

but

x yf x y

xy x y

t x yf tx ty

t xy x y

x yf tx ty

t xy x y

f tx ty t f x y

This is homogeneous function

Now to solve 2

3 3( )

dy x y

dx xy x y

we can’t apply here the homogeneous method

To solve the homogeneous differential equation

),( yxfdx

dy

We use the substitution

x

yv

If ),( yxf is homogeneous of degree zero, which we don’t have there.

Question 4

(b) Solve the following differential equation.

2 10

2 1

dy x y

dx y x

10

Solution

2 10

2 1

dy x y

dx y x

Page 11: Differential Equations - Solved Assignments - Semester Spring 2005

,

the given equation reduces to

2X Y+2h k+10

2Y X+2k h 1

We choose h and k such that

2h k+1=0;2k h 1=0

h+2k 1=0 multiply this with 2 we get

2h+4k-2=0

then adding both

2h k + 1=

now put thevalues

x X h y Y k

dy

dx

0

2h+4k 2=0

3 1 0

k=1 3 2h k+1=0

2h 1 3 1 0

2h+ 2 3 0

2h= 2 3

h= 1 3

h= 1 3,k=1 3

2X Y=

X 2Y

This is a homogenous equation. We substitute Y=VX to obtain

s in

k

soultimately we get

dy

dx

dY dVX V

dX dX

by replacing all value the g

2X

X 2VX

2

1 2V

2 1 2V

1 2V

2 1 2V

1 2V

iven DE

dV VXX V

dX

dV VX V

dX

V VdVX

dX

V VdVX

dX

Page 12: Differential Equations - Solved Assignments - Semester Spring 2005

2

2

2

2

2 2

1 2V

2 2 2

1 2V

1 2V 2

1

2V 1 20

1

dV V V VX

dX

dV V VX

dX

dV dXV V X

dV dXV V X

Integrating both sides

2

2

2 2

22

2

2 2

2

2V 1 20

1

ln | 1| 2 ln | | ln | |

( 1)

now replacing V=Y/X we will get the

X 1

( )

replace the values of X,Y

x=X 1 3, 1 3

X=x+1/3,Y=y-1/3

y 1/3 y 1/3 x+1/3 x+

dV dXV V X

V V X c

X V V c

Y Yc

X X

Y YX X c

now

y Y

2

2 2

2 2

2 2

2 2

2 2

1/3

3 1 3 1 3 1 3 1 9

3 6 1 3 3 3 1 3x 6 1 9

3 3x 2 3 3 6 3 6 1 9

3 3x 3 9 9 3 9

x 3 3 1 3

c

y x y x c

y y xy y x x c

y xy y y x x c

y xy y x c

y xy y x c

Page 13: Differential Equations - Solved Assignments - Semester Spring 2005

Assignment 2 (Spring 2005)

Maximum Marks 60

Due Date 20, April 2005

Assignment Weight age 2%

Question 1

(a) What is the difference between a linear equation (first order) and Bernoulli equation gives at

least two examples of each. 05

Solution

LINEAR DIFFERENTIAL EQUATION (FIRST ORDER)

A linear differential Equation is an Equation of the form ( )dy

a x b x y c xdx

it’s more frequently

used form ( ) ( )dy

f x y g xdx

where f(x) and g(x) are both continuous functions of x. e.g.

3 2dyx y x

dx

dyxy x

dx

BERNOULLI DIFFEREMNTIAL EQUATION

Now the differential equation of the form ndyf x y g x y

dx where n is any real no is called

Bernoulli equation. e.g.

2

2

dy 1

dx

1

y xyx

dyx y

dx y

DIFFERENCE The difference is that linear equation is a special type of Bernoulli’s equation when n is equal to 0 and

1 simply for n=0, 1 the Bernoulli’s equation becomes a linear equation.

Question 1

(b) Solve the following linear differential equation 10

32dy

x y ydx

Page 14: Differential Equations - Solved Assignments - Semester Spring 2005

Solution

3

3

2

2

dyx y y

dx

dy y

dx x y

It is clearly not a linear equation it can be changed to a linear as in first order linear equation

independent and dependent variables can be changed so writing equation as.

3

2

1

1ln

2

12

" "

1 , 2

1

dyy

y

dx x y

dy y

dxx y

dy y

nowit is linear in x

p y y Q y

IF e

IF ey

Multiplying by 1/y on both sides we get

2

2 2

1 12

2

( )

dxx y

y dy y

d xnowit is a differential of

dx y

d xy

dx y

xy c x y y c

y

Question 2

(a) Solve the following initial value problem and mention the type of the equation

3

, (1) 22

dy y xy

dx x y 10

Solution

Page 15: Differential Equations - Solved Assignments - Semester Spring 2005

3

43

4

3 3

2

" " 3

1 2 ,

2

14 4

4

dy y x

dx x y

It is abernaoulli equtionin y where n

p x x Q x x

dy yy x

dx x

put t y

dt dy dt dyy y

dx dx dx dx

Now the equation becomes

22

2ln ln 2

2

2 3

1 1

4 2

24

2 4

dxx xx

dtt x

dx x

dtt x

dx x

now IF will be

IF e e e x

Now

multiplying both sides x

dtx xt x

dx

now

It is the differential of

2dtx

dt

2 3

2 3

2 4

( ) 4

4

dtx x

dx

tx x dx

tx x c

4 2 4y x x c

By applying initial conditions

Now put y=2, x=1

4 2 4

2 1 1

15

c

c

So we get 4 2 4 15y x x

Question 2

(b)Show that 2 24 4y cx c is self orthogonal

Page 16: Differential Equations - Solved Assignments - Semester Spring 2005

Note: A family of curves is said to be self orthogonal if the orthogonal trajectories of the family is

same as the DE of that family. 10

Solution

2 2

2

2

2

2

4 4 (1)

2 42

(1)

4 42 2

2

y cx c

now

dy y dyy c c

dx dx

substituting this valueinequation

we get

y dy y dyy x

dx dx

dy dyy x y y

dx dx

so DE of given faimly is

2

2 2 2dy dy

y xy ydx dx

To get the family of orthogonal trajectories

Replace 1dy dy

bydxdx

2

2

2

2

2

2 2

2

2 2

2

2

2

2 1

2 1

2

2 3

dy dyy x y y

dx dx

by replacing we get

dy dyy x y y

dx dx

dy dyy xy y

dx dx

dy dyy xy y

dx dx

dy dyy xy y

dx dx

If we compare both differential equations (2) and (3) then came to know that both are the same

So the given family of equation is self orthogonal

Question 3

Page 17: Differential Equations - Solved Assignments - Semester Spring 2005

(a)Determine either the following curves are orthogonal to each other or not? If not then find the

condition of orthognality.

2 2 31 , 3y x x y

Note:

Two functions are said to be orthogonal if product of the slopes is -1. 07

Solution

TO check both curves orthogonal or not,

First of all differentiating both the equations w.r.t x 2 2

1 1

1

2 2 0

2

2

( , )

y x

differentiatin wrt x

dyy x

dx

dy x

dx y

Now wecheck at x y

1 1

11

|( , ) 1x y

xdym

dx y

Now considering the second equation

1 1

3

2

1 1

2

2 1|( , )

3

3

( , )

3x y

y x

dyx

dx

now at x y

dym x

dx

Now by applying condition of orthogonal

1 2 1m m

It is not satisfied

Now both the curves are not orthogonal

To find the condition we may proceed as follow and condition for orthognality is

1 2

211

1

3

1

1

3

1 1

3

1 1

1 0

( 3 ) 1 0

31 0

3 0

3 0

m m

putting thevalues

xx

y

x

y

x y

y x

This is the required condition.

Page 18: Differential Equations - Solved Assignments - Semester Spring 2005

Question 3

(b) Solve the following DE by the proper substitution

2 2y ydy sinx

xye +e =dx x

08

Solution

2 2

2

2

2

2

y y

y

y

y

2

2

2

2

2

2

2 y

dy sinxxye +e =

dx x

put e =u

dy du2ye =

dx dx

dy 1 duye =

dx 2 dx

x du sinx+u=

2 dx x

du 2 sinx+ u=2

dx x x

sinxHere p(x)=2/x And Q(x)=

x

2I.F=exp( dx)=x

x

dux +2xu=2sinx

dx

d(x u)=2sinx

dx

Integrate

x u= 2cosx+c

x e = 2cosx+c

Question 4

(a) The population of a town grows at rate proportional to the population at any time. Its initial

population of 50 decreased by 9% in 15 years what will be the population in 25 years? (Just make

the model of the population dynamics as well as just describe the given conditions do not solve

further) 03

Page 19: Differential Equations - Solved Assignments - Semester Spring 2005

Solution

Suppose that P0 is the initial population of the town, as given P0 is 50 and P (t) the population at any

time t. Then population growth can be represented by

dPP

dt

dPkP

dt

dPkdt

P

Integrate both sides

0 0

ln

kt c

kt c

kt c

P kt c

P e

P e e

P P e say P e

0P is the initial population of the town

0P =50 = P (0)

P (15) =50- 9

50100

P (25) =?

Question 4

(b) If a small metal bar whose initial temperature 30C is dropped in to a container of boiling water,

how long will it take for the bar to reach 90C if it is known that its temperature increased 2C in a

second? (Just make the model as well as just describe the given conditions do not solve further)

04

Solution

Small metal bar has the initial temperature T (0) =30 0C=T when dropped into a container of boiling

water (temperature of surrounding matters a lot on the small metal bar) so suppose mT is the

temperature of the boiling water. As we know the boiling temperature of the water is 100c so

100mT

Change in the temperature can be represented as dT

dtand it is proportional to mT T i.e.

propotional to 100dT

Tdt

With conditions T (1) =32as 30+2and T (0) =30

Page 20: Differential Equations - Solved Assignments - Semester Spring 2005

We have too find the time when T=90

( 100)

100

dTk T

dt

dTkT k

dt

Question 4

(c)The initially there were 50 milligrams of the radioactive substance present after 5 hours the mass

increased by 4%. If the rate of decay is proportional to the amount of the substance present at any

time, determine the half- life of the radioactive substance. (Just make the model as well as just

describe the given conditions do not solve further) 03

Solution

Suppose, radioactive substances = R

Then its model would be

proportional to R dR

dt

=kR dR

dt

Where k is the constant of proportionality

With the initial conditions R (0) = 50, R (5) = 52

Since it is given after 5 hours radioactive substances

Increased by 4%, thus 4% of 50 is 2 milligram

That means 52, as we subtract 50+2=52

Then =k R

dRdt

After integration, we get

kt + c

kt c

0 0

ln = kt + c

=e

= e e

R

R

R R where R

Now we have to find half-life of radioactive

Substances, that means t is required if R = R/2 i.e. R=25

Page 21: Differential Equations - Solved Assignments - Semester Spring 2005

Assignment 3 (Spring 2005)

Maximum Marks 60

Due Date 02, May 2005

Assignment Weight age 2%

Question 1

(a) Define uniqueness of solution and condition of its existence? If 1 2, , , ny y y are n solutions, on

an interval I , of the homogeneous linear nth-order differential equation

0011

1

1

yxadx

dyxa

dx

ydxa

dx

ydxa

n

n

nn

n

n on interval I are linearly independent then

is it necessary 1 2, , , nw y y y must be non-zero? 05

Solution

Let )(),(),...,(),( 011 xaxaxaxa nn and )(xg be continuous on an interval I and let Ixxan ,0)( .

If Ixx 0 , then a solution )(xy of the initial-value problem exist on I and is unique

NOT PART OF THE SOLUTION:

Always remember that solution of the differential equation differ by a constant as I told in the

first assignment’s solution we have initial or boundary conditions to determine these unknown

constants in the differential equations

Secondly, If 1 2, , , ny y y are n solutions, on an interval I , of the homogeneous linear nth-order

differential equation 0011

1

1

yxadx

dyxa

dx

ydxa

dx

ydxa

n

n

nn

n

n on interval I are linearly

independent then it is necessary that W ( 1 2, , , ny y y ) must be non-zero since we know

Linear Independence of Solutions:

The solutions

nyyy ,,, 21

are linearly dependent if and only if

IxyyyW n ,0,,2,1

Question 1

(b) Verify that the given two-parameter family of functions is the complimentary solution of the non-

homogeneous differential equation

2 3

1 2 3"' 5 " 6 ' 2sin 8; x xy y y x y c c e c e 05

Solution

Page 22: Differential Equations - Solved Assignments - Semester Spring 2005

Consider the associated homogenous differential equation

"' 5 " 6 ' 0y y y ----------------------- (1)

Given function is

2 3

1 2 3

x xy c c e c e

Taking derivatives of the function 2 3

2 3

2 3

2 3

2 3

2 3

' 2 3

" 4 9

"' 8 27

x x

x x

x x

y c e c e

y c e c e

y c e c e

Putting all these in the given associated homogenous differential equation

2 3 2 3 2 3

2 3 2 3 2 3

2 3 2 3 2 3

2 3 2 3 2 3

"' 5 " 6 '

8 27 5(4 9 ) 6(2 3 )

8 27 20 45 12 18

0

x x x x x x

x x x x x x

LHS y y y

c e c e c e c e c e c e

c e c e c e c e c e c e

Hence "' 5 " 6 ' 0y y y is the complementary solution of the given non-homogeneous

differential equation.

Question 2

(a) Define linearly independence and linearly dependence of the functions?

Comment either the functions given below are linearly independent or dependent

2 3

1 2 3( ) , ( ) , ( )x x xf x xe f x x e f x x e 10

Solution

Linear Dependence A set of functions

)(,),(),( 21 xfxfxf n

is said to be linearly dependent on an interval I if constants nccc ,,, 21 not all zero, such that

Ixxfcxfcxfc nn ,0)(.)()( 2211

Linear Independence

A set of functions

)(,),(),( 21 xfxfxf n

is said to be linearly independent on an interval I if

Ixxfcxfcxfc nn ,0)()()( 2211 ,

only when

.021 nccc

Page 23: Differential Equations - Solved Assignments - Semester Spring 2005

2 3

1 2 3( ) , ( ) , ( )x x xf x xe f x x e f x x e , First of all finding the derivatives of the function

1

1

1

2

2

2

2

2

2

3

3

2 3 2

3

2

3

( )

' ( ) 1

'' ( ) 1 2

( )

' ( ) 2 2

'' ( ) 2 2 4 2

( )

' ( ) 3 3

'' ( ) 3 2 3

x

x x x

x x x

x

x x x

x x x x

x

x x x

x x

f x xe

f x e xe x e

f x e x e x e

f x x e

f x xe x e x xe

f x xe x e x xe x x e

f x x e

f x x e x e x x e

f x x e x xe

2 2

2 3

2 3

2

2 2

2

2

2 2

2

3 2

2 2

3

6 6

, ,

1 2 3

( 2) 4 2 6 6

1

1 2 3

( 2) 4 2 6 6

1

1 2 3

( 2) 4 2 6 6

1

x x

x x x

x x x

x x x

x x x

x x x

x

x

x x e xe x x

Now

w xe x e x e

xe x e x e

x e x xe x x e

x e x x e xe x x

x x

xe e e x x x x x

x x x x x x

x x

xe x x x x x

x x x x x x

x

xe

2

2

2 2

2 2

2 2 2

3

2

2

3 2 2 2 2 2 2

3 3 2

1 2 3

( 2) 4 2 6 6

2 3 1 3

4 2 6 6 ( 2) 6 6

1 2

( 2) 4 2

4 6 2 6 6 9 10 2

8 8 2 0

x

x

x

x

x x x x x

x x x x x x

x x x x x x xx

x x x x x x x x xxe

x x xx

x x x

xe x x x x x x x x x

x e x x

Page 24: Differential Equations - Solved Assignments - Semester Spring 2005

As Wronskian is not equal to zero so all function are linearly independent

Question 2

(b) If functions )(,),( ),( 21 xfxfxf n possess at least n-1 derivatives on an interval I are linearly

independent then is it necessary functions having Wronskian non-zero? If not then give the

example (search a set of linearly independent functions such as their Wronskian is zero). 05

Solution

If functions )(,),( ),( 21 xfxfxf n possess at least n-1 derivatives on an interval I are linearly

independent then it is not necessary that functions having Wronskian non-zero for example

1 2 3( ) 0, ( ) , ( ) xf x f x x f x e at least posses 2 derivatives on interval I

1 2 3

1 2 3

1 2 3

( ) 0, ( ) , ( )

'' ( ) 0, ' ( ) 1, ' ( )

'' ( ) 0, '' ( ) 0, '' ( )

x

x

x

f x f x x f x e

f x f x f x e

f x f x f x e

Then it can easily notice that all above functions are linearly independent since

Firstly, no function is the constant multiple of the other function

Secondly, no one can be written linear combination of remaining two.

Thus they all are linearly independent.

But 0

0, , 0 1 0

0 0

x

x x

x

x e

w x e e

e

This assures us if Wronskian of function is nonzero then there is no doubt set of functions will always

be linearly independent but when Wronskian of functions is zero then set of function may be linearly

dependent or may be linear independent .

Question 2

(c) Find the 2nd

solution of each of Differential equations by reducing order or by using the formula.

2 4 3

1'' 6 ' 4 0 ;x y xy y y x x 05

Solution

For solving this we use the formula

2

( )

1 2

1 ( )

p x dx

ey x y x dx

y x

2

2

'

2

'' 6 ' 4 0

6 4'' 0

x y xy y

nowdividing by x on both sides

we get

y yy

x x

Page 25: Differential Equations - Solved Assignments - Semester Spring 2005

1

4 3

1

6

4 3

2 4 3 2

6ln| |4 3

4 3 2

sec

( )( )

( )( )

dxx

x

nowas y is given and it is given as

y x x

now for ond solution we proceed as follows

ey x x dx

x x

ex x dx

x x

64 3

6 2

4 3

2

( )( 1)

1( )

( 1)

xx x dx

x x

x x dxx

14 3

4 3

3

3

( 1)( )

1

( )

( 1)

( 1)

( 1)

xx x

x x

x

x x

x

x

So this is the required second solution. Question 3

(a) Define fundamental solution set of a differential equation and what is criterion for existence of

fundamental solution set of equation 05

Solution

Any set 1, 2, 3,....... ny y y y of n linearly independent solutions of the homogenous linear differential

equation on an interval are said to be a fundamental set of solutions on the interval.

Criteria for the existence of the fundamental solution set of for a linear nth order homogenous

differential equation is,

There always exists a fundamental set of solution for a linear nth order homogenous differential

equation.1

1 11( ) ( ) ... ( ) ( ) 0

n n

n n on n

d y d y dya x a x a x a x y

dx dx dx

on the interval I Question 3

(b) Check weather 2 2, , lnx x x x in the interval (0, ) form the fundamental set of solution the

differential equation 2"' 6 4 ' 4 0y x y xy y if it is fundamental solution what will be general

solution. 10

Page 26: Differential Equations - Solved Assignments - Semester Spring 2005

Solution

Now for fundamental solution set of 2"' 6 4 ' 4 0y x y xy y differential equation we have to

show that all the three functions are linearly independent.

1 2 3

' "

1 1 1

2 ' 3 " 4

2 2 2

2 ' 3 3 " 4 4 4

3 3 3

2 ' 3 " 4

3 3 3

2 2

2 2

3 3

4 4

, ,

1 0

2 6

ln 2 ln 6 ln 2 3

ln 1 2ln 6ln 5

, , ln

ln

1 2 1 2ln

0 6 6

nowlet y y y are thethree functions

y x y y

y x y x y x

y x x y x x x y x x x x

y x x y x x y x x

w x x x x

x x x x

x x x

x x

2 2 1 1 2 2

2 3

2 2

1 1 1 1

4

2 2 2 2

4 1 2 1

ln 5

1 ln

1 2 1 2ln ,

0 6 6ln 5

2 1 2ln 1 1 2ln 1 21 ln

6 6ln 5 0 6ln 5 0 6

2 1 2ln 1 1 2l

6 6ln 5

x

x x

x x x x x by taking common x and x fromc c respectively

x x x

x x x x x xx x

x x x x x x

xx x x x

x

1

1 1

4 3 2 2

4 3 2

6 1

n 1 12 ln

0 6ln 5 0 3

12ln 10 6 12ln 6ln 5 6 ln

4 5

4 5 0

xx x

x x x

x x x x x x x x

x x x

x x

As not equal to zero so these are independent and so forms a fundamental solution set.

So general solution may be written as

2 2

1 2 3 lnY c x c x c x x

Question 4

(a) Find the solution of the following DE.

''' 9 '' 26 ' 24 0; 0 1, ' 0 2, '' 0 3y y y y y y 10

Solution

To solve this equation

Page 27: Differential Equations - Solved Assignments - Semester Spring 2005

''' 9 '' 26 ' 24 0y y y

Suppose

2

3

'

''

'''

mx

mx

mx

mx

y e

y me

y m e

y m e

Substituting in equation 3 2

3 2

3 2

9 26 24 0

( 9 26 24) 0

0 ( 9 26 24) 0

2

mx mx mx

mx

mx

m e m e me

e m m m

e or m m m

m is the root of equationthenby useof synthatic division we get the roots as

2 1 9 26 24

2 14 24

1 7 12 0

2

2

2 3 4

1 2 3

1 2 3

2 3 4

1 2 3

2 7 12 0

2 3 4 12 0

2 3 4 3 0

2 3 4 0

0 1

1 1

' 2 3 4

x x x

x x x

m m m

m m m m

m m m m

m m m

general solution may be written as

y c e c e c e

applying y

c c c

y c e c e c e

1 2 3

2 3 4

1 2 3

1 2 3

1 2 3

1 2 3

2 3

' 0 2

2 2 3 4 2

'' 4 9 16

' 0 3

3 4 9 16 3

2 2 1

2 2 3 4

2 2 2 2

0 2 4

x x x

applying y

c c c

y c e c e c e

applying y

c c c

EQ EQ

c c c

c c c

c c

3 4 1EQ EQ

Page 28: Differential Equations - Solved Assignments - Semester Spring 2005

1 2 3

1 2 3

2 3

3 4 9 16

4 4 4 4

1 5 12

c c c

c c c

c c

2 3

2 3

3 3

2 2 2

1 2 3 1 1

2 3 4

5 5 4

1 5 12

0 5 10

1 2 1 2

4 0 2 2 2 2 1

1

1 1 1 1/ 2 1/ 2

1 1

2 2

x x x

EQ EQ

c c

c c

c c

then EQ implies c c c

then EQ implies

c c c c c

y e e e

Question 4

(b) The roots of an auxiliary equation are 1 2 3

1, 2 , 2

3m m i m i . What is the corresponding

differential equation? 05

Solution

Given that 1 2 3

12 2

3

10 ( 2) 0 ( 2) 0

3

m m i m i

m m i m i

The corresponding auxiliary equation will be,

2

2

2

23 2

3 2 2

3 2

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

3

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

3

12 2 4 2 2 1 0

3

14 5 0

3

4 54 5 0

3 3 3

3 12 15 4 5 0

3 11 11 5 0

m m i m i

m m m i m i m i

m m m m im i im i

m m m

mm m m m

m m m m m

m m m

Finally the corresponding differential equation will be,

Page 29: Differential Equations - Solved Assignments - Semester Spring 2005

3 23 2

3 23 11 11 5 0 3 11 11 5 0

d y d y dyD D D

dx dx dx

Page 30: Differential Equations - Solved Assignments - Semester Spring 2005

Assignment 4 (Spring 2005)

Maximum Marks 40

Due Date 12, May 2005

Assignment Weight age 2%

Question 1

Solve the differential equation /// //6 1 cosr r .find complimentary cr and particular solution (

pr )

by the undetermined coefficient (superposition approach). 10

Solution /// //6 1 cosr r

Complementary function

To find cr , we solve the associated homogeneous differential equation

/// //6 0r r

Put nr e , 2 3' , r'' , '''n n nr ne n e r n e

Substitute in the given differential equation to obtain the auxiliary equation

/// //

3 2 2

6 0

6 0 6 0

0,0,6

r r

n n n n

n

Hence, the auxiliary equation has complex roots. Hence the complementary function is

1 2 3

6cr c c c e

Particular Integral Corresponding to cos( )g :

1cos sinpr A B

Corresponding to 1f : 2pr C

Therefore, the normal assumption for the particular solution is

1 2p p pr r r

1

cos sinpr A B +C

Clearly there is duplication of

(i) The constant function between cr and2pr .

To remove this duplication, we multiply 2pr with 2 . This duplication can’t be removed by multiplying

with . Hence, the correct assumption for the particular solution 2pr is

2cos sinpr A B C

Page 31: Differential Equations - Solved Assignments - Semester Spring 2005

Then ' sin cos 2pr A B C

'' cos sin 2

''' sin cos

p

p

r A B C

r A B

Therefore

1 1

1 1

''' 6 '' sin cos 6 cos sin 2

''' 6 '' sin 6 cos 6 12

p p

p p

r r A B A B C

r r A B A B C

Substituting the derivatives of pr in the given differential equation and grouping the like terms, we

have

sin 6 cos 6 12 1 cosA B A B C

6 0, 6 1, 12 1 21 1A B A B C C

Solving these equations, we obtain

6 6 6 1 37 1 1 37

6 37

BB

A

A B B B

A particular solution of the equation is

21 37 6cos sin 1 12pr

So

2

1 2 3

6 1 37 6cos sin 12

c pr r r

r c c c e

Question 2

Factor the given differential operator also find the function annihilated by this 4 64D D 10

Solution

4 3

34 3

4 2

2 1

64 64

64 4

64 4 4 16

Since 0

Annihilates each of the functions

1, , , ,

D Annihilates 1

n

n

D D D D

D D D D

D D D D D D

D y

x x x

Since nD )(

Annihilates each of the functions

xnxxx exexxee 1 2 , , , ,

( 4)D Annihilates -4 xe

Since differential operator

nDD 2 222

is the annihilator operator of the functions

Page 32: Differential Equations - Solved Assignments - Semester Spring 2005

2 1cos , cos , cos , , cosx x x n xe x xe x x e x x e x

2 1sin , sin , sin , , sinx x x n xe x xe x x e x x e x

22 2 2 2 2

2 4 2

16 2 16 16 4 12 2 3

2, 2 3

2 4 16D D Annihilates 2 2cos2 3, sin 2 3x xe e

Therefore, the operator 4 264 3 4 16D D D D D D annihilates linear combination 1,-4 xe ,

2 2cos2 3, sin 2 3x xe e

Question 3

Find the annihilator operator of the function 22e e e 10

Solution

Suppose that

2

21 2 3

2

,( ) , y ( ) 2 y ( )

e e e

y e e e

Then

1

2 2

2

3 3 2

2

1 1 0

1 1 2 0

1 1 0

D y D e

D y D e

D y D e

Therefore, the product of two operators

3

1D

Annihilates the given function 2( ) 2f e e e

Question 4

Solve the differential equation // 2 5xy y x e .find complimentary cy and particular solution ( py ) by

the undetermined coefficient (annihilator approach). 10

Solution

Step 1 The given differential equation can be written as

2 2( 1) 5xD y x e

Step 2 The associated homogeneous differential equation is

Page 33: Differential Equations - Solved Assignments - Semester Spring 2005

2( 1) 0

1 1 0

D y

D D

Roots of the auxiliary equation are complex

1, 1m

Therefore, the complementary function is

1 2

x x

cy c e c e

Step 3 Since 3 21 0, 5 0xD x e D

Therefore the operators 3

1D and D annihilate the functions 2 xx e and 5 . We apply 3

1D D to

the non-homogeneous differential equation

3 21 ( 1) 0D D D y .

This is a homogeneous differential equation of order 6.

Step 4 The auxiliary equation of this differential equation is

3 21 ( 1) 0 0

0, 1,1,1,1, 1

m m m y

m

Therefore, the general solution of this equation must be

2 31 2 53 4 6

x x x x xc c e xe e x ey c c x c c e

Step 5 Since the following terms are already present in cy

2 6

x xc e c e

Thus we remove these terms. The remaining ones are

2 31 53 4

x x xc xe e x ec c x c

Step 6 The basic form of the particular solution of the equation is

2 31 53 4

x x xp c xe e x ey c c x c

The constants 1c , 43 ,cc and 5c have been replaced with BA , and C .

Step 7 Since

2 3

2 3

x x x

p

x

p

y A Bxe C e Dx e

y A Bx C Dx e

x

x

Then taking derivative

2 3 2

2 3 2

2 3

'

'

' 2 3

2 3

2 3

x x

p

x

p

x

p

y Bx C Dx e B C Dx e

y Bx C Dx B C Dx e

y B C x C D Dx B e

x x

x x

x

Again taking derivative

Page 34: Differential Equations - Solved Assignments - Semester Spring 2005

2 2 3

2 2 3

2 3

'' 3 2 3

'' 3 2 3

'' 2 2 6 2 3

2 2 3

2 2 3

2 3

x x

p

x

p

x

p

y B C C D e B C x C D Dx B e

y B C C D B C x C D Dx B e

y B C C D B C C D Dx e

x Dx x

x Dx x

x D x

Therefore

2 3 2 3

2 3 2 3

2 3 3

2

2 2 6 2 3

2 2 6 2 3

2 2 6 2 3

2 4 6 6

2 3

2 3

2 3

2

x x

p p

x

p p

x

p p

p p

y y B C C D B C C D Dx e A Bx C Dx e

y y B C C D B C C D Dx Bx C Dx e A

y y B C C D B C B C D C Dx Dx e A

y y B C C D D

x D x x

x D x x

x D x

x x

xe A

Substituting in the given differential equation, we have

2

2 2

5

2 4 6 6 52

x

p p

x x

y y x e

B C C D D e A x ex x

Equating coefficients of 2,x xe x e and the constant terms, we have

2 0, 4 6 0, 6 1, 5

1, 2 3 , , 5

6

1 1 1, , , 5

4 4 6

2B C C D D A

B C C D D A

B C D A

Thus

2 3

2 3

1 1 15

4 4 6

5 3 3 212

x

p

x

p

y x x e

ey x x

x

x

Step 8 Hence, the general solution of the given differential equation is

pc yyy

Or 2 31 2 5 3 3 2

12

xx x e

c e c e x xy x

Page 35: Differential Equations - Solved Assignments - Semester Spring 2005

Assignment 5 (Spring 2005)

Maximum Marks 40

Due Date 21, June 2005

Assignment Weight age 2%

Question 1

State in words a possible physical interpretation of the given initial-value problems

(a)

2

2

15 0

32

(0) 2, '(0) 1

d xx

dt

x x

(b)

2

2

12.5 0

2

(0) 0.5, '(0) 0.25

d xx

dt

x x

(c)

2

2

13.5 0

8

(0) 1.5, '(0) 0

d xx

dt

x x

15

Solution

(a)Comparing 2

2

15 0

32

d xx

dt with

2

20

d xm kx

dt , we got

1m= , k=5

32

W=mg

W=32/32=1

Weight of 1 lb attached to a spring, is released from a point 2 units below the equilibrium position

with initial downward velocity and the spring constant is 5lb/ft.

(b)Comparing 2

2

12.5 0

2

d xx

dt with

2

20

d xm kx

dt , we got

1m= , k=2.5

2

W=mg

W=32/2=16

Weight of 16lb attached to a spring, is released from a point 0.5 units above the equilibrium position

with initial upward velocity and the spring constant is 2.5lb/ft.

(c)Comparing 2

2

13.5 0

8

d xx

dt with

2

20

d xm kx

dt , we got

1m= , k=3.5

8

W=mg

W=32/8=4

Weight of 4lb attached to a spring, body starts from rest from point 1.5 units below the equilibrium,

and spring constant is 3.5 lb/ft.

Page 36: Differential Equations - Solved Assignments - Semester Spring 2005

Question 2

A 24lbweight, attached to a spring, stretches it 4 in. find the equation of motion if the weight is

released from the rest from a point 3 in above the equilibrium point also find the time period(period of

oscillation) and frequency as well as amplitude. 10

Solution

As we know general equation of simple harmonic motion (un-damped free motion) is

kxdt

xdm

2

2

----- (1)

Or 2

2

20

d xx

dt

So basically we needed m, k thus we can find

For consistency of units with the engineering system, we make the following conversions

12 inches 1 foot

11 inches foot

12

4 1

4 inches foot= foot12 3

.

So, 4 1

4 12 3

Stretch s inc ft ft

Further weight of the body is given to be

W 24 lb

But mgW

Therefore W 24 3

32 4m

g

or 3

slugs.4

m

Since 4 1

4 12 3

Stretch s inc ft ft

Therefore by Hook’s Law, we can write

1

243

k

72 lbs/ftk

Equation 1 is becomes

Page 37: Differential Equations - Solved Assignments - Semester Spring 2005

2

2

2

2

372

4

472

3

d xx

dt

d xx

dt

or 2

296 0

d xx

dt .

Since the initial displacement is 3 1

3 f12 4

inches ft t

1

0 , 0 04

x x

The negative sign indicates that body starts from rest above the equilibrium point at the distance of 3

inch from equilibrium point or ¼ foot from equilibrium point

2

296 0

d xx

dt

Subject to 1

0 , 0 04

x x

Putting mtmt emdt

xdex 2

2

2

,

We obtain the auxiliary equation

2 96 0m

or 4 6m i

The general solution of the equation is

1 2cos4 6 sin 4 6x t c t c t

Now, we apply the initial conditions.

1

04

x 1 2

1.1 .0

4c c

Thus 1

1

4c

So that 2

1cos4 6 sin 4 6

4x t t c t

Since

1 26 sin 4 6 4 6 cos4 6x t c t c t .

Therefore

0 0x 20 4 6 .1 0c

Thus

Page 38: Differential Equations - Solved Assignments - Semester Spring 2005

2 0c .

Hence, solution of the initial value problem is

1

cos 4 64

x t t

Time period

2T

Comparing 2

2

20

d xx

dt with

2

296 0

d xx

dt we have

2 96 4 6

It implies

2

4 6 2 6T

Frequency

1 2 6

2f

T

Amplitude

22 2 2

1 2 1 4 0 1 4A c c Question 3

Physically interpret the following differential equations.

(a)

2

2

12 0

16

0 0, ' 0 1.5

d x dxx

dt dt

x x

(b)

2

24 4 0

0 1, ' 0 2.01

d x dxx

dt dt

x x

(c)

2

216 0

0 1, ' 0 0

d x dxx

dt dt

x x

15

Solution

(a)Comparing the given differential equation

2

2

12 0

16

d x dxx

dt dt

With the general equation of the free damped motion

Page 39: Differential Equations - Solved Assignments - Semester Spring 2005

2

2

2

2

22

2

0

0

2 0

d x dxm β kx

dt dt

d x β dx kx

dt m dt m

d x dxλ x

dt dt

So here

m =1/16, β =2, k =1

W=mg

W=32/16=2lb

we see that

22 12 16

1 16 1 16

16, 4

β kλ ,

m m

λ

So that 022 λ

System is Over-damped motion.

The problem represents

“A 2lb weight is attached to a spring whose spring constant is 1lb/ft i.e. k=1lb/ft. the system is Over-

damped with resisting force numerically equal to 2 times of the instantaneous velocity i.e.

2dx dx

Damping force -βdt dt

Inspection of the boundary conditions:

0 0, ' 0 1.5x x

Reveals that the weight starts from equilibrium position with upward velocity of 1.5 ft/sec

Solution

(b)Comparing the given differential equation

2

24 4 0

d x dxx

dt dt

With the general equation of the free damped motion

2

20

d x dxm β kx

dt dt

2

20

d x β dx kx

dt m dt m

Page 40: Differential Equations - Solved Assignments - Semester Spring 2005

2

2

22 0

d x dxλ x

dt dt

So here

m =1, β =4, k =4

W=mg

W=32=32lb

we see that

24 42 4

1 1

2, 2

β kλ ,

m m

λ

So that 2 2 0λ

System is critically -damped motion.

The problem represents

“A 32lb weight is attached to a spring whose spring constant is 4lb/ft i.e. k=4lb/ft. the system is

critically damped with resisting force numerically equal to 4 times of the instantaneous velocity i.e.

4dx dx

Damping force -βdt dt

Inspection of the boundary conditions:

0 1, ' 0 2.01x x

Reveals that the weight starts from1 unit above the equilibrium position with a downward velocity of

2.01 ft/sec

Solution

(c)Comparing the given differential equation

2

216 0

d x dxx

dt dt

With the general equation of the free damped motion

2

20

d x dxm β kx

dt dt

2

20

d x β dx kx

dt m dt m

2

2

22 0

d x dxλ x

dt dt

Page 41: Differential Equations - Solved Assignments - Semester Spring 2005

So here

m =1, β =1, k =16

W=mg

W=32=32lb

we see that

21 162 16

1 1

1, 4

2

β kλ ,

m m

λ

So that 2 2 0λ

System under -damped motion

The problem represents

“A 32lb weight is attached to a spring whose spring constant is 16 lb/ft i.e. k=16 lb/ft. the system is

under -damped with resisting force numerically equal to 1 times of the instantaneous velocity i.e.

1dx dx

Damping force -βdt dt

Inspection of the boundary conditions:

0 1, ' 0 0x x

Reveals that the weight starts from rest 1 unit below the equilibrium position

Question 4

(a) A 16-lb weight stretches a spring 8/3 ft. Initially the weight starts from rest 2-ft below the

equilibrium position and the subsequent motion takes place in a medium that offers a damping force

numerically equal to ½ the instantaneous velocity. Find the equation of motion, if the weight is driven

by an external force equal to .3cos10 ttf 10

(b)Solve the following differential equation.

2 32 2 lnx y xy y x x

10

Solution

(a) Given w=16 lb and s=8/3 ft , β = ½ , .3cos10 ttf Initially the weight starts from rest 2-ft

below the equilibrium position and the subsequent motion takes place in a medium i.e

0 2, ' 0 0x x then differential equation of forced motion

Page 42: Differential Equations - Solved Assignments - Semester Spring 2005

2

2

d x dxm kx f t

dt dt

16 1

32 2

WW mg m

g

166

8 3

FF ks k

s

2

2

1 16 10cos3

2 2

d x dxx t

dt dt

or 2

212 20cos3

d x dxx t

dt dt

First consider the associated homogeneous differential equation.

2

212 0

d x dxx

dt dt

Put mtmtmt emdt

xd, me

dt

dx, ex 2

2

2

Then the auxiliary equation is:

2

2

2

2

12 0

1 112 0

4 4

1 48 10

2 4

1 47

2 4

1 47

2 2

m m

m m

m

m

m i

1 47

2 2m i

Thus the auxiliary equation has complex roots

1 2

1 47 1 47,

2 2 2 2m m i m m i

So that the complementary function of the equation is

47 472

cos sins1 22 2

tx e c t c tc

To find a particular integral of non-homogeneous differential equation we use the undetermined

coefficients, we assume that

cos3 sin3px t A t B t

Then 3 sin 4 3 cos4px t A t B t

9 cos4 9 sin 4px t A t B t

So that

Page 43: Differential Equations - Solved Assignments - Semester Spring 2005

12 9 cos 4 9 sin 4 3 sin 4 3 cos 4 12 cos3 12 sin3

3 3 cos 4 3 3 sin 4

p p px x x A t B t A t B t A t B t

A B t A B t

Substituting in the given non-homogeneous differential equation, we obtain

3 3 cos4 3 3 sin 4 20cos3A B t A B t t

Equating coefficients, we have

3 3 20A B

3 3 0A B

Solving these equations, we obtain

10 10

3 3A , B

Thus 10

cos3 sin33

px t t t

Hence the general solution of the differential equation is:

47 47 102

cos sins cos3 sin31 22 2 3

tx t e c t c t t t

1 47 472cos sins

1 22 2 2

47 47 47 102sin cos 3 sin 3 cos3

1 22 2 2 3

tx t x t e c t c t

te c t c t t t

Now 0 2x gives

1 1

10 10 4.1 2 2

3 3 3c c

Also 00 x gives

2

2

2

47 10 1 40 3

2 3 2 3

47 20 10

2 3

64

3 47

c

c

c

or

Hence the solution of the initial value problem is:

4 47 64 47 102

cos sins cos3 sin33 2 2 33 47

tx t e t t t t

Page 44: Differential Equations - Solved Assignments - Semester Spring 2005

Solution

(b)

2 3

1

2

2

2

1 2

2 2 ln

,

1

2 2 0

2 1 0

1,2

m m

m

m

c

x y xy y x x

put

y x y mx

y m m x

x m m m

m m

m

y c x c x

2

1 2

1 1 2 2

2

2 2 2

2

5

1 3

4

2 3

,

. .

21 2

0ln

ln 2

0ln

1 ln

p

y x y x

By V D P y u y u y

x xW x x x

x

xW x x

x x x

xW x x

x x

311

4

1

ln

int

1ln

4 4

Wu x x

W

On egrating

xu x

222

3

2

ln

int

1ln

3 3

Wu x x

W

On egrating

xu x

2

1 2

5 5

5

5 2

1 2

1 1ln ln

4 4 3 3

ln 1 ln 1

4 16 3 9

ln 7

12 16 9

p

p

p

y u x u x

x xy x x

x xy x

xy x c x c x

Page 45: Differential Equations - Solved Assignments - Semester Spring 2005

Assignment 6 (Spring 2005)

Maximum Marks 60

Due Date 05, July 2005

Assignment Weight age 2%

Question 1

(a) Define sequence and series define both with at least two examples, can every sequence be

represented as a series or not? Consider the series 2 3

0

1 ...! 2! 3!

n

n

x x xx

n

find the derived

series by differentiating the series. Show that both have same radius of convergence. 10

NOTE:

A series is said to be derived series which is obtained by integrating or differentiating term by term a

given series.

Solution

Sequence:

The sequence is defined as the function having the natural numbers as its domain N= {1, 2, 3…}.

Series:

The series is defined sum of the terms of a sequence.

Examples

1

2

1

1 2

1

1 2 3{ } , , ,... (1)

2 3 4 5

1 4 9{ } , , ... (2)

2 1 3 5 7

,

n

n

n

n

n

n

n

n

now for series series are the sum of the terms of a sequence

ar a ar ar

3

1

...

1 1 1 1...

( 1) 1.2 2.3 3.4k

ar

k k

Every sequence definitely can be represented as a series by simply adding the terms of that sequence.

Both the sequence 1 and 2 can be represented as a series like

1

2

1

1 2 3...

2 3 4 5

1 4 9...

2 1 3 5 7

n

n

n

n

n

n

Now we are going to find the derived series

The given series is

Page 46: Differential Equations - Solved Assignments - Semester Spring 2005

2 3

0

2 3

2 3

0

1 ...! 2! 3!

term by term

0 1 ...2 6

1 ...! 2! 3!

n

n

n

n

x x xx

n

now to get derived series we differentiate

x xx

we will get this series which is same as the above series

x x xx

n

so both the se

1

n n n n

1 1 ! ! !lim lim lim lim 0

1 ! 1 ! ! 1

n

n

ries will have the same radius of convergence but for is series

radius of convergence is

nc n n

c n n n n

so radius of convergence of orignal and derived series

.are same

Question 1

(b)If 3 5 7

...3 5 7

x x xSinx x the find the few terms of the power series of Cscx 05

Show all the steps involved each step has its own importance.

Solution

3 5 7 9

3 5 7 9

2 4 6 8

1sec

x x x xx-

3 5 7 9

x x x xx-

3 5 7 9

x x x x1-(

x 3 5 7 9

Co xSinx

Page 47: Differential Equations - Solved Assignments - Semester Spring 2005

I Now as we have that 3 5 7

...3 5 7

x x xSinx x

2

pplying the boinomial theorem i.e. (1+x)n =1+nX+n(n-1)/2! X2+n(n-1)(n-2)/3! X3...

When n is a ve integer or it is in the form of fraction then we use the following formula .

x1+(-1)

1=

x

4 6 8 2 4 6 8

2 4 6 8

2 4 6 8 2 4 6 8

2 4 6 8

x x x x x x x

3 5 7 9 3 5 7 9

x x x x...

3 5 7 9

x x x x x x x x1+(-1)( (

1 3 5 7 9 3 5 7 9

x x x x(

3 5 7 9

x

2 4 6 8 4 8 6 8

6 8

2 4 6 8 4

...

x x x x x x x x1+ neglecting higher tems neglecting higher terms

3 5 7 9 4 25 15 211

x xneglecting higher terms ...

27 45

x x x x x1

1 3 5 7 9 4

x

x

8 6 8

6 8

2 4 4 6 6 6 8 8 8

6 8

2 4 6 8

x x x

25 15 21

x x...

27 45

x x x x x x x x x1-

1 3 5 4 7 15 27 9 25 21

x x...

27 45

1 x 9x 296x 206x1- ...

3 20 945 945

x

x

So it is the required power series for the given function

Question 1

(c) Find the radius of convergence of 1

1

( 1) ( 2)n n

n

n x

05

.

Solution

1

0

1 1

1 2

1

( 1) ( 2) ( )

( 1) ( 1) 1

n n n

n

n n

n n

n n

compareit n x with c x x

c n c n

Page 48: Differential Equations - Solved Assignments - Semester Spring 2005

1

2

1

lim

( 1) ( 1)lim

( 1)

1lim 1

nn

n

n

n n

n

cR

c

nR

n

nR

n

So the radius of convergence is 1

Question 2

(a) Find the function represented by the following power series 0

n

n

x

05

Solution

2 3 4

1

0,1,2,3...

1 ...

inf

S1

1 and r=x

1so

1-x

n

n

now for n

we can get the series

x x x x x

so it is an inite geometric series so finding it sum

a

r

here a

Question 2

(b) Show that the given DE has a regular singular point at x=0.Determine the indicial equation, the

recurrence relation and the roots of the indicial equation. 10

" 0xy y

Solution

2

" 0

tan

" 0

sin int x=0 is a singular point

and p(x)=0 for which x-0) is analytic

1and q(x)= for which x-0) is analytic

x x

xy y

in s dard form it can be written as

yy

x

now by defination of gular po

Page 49: Differential Equations - Solved Assignments - Semester Spring 2005

so the point x=0 is regular singular point.

0

1

0

2

0

0

" 0 (1)

sup

' ( )

'' ( )( 1)

1

( )(

r n

n

n

r n

n

n

r n

n

n

n

n

xy y

now we pose there is a solution of the form

y a x

y a r n x

y a r n r n x

now substitutiong all theses values in equation no

x a r n

2

0

1

0 0

1 1

0

1 0

1

1

0 1

1) 0

( )( 1) 0

( 1) ( )( 1) 0

putting this value in the equation

r n r n

n

n

r n r n

n n

n n

r r n r n

n n

n n

r n r n

n n

n n

r n x a x

a r n r n x a x

a r r x a r n r n x a x

now we can write a x a x

now

1 1 1

0 1

1 1

1 1 1

0 1

1 1

1 1

0 1

1

( 1) ( )( 1) 0

( 1) ( )( 1) 0

( 1) [ ( )( 1) ]

( 1)

r r n r n

n n

n n

r r n r n

n n

n n

r r n

n n

n

a r r x a r n r n x a x

a r r x a r n r n x a x

a r r x a r n r n a x

now we have got indicial equation as r r

0

1

1

1

1

0 1

comparing the cofficent of

( )( 1) 0

( )( 1)

r 1

(1 )( )

r n

n n

nn

nn

r r

now x

we get a r n r n a

aa

r n r n

so for the greater root

we get the relation

aa

n n

Page 50: Differential Equations - Solved Assignments - Semester Spring 2005

01

012

023

0

1,2,3...

11.2

n=22.3 2.3!

33.4 3.4!

( 1)

!( 1)!

n

n

now replacing values of n

afor n a

aafor a

aafor n a

so continuing in this manner we will get

aa

n n

so for the series sol

2

01

( 1)( ) [1 ... ...]

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

n

ution we will get the following series solution

ax xy x x

n n

Question 3

(a) Define the ordinary and singular point. Is it necessary that ordinary and singular points are always

real? Find the singular point for the following differential equations.

(1) 2( 4) 2 6 0x y xy y

(2) 1

3 01

y xyx

08

Solution

Ordinary point:

A point 0x is said to be a ordinary point of a differential equation 0)()()( 012 yxayxayxa if

both 1

2

( )P(x) =

( )

a x

a x and) 0

2

( )Q(x)=

( )

a x

a x are analytic at 0x .

Singular point:

A point that is not an ordinary point is said to be singular point of the equation

Or

A point 0x is said to be a singular point of a differential equation 0)()()( 012 yxayxayxa if

both 1

2

( )P(x) =

( )

a x

a x and) 0

2

( )Q(x)=

( )

a x

a x or either one of these are not analytic at 0x .

Both singular and ordinary points need not be real numbers

(1) 2( 4) 2 6 0x y xy y

The equation 2( 4) 2 6 0x y xy y has the singular points at the solutions of 2 4 0x , namely,

2x i .

All other finite values, real or complex, are ordinary points.

Page 51: Differential Equations - Solved Assignments - Semester Spring 2005

(2) 1

3 01

y xyx

The equation 1

3 01

y xyx

has no singular point. But having ordinary point at the solutions of

1 0x , namely, 1x .

Question 3

(b) If 0xx is an ordinary point of the differential equation 0)()( yxQyxPy , can we always

find two linearly independent solutions in the form of power series centered at 0x :

.)(0

0

n

n

n xxcy

Give the reason? 02

Solution

Yes! We can always find the two linearly independent solutions and THEOREM (Existence of Power

Series Solution) gives us authentication .i.e.

“If 0xx is an ordinary point of the differential equation 0)()( yxQyxPy , we can always

find two linearly independent solutions in the form of power series centered at 0x :

.)(0

0

n

n

n xxcy

A series solution converges at least for Rxx 0 , where R is the distance from 0x to the closest

singular point (real or complex)”

Question 4

(a) Define the regular singular and irregular singular point. Find the singular point for the following

differential equations.

(1) 2( 4) 2 30 0x y xy y

(2) 1 2 1 5 0x y x y y 06

Solution

Regular singular point

A Singular point x x 0 of the given equation 2 1 0( ) ( ) ( ) 0a x y a x y a x y is said to be a regular

singular point if both 10 0

2

( )( ) ( ) ( )

( )

a xx x P x x x

a x and 2 2 0

0 02

( )( ) ( ) ( )

( )

a xx x Q x x x

a x are analytic

at 0x .

Irregular singular point

A singular point that is not regular is said to be an irregular singular point of the equation

Or

Page 52: Differential Equations - Solved Assignments - Semester Spring 2005

A Singular point x x 0 of the given equation 2 1 0( ) ( ) ( ) 0a x y a x y a x y is said to be an irregular

singular point if both 10 0

2

( )( ) ( ) ( )

( )

a xx x P x x x

a x and 2 2 0

0 02

( )( ) ( ) ( )

( )

a xx x Q x x x

a x or either one

of them are not analytic at 0x .

(1) 2( 4) 2 30 0x y xy y

2x are singular points of the equation

Because 2 4 0x or 2x .

Now write the equation in the form

2 2

2 300

( 4) 4

xy y y

x x

or 2 30

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

xy y y

x x x x

2( )

( 2)( 2)

xP x

x x

and

30( )

( 2)( 2)Q x

x x

Then

Clearly 2x are regular singular points.

The factor ( 2)x appears at most to the first powers in the denominator of )(xP i.e.1 and at most to

the second power in the denominator of ( ) . .1Q x i e then 2x is a regular singular point.

Similarly, for x=-2.

(2) 1 2 1 5 0x y x y y

1x are singular point of the equation

Because 1 0x or 1x .

Now write the equation in the form

1 52 0

1 1

xy y y

x x

or 0

2 50

11y y y

xx

0

2( )

1P x

x

and

5

( )1

Q xx

Then

Clearly 1x are regular singular points.

The factor ( 1)x appears at most to the first powers in the denominator of )(xP i.e.0 and at most to

the second power in the denominator of ( ) . .1Q x i e then 1x is a regular singular point.

Question 4

Page 53: Differential Equations - Solved Assignments - Semester Spring 2005

(b) If x x 0 is a regular singular point of equation 2 1 0( ) ( ) ( ) 0a x y a x y a x y , then is it

necessary there exists always two series solution of the form if not then Give the reason? 02

Solution

No it is not necessary there exist two solutions Frobenius’ Theorem gives us authentication that there

exist just one power series solution .i.e.

If x x 0 is a regular singular point of equation 2 1 0( ) ( ) ( ) 0a x y a x y a x y , then there exists at

least one series solution of the form

0 0 00 0

( ) ( ) ( )r n n rn n

n n

y x x c x x c x x

where the number r is a constant that must be determined. The series will converge at least on some

interval .0 0 Rxx

Question 4

(c) Define second order Bessel differential equation and Bessel function. Find 3/ 2J x 07

Solution

Bessel differential equation:

A second order linear differential equation of the form

022

2

22 yvx

dx

dyx

dx

ydx

is called Bessel’s differential equation.

Bessel function:

Solution of the 022

2

22 yvx

dx

dyx

dx

ydx differential equation is usually denoted by xJv and is

known as Bessel’s function.

And now to find 3/ 2J x

Consider

1 12

v v vv

J x J x J xx

For 1

2v

1 1 11 12 2 2

122J x J x J x

x

Page 54: Differential Equations - Solved Assignments - Semester Spring 2005

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 x

32

32

1 2 2cos sin

2 cossin

J x x xx x x

xJ x x

x x

Page 55: Differential Equations - Solved Assignments - Semester Spring 2005

Assignment 7 (Spring 2005)

Maximum Marks 60

Due Date 13, July 2005

Assignment Weight age 2%

Question 1

(a) write the give DE as a system in normal form 4 3

4 33 3 6 10

d y d yy

dt dt 07

Solution

1 2 3 4

' ' ' '

1 1 2

'

1 2

'

2

'

2 3

3

'

3

'

3 4

'

4

'

4 4 1

'

1 2

'

2 3

3

, ' , " "'

"

"

"'

""

103 2

4

y x y x y x y x

now x y so x y x

x x

now y x

so x x

asy x now differentiating both sides

y x

sox x

x y

x x x

so the required normal form is

x x

x x

x

'

4

'

4 4 1

103 2

4

x

x x x

Question 1

(b)What is degenerate system Explain in your own words with example( no definition is required from

books)

Write the following system in the normal form if possible

3 32

3 3

3 2 3

3 2 3

2 10 4 6

2 5 3 4 4

d y d x dx dyt

dt dt dt dt

d x d y d y dyx

dt dt dt dt

08

Solution

Page 56: Differential Equations - Solved Assignments - Semester Spring 2005

The system of DE is said to be degenerate system If it cannot be reduced to a linear system in the form

of normal form .Some time the situation arises that when we write the highest derivative in the form of

remaining variables then we try to eliminate the both the highest derivatives eliminate at same time so

it is not possible to write it in the form of linear normal from this type of systems are known as

degenerate systems.

The solution of the given system is not possible it is a degenerate system because

3 32

3 3

3 2 3

3 2 3

3 32

3 3

3 3 2

3 3 2

2 10 4 6

2 5 3 4 4

2 10 4 6 (1)

4 2 5 3 4 (2)

(1)

d y d x dx dyt

dt dt dt dt

d x d y d y dyx

dt dt dt dt

so writinging again

d y d x dx dyt

dt dt dt dt

d y d x d y dyx

dt dt dt dt

multiplying

3 32

3 3

3 3 2

3 3 2

2 by1

4 2 20 8 12

4 2 5 3 4

by and

d y d x dx dyt

dt dt dt dt

d y d x d y dyx

dt dt dt dt

It is clear that it is a degenerate system

Question 2

(a)Explain the difference between the echelon form and reduced echelon form. 05

Solution

Explain the difference between the echelon form and reduced echelon form

The difference in the reduced echelon and echelon form is that for each pivot element

ija in the echelon form ija =1 and for each element in that column in which pivot exists for

Which i is less than the pivot i are non zero and reaming all element are zero. Now in the case of

reduced echelon form each pivot element ija =1 and all element in that column should be zero in

which pivot exists, the best example of reduced echelon form is identity matrix. Question 2

(b)Solve the following system of equations by using gauss elimination method

1 2 3

1 2 3

1 2 3

6 6 6 6

2 4 6 12

10 5 5 30

x x x

x x x

x x x

10

Solution

The augmented Matrix may be written as

Page 57: Differential Equations - Solved Assignments - Semester Spring 2005

1

2

3

2 1

6 6 6 6 1 1/ 6

2 4 6 12 2 3 1/ 2

10 5 5 30 1 R

1R

1 4R

1

R

R R

RR

3

1 2

2

2

2

0R

0 5R

1

0

0 R

1

0

0

1

R

RR

R

R

R

3 2

23

R

0

R

R R

now by backward substitution we will get the values of all variables

1 2 3 11/5x x x Question 3

(a) Define are Eigen values and vectors.

Solution

The equation Ax y can be viewed as a linear transformation that maps a given vector x into a new

vector y .to find such a vector we set y= x, where is a scalar proportionality factor, and seek

solutions of the equations

Ax= x

(A- I)x=0 ---------------(1)

Where satisfy the equation det (A- I)=0and are called eigenvalues of matrix A

And non zero solution of equation 1 obtained by using the Eigen values are known as Eigen vectors.

05 Question 3

(b) Find the Eigen values and vectors of the following

5 2 4

1 1 1

4 3 3

10

Solution

Page 58: Differential Equations - Solved Assignments - Semester Spring 2005

Let

5 2 4

1 1 1

4 3 3

A

then Eigen values can be obtained as

The characteristic equation of the matrix A is

5 2 4

det 1 1 1 0

4 3 3

A I

Expanding the determinant by the cofactors of the second row, we obtain

2

2

2 2 3

2 3

3 2

5 1 3 3 2 3 4 4 3 4 1 0

5 3 3 3 2 3 4 4 3 4 1 0

2 5 2 7 4 7 4 0

10 5 2 14 2 28 16 0

10 3 14 2 28 16 0

3 24 42 0

1

2

3

4.4704

5.0697

1.6006

haveeigenvalues

But its have complex Eigen vectors corresponding to Each Eigen value.

Question 4

Solve the following system of DE by the method of undetermined coefficients.

6 2 1

4 3 5 7

t

t

dxx y e

dt

dyx y e t

dt

Solution

Page 59: Differential Equations - Solved Assignments - Semester Spring 2005

2

6 1' ( )

4 3

2 0 1' and F(t)=

1 5 7

the homogenous system

6 1X'=

4 3

6 1det( ) 0

4 3

(6 )(3 ) 4 0

9

t

X X F t

dx

xdtX X e t

dy y

dt

now first solve the

X

now A I

1 2

1

1

2

1

2

1 2

1 2

1 2

1

1

2

14 0

2, 7

for 2

6 2 1 0

4 3 2 0

4 1 0

4 1 0

4 0

4 0

1, 4

1

4

7

1 1

4 4

now

k

k

k

k

k k

k k

so solving we get

k k

K

now for

k

k

1 2

2

2 7

1 2

0

0

0

sin values

1K

1

1 1

4 1

t t

c

k k

now cho g arditrary

now as X c e c e

Page 60: Differential Equations - Solved Assignments - Semester Spring 2005

1

3 2 1

3 2 1

2t

2 0 1( )

1 5 7

F(t) on e is an eigen value so correct form of parti

t

t

p

t

now F t e t

so the particular solution is

a a aX e t

b b b

as by replacing the e in the

34 2 12 2

34 2 1

cular solution is

t t

p

aa a aX te e t

bb b b