Differential Equations - Solved Assignments - Semester Spring 2008

8/9/2019 Differential Equations - Solved Assignments - Semester Spring 2008

1/27

ASSIGNMENT 01Maximum Marks: 30

Q#1:

Solution:4 2

2 4

2 4

2 2 2

1 1 2

1 2

(1 ) (1 ) 0

1

1 1

1

1 1

1 1 21 2 1 ( )

1tan ( ) tan ( )

2

tan ( )tan[ ]

2

.

x dy x y dx

xdy dx

y x

Integrating both sides we get

xdy dx

y x

xdy dxy x

y x c

xy c

isthe general solution

Q#2:


2/27

2

2 2 2

2 2

2

2 2

2

2

2

2 2

2 2 2

2

( 1) ( )

1 ( ) ( )

/ hom .

dy x y x y

dx

dy x x yx y

dx

dy x yx y

dx x

x yx y

dy x

xdx

x

dy x xy y

dx x x x

dy y y

dx x x

As it is a function of y x so it is ogeneous


3/27

2

2

2

2

1

1

1

1

1

1

tan ( ) ln ln

tan ( ) ln

tan[ ln ]

tan[ ln ]

.

To solve it put y vx

yv

x

Then above equation becomes

dvv x v v

dx

dvx v

dx

dv dx

v x

Integrating both sides we get

dv dx

v x

v x c

v cx

v cx

y x cx

isthe general solution


4/27

Q#3:In the given problem, determine whether the given equation is exact. If exact, solve it

2( 2 tan ) (sec ) 0 ; ( )4

Sin x y dx x y dy y

Solution:

2

2 2

2

Here

2 tan & sec

sec sec

.

2 tan 1 & sec 2

int 1 . . ,

cos2( , ) tan ( ) 3

2

M Sin x y N x y

NMy y

xy

M NClearly soit is an exact differential equaiton

y x

Nowtake

f f M Sin x y N x y

x y

On egrating no w r t x we get

x f x y x y g y

Par

. .tially differentiate w r t y we get


5/27

:

2 /

1

2 1

1 2

sec ( )

( )

cos2( , ) tan

2

cos2tan

2

cos2( )tan

2

( )4

1

2

cos 2 1tan

2 2

f x y g y

y

g y c

then above equation becomes

x f x y x y c

xc x y c

xc c c x y c

Given y

Aboveequationbecomes

c

xThen the particular solution is x y

we ge

2 2 tan 1 2t Cos x x y


6/27

ASSIGNMENT 02

Maximum Marks: 30

Due Date: 3rd May, 2008

Question 1: Marks=10

Tell that which kind of differential equation is that, also solve that D.E

( ) 1dy

x y ydx

1(1 )( ) ln

( ) 1

1 1(1 ) (1)

.

1 1( ) 1 , ( )

.

(1),

( 1)

( )

int ,

dxP x d x x x xx

x x x

xx

x x

x x

dy x y y

dx

dyy

dx x x

It is a linear D E

Here P x Q xx x

I F e e e xe

Multiply it by no we get

dy xe e x y e

dx

d xe ye

dx

On egrating we get

xe y e dx

xe y e c



7/27

Solve the following D.E by using an appropriate substitution

sec( ) x y dy dx Solution:

2

2

sec ( )

. . ,

1

1

sec( ) 1

sec ( 1) 1

1sec sec

sec

1 sec

1

1 cos

1

2cos2

1s

2 2

tan2

(tan

x y dy dx

Put u x y

Differentiate w r t x we get

du dy

dx dx

dy du

dx dx

dy x y can be written as

dx

duu

dx

duu udx

udu dx

u

du dxu

du x cu

uec du x c

ux c

x y

)

2x c


Find an equation of orthogonal trajectory of the curve2 2

x y C Solution:


8/27

2 2

2 2

. . ,

2 2 0

.

1

int ,

ln ln ln

x y C

Differentiate w r t x we get

dyx y

dxdy x

dx y

Now we write down the D E for theorthogonal family

dy y

dydx x

dx

dy dx

y xon grating we get

y x c

cy

x

xy c

Hencethe familyof concentriccircles x y c and the f

.

amily

xy c are orthogonal trajectries


9/27


Upload Date: May 09, 2008

Due Date: May 16, 2008


By using Wronskian, check whether the following functions given below are linearly independent. If not, show its linearly

dependence.

2 2

2 31

21 2 3

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

( ) ( ) cos 2 , ( ) 1, ( ) cos

i t f t t f t t t f t

ii f x x f x f x x

Solution:

(i) / //

1 1 1

2 / //

2 2 2

2 / //

3 3 3

1 2 3

/ // /// 2

1 1 1

2 / // ///

2 2 2

2 3; 2 , 0( ) ( ) ( )

( ) 1; ( ) 2 , ( ) 2

( ) 2 ; ( ) 4 1, ( ) 4

det min

2 3 2 0

1 2 2 14 0

2 4 1 4

.

t f t f t f t

f t t f t t f t

f t t t f t t f t

er ant is

f f f t

W t t f f f

t t t f f f

so it islinearlyindependent


10/27

21 2 3

/ // ///

1 1 1

/ // ///

2 2 2

2

( )

2 1

0 2 22 2

4 2 0 2 2

( )

22 2(4 2 2 4sin 2 2 ) 0

4 2 2 2

ii

f f f Cos x Cos x

W CosxSinx Sin x f f f Sin x

Cos x Cos x f f f

Expanding C IInd column

Sin xSin xSin xCos x xCos x

Cos x Cos x

But vanishing of wroskiandoes not gu

2

3

2 13

3 1 2

.

socheck .

( ) cos

1 cos2

2

2

& .

areenteethelinear dependence

itslinear dependence by that method

Since f x x

x

f ff

It shows that f isa linear combinationof f f soit islinearlydependent


Solve the given D.E subject to the indicated initial conditions.

Solution:

4 / // ///

4

4

4

2 2

2

2

1 2 3 4

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

( 1) 0

1 0

( 1)( 1) 01 0 1,1

1 0 ,

x x

d y y y y y y

dx

Above differential equation canbe writtenas

D y

Its auxiliary equationis

m

m mm gives m

m gives m i i

we get

y c e c e c Cosx c Sinx


11/27

/

1 2 3 4

//

1 2 3 4

///

1 2 3 4

1 2 3

/

1 2 4

//

1 2 3

///

1 2 4

sin cos

sin cos

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

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

(0) 0 0 .....(3)

(0) 1 1 ..

x x

x x

x x

y c e c e c x c x

y c e c e c Cosx c Sinx

y c e c e c x c x

y givesc c c

y gives c c c

y gives c c c

y gives c c c

1 2 1 2 1 2

1 2 1 2 2 2 2 1

4 4

3 3

...(4)

2 2 0 0(1) & (3)

1 1 1 12 4 2 2 1 &( ) & ( )

2 2 4 4

1 1 12 0

4 4 2

1 1(3) 0 0

4 4

1 1 1

4 4 2

x x

Add c c c c c c

Add c c c c c c c c

From c c

From c c

Thereforethe general solution is y e e Sinx


Suppose that an object of mass m is falling downwards from a heighth with some velocity m/sv (velocity of the ball changes

w. r .t the time t).

As the ball moving downwards, there is some air resistance (some kind of force) say v where is the constant.

Deduce the differential equation that describes the rate of change of speed m/sv w.r.t the time t from above information.

Also solve that differential equation for m=10 kg and 5 .

Hint: In that Question, you have to use Newtons 2nd

law of motion. Here you can also say that the ball moves downward with

acceleration a.

Solution:Since the ball is moving downward so the force applied is the weight W of the object where W = mg (g =9.8 ms

-2or approx 10 ms

-2)

Also air resistance = R = v According to theNewtons 2

ndlaw of motion


12/27

2

2

2

Net force = m a

W-R =ma

mg- v =ma

dv d xmg- v = m m

dt dtdv

mg- v =mdt

mg-5vdv

mdt

dv 5v( 10 , 10)g -

dt

.

but g ms mm

It is the required differential equation

1

5

dv 5v 100 5v10-

dt 10 10

var ,

dv 1

100-5v 10

int

ln |100-5v| tln ...(1)

5 10

10; 0

100

(1) .

Separating the iables we get

dt

while egrating we get

c

when t v then c

put it in eqaution no we get the required differential equation


13/27


Upload Date: May 17, 2008

Due Date: May 22, 2008


Solve the Differential Equation by the variation of parameters

// tan y y x Solution:

// tan y y x is a non-homogeneous differential equation.

Its particular solution is yp = u1(x) y1(x)

Step1:To find out the complementary function

//

2

1 2

0

,

1 0

mx

c

y y

Put y e we get

m

m i

so y c Cosx c Sinx

Step2:

From the above complementary solution, we identify

1 2

1 2

,

hom .

( , ) (cos ,sin ) 1 0os

y Cosx y Sinx aretwolinearly independent solutions of associated

ogeneous equationThereforeCosx Sinx

W y y W x xSinx C x

Step 3:

The given D.E is// tan

( ) tan

y y x

where f x x

Step 4:

Now we construct the determinants W1 and W2 given by

2

21 /

2

1

2 /

1

00 sintantan cos cos( )

0 cos 0

sin tan( )

ySinx xW sinx x

x x x f x y

y xW sinx

x x y f x

Step 5:


14/27

Next we determine the derivatives of the unknown variables u1 and u2 through the relations

/ 11

/ 22

sin tansin tan

1

sin sin1

W x xu x x

W

W xu xW

Step 6:

Now integrate the above following we get

2

1

2

2

sinsin tan

cos

1 cos

cos

(cos sec )

sin ln sec tan

sin cos

xu x x dx dx

x

xdx

x

x x dx

x x x

u x dx x

Step 7:

Therefore, the particular solution of the given differential equation is

1 1 2 2

(sin ln sec tan ) cos cos sin

ln sec tan

py u y u y

x x x x x x

x x

Step 8:

Now the general solution of the differential equation is

1 2 ln sec tanc p y y y c Cosx c Sinx x x


Solve the differential equation

Solution:

-------------------(1)

It can be re-written as

For complementary Solution

2 cosx y y y e x

2 cosx y y y e x

2 2 cosx D y Dy y e x

2 2 1 cosx D D y e x


15/27

Where

Its characteristic or auxiliary equation is

So ,

For a particular solution we assume

Putting values in (1)

Equating co-efficient of like terms

---------------------(2)

--------------------(3)

Multiply (2) by 4 and (3) by 3

2 2 1 0 D D y 0F D y

2 2 1F D D D

2 2 1 0m m

2

1 0m

1, 1m

1 2 xcy c c x e

cos sinxpy e A x B x cos sin sin cosx xpy e A x B x e A x B x

cos sinxpy e A B x A B x cos sin sin cosx xpy e A B x A B x e A B x A B x cos sinxe A B A B x A B A B x 2 cos 2 sinxe B x A x

2 cos 2 sin 2 cos sin

cos sin cos

x x

x x

e B x A x e A B x A B x

e A x B x e x

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

sin cos

B x A x A B x A B x A x

B x x

(3 4 )cos ( 4 3 )sin cos A B x A B x x

3 4 1A B 4 3 0A B

12 16 4A B 12 9 0A B


16/27

So

3

25A

So general solution is

1 2

3 4

( ) ( )25 25

c p

x x

y y y

y c c x e e Cosx Sinx

4

25B

43 4 1

25A


17/27


Upload Date: June 25, 2008

Due Date: July 01, 2008


A 4-lb weight is attached to a spring whose constant is 2lb /ft. The medium offers a resistance to the motion of the weight numerically

equal to the instantaneous velocity. If the weight is released from a point 1 ft above the equilibrium position with a downward velocity

of 8ft /s, determine the time that the weight passes through the equilibrium position. Find the time for which the weight attains its

extreme displacement from the equilibrium position. What is the position of the weight at this instant?

Solution:Since Weight = 4 lbs and constant = k = 2lb/ft

Therefore, by Hooks law

4 = 2s

S = Stretch = 2ft & Damping force = dx/dt

Where

1

14 32

8Since W mg m m slugs

Thus Differential equation of motion of free damped motion is given by2

2

2

2

2

2

/

0

8 16 0

the weight is released from a point 1 ft above

the equilibrium position with a downward velocity

of 8ft (0) 1,/ (0) 8s

d x dxm kx

dt dt

d x dxm kx

dt dt

d x dxx

dt dt

As

with initial conditions x x

dAs

2

2

2

2

8 16 0

,

8 16 0

( 4) 0

4, 4

mt

x dxx

dt dt

Put x e we get

m m

m

m


18/27

4

1 2

/ 4 4

1 2 2

1 2

4 4

1 2

4

/ 4 4

/ 4

/

/

( ) ( )

( ) 4( )

1, 4

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

( ) ( 1 4 )

( ) 4 4 ( 1 4 )

( ) (4 4 16 )

( )

(0) 1, (0) 8

t

t t

t t

t

t t

t

x t c c t e

x t c c t e c e

c c

x t c c t e becomes x t t e

For extreme values x t t e

x t e t e

x x giv

x t t e

es

x t

4

/ 4

// 4 4 4 4

/

4

4

1 1

// 3

(8 16 )

( ) 8(1 2 )

( ) 16 32 (1 2 ) ( 16 32 64 ) ( 48 64 ) ...(2)

( ) 0

(8 16 ) 0

10 8 16 0, (2),

2

1( ) ( 48 32) 16

2

t

t

t t t t

t

t

t e

x t t e

x t e t e t e t e

For critical value put x t

t e

but e so t t put it in no we get

x e e

3

4 2

max

10 ( ) max

2

1 1, ( ) (1 12 ) ( ) 7

2 2

t

ve so x t has imum value at t

At t x t t e becomes x e

//

4

det min " " ,

. ( ) 0

( 48 64 ) 0 ( sin (2))

48 64 0

3

4

t

To er e time t wheretheweight passes throughtheequilirium position

its accelarationbecomes zero That is x t

t e by u g no

t

t


Solve the non homogeneous differential equation

3 23 2 3

3 23 6 6 3 ln

d y d y dy x x x y x

dx dx dx

Solution:

For yc


19/27

Put

2 31 2 3

2 3

3 3 2 2 1

3 2

2

, ( 1) , ( 1)( 2)

( 1)( 2) 3 ( 1) 6 6 0

( 1)( 2) 3 ( 1) 6 0

11 6 06

1 5 6 0

1 2

m

m m m

m m m m

y x

dy d y d ymx m m x m m m x

dx dx dx

we get x m m m x x m m x x mx x

Auxiliary equationbecomes

m m m m m

mm m

either m or m m

either m or m

2 3

1 2 3

,3

c

Therefore

y c x c x c x

Lets find the particular solution

2 3

1 2 2

2 3 / / /

1 2 3

2 / / /

1 2 3

3 / / /

1 2 3

, ,

0

(1) (2 ) (3 ) 0

(0) (2) (6 ) 3 ln 3(1 ln )

Suppose

y x y x y x

Now

u x u x u x

u u x u x

u u u x x x

To solve these above equations, we can write them in matrix form as


20/27

2 3

2

3

1 2

2 3

2

1 2

2

2 3

0

1 2 3 0

0 2 6 3(1 ln )

2

0 2 0

1 2 3 0

3(1 ln )0 1 3

2

1 2 3 0

0 2 0

3(1 ln )0 1 3

2

x x x

x x

x x

R

Apply R xR and

x x

x x

xx

Interchange R and R

x x

x x

xx

From these above matrix, we write it as2 / / /

1 2 3

2 3/ /

2 3

/ /

2 3

// /2 3

/

3

/

3

2

3

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

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

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

2

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

3(1 ln )

2

3(1 ln )

2

int ,

3 (ln )ln

2 2

u u x u x

x u x u

xu xu

From u xu put it in no

xwe get xu

xu

x

After egaration we get

xu x

/

/

2

2

(2 )

3(1 ln )

int ,

3 ln

From

we get u x

after egaration

u x x


21/27


22/27


Upload Date: July 10, 2008



Find the interval of convergence of the given power series

(i)

1

( 7)n

n

x

n

(ii)

1

( 1) ( 5)

10

n n

nn

x

Solution:

(i)

1

1

1

1

1

( 7)

( 7) ( 7)

1

( 7)lim lim lim 7 7

( 7) 1 1

7 1

. . 1 7 1 8 6 ( 8, 6)

int

n

n

n n

n n

n

n

nn n nn

x

n

x x

Here a an n

a x n n R x x

a x n n

Theabove power seriesconverges for x

i e x x or x

It istherequired erval of converge

.nce

(ii)

1

1 1

1 1

1 1

1

1

( 1) ( 5)

10

( 1) ( 5) ( 1) ( 5)

10 10

5( 1) ( 5) 10 1lim lim lim 7

10 ( 1) ( 5) 10 10

5 10

. . 10 5 10 5 15

n n

nn

n n n n

n nn n

n n n

n

n n nn n nn

x

x x Here a a

xa xR x

a x

Theabove power series converges for x

i e x x o

( 5,15)

int .

r x

It is the required erval of convergence

Question 2: Marks = 10

Solve the Differential equation // 2 0 y x y

Note: Mention each and every step

Solution:


23/27

// 2

2

2 1 0

2

0

0

0 ...(1)

( ) 1, ( ) 0, ( )

& ( ) 1 0, ( ) 0, ( )

0 int . sin

int

.

( )

n

n

n

n

n

n

y x y

Here a x a x a x x

P x Q x R x x

So x x is an ordinary po As there are no finite gular

po so there exist two solutions of the form y c x which

converges for x

y x c x

/ 1 // 2

0 1 2

2 2

2 0

2 2

2 0

2 2

2 3

4 0

, ( ) , ( ) ( 1)

(1) ( 1) 0

( 1) 0

2 6 ( 1) 0

2 2 2

n n

n n

n n

n n

n n

n n

n n

n n

n n

n n

n n

n n

y x nc x y x n n c x

Therefore eq no becomes n n c x x c x

n n c x c x

c c x n n c x c x

put k n and k n in Ist and nd

2 3 2 2

2 2

2 3 2 2

2

2 3 2 2

2

3

2 2

.

2 6 ( 2)( 1) 0

2 6 [( 2)( 1) ] 0

0, 0 & ( 2)( 1) 0

0

0

12,3,4,...

( 2)( 1)

k k

k k

k k

k

k k

k

k k

k k

summation respectively

c c x k k c x c x

c c x k k c c x

Thus c c k k c c

Thereforec

c

c c for k k k


24/27

2 2

4 0 0

5 1 1

6 2

7 3

2

8 4 0 0

9 5 1

1

( 2)( 1)

1 22,

4 3 4!

1 63,5 4 5!

1 14, (0) 0

6 5 6 5

1 15, (0) 0

7 6 7 6

1 1 2 ( 1) 606, ( )

8 7 8 7 4! 8!

1 1 67, (

9 8 9 8 5!

k k As c ck k

For k c c c

For k c c c

For k c c

For k c c

For k c c c c

For k c c c

2 2

1 1

2 3

10 8 0 0

11 7

2 3 3

12 8 0 0 0

0

0

( 1) 6.6.7 ( 1) 252)

9! 9!

1 1 ( 1) 60 ( 1) 608, ( )

10 9 10 9 8! 10!

19, 0

11 10

1 1 ( 1) 60 ( 1) 90 60 ( 1) 540010, ( )

12 11 12 11 8! 12! 12!

n

n

n

c c

For k c c c c

For k c c

For k c c c c c

Therefore y c x c x

0 1 2 3 4 5 6 7 8 9 101 2 3 4 5 6 7 8 9 10 ...c x c x c x c x c x c x c x c x c x c x

Put the values of constants c0, c1, and get the answer


25/27


Upload Date: July 22, 2008


Question 1: Marks=10Express the given Bessel function in terms of Sin(x) and Cos(x) and power of x.

7

2

1 1

sin

2( ) ( ) ( )v v v

J

U g the technique

v J x J x J x

x

Answer:

1 1

5

2

5 5 51 12 2 2

3 7 52 2 2

7 5 32 2 2

52

3 / 2

72

Consider that

2

522

5

5 (1)

3 s in 2 2 2( . cos ) sin

sin 2 2( ) . cos

(1)

5

v v v

t ak in g v

v J x J x J xx

A s f o r

J x J x J xx

J x J x J xx

J x J x J xx

w e k n o w

x J x x x

x x x x x

x J x x

x x x

T he n e qu at io n b ec om es

J x

2

2 3 2 3 2 s in[ 1 s in .( cos )] cos

x x x x

x x x x x xx

Question 2: Marks=5


26/27

Find the general solution of the differential equation2 '' ' 24 4 (4 25) 0 (0, ) x y xy x y on

Solution:The general form of Bessel differential equation is

22 2 2

2( ) 0

d y dy x x x v y

dx dx -----(1)

2 '' ' 2

22 2

2

2

2 2 22

4 4 (4 25) 0

4,

25( ) 0

4

5( ( ) ) 0 (2)2

x y xy x y

Dividing by we get

d y dy x x x y

dx dx

d y dy x x x ydx dx

Comparing (1) and (2) , we get

2 25 5

4 2v v

so the general solution is

1 5 2 5

2 2

( ) ( )y c J x c J x

Question 3 : Marks=5

Reduce the third order equation / / / / / / 23 6 10 1 y y y y t to the normal form.

Solution: / / / / / / 2

/ / / / / / 2

1

/

2

/ /

3/ /

1 2

/ / /

2 3

/ / / /

3

/ 2

3 1 2 3

3 6 10 1

10 6 3 ( 1)

int var

10 6 3 1 .

Given equationis y y y y t

y y y y t

Now roduce the iables

y x

y x

y x x y x

x y x

x y

x x x x t isthe required normal form


27/27

1

/

2

/ /

3

/ /

1 2

/ / /

2 3

/ / / /

3

/

3 1 2 3

1 13 2 cos

3 3

y x

y x

y x

x y x

x y x

x y

x x x x t

Differential Equations - Solved Assignments - Semester Spring 2008

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