Differential Equations - Solved Assignments - Semester Summer 2007

8/9/2019 Differential Equations - Solved Assignments - Semester Summer 2007

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Assignment No 1 Question #1 :Answer

Given D.E is(1 ) (1) y x

dye xe

dx

It is not separable.By dividing both sides by ye

1 (2) x ydy

xe edx

1 (2) x ydy

xedx

Suppose thatz=x+y Diff w.r.t.x

1+dz dydx dx

dzdx

z xe

Equation (2) becomes

. ze dz x dx It is separable eq.

Taking integrating both side

. ze dz x dx


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sin( )( , )

sin( )

y y x

x f x y y

x x

For homogenous function

sin( )( , )

sin( )

y y x

x f x y y

x x

For substituting

(3) y vx Eq (1) can be written as

sin (2)sin

dy y v xdx x v

Differentiating Equation (3) with respect to x. we have

(4)dy dv

v xdx dx

From Eq (2) & Eq (4) we have

sinsin

dv y v xv x

dx x v

sinsin

dv vx v xv xdx x v

( sin 1)sin

dv x v vv x

dx x v

sin 1sin

dv v vv x

dx v

sin 1sin

dv v v x v

dx v

1sin

dv xdx v

1sin v dv dx

x

By integrating we have1

sin v dv dx x


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By using formula we havesin cos xdx x

cos lnv x c

cos( ) ln y y

x c wherev

x x

cos( ) ln y

x c x

Which is the required result

Question #3 :

2 2(xy -1)dx-(1-xy )dy (1)

Answer.2 2M=xy -1 & N=-(1-xy )

2 2(xy -1) & -(1-xy )

2 & 2

M N y y y y

M N xy xy

y y

2

2

(xy -1)

-1+xy =N

f M

x

f y

2

2 2

f(x,y)= (xy -1)

1f(x,y)= y ( )

2

dx

x x h y

Where h(y) is constant of integration

2 2

2

2 2

Differentiating w.r.t y

y( ( ))2

'( )

'( ) 1

f x x h y y y

x y h y

x y h y x y N


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2 2

2 2

2 2

2 2

By comparing sides

=h'(y) = -1 is independent of x

or integerating we have h(y)= -y

yf(x,y)= ( )

2y

f(x,y)=2

Hence the general solution of equation is given by

f(x.y)=c

y

2

y 2( )

x x y

x x y

x x y c

or x x y

c

Question#04:

Solve (by finding I.F) ( ) 1dy

x y ydx

Solution:

1

y ydxdy

x

dividing by x


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1(1 )

1

1 1(1 ) (1)

.

.

. (1)

1 1(1 ) ( )

( . )

int .

.

( 1)

dx x

x

x x x

x x

x x

x

dy y y

dx x xdy

ydx x x

It is linear in y

I F e

xe

I F multiplybyeq

dy xe xe y xe

dx x xd xe y e dx

Taking egrate both sides

xe y e c

e xy c


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Assignment 2

Question#01Solve the differential equation and mention the name of type of this D.E

3 23 (1) 2dy x y x y with ydx

solution:

3 2

2 2

3

' ',

3

First order Ordinary differential equation

dy x y x y

dx Nowbydividing the equationby x we get

dy y x y

dx x

2

2 2

' .' ',

3

It is a Bernoulli s equation By dividing the above equationby y then we get

dy y x

dx xy

1 2

1

2

2

2

. . . ' '

1

3

( 1) .

3

Putting t y

t y

Diff w r t x

dt dy y

dx dxThen equationbecomes

dt t x

dx xmultiply by above equation

dt t x

dx x

3

33ln

ln

33

int

.

.

1.

dx x x

x

egrating factor will be

I F e e

I F e

I F x x


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3 4

3

3

3

int

1 3 1

1

int ,

ln

1 1ln

Nowmultiplying theequationby egrating factor

dt t dx x x xd t

dx x x By egrating the above equation we get

t x C

x

x C where t x y y

3

1ln1 ( ln1 0)

21

0212

1 1ln

2

Applying initial condition

C

C

C

Therequired result is

x yx

Question#02 : Find an equation of orthogonal trajectory of the curve2( ) y x c

Solution.


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1 2

1 2

1

2

sin (0) 0,

0 cos 0 sin 0

0 .1 .0

0

sin4

Now u g boundary equation y we get

c c

c c

c

Hence the solutionbecomes y c x

2

2

2

sin 1,2

1 sin 42

1 sin 2

1 .0

1 0

.

, .

Nowu gtheboundarycondition y we get

c

c

c

This is a clear contradiction

Therefore theboundary value problemhas no solution


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Assignment 3

Question#01: Solve by the method of U.C.

22

2 3 2 2 2xd y dy

y x xedx dx

Solution:For complementary function,

2

2

2

3 2 0

the auxiliary equation is

3 2 0

2 2 0

( 2) 1( 2) 0

( 1)( 2) 0

( 1) 0 ,( 2) 0

1, 2

D D

m m

m m m

m m m

m m

m m

The roots are

The complementary solution is

x xc ecec y

221

For the particular solution of the given equation, well find the particular solution of :2223 x y y y .(2)

e x y y y 223 .(3) For (2), we assume P.I. as

2Cx Bx A y p

Cx B y p 2

C y p 2

Substituting in (2), we get22 2222632 xCx Bx ACx BCx

Or 22 2232622 x A BC xC BCx Equating the co-efficient

Therefore,C = 1

062 C B or 3 B

0232 A BC or27

A


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27

. 32

P I x x

For (3) we assume P.I.as:

2

x p

x

y x D Ex e

Dx Ex e

x x p e Ex Dxe Ex D y 22 xe Ex x E D D 22

x x p e Ex E De Ex x E D D y 222 2 xe Ex x E D E D 2422

Substituting in (3) we have:

x x x x

xee Ex Dxe Ex x E D De Ex x E D E D 2223422222

Or x x E E D 222

Therefore:02E D and 22E

Thus:1 E and 2 D

P.I. of (3) is: xe x x 22

The general solution of the given equation is:

222

21 2327

x xe x xecec yx x x

Question 2:Solve the D.E. by variation of Parameters

y"' y' cosec x Solution:

y"' y' cosec x

The function f(x) can be identified as


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f x cosec x

To find y c, at first the associated homogenous equation will be solvedmx mx 2 mx 3 mxy e ,y ' m e ,y '' m e ,y "' m e

Then the associated homogenous equation gives

3 mxm m e 0

Therefore, the auxiliary equation is3

mx

m m 0

as

e 0, x

Using the quadratic formula, roots of the auxiliary equation are

2m m 1 0

m 0, i

Hence, the complementary function y c is

c 1 2 3y c c cos x c sin x

Here

1

2

3

y 1

y cos x

y sinx

So

1 2 3

2 2

1 cos x sin x

W y ,y ,y 0 sin x cos x

0 cos x sinx

After expanding.

1 sin x cos x cos x 0 0 sin x 0 0 1

1

2 2

2 2

2 2

0 cos x sinx

W 0 sinx cos x

cosec x cosx sinx

0 sin x cos x cosx 0 cosx.cosec x sinx 0 sinx.cosec x

cos x.cosec x sin x.cosec x

cosec x cos x. sin x cosec x


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2

1 0 sinx

W 0 0 cos x

0 cosec x sinx

1 0 cosx.cosec x 0 0 0 sinx 0 0

cosx.cosec x cot x

3

1 cos x 0

W 0 sinx 0

0 cosx cosec x

1 sin x.cosec x 0 cos x 0 0 0 0 0

1

Integrating

11

22

33

Wu' cosec x

W

Wu' cot xWW

u' 1W

Gives

1

2

3

u cosec x dx

ln cosec x cot x

u cot x dx

cosxdx ln sinx

sinxu 1dx

x

The particular solution y p of non -homogenous equation will be py ln cosec x cotx ln sinx cosx xsinx

Hence the general solution of the given equation is

c p y y y

1 2 3c c cosx c sinx ln cosec x cotx ln sinx .cosx xsinx


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p

Substituting in the given differential equation , then

(A+2B) cos ( 2 ) sin 4 cos 2sin

quating coefficients , we obtain

A+2B=4

2 2these equation

A=0, B=2

thus X 2sin

t A B t t t

E

A Bsolving

1 2

'1 2 1 2

1 2

1

'1 2

2

general solution is

( ) ( cos sin ) 2 sin

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

(0) 0 .1 .0 0 0

0

(0) 3 .1 .1 2 3

c p

t

t t

t

hence

x x x

x t e c t c t t

x t e c t c t e c t c t t

now we apply the boundary conditions x c c

c

x c c

c

1

sin 2sin

sin 0 t 0

sin , 2sin

sin 2sin

t

t

t

t

Steady StateTransient

the solution of the initial value problem is

x e t t

e t as

therefore

e t Transient Term t Steady State

hence

x e t t

the effect ofte transient ter

:

2

m becomes negligible for

t

-----------------------------------------------------------------------------------------------------Question#02

Solve by using Cauchy Euler method3 ' ' ' 2 ' ' ' 44 5 15 x y x y xy y x

Solution:


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3 3 2 2 4

t

2 2 2

3 3 3 2

3 2 2 4

3 2 4

( 4 5 15) (1)

x = e or t = Inx so that

( 1)( 1)( 2) 3 2

in eq(1), then we get

( 3 2 4 4 5 15)

( 7 15) (2)

t

t

x D x D xD y x

Let

xD

x D x D

Substituting

y e

or

y e

T

3 2

2

3 21 2 3

4 4 4

3 2

eq of (2) is

7 15 0

( 3)( 4 5) 0

3, 2

( cos sin ).

7 15 64 16 28 15 37The general solution of (2)is

t t c

t t t

p

hecharacteristic

or

so i

The complementry solution is

y c e e c t c t

e e e y

y

43 2

1 2 3

t

43 2

1 2 3

( cos sin ) 37Where t= Inx or x = e

[ cos( ) sin( )37

t t t e

c e e c t c t

x y c x x c Inx c Inx

-----------------------------------------------------------------------------------------------------------

Question#03Find the first four terms of a power series in x for the function secxSolution:


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2 4 6

2 4 6 2 4 6

2 4 6

2 4 6

4 6

1sec , cos 1 ...

cos 2 24 720using

1 12 24 720 2 24 720

_________________

...2 24 720

...2 4 48

______________5 7

+.24 360

As we know that

x x x x where x

xTherefore long division

x x x x x x

x x x

x x x

x x

2 4 6

4 6

6

2 4 6

5 611 ...2 24 720

..

5 5+...

24 48__________________

61...

720( ) sec

5 61sec 1 ...2 24 720

the interval of converge

x x x

x x

x

hence the power series for the function f x x is

x x x x

nce of this series is ( , ).2 2


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Assignment# 05

Question 1: Marks=10

The D.E

'' '3 0 x y y y has regular singular point at x=0. Justify your answer.

Solution:n

n = 0

' 1

nn = 0

'' 2n

n = 0

'' ' r 10

try a solution of the form y = c

Therefor y = (n + r) c

y = (n + r)(n + r - 1) c So that

xy + 3y - y = x r(r + 2)c + (k + r + 1

n r

n r

n r

We x

x

x

x

k k + 1 k

k = 0

1 2

k + 1 k

)(k + r + 3) c c x = 0

So that the indicial equation and exponent are r(r + 2) = 0 andr = 0, r = -2, respectively.Since (k + r + 1)(k + r + 3)c - c = 0,

1

k k + 1

01

012

023

3 04

1

n

k = 0, 1, 2,...It follows that when r = 0,

cc =(k + 1)(k + 3)

cc =

1.32cc

c = =2.4 2!4!

2ccc = =

3.5 3!5!c 2c

c = =4.6 4!6!

cc = , n = 1, 2 ,...

n!(n + 2)!


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01 0

n = 1

0= 0

2

k + 1 k

one series solution is

2y = c x 1 +n! (n + 2)!

2= cn!(n + 2)!when r = -2, Eq.(1) becomes.

(k - 1)(k + 1)c - c = 0 (2)But not here that we don't divi

n

n

n

Thus

x

x

Now

1 0 2 1

ded by (k -1)(k +1) immediatelysince this term is zero for k = 1. However, we use the recurrencerelation (2) for the cases k = 0 and k =1-1.1c - c = 0 and 0.2c - c = 0The latter equa 1

0

k k + 1

23

3 24

4 25

n

tion implies that c = 0 and so the former equationimplies that c = 0. We find,

cc = k = 2, 3, ...(k - 1)(k + 1)

cc =

1.3c 2c

c = =2.4 2!4!c 2c

c = =3.5 3!5!

2cIn general c = 2

-2 22 2

n = 3

, n = 3,4,5...n!(n - 2)!

2Thus y = c x x + (3)n!(n - 2)!

n x

2

1

, close inspection of (3) reveals that y is simplyconstant multiple of y . To see this, let k = n - 2 in (3). Weconclude that the method of Forbenius gives only one seriessolution of the gi

However

ven differential equation.

Question 2: Marks=5Find the general solution of the equation

2 '' ' 2 1( ) 0 (0, )25

x y xy x y on

Solution:


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2 '' ' 2 2

2 '' ' 2

2

Bessel Differential Equation isx + xy + (x - v )y = 0 (1)

1x + xy + (x - )y = 0 (2)25

Comparing eq (1) and (2), then we get1 1

v = , therefore v = 525So the general solution of eq

The y

y

1 1 2 15 5

(2) isy = C J (x) + C J (x)

Question 3: Marks=10

Solve the Bessel function in terms of 32

sin x of the J

Solution:

1 1

12

1 1 11 12 2 2

1 3 12 22

3 1 12 22

1/ 2 1/ 2

32

32

C o ns id er t ha t2

12 2

1

1

2 2( ) s in , ( ) cos

1 2 2s in cos

2 s incos

v v v

t ak in g v

v J x J x J x

x A s f o r

J x J x J x x

J x J x J x x

J x J x J x x

w e k n o w

J x x J x x x x

J x x x x x x

x J x x x x


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Assignment 6

Question#01:(2 1) 2 4

t

D x Dy

Dx Dy e

Solution.

The given differential equation is

2 2 4

-

To solve the highest derivative Dy, we eliminate the highest

derivative Dx.

Therefore, multiply the second equation with 2and then

su

t

Dx x Dy

Dx Dy e

btract from the first equation to have

2 2 4

2 2 2

----------------------------------

= 4 2

t

t

Dx x Dy

Dx Dy e

x e

Therefore, it is impossible to solve the system for the highest derivatives

of each variable.

Thus the system cannot be reduced to the linear normal form.

Hence, the system is a degenerate formation.

Question#02:Use the Gauss Jordan elimination method to solve the following system of linearalgebraic equations

1 3 4

1 2 3 4

1 2 3

1 2 3 4

5x 4x 2x 3

x x 2x x 1

4x x 2x 1

x x x x 0


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Solution:The augmented matrix of given system of linear algebraic equation is

5 0 4 2 3

1 1 2 1 1

4 1 2 0 1

1 1 1 1 0

We shall perform row operations on augmented matrix.Interchanging 1 st and 2 nd row, then we have

1 1 2 1 1

5 0 4 2 3~

4 1 2 0 1

1 1 1 1 0

2 1

1 1 2 1 10 5 6 3 2

~ R 5R4 1 2 0 1

1 1 1 1 0

3 1

1 1 2 1 1

0 5 6 3 2~ R 4R

0 5 6 4 3

1 1 1 1 0

4 1

1 1 2 1 1

0 5 6 3 2

~ R R 0 5 6 4 30 2 1 0 1

2

1 1 2 1 1

6 3 20 11

~ R 5 5 55 0 5 6 4 3

0 2 1 0 1

3 2

1 1 2 1 1

6 3 20 1

~ R 5R 5 5 50 0 0 1 1

0 2 1 0 1

Interchanging 3 rd and 4 th row


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1 1 2 1 1

6 3 20 1

~ 5 5 50 2 1 0 1

0 0 0 1 1

3 2

1 1 2 1 1

6 3 20 1

5 5 5~ R 2R7 6 1

0 05 5 5

0 0 0 1 1

3

1 1 2 1 1

6 3 20 1

5 5 5 5~ R6 17 0 0 17 7

0 0 0 1 1

4

1 1 2 1 1

6 3 20 1

5 5 5~ R6 1

0 0 17 7

0 0 0 1 1

3 4

1 1 2 1 1

6 3 20 16

~ R R 5 5 57 0 0 1 0 1

0 0 0 1 1

2 3

1 1 2 1 1

3 80 1 06

~ R R 5 55 0 0 1 0 1

0 0 0 1 1

2 4

1 1 2 1 1

0 1 0 0 13~ R R

0 0 1 0 150 0 0 1 1

1 2

1 0 2 1 0

0 1 0 0 1~ R R

0 0 1 0 1

0 0 0 1 1


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1 3

1 0 0 1 2

0 1 0 0 1~ R 2R

0 0 1 0 1

0 0 0 1 1

1 4

1 0 0 0 1

0 1 0 0 1~ R R

0 0 1 0 1

0 0 0 1 1

solution becomes

1

2

3

4

x 1x 1x 1x 1

Question#03:Find the general solution of the given system of D.E

dxx 2y

dtdy

2x ydt

Solution:

The given homogenous system of differential equations is

dx dt 1 2 xdy dt 2 1 y

So, the coefficient matrix will be

1 2A

2 1

Then, we shall find the eigenvalues and eigenvectors of the coefficient A. For that, thecharacteristics equation will be

2

2

1 2det A I 2 1

1 1 2 2

1 2 4

2 3

The characteristic equation is


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2

det A I 0

2 3 0

And for the roots of characteristic equation

2

22 3 0

3 3

1 3 0

1,3

For 1

1 1 2

2 1 2

1 1 2 k 2k 2kA I K

2 1 1 k 2k 2k

1 2 1 2

1 2 1 2

2k 2k 2k 2k 0A I K 0 0

2k 2k 2k 2k 0

As both equations represents the same equation, so

1 2

1 2

k k 0

k k

or

If we select an arbitrary value for k 2 as k 2 = -1 and subsequently k 1 = 1, then the correspondingeigenvector is

11

2

k 1K

k 1

For 3

1 1 2

2 1 2

1 3 2 k 2k 2kA I K

2 1 3 k 2k 2k

1 2 1 2

1 2 1 2

2k 2k 2k 2k 0A I K 0 0

2k 2k 2k 2k 0

Now too, as both equations represents the same equation, so 1 2

1 2

k k 0

k kor

If we select an arbitrary value for K2 as K2 = 1 and subsequently k 1 = 1, then the correspondingeigenvector is

1

22

k 1K

k 1


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Hence weve obtained 2 linearly independent solution vectors of the given homog enous systemas

t 3t

1 21 1

X e ,X e1 1

Thus the general solution of the system is

1 1 2 2

t 3t1 2

t 3t1 2

t 3t1 2

X c X c X

1 1X c e c e

1 1

c e c ex t

y t c e c e

The solution of the system is

t 3t1 2

t 3t

1 2

x t c e c e

y t c e c e

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Differential Equations - Solved Assignments - Semester Summer 2007