Differential Equations - Solved Assignments - Semester Spring 2007

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

1/49

ASSIGNMENT 01

Question 1:

Marks=10

Solve (1 )x ydye x e

dx

The equation is not separable

( )

2

2

2

1

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

1

. (1)

int

2

0

2

ln2

x y

x y

z

z

z

x y

dy xe e

dx

dyxe

dx

put z x y then

dz dy

dx dx

Equation no becomes

dzxe

dx

e dz xdx

egrating we have

xe c

r

xe c where z x y

x x y c

Question 2:

Marks=10

Solve2 2 2( ) 0 x xy y dx x dy


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2 2 2

2 2

2

2 2 2 2 2 2 2 2

2 2 2

2

2 2

2

( )

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

(1)

( ) ( ) (1 )

1

1 1

1

x dy x xy y dx

dy x xy y

dx x

put y vx

dy dvv xdx dx

equation becomes

dv x x vx vx x x v v x x v vv x

dx x x x

v v

dv x v v v v

dx

dv v

dx xd

2

2

1

1

1

int

1

tan ln

tan ln

v dx

v x

egrating we get

dv dx

v x

v x c

yx c

x

is the required general solution

Question 3:

Marks=10

Determine whether the given equation is Exact, if so please

2 2 2(2 tan ) ( tan ) 0 xy y y dx x x y Sec y dy


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2 2 2

2 2

2

2 2 2

2

2 tan

tan sec

2 1 sec 2 tan

2 tan

( , )

2 tan (1)

tan sec (2)

int (1) . .

( , ) tan

M xy y y

N x x y y

M x y x y

Y

N x yX

the eqis exact thereforethe function f x y is

f xy y y

x

f x x y y

y

egrating w r t x

f x y x y xy x y h

2 2 / 2 2 2

( ) (3)

( ) tan int. .

sec ( ) tan sec

y

whereh y isthecons t of egrationdiff the above equation w r t y

fx x x y h y x x y y

y

/ 2

2

( ) sec

( ) tan

(3)( , ) tan tan

h y y

h y y

therefore eq becomes f x y x y xy x y y c

Question 4:

Marks=10

Solve (by finding I.F)

1ydy

dx e x

the equation may be written as


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2

2

2

(1)

.

. (1)

2

2

y

y

dy y

y y

y y

yy

yy

dxe x

dy

or

dxx e

dyIt is a linear in x

I F e e multiply by eq

therefore

d xe e

dy

or

xe e dy c

e xe c

or

e x ce


5/49

ASSIGNMENT 02

uestion 1: Marks=10

olve the differential equation and mention the name of type of this D.E

3dy y xy

dx

3

3 2

2

3

2

3

.

1 1

1

. . .

2

olution

dy y xy

dx

t is known asa

dyx

y dx y

dyy x

y d

bernoulli different

x

putting y z

Diff w r t x

dy dzy

dx dx

ial equation

3

3

2

1

2

2

2 2

.

2, 2

dy dz

y dx dx

dy dz

y dx dx

dzz xdx

dzz x

dx

is linera differential equation of the first order

p q x


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2

2

2 2

2

2 2

22 2

22 2

2

,

.

( . ) ( . )

( ) ( ) 2

. ( 2)

. 1.

( )2

1

2

pdx

dx

x

x x

x

x x

x

x x

x

x x

sweknowthat Integrating factoris e so

F e

I F I F q dx

z e e x dx

x e dx

x e e dx

ez e x e c

ee x e c

y

uestion 2: Marks=10

nd an equation of orthogonal trajectories of the curve2

x cy


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2

2

2

2 2

22

2 2

(1)

2

(1)

2 . .

2.

2

,

2

2

nt

2

22 2

.

22 .

x cy

dycy

dx

xFromequation wehave c

y x dy

y y dx

x dy

y dx

dy y

dx x

Bythe rule of orthogonal trajectories we get

dy x

dx yy dy x dx

egrating both sides

y dy x dx

y xc

yx c

y x c

Question 3: Marks=10

fossilized bone is found to contain1

1000of the original amount of C-14. Determine the age of the fissile.

et A(t)be the amount present at any time tand A0the original amount of C14.herefore, the process is governed by the initial value problem.

0

A, (0)

dtkA A A

We know that the solution of the problem is

0( )ktA t A e


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nce half life of the carbon isotope is 5600 years, Therefore

0A(5600)=2

A

o that

560000

5600 ln 22

kA A e or k

= -0.00012378

ence(0.00012378)

0( ) tA t A e

tdenotes the time when fossilized bone was found then

0( )1000

AA t

(0.00012378)0

0000

tAA e

0.00012378 ln1000t

herefore

ln100055,800 .

0.00012378 years


9/49

ASSIGNMENT 03

Maximum Marks: 30

Upload Date: April 11, 2007

uestion 1: Marks=10olve the initial value problem

2

2

'

6 9 0

(0) 2 , (0) 3

d y dyy

dx dx

with y y

olution:

iven equation is

2

26 9 0

d y dyy

dx dx

xpected solution of this equation is y = emx

ubstituting it in the differential equation, we get

2

26 9 0

mx mx mxd de e edx dx

fter differentiation, we get2 6 9 0mx mx mxm e me e

We factorize it

2 6 9 0mx m m nce 0

mxe ,we conclude that

2 6 9 0m m 23) 0

3 , 3

m

m

neral solution is3 3

1 2

x xy C e C xe

iven the initial conditions y(0) = 2 and y(0) = -3


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3(0) 3(0)

1 2

0 3(0)

1 2

1

(0) 2 (0)

2 (0)

2

y C e C e

C e C e

C

3 3 3

1 2

3 3 3

1 2 2

3

1 2 2

'( ) ( 3) ( 3)

3 3

3 3

x x x

x x x

x

y x C e C xe e

C e C xe C e

e C C x C

3(0)

1 2 2

0

1 2

1 2

'(0) 3 3 3 (0)

3 3

3 3

y e C C C

e C C

C C

ut from first initial value of y(0), we got C1 = 2erefore,

2

2

2

3 3(2)

3 6

3

C

C

C

ubstituting the values of C1 and C2 in3 3

1 2x x

y C e C xe , we can get

3 32 3

x xy e xe

uestion 2: Marks=10

olve

' ' ' ' ' '6 3 10 0 y y y y


11/49

olution:

' ' ' ' ' '6 3 10 0 ( )

2 3, ,

( )

3 26 3 10 0

3 2( 6 3 10) 0

0

3 26 3 10 0

by synth

y y y y i

mxPut y e

mx mx mx y me y m e y m e

Putting these values in the eq i

mx mx mx mxm e m e me e

mxm m m e

mxSince e

m m m

Which can be factorized

etic division we find

1 6 3 101

1 7 10

____________________

1 7 10 0

-1 is a root.he coefficients of the quotient are-7, 10

hus we can write the auxiliary equation as:21)( 7 10) 0m m m

1m or 2 7 10 0m m ( 5)( 2) 0m m

5m or 2m he roots are -1, 2 and 5.herefore solution of the differential equation is

2 5

1 2 2

x x xy C e C e C e

uestion 3: Marks=10

iven that2

1y x is a solution of2 ' ' '3 4 0 x y xy y

nd the general solution of the differential equation on the interval (0, ) .olution:he equation can be written as


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2

3 40

2

2

2

( )2 1 2

1

32

2 4

3ln2

2 4

122

2ln

2

(0, )

1 1 2 2

1

Dividing by x

y y yx x

he nd solution y is given by

Pdxe

y y x dxy

dxde

y x dxx

xey x dx

x

y x dxx

y x x

Hencethe gernal solutionof thedifferential equation

n is given by

c y c y

c

2 2 ln2

x c x x


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Assignment 4

Answer# (1):-

22

2

22

2

2 3 3

:

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

Associated homogeneous differential equation is:

2 '' 3 ' 0.....................(i)

Now solution to the equation (i) is det

d y dy y x Sinxdx dx

solution

d y dy y x Sinx A

dx dx

y y y

2

2

1 2

ermined by the solution

of associated auxillary equation given by:2 3 1 0.....................(ii)

2 2 1 0

2 ( 1) 1( 1) 0

( 1)(2 1) 0

Therefore the roots of the given equation(ii) are

1,

m m

m m m

m m m

m m

m m

1

21 2

1

2

Hence, the complimentary solution to the

associated homogeneous equation (i) is given as

xx

cy c e c e


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2

Now we assume that the particular integral

cos sin

. . .' 2 sin cos

. . .

'' 2 cos sin

Substituting the values of , ', '' in the equation (A)we

will ha

p

p

p

p p p

y Ax Bx C D x E x

Diff w r t x

y Ax B D x E x

Again diff w r t x

y A D x E x

y y y

2 2

2 2

ve:

4 2 cos 2 sin 6 3 3 sin 3 cos + x+ + cosx+ sin +3sin

(6 ) (3 )cos (3 )sin (4 3 ) +3sin .................(iii)

Equating like coefficient on both sides of equation (

A D x E x Ax B D x E x Ax B C D E x x x

Ax A B x E D x D E x A B C x x

iii), we have:1

6 0 6

4 3 0 14

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

3 3 3 3 .............(v)

A

A B B

A B C C

E D iv

and

E D D E


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2

Solving equation (iv) and (v) simultaneously we have:

3E-D=0

3E+9D=-9

......................

-10D =99

D=-10

Now,

93E+ 0

10

93

10

9 1 3*

10 3 10Hence the particular integral :

9 36 14 cos sin

10 10

an

p

p

E

E

y is

y x x x x

1

221 2

d the general solution to the given non-homogeneous differential equation is:

9 36 14 cos sin .

10 10

c p

xx

y y y

y c e c e x x x x

Answer# (2):-


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' '

' '

' '

2

4 36 3

To find the complementary function we solve the associated homogeneous

differential equation:

4 36 0

4

9 0

The auxillary equation is

9 0 3

roots of the auxillary equ

y y Csc x

y y

Dividing by

y y

m m i

1 2

1 2

ation are complex. Therefore, the complementary

function is

cos3 sin 3

From the complementary function, we have

y cos3 , sin 3cos3 sin3

(cos3 ,sin 3 ) 3-3sin3 3cos3

Now by

c y c x c x

x y xx x

W x xx x

1 2

dividing with 4, we put the given equation in the following

standard form:

1'' 9 csc3

4

1So we identify the function ( ) csc3

4

We shall now construct the determinants and

y y x

f x x

W W


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1

2

1 2

1 21 2

0 sin31 1

csc3 .sin314 4csc3 3cos3

4

cos 3 01 cos3

1 4 sin33sin3 csc34

Therefore the derivatives ' and ' are given by

1 1 cos3' , '

12 12 sin3

Now i

x

W x xx x

xx

W xx x

u u

W W xu u

W W x

1 2

ntegrating the last two equations . . , we obtain

1 1' , ' ln sin3

12 36

The particular solution of the non-homogeneous equation is:

1 1cos3 (sin3 ) ln sin 312 36

Hence the general solution of the

p

w r t x

u ux x

y x x x x

1 2

given differential equation is:

1 1cos3 sin3 cos3 (sin 3 ) ln sin3

12 36c py y y c x c x x x x x

Answer #(3):-


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' ' ' '

' ' '

3 2

tan

To find the complementary function we solve the associated homogeneous

differential equation

' 0The auxillary equation is

0 ( 1) 0

0,

Therefore, the complementary funct

y y x

y y

m m m m

m m i

1 2 3

1 2 3

1 2 3

1 2 3

ion is

cos3 sin 3

,

1, cos , sin

Now we compute the wronskian of , ,

1 cos sin

( , , ) 0 - sin cos

0 -cos - sin

c y c c x c x

Therefore

y y x y x

y y y

x x

W y y y x x

x x

1 3

2 2

By the elementary row operation , we have

1 0 0

0 - sin cos

0 -cos - sin

(sin cos ) 1 0

The given differential equation is alrea

R R

x x

x x

x x

st nd rd

1 2 3

dy in the required standard form

''' 0. '' ' 0. tan

The determinants , and are found by replacing 1 ,2 ,3 column of by the column

y y y y x

W W W W


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2 2

1

2

0 cos sin

0 -sin cos tan (cos sin ) tan

tan - cos - sin

1 0 sin

0 0 cos 1(0 cos tan ) s

0 tan -sin

x x

W x x x x x x

x x x

x

W x x x

x x

3

1 2 3

31 21 2 3

in

1 cosx 0

0 sin 0 1( sin tan ) 0 sin tan

0 cos tan

Therefore the derivatives ' , ' , ' are given by

' tan , ' sin , ' sin tan

N

x

and

W x x x x x

x x

u u u

WW Wu x u x u x x

W W W

1 2 3ow integrating these three derivatives of the function ' , ' , 'u u u

11

22

33

2

2 2

sintan ln cos

cos

sin cos

sin tan

sinsin sin seccos

(cos 1)sec (cos sec sec )

(cos sec ) cos sec

sin ln sec tan

W xu dx xdx dx x

W x

Wu dx xdx x

W

Wu dx x xdx

W

x x dx x xdxx

x xdx x x x dx

x x dx xdx xdx

x x x

2 2

Thus, the particular solution of the non-homogeneous equation

ln cos cos cos (sin ln sec tan )(sin )

ln cos cos sin sin ln sec tan

ln cos 1 sin ln sec tan

Hence, the general equation of th

p y x x x x x x x

x x x x x x

x x x x

1 2 3

e given differential equation is:

cos sin - ln cos 1 sin ln sec tan y c c x c x x x x x


20/49

ASSIGNMENT 05

Maximum Marks: 30

uestion 1: Marks=10

olve the initial value problem

2

2

/

2 2 4cos 2 int

(0) 0 , (0) 3

d x dx x t S

dt dt

x x

Solution:

2

2

mx 2

2

associated homogeneous equation

2 2 0

x=e , ,

the auxiliary equation is

m 2 2 0.

2 4 81

2

mx mx

t

c

First of all

d x dxx

dt dt

By putting the

x me x m e

Then

mfor finding roots

m i

Thus the complementry function is

x e

1 2cos sinc t c t


21/49

p

2

2 2

2

2 2

the particular integer

x cos sin

sin cos

cos sin

2 2 cos sin 2 sin 2 cos 2 cos 2 sin

2 2 2 cos 2 sin

in the give

p

p

p p

p

p p

p

For

A t B t

x A t B t

x A t B t

d x dxx A t B t A t B t A t B t

dt dt

OR

d x dx x A B t A B t

dt dt

By substituting

n differencial equation,we have

2 cos 2 sin 4cos 2sin

coefficient,we obtain

A+2B=42A+B=2

A B t A B t t t

The equation

p

c

1 2

1 2 1

By solving these two equation,we get

A=0,B=2

Thus

2sin

of the differencial equation is as given below,

(t) = x

cos sin 2sin

. . .

cos sin sin

p

t

t t

t

Hence general solution

x

t e c t c t t

Diff w r t x

t e c t c t e c t

2

1 2

1 2

1

2

-t

cos 2cos

Now we apply the boundry conditions

0 0 , .1 .0 0 0 ,

0 3 , .1 .1 2 3 ,

solution of the initial value problem is

=e sin 2

0

sin

1

c t t

c c

c

c

cc

hus

t t

uestion 2: Marks=10


22/49

olve by using (tx e )

22 2

23 5 sin(ln )

d y dy x x y x x

dx dx

Solution:

22 2

2

22

2

2 2

t

2 2

3 5 sin(ln )

First of all associated homegeneous differential equation is.

3 5 0

3 5) 0

With the substitution of x=e , We have:

xD= , 1

So, the homegeneous di

d y dyx x y x x

dx dx

d y dyx x y

dx dx

x D xD y

x D

fferential equation becomes,

2

2

2

2

2

2

1 3 5 0

3 5 0

4 5 0

4 5 0

4 5 0

y

y

y

OR

d dy

dt dt

d y dyy

dt dt

mt

2 mt mt mt

2 mt

mt 2

22

By putting the y=e , We get the auxiliary equation as ,

e 4 e 5e 0

4 5 e 0

0 , 4 5 0

4 4 4 54 4 16 20

2 2 2

m m

m m

m m

b b acm

a


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2

1 2

2

1 2

2 2

22

2

4 4 4 22

2 2

cos sin

ln

cos(ln ) sin(ln )

Now the non-homogeneous equation becomes,

So equation becomes

2 5 sin

By the method

t

c

t

t

t

ii

y e c t c t

where t x

yc x c x c x

according to condition

x e

so

x e

d y dy y e t dt dt

2

2

2

2

2

2

2

2

2

1

of undetermined coefficient we try a particular

olution of the form

1( sin )

4 5

1(sin )

2 4 2 5

1(sin )

1

(sin )2

cos2

, ,

( cos

t

p

t

p

t

p

t

p

t

p

c p

t

y e t

y e t

y e t

ty e t

ty e t

So the General solution

y y y

y e c t c

2

2

22

1 2

sin ) cos2

lncos ln sin ln cos ln , ln2

ttt e t

x xy x c x c x x Where t x

Question 3: Marks=10

nd the interval of convergence of the power series


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0 (2 )!

n

n

x

n

Solution:or finding interval of the series we use ratio test

1

1

1

1

1

2 !lim lim .

2 1 !

2 !.lim lim .

2 1 !

. 2 !

lim lim 2 2 !

2 !lim lim

2 2 !

2 !lim

2 2 (2 1) (2 )!

1lim 0 1

2 2 (2 1)

n

n

nn nn

n

n

nn nn

n

n nn

n

n nn

n

n

na x

a n x

na x x

a xn

x na

a n

nax

a n

nx

n n n

xn n

Thus the power series converge

.s x

hus the interval of the convergence of the given series is ,


25/49

Assignment 6

uestion 1: Marks=10

ind solution of the differential equation

/ /4 0y y

in the form of a powers series in x.

Solution:

We assume that a solution of the given differential equation is

0

0 1

n n

n n

n n

y c x c c x

Then term by term differentiation

1 1

1

1 2

n n

n n

n n y nc x c nc x

22

1n

n

n

y n n c x

Substituting the expression for/ /

y and y .

we get

2

2 04 4 1

n n

n nn ny y n n c x c x

Notice that both series start with 0x . If we, respectively, substitute

k= n - 2, k= n, k= 0,1, 2,K

In the first series and second series on the right hand side of the last equation.


26/49

Then we after using, in turn, n = k+ 2 and n k, we get

20 0

4 4 2 1k k

k k

k k

y y k k c x c x

20

4 4 2 1k

k k

k

y y k k c c x

Substituting in the given differential equation, we obtain

20

4 2 1 0k

k k

k

k k c c x

24 2 1 0

k kk k c c

2, k=0,1,2,k

4 2 1

kk

cc

k k

From iteration of this recurrence relation it follows that

0 02 24.2.1 2 .2!

c cc

1 13 24.3.2 2 .3!

c cc

024 4

3 15 4

4.4.3 2 .4!

4.5.4 2 .5!

cc

c c

046 6

5 17 6

4.6.5 2 .6!

4.7.6 2 .7!

cc

c c

o on


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his iteration leaves both0

c and1

c arbitrary.

We have

2 3 4 5 6 7

0 1 2 3 4 5 6 7 c c x c x c x c x c x c x c x L

2 3 4 5 6 70 0 01 1 10 1 2 2 4 4 6 62 .2! 2 .3! 2 .4! 2 .5! 2 .6! 2 .7!

c c cc c cy c c x x x x x x x L

r

2 4 6 3 5 710 12 4 6 2 4 6

1 1 1 1 11

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

cy c x x x L c x x x x L

a general solution.

When the series are written in summation notation,

2 2 1

1 0 2 1

0 0

1 1and 2

2 ! 2 2 1 ! 2

k kk k

k k

x xy x c y x c

k k

he ratio test can be applied to show that both series converges for allx.

Maclaurin series as 1 0 cos / 2 y x c x and 2 12 sin / 2 y x c x

uestion 2: Marks=10

Whether 3x are singular points of the equation

2 2 / / / 9) ( 3) 0x y x y y

Solution:


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2 2 / / / ( 9) ( 3) 0 x y x y y

Dividing the equation by 2 2( 9)x = 2 2( 3) ( 3)x x

/ / /

2 2 2 2

/ / /

2 2 2

2 2 2

( 3) 10

( 3) ( 3) ( 3) ( 3)

1 10

( 3)( 3) ( 3) ( 3)

1 1and q x

( 3)( 3) ( 3) ( 3)

x y y y

x x x x

y y y x x x x

Where

p x x x x x

X=3 is regular singular points because power of x-3 in P (x) is 1 and in Q (x) is 2

X= -3 is an irregular singular point because power of x+3 in P (x) is 2.

Question 3(a) : Marks=5

ind the general solution of the equation

2 / / / 2 1( ) 016

x y x y x y

Solution:

The Bessel differential equation is


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2 / / / 2 2

2 / / / 2

( ) 0 1

1( ) 0 2

16

x y x y x v y

x y x y x y

Comparing 1 and 2 we get

2 1 1, therefore v=16 4

v

So, the general solution of 1 is

1 1/ 4 2 1/ 4y C J x C J x

Question 3(b) : Marks=10

erive the expression of3

( )2

n J x for n

Solution:

Consider

1 12

v v vv

J x J x J xx

For1

2v

1 1 11 12 2 2

122

J x J x J xx


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3 1 12 22

3 1 12 22

1

1

J x J x J xx

J x J x J xx

we know that

1/ 2 1/ 22 2

( ) sin , ( ) cos J x x J x xx x

3

2

32

1 2 2sin cos

2 sincos

J x x x x x x

x J x x

x x


31/49

Assignment 7

Q No.1Solve the system of equation by systematic elimination

1 1 2

3 2 1

D x D y

x D y

Solution:

1 1 2 (1)

3 2 1 (2)

D x D y

x D y

From equation (2)

1 2

3

D yx

Putting in equation (1), then we have

1 2

1 1 23

D y D D y


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2 5 7D y

The associated homogenous equation is

2 5 0D y

The auxiliary equation is

2 5 0m

Roots are 5m i

1 2cos 5 sin 5

c y c t c t

Now to determine the particular solutionp

y

'

''

0

0

p

p

p

y A

y

y

Thus substituting in the given D.E, we have

'' 5 7p py y

Equating the coefficients and constants terms gives.


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5 7

7

5

A

A

Hence the solution of variable y is given by

c p y y y

1 2

7cos 5 sin 5

5 y c t c t

From equation (1) we have

1 3

2

xy

D

Putting in equation (2)

2 1 1 1 3 2 2 D D x D x D

2 1 1 3 1 2 2 D D x D x D D

2 5 3D x

The auxiliary equation of the differential equation for x is

2 5 0m

Roots are 5m i


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The roots of the auxiliary equation are complex. Therefore

3 4cos 5 sin 5

c x c t c t

We assume that

px A

'

''

0

0

p

p

x

x

Substituting in the differential equation we have.

'' 5 3p p

x x

Equating the coefficients we have

5 3A 3

5A

so that

c p x x x

3 4

3

cos 5 sin 5 5 x c t c t

Thus


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3 43

( ) cos 5 sin 55

x t c t c t

1 2

7( ) cos 5 sin 5

5 y t c t c t

Q No. 2: If possible rewrite the following equations in the canonical form

'' '

'' '

2

3

x y

x y

SOLUTION:

Writing the system in the operator form

2

2

2

3

D x Dy

D x Dy

2

2

2

3

0 1

D x Dy

D x Dy

This is absurd. Thus the given system cannot be reduced to a canonical form.Hence the system is a degenerate system.

Q No. 3: Solve the following system of linear equations by Gaussian elimination method


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1 3

1 2 3

1 2 3

2 1

2 4 2

8 3 2

x x

x x x

x x x

Solution

2 0 1 1

2 4 1 2

1 8 3 2

Theaugmented matrixof the systemis

By interchanging 13R

1 8 3 2

2 4 1 2

2 0 1 1

1 2

1 8 3 2

0 20 5 6 2

2 0 1 1

R R

1 3

1 8 3 2

0 20 5 6 2

0 16 7 3

R R


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2

1 8 3 2

5 60 1

20 20 20

0 16 7 3

R

2 3

1 8 3 2

5 60 1 16

20 20

90 0 3

5

R R

3

1 8 3 2

5 60 1

20 20 3

30 0 1

5

R

1 8 3 2

5 60 1

20 20

30 0 1

5


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1 2 3

2 3

3

8 3 2

5 6

20 203

5

Thelast matrix is in row echelon form and representsthe system

x x x

x x

x

2

2

1

1

1 2 3

5 3 6

20 5 20

9

20

9 38 3 2

20 5

1

5

lg

1 9 3, ,

5 20 5

x

x

x

x

Solution set of given system of linear a ebraic equation

x x x


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Assignment 8

Question # 1:- Find the Eigen values and Eigen vectors of the following matrix

7 2 1

2 15 2

1 2 7

A

Solution:

For Eigenvalues

det 0

7 2 1 1 0 0

2 15 2 0 1 0 0

1 2 7 0 0 1

A I

7 2 1

det 2 15 2 0

1 2 7

A I

Expanding the determinant

det 1( 2( 2) 1 15 ) 2 2 7 2

7 15 7 2 2

A I


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We have

27 ( 22 96) 0

7 16 6 0

Thus the Eigen values of the matrix are

1 2 37, 16, 6

For Eigenvectors are:

For 1 7 we have

=>

0

0

0

=>By row reducing the augmented matrix


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1 3

2 1

2

2 3 1 2

2

&

0

0

0

2

0

0

0

2

0

0

0

&

0

0

0

2

0

0

0

ExchangingR R

R R

R

R R R R

R

Thus we have the following equations in k1, k2, and k3. The number k3 can be chosen

arbitrarily.

1 3, 2 3

1

2k k k k

Choosing 3K = 1, we get k1=1 and k2 = , hence the eigenvector corresponding 1 7 is


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1

1

4

1

K

1

1

1

2

1

K

For2

16 we have

0

16 0 0

0

A

]

By the row reducing the augmented matrix


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1 3

2 1 3 1

2

3 2

0

0

0

&

0

0

0

2 & 9

0

0

0

50

0

0

20

0

0

0

Exchanging R R

R R R R

R

R R

1 22

0

0

0

R R

Thus we have the following equations in k1, k2, and k3. The number k3 can be

chosen arbitrarily.

1 3, 2 34k k k k


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Choosing k3 = 1, we have 1K =1 and k2 = -4, hence the eigenvector corresponding

216 is

2

1

41

K

For2

6 we have

0

6 0 0

0

A

2 1 3 1

2

1 2

0

0

0

2 &

0

0

0

5

0

0

0

2

0

0

0

R R R R

R

R R

Now, we have the following equations in k1, k2, and k3. The number k3 can be chosen

arbitrarily.

=> 1 3, 2 0k k k


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26

Choosing k3 = 1, we get k1= -1 and k2 = 0,

Hence the eigenvector corresponding

2 6

3

1

0

1

K

Question # 2:

Solve the homogeneous system of differential equations

2 3

2

dxx y

dt

dyx y

dt

Solution:


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The given system can be written in the matrix form as

2 32 1

dx

xdtdy y

dt

Therefore, the coefficient matrix

2 3

2 1A

Now se find the eigenvalues and eigenvectors of the coefficient A. The characteristics equation

is

2

det 0

2 30

2 1

3 4 0

A I

Therefore, the characteristic equations

1 4 0 1,4

Therefore roots of the characteristic equation are real and distinct and so ate the eigenvalues,

For 1 , we have

1

2

1 2

1 2

1 2

1 2

2 1 3

2 1 1

3 3

2

0

3 3 0

2 0

k A I K

k

k k A I K

k k

A I K

k k

k k

These two equations are nod different and represent the equation.


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Thus we can choose value of the constant k2 arbitrarily, If we choose2

K = -1 then

1K = 1.

Thus the corresponding eigenvector is

1

1

1K

For =4 we have

1

2

1 2

1 2

1 2

1 2

2 4 3

2 1 4

2 3

2 3

0

2 3 0

2 3 0

k A I K

k

k k A I K

k k

A I K

k k

k k

Again the above two equations are not different and represent the equation

21 2

32 3 0

2

kk k

Again m the constant k2 can be chosen arbitrarily. Let us choose k2 = 2 then k1 = 3.

Thus the corresponding eigenvector is

4

1 2

1 3,

1 2

t t X e X e

Hence the general solution of the system is the following


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1 1 2 2

4

1 2

4

1 2

41 2

1 3,

1 2

( ) 3

( ) 2

t t

t t

t t

X c X c X

X c e c e

x t c e c e

y t c e c e

4

1 2

4

1 2

( ) 3

( ) 2

t t

t t

x t c e c e

y t c e c e

Question # 3: Write the given system in matrix form

2

2

1

2 3

2

dxx y z t

dt

dy x y z t

dt

dzx y z t t

dt


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Solution:-

In the matrix notation

' 2

1 1 1 0 1 1

2 1 1 3 0 0

1 1 1 1 1 2

X X t t

'

2

1 1 1

2 1 1

1 1 1

0 1 1

3 0 0

1 1 2

X X g t

Where g t

t t

Documents

Differential Equations - Solved Assignments - Semester Spring 2007