Assignment # 01 Question 1: Marks=10

Solve the initial value problem 2( ) 2 (0) 2

dyy x y x with y

dx+ = = −

Solution:

( )

( )

( ) ( )

2

2

2

22

2

22

2 2

22 2

22

1 2

.

2

1int

2

1

ln 1 (1)2

0 2

2 0 (1)

2ln 0 1

22 (1)

ln 1 22

2ln 1 4

ln 1 4

ln 1 4

y x dy xdx

It can be seperable form

xydy dx

xTaking egral both sides

xydy dx

x

yx c

y

putting y and x in eq

c

c put in

yx

y x

y x

y x

+ =

=+

=+

= + + →

= −= − =

−= + +

=

= + +

= + +

= + +

= + +

∫ ∫

Question 2: Marks=10 Solve the D.E ( 1) ( 3 5)x y dy x y dx− − = + − Solution:

3 5

1

omogeneous

omogeneous

3 3 5

13 5 0 1 0

1 2

dy x y

dx x y

The equation is neither separable nor h

The equation is reduce to h

putting

x X h dx dX

y Y k dy dY

dy dY

dx dXdY X Y h k

dX X Y h know h k and h k

solving both k and h

dY

+ −=− −

= + ⇒ == + ⇒ =

=

+ + + −=− + − −

+ − = − − == =

∴

2 2

2

3(1)

omogeneous

(1)

3 1 3

11 3

1

1 3 1 2

1 1var

1

2 1

X Y

dX X Ythe aboveequation is h

putting Y vX in

dY dvv X

dX dXdv X vX v

v XdX X vX v

dv vX v

dX v

dv v v v v vX

dX v vSeparable iable

v dXdv

v v X

+= →−

=

= +

+ ++ = =− −

+= −−+ − + + += =

− −

− =+ +

( )( )( )( )

( )

2

2

22

2

2 2 21

2 2 1

2 2 21

2 2 1

1 2 2

2 2 1 1

1 1ln 2 1 ln

2 11

ln 1 ln11

ln 1 ln1

ln ln

ln ln

ln ln ln

ln

v dXdv

Xv v

v dXdv

Xv v

v dv dXdv

v v Xv

v v X cv

v X cv

YX c

YXX

Y X XX c

X Y X

X XX c

X Y X Y

XX X Y X c

X YX

XX Y

− + −− =

+ +

− + −− =

+ +

+− + =+ + +

− + + − = ++

− + − = ++

− + − = ++

+− − = ++

− = ++ +

− + − = ++

++

∫ ∫

∫ ∫ ∫

( 2)ln 2 1

2 1

2 4ln 3

3

Y c

xx y c

x y

xx y c

x y

+ =

− + − + − =− + −

− + + − =+ −

Question 3: Marks=10 In the given problem determine, whether the given equation is exact. If exact, solve

3 2 26 (4 9 ) 0xy dx y y x dy+ + = Solution:

( )3 2 2

3 3 2 2

2 2

3 2 3

6 4 9 0

6 , 4 9

18 , 18

int tan

6 3

t

xy dx y y x dy

M xy N y y x

M Nxy xy

y x

M N

y x

Sothe differentail equationis exact

Now egrating M with respect to x and taking y terms as cons t

Mdx xy dx x y

In egrate N with respect to y for te

+ + =

= = +∂ ∂= =∂ ∂

∂ ∂=∂ ∂

= =∫ ∫

3 4

2 3 4

,

4

3

rms involving y alone

y dy y

Hencethe general solution is

x y y c

=

+ =

∫

Question 4: Marks=10 Solve (by using I.F method)

4 3 2 4 2 2(2 2 ) ( 3 ) 0y yxy e xy y dx x y e x y x dy+ + + − − =

( )( )

4

4 3

4 3 2

2 4 2 2

4 2

4 2 4 3 2

4 3

3 2

3 2

14 ln

:

2 2

2 8 6 1

3

2 2 3

2 2 3 2 8 6 1

2 2

4 2 2 1 4

2 2 1

. .

y

y yy

y

yx

y x

y y yx y

y

y

y

dyyy

Solution

M xy e xy y

M xy e xy e xy

N x y e x y x

N xy e xy

M N

N M xy e xy xy e xy e xy

M xy e xy y

xy e xy

yy xy e xy

I F e e

−

= + += + + +

= − −= − −≠

− − − − − − −=+ +

− + + −= =+ +

∫= = =4

1

y

Multiplying I.F. by both side of given equation

( ) ( )4 3 2 4 2 24 4

22

3 2 4

3

22

3

1 12 2 3 0

2 1 32 0

2 12 0

y y

y y

y

y

xy e xy y dx x y e x y x dyy y

x x xxe dx x e dy

y y y y

Theequation is exact

xxe dx

y y

x xx e c

y y

+ + + − − =

+ + + − − =

+ + =

+ + =

∫

ASSIGNMENT 02

Question 1: Solve the differential equation and mention the name of type of this D.E

2ln .dy

x y x ydx

+ =

:

This is a "Bernoulli"equation

Solution

2

Given eq.can be written as

ln(1)

Comparing with general Bernoulli Eq.

( ) ( ) n

dy y xy

dx x x

dyP x y Q x y

dx

+ = − − − − − −

+ =

1 1

weget

1 ln( ) , ( )

2

thus wesubstituten

xP x Q x

x xn

v y y− −

= =

=

= =

2

Eq.(1) becomes

ln(2)

dv dyy

dx dx

dv v x

dx x x

−= −

− = − − − − − − −

1

ln

ln 1

2 2

Now this Eq. is linear Eq.

Integrating factor of this linear D.E is

( )

( )

Multiply Eq.(2) by U(x)

1 ln

dxxx

x

U x e e

U x e x

dv v x

x dx x x

−

− −

−

∫= =

= =

− = −

12

12

ln( )

integrating both side w.r.t x.

ln

integrating by parts

d xx v

dx x

xx v dx

x

−

−

= −

= −∫

12

1

1

ln 1[ ]

ln 1

ln 1

Re var

ln 1

1

ln 1

xx v dx C

x xx

x v Cx x

v x Cx

verting back to original iable

y x Cx

yx Cx

−

−

−

= − − + +

= + +

= + +

= + +

=+ +

∫

Question 2:

Solve the D.E by using an appropriate substitution cos( )x y dy dx+ =

:

taking derivative w.r.t x.

1 1

sogiven D.E becomes

cos [ 1] 1

cos cos 1

cos 1 cos

cos

1 cos1

11 cos

solution

Let x y u

dy du dy du

dx dx dx dx

duu

dxdu

u udxdu

u udxu

du dxu

du dxu

+ =

+ = => = −

− =

− =

= +

=+ − = +

2

2

2

Since, 1 cos =2cos 2

So above Eq. becomes

11-

2cos2

11 sec

2 2

integrating

tan2

Reverting back tooriginal variable

tan2

tan( )2

uu

du dxu

udu dx

uu x c

u x y

x yx y x c

or

x yy c

+

=

− =

− = +

= +++ − = +

+= +

Question 3:

Find an equation of orthogonal trajectory of the curve 2 2 2x y c+ =

2 2 2

:

Differentiate it w.r.t 'x' to find theDE.

2 2 0

2

2

write down the DE for the orthogonal family

1

( )

this is linear as well as a separable DE. Find the integrating factor.

(

x y c

solution

dyy x

dxdy x

dx y

dy x

dx y

dy yxdx xy

u

+ =

+ =

−=

= −

= − =−

1 1)

which gives thesolution

. ( )

( )

This represents family of straight lines through origin.

dxxx e

x

y u x m

or

my mx

u x

y mx

−∫= =

=

= =

=

ASSIGNMENT NO. 3

Question 1: Marks=10 Determine whether the following functions are linearly dependent or independent using Worskian determinant.

2 2

( ) 9cos(2 )

( ) 2cos 2sin

f x x

g x x x

== −

Solution:

2 2

2 2

( ) 9cos(2 )

( ) 18sin(2 )

( ) 2cos 2sin

2(cos sin )

2cos(2 )

( ) 4sin(2 )

9cos(2 ) 2cos(2 )( ( ), ( ))

18sin(2 ) 4sin(2 )

f x x

f x x

g x x x

x x

x

g x x

x xW f x g x

x x

=′ = −

= −= −=

′ = −

=− −

( ) ( )36 cos(2 )sin(2 ) 36 cos(2 )sin(2 )

0

x x x x= − +=

Thus functions f(x) and g(x) are linearly dependent functions. Question 2: Marks=10 a) Suppose that

1 2 1 2

sin(2 )

cos(3 ) sin(3 ) , c and c are constants

p

c

y t

y c t c t

=

= +

Find whether yp is a particular solution and yc is the complementary function of following non-homogenous differential equation. 9 5sin(2 )y y t′′ + =

Solution:

sin(2 )

2cos(2 )

4sin(2 )

Then

9 4sin(2 ) 9sin(2 ) 5sin(2 )

Hence

sin(2 ) is a particular solution of given non-homogenous equation

p

p

p

p p

p

y t

y t

y t

y y t t t

y t

=′ =′′ = −

′′ + = − + =

=

Now

1 2 1 2

1 2

1 2

cos(3 ) sin(3 ) , c and c are constants

3 sin(3 ) 3 cos(3 )

9 cos(3 ) 9 sin(3 )

c

c

c

y c t c t

y c t c t

y c t c t

= +′ = − +′′ = − −

( )1 2 1 2

1 2 1 2

Then

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

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

0

c cy y c t c t c t c t

c t c t c t c t

′′ + = − − + += − − + +=

Thus yc is the complementary function of given non-homogenous differential equation.

b) Let 1xy e= is a solution of the following differential equation

2 0y y y′′ ′− + =

What is its second solution and general solution? Solution:

1xy e=

2 0y y y′′ ′− + =

Compare it with ( ) ( ) 0y P x y Q x y′′ ′+ + =

Then ( ) 2

( ) 1

P x

Q x

= −=

The second solution is given by ( )

2 1 21

P x dxe

y y dxy

−⌠⌡

∫=

Put values ( 2)

2 2( )

dx

xx

ey e dx

e

− −⌠⌡

∫=

2

2

dx

xx

ee dx

e

⌠⌡

∫=

2

2

xx

x

ee dx

e

⌠⌡

=

xe dx= ∫

xxe= Hence general solution of given differential equation is

1 1 2 2y c y c y= +

1 2x xy c e c xe= +

Question 3: Marks=10

Let 1y x= is a solution of the following differential equation 2 0x y xy y′′ ′− + =

Find its second solution using REDUCING ORDER METHOD. Solution:

1y x=

Let

2 1( ) ( )y x u x y=

Then

2y ux=

2y u x u′ ′= +

2 2y u x u′′ ′′ ′= +

Put in given differential equation

22 2 2 0x y xy y′′ ′− + =

2( 2 ) ( ) 0x u x u x xu u xu′′ ′ ′+ − + + = 3 2 0u x x u′′ ′+ =

10u u

x′′ ′+ =

If we take w u′= then

10w w

x′ + = ------------(1)

It is first order linear equation, whose integrating factor is 1

lndx xxe e x∫ = = Multiply equation (1) by integrating factor

0xw w′ + =

( )0

d xw

dx=

1 1, where is constant of integrationxw c c=

Put value of w

1xu c′ =

1cu

x′ =

Integrate with respect to x

1 2 2= ln , where is constant of integrationu c x c c+

Since 2y ux= , so

( )2 1 2lny x c x c= +

Choosing c1 =1 and c2 = 0

2 lny x x=

Hence general solution is

1 2 , where A and B are constantsy Ay By= +

lny Ax Bx x= +

Question 4: Marks=10

Solve 33 4 xy y e−′′ ′+ =

Solution: Given differential equation is

33 4 xy y e−′′ ′+ =

Associated homogenous differential equation is 3 0y y′′ ′+ =

Put mxy e=

mxy me′ = 2 mxy m e′′ =

Substituting in the give differential equation, we have

2 3 0mx mxm e me+ = 2( 3 ) 0mxm m e+ =

Since xemx 0 ∀≠ , the auxiliary equation is 2 3 0m m+ =

( 3) 0m m + =

0, 3m = − Thus complementary function, yc , is

31 2

xcy c c e−= +

Next we find a particular solution of the non-homogeneous differential equation. Particular Integral The input function

3( ) xg x e−=

Since 3xAe− is present in cy . Therefore, it is a solution of the associated homogeneous differential equation

3 0y y′′ ′+ =

To avoid this we find a particular solution of the form 3x

py Axe−=

We notice that there is no duplication between cy and this new assumption for py

3 3

3

3

( 3 1)

x xp

x

y Axe Ae

Ae x

− −

−

′ = − +

= − +

3 3 3

3 3

3

y 9 3 3

9 6

(9 6)

x x xp

x x

x

Axe Ae Ae

Axe Ae

Ae x

− − −

− −

−

′′ = − −

= −= −

Substituting in the given differential equation

3

3 3 3

3 4

(9 6) 3 ( 3 1) 4

9 6 9 3 4

3 4

4

3

xp p

x x x

y y e

Ae x Ae x e

Ax A Ax A

A

A

−

− − −

′′ ′+ =

− + − + =− − + =

− =−=

So particular solution of given differential equation is

3 34

3x x

py Axe xe− −−= =

Hence, general solution is

3 31 2

4

3

c p

x x

y y y

c c e xe− −

= +

= + −

ASSIGNMENT 04

Question 1: Marks=10 a) Find the differential operator that annihilates the following function. ( ) 1 sinf x x= +

Also check your answer by applying that operator. Solution:

1 2

1

1

1

1

( ) 1 sin

Let ( ) 1 , ( ) sin

Since ( ) is a constant so it vanishes after first differentiation. i-e

( ) (1) 0

So D is the annihilator operator of ( ).

As we know that for functions

con x

f x x

y x y x x

y x

Dy x D

y x

x eα−

= += =

= =

( )( )

1

2 2 2

12

s

or

sin

the annihilator operator is

2

where , are real numbers and n is non-negative integer.

So for ( ) sin , compare it with sin

0 , 1 , 1 0 1

So annihilator oper

n x

n x

x

x e x

D D

y x x x e x

n n

n

α

α

β

β

α α β

α ββ

α β

−

−

− + +

== = − = ⇒ =

( )( )

2

12 2 2

2

ator of ( ) is

1 1

Therefore 1 annihilates the function ( ) 1 sin

y x

D D

D D f x x

+ = +

+ = +

Check:

( ) ( ) ( )( )( ) ( )

( )( ) ( )( ) ( )

( ) ( )

2 3

3

2

3 2

2

1 1 sin 1 sin

1 sin 1 sin (1)

Now

1 sin (1) (sin ) cos

1 sin 1 sin (cos ) sin

1 sin 1 sin ( sin ) cos

Putting values in equation (1)

1 1 sin

D D x D D x

D x D x

D x D D x x

D x D D x D x x

D x D D x D x x

D D x D

+ + = + +

= + + + − − − − − − −

+ = + =

+ = + = = −

+ = + = − = −

+ + = ( ) ( )3 1 sin 1 sin

cos cos

0

x D x

x x

+ + +

= − +=

NON-HOMGENOUS LINEAR DIFFERENTIAL EQUATION: A non-homogeneous linear differential equation of order n is an equation of the form

)(011

1

1 xgyadx

dya

dx

yda

dx

yda

n

n

nn

n

n =++++−

−− ⋯

The coefficients naaa ,,, 10 … can be functions ofx or constants. ( )g x is non zero function of x.

A second order, linear non-homogeneous differential equation is 2

2 1 02( )

d y dya a a y g x

dx dx+ + =

There are two common methods for finding particular solutions of linear non-homogeneous differential equations:

1. Undetermined Coefficients 2. Variation of Parameters.

Undetermined Coefficients: The method of undetermined coefficients developed here is limited to non-homogeneous linear differential equations

That have constant coefficients, and Where the function )(xg has a specific form.

So either we use

• Undetermined Coefficients-Superposition approach or • Undetermined Coefficients-Annihilator Operator Approach

two things must be kept in mind. 1. Non-homogeneous linear differential equation has constant coefficients. That is, in following equation

1

1 1 01( ) (*)

n n

n nn n

d y d y dya a a a y g x

dx dx dx

−

− −+ + + + = − − − − −⋯

The coefficients naaa ,,, 10 … are constants.

2. This method is applicable for only those values of ( )g x , which have finite family of derivatives. That is, functions

with the property that all their derivatives can be written in terms of just a finite number of other functions. For

example

( ) sin

( ) cos

( ) sin

( ) cos

( ) siniv

g x x

g x x

g x x

g x x

g x x

=′ =′′ = −′′′ = −

=

And the cycle repeats. Notice that all derivatives of ( )g x can be written in terms of a finite number of functions. [In this case,

they are sin x and cos x, and the set {sin x, cos x} is called the family (of derivatives) of ( ) sing x x= .]

This is the criterion that make equation (*) susceptible to the method of undetermined coefficients. Here's an example of a function that does not have a finite family of derivatives Its first four derivatives are

2

2

4 2 2

4 2 3

( ) tan

( ) sec

( ) 2sec tan

( ) 2sec 4sec tan

( ) 16sec tan 8sec taniv

g x x

g x x

g x x x

g x x x x

g x x x x x

=′ =′′ =′′′ = +

= +

Notice that the nth derivative ( n ≥ 1) contains a term involving tan n-1 x, so as higher and higher derivatives are taken, each one will contain a higher and higher power of tan x, so there is no way that all derivatives can be written in terms of a finite number of functions. The method of undetermined coefficients could not be applied if ( ) tang x x= . So just what are the Following

table gives the list of functions ( )g x whose derivative families are finite.

Nonzero Functions with a Finite Family of Derivatives

Function Family

k {1}

( k: a constant)

x n { x n , x n−1 ,… x, 1}

( n: a nonnegative integer)

e kx { e kx }

sin kx {sin kx, cos kx}

cos kx {sin kx, cos kx}

a finite product of any of the preceding types {all products of the individual family members}

If these restrictions do not apply to a given nonhomogeneous linear differential equation, then a more powerful method of determining a particular solution is needed: the method known as variation of parameters. In view of above discussion, we can easily answer the given questions.

b) Decide whether the given equations are linear non-homogenous differential equations and the method of undetermined coefficients can be applied to find a particular solution of the given equations. Do not solve the equations, just give a short reason if your answer is no.

• 22 5 cost ty y yt te t e′′ ′− + = −

This is a non-homogeneous linear differential equation . Since the coefficient of y is not constant, the method of undetermined coefficients cannot be applied to find a particular solution of the given equation.

• 23 4sin

tty y y t e

t′′ ′+ − = −

Here

2

2

( )sin

( ) cos

t

t

tg t t e

t

g t t ec t t e

= −

= −

This is a non-homogeneous linear differential equation . Since cosec(t) does not have finite derivative family, so the method of undetermined coefficients cannot be applied to find a particular solution of the given equation.

• 2

2 2 tt

ty y y te

e′′ ′+ + = −

This is a non-homogeneous linear differential equation. The method of undetermined coefficients can be applied to find a particular solution of the given equation.

c) Does the differential operator 4( 1)D − annihilate the function3 xx e ? Justify your answer.

As we know, the differential operator ( )nD α− annihilates the function1 n xx eα−

.

In the function 3 xx e ,

n -1 = 3 , n = 4 α = 1

Thus 4( 1)D − annihilates the function3 xx e .

Question 2: Marks=10 Solve the following differential equation using UNDETERMINED COEFFICIENT-Annihilator Operator Approach.

34 12 2 3y y y x x′′ ′− − = − +

Solution:

( )3

2 3

2

2

2 61 2

3

4 12 2 3

4 12 2 3 (1)

Auxillary equation is

4 12 0

6 2 12 0

( 6) 2( 6) 0

( 2)( 6) 0

2,6

(2)

Now input function,g(x), is given by

g(x) = 2 3

x xc

y y y x x

D D y x x

m m

m m m

m m m

m m

m

y c e c e

x x

−

′′ ′− − = − +

− − = − + − − − − − − − − − −

− − =− + − =

− + − =+ − =

= −= + − − − − − − −

− +

As we know that 1. The polynomial function

1110

−−+++ n

n xcxcc ⋯

can be annihilated by finding an operator that annihilates the highest power of .x 2. The differential operators

D , 2D , 3D , … ,4D

are respective annihilator operators of the following functions

… , , , ),constant a( 32 xxxk

Use these results. Since in g(x) highest power of x is 3, so 4D is the annihilator operator. That is,

[ ] ( )4 4 3g(x) = 2 3 0D D x x− + =

Thus we can write equation (1) as

( ) ( )4 2 4 34 12 = 2 3 0D D D y D x x− − − + =

( )4 2 4 12 0D D D y− − =

( )6 5 44 12 0 (3)D D D y− − = − − − − − − −

This is homogenous differential equation of order 6. Its auxillary equation is

6 5 44 12 0m m m− − =

( )4 2 4 12 0m m m− − =

4( 2)( 6) 0m m m+ − =

0, 0, 0, 0, 2, 6m = −

Thus the general equation of homogenous equation (3) is 2 3 2 6x xy A Bx Cx Dx Ee Fe−= + + + + +

Compare the right hand side with equation (2) 2 6x xEe Fe− + are in cy

So particular integral of given non-homogenous differential equation can be written as 2 3

py A Bx Cx Dx= + + +

22 3py B Cx Dx′ = + +

2 6py C Dx′′ = +

( ) ( )

( ) ( )

2 2 3 3

2 2 3 3

2 3 3

4 12 2 6 4 2 3 12 2 3

2 6 4 6 12 12 12 12 12 2 3

12 4 2 12 8 6 12 12 12 2 3

p p py y y C Dx B Cx Dx A Bx Cx Dx x x

C Dx B Cx Dx A Bx Cx Dx x x

A B C B C D x C D x x x x

′′ ′− − = + − + + − + + + = − +

= + − − − − − − − = − +

= − − + + − − + + − − − = − +

Compare coefficients. 12 2

1

6

D

D

− =

= −

12 12 0

0

1

6

C D

C D

C

− − =+ =

=

12 8 6 1

Put values of C and D

1

9

B C D

B

− − + = −

= −

12 4 2 3

Put values of B and C

5

27

A B C

A

− − + =

= −

Thus

2 35 1 1 1

27 9 6 6py x x x= − − + −

General solution is

2 6 2 31 2

5 1 1 1

27 9 6 6x xy c e c e x x x−= + − − + −

Question 3: Marks=10

Solve the following differential equation using VARIATION OF PARAMETERS method.

22

1

xey y y

x′′ ′− + =

+

Solution:

( )

2

2

2

1 2

21

Auxillary equation is

m 2 1 0

1 0

1, 1

x

x xc

ey y y

x

m

m

m

y c e c xe

′′ ′− + =+

− + =

− === +

From the complementary function we consider,

1 2( ) , ( )x xy x e y x xe= =

2

2

( , ) ( )x x

x x x x x x

x x x

x

e xeW e xe e e xe xe

e e xe

e

= = + −+

=

Thus 1( )y x and 2( )y x are two linearly independent solutions.

Now compare given differential equation with the following equation. ( ) ( ) ( )y P x y Q x y f x′′ ′+ + =

Then

2( )

1

xef x

x=

+

Now

1

2

0

1

x

xx x

xeW e

e xex

=+

+

2

2 1

xxe

x

−=+

( ) ( )

( )

21

1 2 2 2

21

1 1

Integrate to get

1ln 1

2

x

x

W xe xu

W x e x

u x

− −′ = = =+ +

= − +

2

2

0

1

x

xx

eW e

ex

=

+

2

2 1

xe

x=

+

( ) ( )2

22 2 2 2

1

1 1

x

x

W eu

W x e x′ = = =

+ +

12

Integrate to get

tan ( )u x−=

So,

( )1 1 2 2

2 1ln 1 tan ( )2

p

xx

y u y u y

ex xe x−

= +

= − + +

Thus, general solution is

( )2 11 2 ln 1 tan ( )

2

c p

xx x x

y y y

ec e c xe x xe x−

= +

= + − + +

Question 4: Marks=10

A 16 lb object stretches a spring 8

9 feet . There is no damping and external forces acting on the system. The spring is initially

displaced 6 inches upwards from its equilibrium position and given an initial velocity of 1 ft/sec downward. Find the displacement at any time t. (NOTE: Use feet as the unit of measurement.) Solution: Weight of an object = 16 lb Acceleration due to gravity = g = 32ft/sec2

Spring stretch = 8

9 ft

We can find the mass of an object, m, by following equation

16 1slugs

32 2

W mg

Wm

g

=

=

= =

Spring stretch = 8

9 ft

Find value of spring constant ,k , using Hooke’s Law

k =

16 9 16 14418 lb/ft

8 8 89

W

s×= = = =

Let x(t) is the displacement at any time t.

Then the equation of Simple Harmonic Motion is 2

2

2

2

2

2

2

2

2

2

118

2

118 0

2

36 0 (1)

Auxillary equation is

m 36 0

36

6

d xm kx

dt

d xx

dt

d xx

dt

d xx

dt

m

m i

= −

= −

+ =

+ = − − − − − − − − − −

+ == −

= ±

General solution of homogenous differential equation (1) is

1 2( ) cos(6 ) sin(6 ) (2)x t c t c t= + − − − − − − −

Since the initial displacement is 6 inches, that is -0.5 ft, and the initial velocity is 1 ft/sec, the initial conditions are (0) 0.5 ft , (0) 1 ft/secx x′= − =

At t = 0, equation (2) becomes

1 2 1(0) cos(0) sin(0)x c c c= + =

Apply initial condition, (0) 0.5 ftx = − , to get value of 1c

1 0.5c = −

Differentiate equation (2)

1 2

1

2

2 2

( ) 6 sin(6 ) 6 cos(6 )

Since 0.5

( ) 3sin(6 ) 6 cos(6 )

For t =0

(0) 3sin(0) 6 cos(0) 6

x t c t c t

c

x t t c t

x c c

′ = − += −

′ = +

′ = + =

Apply initial condition, (0) 1 ft/secx′ = , to get value of 2c

2

1

6c =

Put values of 1c and 2c in equation (2)

1( ) 0.5cos(6 ) sin(6 )

6x t t t= − +

Assignment # 05

Question#01: Interpret and solve the initial value problem

2

2

/

5 4 0

(0) 1 , (0) 1

d x dxx

dt dt

x x

+ + =

= =

Solution.

2

2

22

2

2

2 2

Comparing the given differential equation (1) and (2).

5 4 0 (1)

The general equation of the free damped motion

2 0 (2)

see that

5, 4

2so that,

0

Therefore, the probl

d x dxx

dt dt

d x dxw x

dt dtwe

w

w

λ

λ

λ

+ + = →

+ + = →

= =

− >em represents the over-damped motion of a mass on a spring.

Inspection of the boundary conditions.

x(0) = 1, x (0) = 1

reveals that the mass starts 1 unit below the equilibrium position with a downward

ve

′

locity of 1 ft/sec.

( )( )

( )

2

2

2mt mt 2 mt

2

2

41 2

Now we have o solve the D.E

5 4 0

put,

x=e , e , e

Then the auxiliary equation is

m 5 4 0

4 1 0

1, 4

Thus the solution of the differential equation is:

x t t

t

d x dxx

dt dtwe

dx d xm m

dt dt

m

m m

m m

c e c e− −

+ + =

= =

+ + =+ + =

= − = −

= +

( )

( )( )

( )

41 2

1 2

1 2

1 2

1 2

1 2

x t 4

apply the boundary conditions.

x 0 1 .1 .1 1

x 0 1 4 1

,

1

4 1

,

5 2,

3 3Therefore, solution of the initial value problem is

x t

t

t tc e c e

c c

c c

thus

c c

c c

solving above two equations we can get

c c

− −′ = − −

= ⇒ + =′ = ⇒ − − =

+ =− − =

= = −

= 45 2

3 3t te e− −−

Question#02 Solve the initial value problem

22

2

/

0

cos

(0) 0 , (0) 0

.

o

d xw x F t

dt

with x x

where F is constt

γ+ =

= =

Solution.

22

2

22

2

2

2 2

1 2

0

,

0

( )

mt mt

c

d xw x F Cos t

dtThe complementary function is

Associated equation is

d xw x

dt

x e x m e

Then auxiliary equation is

m w

m wi

Thus complementary solution is

x t C Coswt C Sinwt

For particular solution we ass

γ°+ =

+ =

′′= =

+ == ±

= +

2 2

2 2 2 2 2

2 2 2 2 2

2 2

( )

( )

( )

( ) ( )

int

( )

p

p

p

p p

p p

ume

x t ACos t BSin t

x t A Sin t B Cos t

x t A Cos t B Sin t

x w x A Cos t B Sin t Aw Cos t Bw Sin t

x w x A w Cos t B w r Sin t

Putting o ginven differential equation

A w C

γ γγ γ γ γ

γ γ γ γ

γ γ γ γ γ γ

γ γ γ

γ

= +′ = − +

′′ = − −

′′ + = − − + +

′′ + = − + −

− 2 2

2 2 2 2

2 2

1 2 2 2

1

2 2

( )

( ) ; ( ) 0

( ) ( )

( )

0

( )

( )

p

os t B w Sin t F Cos t

Comparing Coefficients

A w F B w

Fx t Cos t

w

Hence the general solution of the differential equation is

Fx t C Coswt C Sinwt Cos t

w

x t C wSin

FA B

w

ο

ο

ο

γ γ γ γ

γ γ

γ

γ

γ

γγ

°

°

+ − =

− = − =

=−

= + +−

′ = −

= =−

∵

2 2 2

1 2 2 2

1 2 2

1 2 2

( )

(0) 0

(0) .1 .0 .1

0

Fwt C wCoswt Sin t

w

Now we apply boundary value condition

x

Fx C C

w

FC

w

FC

w

ο

ο

ο

ο

λ γγ

γ

γ

γ

+ −−

=

= + +−

= +−

= −−

Now x′ (0) = 1 2 2 2.(0) .1 .(0)

FC w C w

w γ°− + −

−

0 = wC2

02 =⇒ C

2 2 2 2( ) ( )

The solution of initial value is

F Fx t Coswt Cos t

w wγ

γ γ° °= − +

− −

∵

Question#03 Solve by using (tx e= )

22 2

24 4 3 sin(ln( )) , 0

d y dyx x y x x where x

dx dx− + = − <

Solution. Consider the associated homogeneous equation

22

24 4 3 0

d y dyx x y

dx dx− + =

The differential equation can be written as:

2

2 2

4 4 3 0

(4 4 3) 0

x y xy y

x D xD y

′′ ′− + =− + =

Where

( )

22

2

2 2

,

ln( )

1

t

d dD D

dx dxPut

x e then t x

Let

xD

xD

= =

− = = −

= ∆

= ∆ ∆ − = ∆ − ∆

2

2 2

2 2

(4 ( 1) 4 3) sin

(4 4 4 3) sin

(4 8 3) sin

t

t

t

y e t

y e t

y e t

∆ ∆ − − ∆ + =∆ − ∆ − ∆ + =∆ − ∆ + =

( ) ( )( )( )( ) ( )

2

2

Hence the auxiliary equation is

4 8 3 0

4 6 2 3 0

2 2 3 1 2 3 0

2 3 2 1 0

2 3 0 2 1 0

2 3 2 1

3 1

2 23 1

,2 2

∆ − ∆ + =∆ − ∆ − ∆ + =∆ ∆ − − ∆ − =

∆ − ∆ − =

∆ − = ∆ − =∆ = ∆ =

∆ = ∆ =

∆ =

1 32 2

1 21 32 2

1 21 32 2

1 2

( (

( (

) )

) )

c

t tc

c x x

t ty c e c e

y c e c e

y c c

+

+

− + −

=

=

=

Now for particular solution.

( ) ( )

( )( )( )

( )

( )

22

2

2

22

22

2

2

2 2 22

2

1sin

4 8 3sin

.sin4 2 8 2 3

1. sin4 16 16 8 16 3

1. sin4 8 3

1

sin.

4 8 38 1 sin 8 1 sinsin

. . .8 1 8 1 8 1 64 1

1 sin.

tp

t

p

t

t

t

t t t

t

y e t

u g shifting theorm

ey t

e t

e t

putting

te

t tte e e

te

=∆ − ∆ +

=∆ + − ∆ + +

=∆ + ∆ + − ∆ − +

=∆ + ∆ +

∆ = −

=− + ∆ +

∆ + ∆ += = =

∆ − ∆ − ∆ + ∆ −

∆ += ( ) ( ) ( )

( )

( ) ( )( )

2 22

2

1 3 22 2

1 2

8 1 sin. . 8 1 sin 8cos sin

64 1 65 65 65

8cos ln( ) sin ln( ) , ln( )65

( ( 8cos ln sin ln65

) )

t tt

c p

t e ee t t t

xx x t x

y y y

xx x x xy c c

∆ += = − ∆ + = − +

− − −

= − − + − = −

= +

− + − − − + −=

Assignment No. 6

Q#01:- Find solution of the D.E(differential equation)

/ 1y y+ =

in the form of a powers series in x.

Answer.

/

nn

n=0

' n-1n

n=1

'

n-1 nn n

n=1 n=0

n+1 nn=0 n=0

1 0

1 0

n+1 n

1

y = c x

. . .

y c nx

1

c nx c x =1

Put n=n+1

(n+1)c c 1

c +c 1

1 c

(n+1) c +c

n n

y y

Let

Diff w r t x

y y

or x x

Equating the co efficient liketerms

c

and

∞

∞

∞ ∞

∞ ∞

+ =

=

+ =

+

+ =

−=

= −=

∑

∑

∑ ∑

∑ ∑

n+1 n

0

1 c c

n+1 n=1,2,3...put

= −

0 012 1 0

023

04

230

0 0 0

(1 ) 1c 1

2 2 2!1

3 3!1

4!...

...

...

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

2! 3!

c ccwhere c c

ccc

cc

Thus the required solution is

cxy c c x c x

− −= − = − = = −

−−= = −

−=

−= − − + − − +

Q#02:- Whether 5x = ± are singular points of the equation? 2 2 / / /( 25) ( 5) 0x y x y y− + − + =

Answer.

( ) ( )

( ) ( ) ( )

( ) ( ) ( ) ( )

( )( ) ( ) ( )

( )( ) ( )

( )( ) ( )

( )

22

2 2 22

2 2 2 2

2 2 2

2

2 2

- 25 -5 0

- 25 -5 5

-5 10

-5 5 -5 5

1 10

-5 5 -5 5

1

-5 5

1

-5 5

5 sin int -5

x y x y y

Dividing the equation by x x x

xy y y

x x x x

y y yx x x x

where

P xx x

Q xx x

x is regular gular po beacuse power of x in P x

′′ ′+ + =

= +

′′ ′+ + =+ +

′′ ′+ + =+ +

=+

=+

= ( )( )

1 2.

-5 sin int 5 2.

is and in q x is

x is an irregular gular po beacuse power of x in P x is= +

Q#03:- Find the general solution of the equation / / / 4 0x y y y+ − =

( By using Frobenius method.)

Answer.

/ / / 4 0x y y y+ − = Let

0

n rn

n

y c x∞

+

=

=∑

( ) ( ) ( )

/ / /

1 1

0 0 0

4 0

1 4 0n r n r n rn n n

n n n

x y y y

n r n r c x n r c x c x∞ ∞ ∞

+ − + − +

= = =

+ − =

+ + − + + − =∑ ∑ ∑

( )2 1

0 0

4 0n r n rn n

n n

n r c x c x∞ ∞

+ − +

= =

+ − =∑ ∑

( )22 1 10

0 0

4 0r n nn n

k n

x r c x n r c x c x∞ ∞

− −

= =

+ + − =

∑ ∑

( ){ }22 10 1

0

1 4 0r kk k

k

x r c x k r c c x∞

−+

=

+ + + − =

∑

2 0r = , so the indicial roots are equal

&

1 2 0r r= =

( )2

11 4 0 0,1,2,....k kk r c c k++ + − = =

( )1 2

40,1,2,...

1k

k

cc k

k+ = =

+

( )1 0 20

4

!

nn

n

y c xn

∞

=

= ∑

To obtain the 2nd linearly independent solution we set:

0 1c =

( ) ( ) ( )1

2 1 1 221 2 3

1

161 4 4 ...

9

dxxe

y y x dx y x dxy x

x x x x

− ∫

= = = + + + +

∫ ∫

( )12 316

1 8 24 ...9

dxy x

x x x x=

+ + + +

∫

( ) 2 31

1 14721 8 40 ...

9y x x x x dx

x = − + − + ∫

( ) 21

1 14728 40 ...

9y x x x dx

x = − + − + ∫

( ) 2 31

1472ln 8 20 ...

27y x x x x x

= − + − +

( ) ( ) ( ) 2 31 1 2 1 1

1472ln 8 20 ...

27c y x c y x x y x x x x

= + + − + − +

Assignment # 07 Q#01 :- Solve the Bessel function in terms of 7

2

sin cosx and x of the J

Solution:

( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( )

( )

( )

1 1

52

5 5 51 12 2 2

3 7 52 2 2

7 5 32 2 2

52

3 / 2

72

C o n s i d e r t h a t

2

52 2

5

5(1)

3 s i n 2 2 2( . c o s ) s i n

s i n 2 2( ) . c o s

(1)

5

v v v

t a k i n g v

vJ x J x J x

xA 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

xJ x x x

x x x x x

xJ x x

x x xT h e n e q u a t i o n b e c o m e s

J x

π π π

π π

− +

=

− +

+ =

+ =

+ =

= − − − − −

= − −

= −

=2

2 3 2 3 2 s i n[ 1 s i n . ( c o s ) ] c o s

xx x x

x x x x x xxπ π π − − − −

Q#02 :- Find the general solution of the given differential equation on(0,∞ )

22 2

216 16 (16 1) 0

d y dyx x x y

dx dx+ + − =

Solution: The Bessel differential equation is

22 2 2

2( ) 0

d y dyx x x v y

dx dx+ + − = -----(1)

2

2 22

22 2

2

22 2 2

2

16 16 (16 1) 0

16.

1( ) 0

16

1( ( ) ) 0 (2)

4

d y dyx x x y

dx dxDividing by

d y dyx x x y

dx dx

d y dyx x x y

dx dx

+ + − =

+ + − =

+ + − = − − − −

Comparing (1) and (2) , we get

2 1 1

16 4v v= ⇒ = ±

so the general solution is

1 1 2 1

4 4

( ) ( )y c J x c J x−

= +

Q#03 :- Reduce the third order equation

/ / / / / /3 9 6 cosy y y y t= − − + + to the normal form solution:-

/ / / / / /

/ / / / / /

3 9 6 cos

3

1 13 2 cos

3 3int var

y y y y t

Dividing by

y y y y t

Now roduce the iables

= − − + +

= − − + +

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

=

=

=

= =

= =

=

= − − + +

Assignment No. 8 Question#01: Find the eigen values and eigen vectors of the following matrix

1 2 2

2 1 2

2 2 1

A

− = − − −

solution.

3 2

2

12

2

1 2 2

2 1 2 0

2 2 1

exp .

3 9 5 0

sin .

1, 4 5 0

1, 1, 5

5

4 2 2

5 2 4 2

2 2 4

2 4 2

4 2 2

2 2 4

2 4 2

0 6 6

0 6 6

A I

After ansion

By u g synthetic division

For

A I

R

R

λλ λ

λ

λ λ λ

λ λ λλ λ

λ

− −− = − − =

− − −

− + + + =

= − − − == − = −

=− − − = − − − − −

− − = − − − − −

− − = − − − −

1 3 1

1 2 3

3 2

2 ,

1 2 11 1 1

0 1 1 , ,2 6 6

0 1 1

1 2 1

0 1 1

0 0 0

R R R

R R R

R R

+ +

− − = − −

− − = −

1

2

3

1 2 3

2 3 2 3

2 3

1

1

( 5 ) 0

1 2 1 0

0 1 1 0

0 0 0 0

2 0

0

,

0

1

1

1

A I v

x

x

x

x x x

x x x x

x a x a

x

a

v a a

a

− =− −

=

− − =+ = ⇒ − == = −=

= − −

2 1 3 1

1 2 3

1 2 3

1

2 2 2

( 1) 2 2 2

2 2 2

2 2 2

0 0 0 ,

0 0 0

2 2 2 0

, ,

For

A I

R R R R

x x x

x a x b x a b

λ = −−

− − = − − −

− = − +

+ − =

= = = +

1

2

3

1 0

0 1

1 1

x a

x b

a bx

a b

= +

= +

Question#02 Solve the homogeneous system of differential equations

2

dxx y

dtdy

x ydt

= +

= − −

solution. The given system can be written in matrix form

1

1

1 2 2 1

1 2

1

1 1

2 1

1 10

2 1

exp

1 1( ) 0

2 1

(1 ) 0 (1 )

1 (1 )

1

1

dxxdt

dy y

dt

A I

After ension

i

For i

kiA I K

i k

i k k k i k

choose k then k i

ki

λλ

λ

λλ

λ

= − −

−− = =

− − −

= ±= −

+ − = = − − +

= + + = ⇒ = − += = − +

= − −

1

1

1 2 2 1

1 2

2

1 2

1 2

1 1( ) 0

2 1

(1 ) 0 ( 1)

1 ( 1)

1

1

1 1,

1 1

1 1

1 1

it it

it it

For i

kiA I K

i k

i k k k i k

choose k then k i

ki

X e X ei i

X C e C ei i

λ

λ

−

−

=−

− = = − − −

= − + = ⇒ = −= = −

= −

= = − − −

= + − − −