question bank engineering mathematics- ii

36
JAHANGIRABAD INSTITUTE OF TECHNOLOGY ENGINEERING MATHEMTICS-II (QUESTION BANK) NUMERICAL PROBLEMS WITH THEORY PREPRAED BY MOHAMMAD IMRAN (ASSISTANT PROFESSOR, JIT) E-mail: [email protected] Website: www.jit.edu.in Mobile no 9648588546

Upload: mohammad-imran

Post on 21-Apr-2017

153 views

Category:

Engineering


21 download

TRANSCRIPT

Page 1: Question bank Engineering Mathematics- ii

JAHANGIRABAD INSTITUTE

OF

TECHNOLOGY

ENGINEERING

MATHEMTICS-II

(QUESTION BANK)

NUMERICAL PROBLEMS WITH

THEORY

PREPRAED BY

MOHAMMAD IMRAN

(ASSISTANT PROFESSOR, JIT)

E-mail: [email protected]

Website: www.jit.edu.in

Mobile no 9648588546

Page 2: Question bank Engineering Mathematics- ii

DIFFERENTIAL EQUATION

Linear Differential Equations

Definition: A differential equation in which independent variable and only its first derivative involved

Called “Linear Differential Equation”.

1 2 3

1 2 3 11 2 3

n n n n

n nn n n n

d y d y d y d y dya a a a a y Q

dx dx dx dx dx

− − −

−− − −+ + + + − − − − − − − + + =

Above equation is an example of linear Differential equation of order n. Where 1 2 1, ,............ na a a −

all are constant and Q is any function of x which is called Differential equation with constant coefficients.

Different notations of Differential operator:

First derivative : ', ,

dD y

d x

Second Derivative: 2

2 ''

2, ,

dD y

d x

Third Derivative :

33 ' ' '

3, ,

dD y

d x

.

.

.

.

Nth

Derivative : , ,n

n n

n

dD y

dx

Order and Degree of A differential Equation :

• The highest derivative in the given differential equation called the “Order of the

Differential Equation”.

For Example consider a differential equation

32

22 0

d y dyy

dx dx

+ + =

the order of the differential equation is 2.

• In a differential Equation After made free from the radicals and fraction the degree of

the highest derivative called “Degree of the Differential equation”.

Page 3: Question bank Engineering Mathematics- ii

For Example consider a differential equation

32

22 0

d y dyy

dx dx

+ + =

the Degree of the differential equation is 1.

Ordinary and Partial Differential Equations :

Differential equation in which unknown function is depends on only one independent

variable called “Ordinary Differential Equation”.

Differential equations in which two or more independent variables are involved called “Partial

Differential Equation”

Linear Differential Equation

Differential equation in which dependent variables and all its first degree derivatives involved called

“Linear Differential Equation”

dy dyP Qy R

dx dx+ + =

Where P, Q are constant and R is the function of x only.

Linear Differential Equation of second order with constant coefficients :

2

2

d y dyP Qy R

dx dx+ + =

Where P, Q are constant and R is the function of x only.

Non-Linear differential Equation :

A Differential Equation in which dependent variable and its derivative Having degree more

than 1 called “Non-Linear differential Equation”.

Auxiliary Equation :

Suppose we have a Differential Equation as

2

23 4 0

d y dyy

dx dx− − = ----------------------------------------------------- (1)

Page 4: Question bank Engineering Mathematics- ii

Equation (1) we can rewrite as

2 3 4 0D y Dy y− − =

� ( )2 3 4 0D D y− − =

On equating the coefficient of y with 0 we will get 2 3 4 0D D− − = Which is called “Auxiliary

Equation”.

Solution of different type of Differential Equations :

When given differential equation not having any function ( having “0” ) in right hand side

1

1 210

n n

n n

d y d ya a y

dx dx

−− − =

Case –I : when all the roots of Auxiliary Equation are distinct

Let the roots of Auxiliary Equation are 1 2 3, , .....................

nm m m m than the general solution will be

1 2 3

1 2 3 ............... nm x m x m x m x

ny c e c e c e c e= + + + +

Case –II : when all the roots of Auxiliary Equation are equal than the general solution will be

( )2 1

1 2 3 ...............m xn

ny c c x c x c x e−= + + + +

Where m is equal roots of Auxiliary Equation

Case –III: when some roots of Auxiliary Equation are equal and some are different than the

General Solution will be

( ) ( )32 42 1

1 2 3 1 2 3............... ............... nm x m xm xm x m xn

n n n n n ny c c x c x c x e c e c e c e c e

−+ + + += + + + + + + + + +

Case –IV : when roots of Auxiliary Equation are imaginary as iα β± than

( )1 2cos sinxy e c x c xα β β= +

Case –V : when roots of Auxiliary Equation are imaginary and repeated twice than

{ } { }( )1 2 3 4cos sinxy e c c x x c c x x

α β β= + + +

Case –VI : when roots of Auxiliary Equation are Irrational as α β± where β is positive than

Page 5: Question bank Engineering Mathematics- ii

( )1 2cosh sinhxy e c x c xα β β= +

Case – V: if ,i iα β α β± ± be the two equal pairs of imaginary roots and all other roots real and

Different.

( ) 5

1 2 3 4 5cos ( )sin ..................... nm x m xax

ny e c c x x c c x x c cβ β = + + + + + +

Some useful expansions :

(i) 1 2 3 4 5(1 ) 1 ..............................D D D D D D−− = + + + + + + ∞

(ii) 1 2 3 4 5(1 ) 1 ..............................D D D D D D−+ = − + − + − + ∞

(iii) 2 2 3 4(1 ) 1 2 3 4 5 .............................D D D D D−+ = − + − + − ∞

(iv) 2 2 3 4(1 ) 1 2 3 4 5 .............................D D D D D−− = + + + + + ∞

(v) 3 2 3(1 ) 1 3 6 10 .............................D D D D−− = + + + + ∞

(vi) 3 2 3(1 ) 1 3 6 10 .............................D D D D−+ = − + − + ∞

� Remark :

When given differential equation not having any function ( having “0” ) in

right hand side then there is no need to find out particular integral (P.I) so general

solution will be � = ����������������(�. �)

Question : find the general solution of the differential equation 5 3

5 30

d y d y

dx dx− = (2009)

2

1 2 3 4 5( ) x xy c c x c x c e c e

−= + + + + Ans

Question : find the particular integral of ( )2 4 4 sin 2D D y x− − = (UPTU2010-11

cos 2.

8

xP I = Ans

Question : Find the particular integral of

22

22 1

d yx x

dx= + − (uptu 2013-14

Page 6: Question bank Engineering Mathematics- ii

4 3 22.

12 6 2

x x xP I

= + −

Ans

Question : Find the particular integral of

22

2

d yy x

dx− = (UPTU 2012

2

1 2. 2x x

C S C e C e x− = + − + Ans

Question: solve the differential equation (UPTU 2009-10

2

25 6 sin 3 cos 2

d y dyy x x

dx dx+ − = +

Question : solve

22 3

22 xd y dy

y x edx dx

− + = UPTU 203-14

( )3

2

1 2. 2 16 12816

xx e

C S C C x e x x x = + + + − + Ans

Question : solve

3 22

3 26 1

d y d y dyx

dx dx dx− − = + UPTU 2001

3 23 2

1 2 3

1 25

6 18 3 6

x x x xCS c c e c e x

− = + + − + −

Ans

Question : if ( )2

20; ,

d x gx a a b and g

dt b+ − = are positive numbers and

' ; 0 0dx

x a whentdt

= = = , show that

( )' cosg

x a a a tb

= + − (UPTU 2007

1 2cos sing g ag b

CS c x c xb b b g

= + +

Ans

Question : if

3 2

3 23 4 2 0

d y d y dyy

dx dx dx− + − =

(uptu 2001

Question : obtain the general solution of the differential equation'' '2 2 cosxy y y x e x− + = + uptu 2002

Ans ( ) [ ]1 2

1. cos sin 1 sin

2 2

xx xe

C S e c x c x x x= + + + +

[ ] [ ]6

1 2

1 1. cos3 sin 3 sin 2 cos 2

30 20

x xC S C e C e x x Ax x ns

−= + − + + −

Page 7: Question bank Engineering Mathematics- ii

Question : find the complete solution of

23

23 2 sin 2

xd y dyy xe x

dx dx− + = + (uptu 2003

( )3

2

1 2

3 1. 3cos 2 sin 2

2 2 20

xx x e

C S c e c e x x x

= + + + + −

Question : solve the differential equation ( )2 21 cos ( 2007xD y xe x uptu− = +

Question : find the complete solution of

2

23 2 sin 2

xd y dyy xe x

dx dx− + = + (uptu 2001

Question : solve

22

2sin

d ya y ax

dx+ = ( UPTU 2008

1 2

cos. cos sin

2

x axC S c ax c ax

a= + −

Question : solve

3 2

3 23 4 2 cos

xd y d y dyy e x

dx dx dx− + − = + (UPTU 2001, 2006

Ans ( )1 2 3

3sin cos. cos sin

10

x x x xC S e c x c x c e

= + + +

Question : solve ( )24 sin 3 cos 2D x x+ = + ( UPTU 2008

Ans ( ) ( )1 2

1 sin 2. cos 2 sin 2 sin 3

5 4

x xC S c x c x x

= + − +

Question : solve the differential equation ( )4 3 2 22 3 3 4sinxD D D y e x+ − = + (UPTU2008

Ans ( ) ( ) [ ]0 3 2

1 2 3 3

3 2. cos 2sin

20 5

x x x xC S c c x e c e c e e x x

−= + + + + − −

Question : solve

3 23 2

3 23 log

d y d y dyx x x y x x

dx dx dx+ + + = + UPTU 2001

1

1 21 2 3

3 3 1cos log sin log log

2 2 2y c x x c x c x x x

= + + + + Ans

Question : solve ( )2

2

2log sin log

d y dyx x y x x

dx dx+ + = UPTU 2002

Page 8: Question bank Engineering Mathematics- ii

Ans ( ) ( ) ( ) ( ) ( ) ( )2

1 2

1 1. cos log sin log log sin log log cos log

4 4C S c x c x x x x x= + + −

Question : solve the following simultaneous differential equation (UPTU 2011-12, 2013-14

( )4 , 4 4 ,dx dx dy

x y ydt dt dt

= − + + = − with condition (0) 1x = (0) 0y = Ans ( ) 2

1 2

tx c c x e−= + 2( ) ty t te−=

Question: solve the simultaneous equations UPTU 2008

3 2 , 5 3dx dy

x y x ydt dt

= + = + ( ) ( )3 10 3 10

1 2

t tAns x c e c e

+ −= +

( ) ( )( )3 10 3 10

1 2

110 10

2

t ty c e c e

+ += −

QUESTION: Question: solve the simultaneous equations UPTU 2008

5 2 , 2 0dx dy

x y t x ydt dt

+ − = + + =

Ans ( ) 3

1 2

1

9 27

t tx c c t e

−= + + + 3 3 321 2

4 2

2 27 9

t t tc ty c e c te e

− − − = + + + −

1

1

27c = − 2

6 2

27 9c = − = −

Question : The equations of the motions of a particle are given by UPTU 2009

0, 0dx dy

wy wxdt dt

+ = − =

Find the path of the particle and show that it is a circle.

Ans 1 2cos sinx c wt c wt= + ,

1 2sin cosy c wt c wt= − , 2 2 2x y r+ = Which is a equation of circle

Question : solve the following differential equation (uptu 2009

1, 1dx dy

y xdt dt

= + = +

Ans 1 2 1t t

x c e c e−= + − ,

1 2 1t ty c e c e

−= − −

Question: Solve the simultaneous differential equations uptu 2005

2 2

2 23 4 0, 0

d x d yx y x y

dt dt− − = + + =

( ) ( )

( ) ( )

1 2 2 3 4 4

1 2 3 4

2 2 2 2 2 2t t

t t

x c c c t e c c c t e

y c c t e c c t e

= − + + + + +

= + + + Ans.

Question: Solve the simultaneous differential equations uptu 2001

Page 9: Question bank Engineering Mathematics- ii

2 2

2 23 , 4 3 sin 2td x dx d y dx

x e y tdt dt dt dt

+ + = − + =

1 2 3 4

1 2 1

2 3 4

2cos sin cos3 sin 3 cos 2

5 15

1cos 3 sin 3 sin 2 2 sin

15

12 cos 2 sin 3 2 cos3

5

t

t

ex c t c t c t c t t

y c t c t t c t

c t c t c t e

= + + + + +

= + + −

+ + − −

Question: Solve the simultaneous differential equations uptu 2008

4 3 , 2 5tdx dy

x y t x y edt dt

+ + = + + =

2 7

1 2

2 7

1 2

5 31 1

14 196 8

2 1 9 5

3 7 98 24

t t t

t t t

x c e c e t e

y c e c e t e

− −

− −

= + + − −

= − + − + +

Ans.

Question: Solve the simultaneous differential equations uptu 2003

3 sin , cosDx Dy x t Dx y x t+ + = + − =

( )

( )

3

1 2

3

1 2

12sin cos

5

12cos sin 2 2

5

t t

t t

x c e c e t t

y t t c e c e

= + + −

= + − +

Ans.

Question: Solve the simultaneous differential equations uptu 2001

2 2

2 24 4 , 4 4 25 16 td x dx d y dy

x y y x edt dt dt dt

− + = + + = +

Ans

3 3

1 2 3 4 3 4

3 3

1 2 3 4

25 (3 4 )cos (4 3 )sin

cos sin

t t t

t t t

y c e c e c c t c c t e

x c e c e c t c t e

= + + − + + + −

= + + + −

Question solve

1/222

21

d y dy

dx dx

= −

uptu 2002

Ans 1cos( )y x c c= − + +

Question : Solve

22

20

d y dy dyx x

dx dx dx

+ − =

uptu 2010

Ans 2

1 2log( 2 )y x c c= + +

Page 10: Question bank Engineering Mathematics- ii

Question : solve 2" 4 ' (4 2) 0y xy x y− + − = given that 2

xy e= is an integral included in the complementary

function . uptu 2004

Ans 2

1 2( )xy e c x c= +

Page 11: Question bank Engineering Mathematics- ii

DEFINITION:

Infinite Series of Trigonometric functions

periodic function, used in Fourier analysis.

EVEN FUNCTION:

Let f(x) be a real-valued function of a real variable.

Then f is even if the following equation holds for all

x and -x in the domain of f.

( ) ( )f x f x− =

or ( ) ( ) 0f x f x− − =

Geometrically speaking, the graph face of an even function

is symmetric with respect to the

graph remains unchanged after

about the y-axis.

Examples of even functions are

or any linear combination of these.

ODD FUNCTIONS

Again, let f(x) be a real-valued function of a real variable.

Then f is odd if the following equation holds for all

x and -x in the domain of f:

( ) ( )f x f x− = −

or

( ) ( ) 0f x f x− + =

Geometrically, the graph of an odd function has rotational

symmetry with respect to the origin

remains unchanged after rotation

origin.

Examples of odd functions are x

any linear combination of these.

FOURER SERIES

Trigonometric functions which represents an expansion or approximation of a

periodic function, used in Fourier analysis.

valued function of a real variable.

if the following equation holds for all

Geometrically speaking, the graph face of an even function

with respect to the y-axis, meaning that its

remains unchanged after reflection

Examples of even functions are |x|, x2, x4, cos(x), cosh(x),

y linear combination of these.

valued function of a real variable.

if the following equation holds for all

an odd function has rotational

origin, meaning that its graph

rotation of 180 degrees about the

x, x3, sin(x), sinh(x), erf(x), or

any linear combination of these.

which represents an expansion or approximation of a

Page 12: Question bank Engineering Mathematics- ii

Properties involving addition and subtraction

• The sum of two even functions is even, and any constant multiple of an even function is even.

• The sum of two odd functions is odd, and any constant multiple of an odd function is odd.

• The difference between two odd functions is odd.

• The difference between two even functions is even.

• The sum of an even and odd function is neither even nor odd, unless one of the functions is equal to zero

over the given domain.

Properties involving multiplication and division

• The product of two even functions is an even function.

• The product of two odd functions is an even function.

• The product of an even function and an odd function is an odd function.

• The quotient of two even functions is an even function.

• The quotient of two odd functions is an even function.

• The quotient of an even function and an odd function is an odd function.

Properties involving composition

• The composition of two even functions is even.

• The composition of two odd functions is odd.

• The composition of an even function and an odd function is even.

• The composition of any function with an even function is even (but not vice versa).

Other algebraic properties

• Any linear combination of even functions is even, and the even functions form a vector space over the reals.

Similarly, any linear combination of odd functions is odd, and the odd functions also form a vector space

over the reals. In fact, the vector space of all real-valued functions is the direct sum of the subspaces of even

and odd functions. In other words, every function f(x) can be written uniquely as the sum of an even function

and an odd function:

( ) ( ) ( )e of x f x f x= +

where

[ ]1

( ) ( ) ( )2

ef x f x f x= + −

is even and

[ ]1

( ) ( ) ( )2

of x f x f x= − −

is odd. For example, if f is exp, then fe is cosh and fo is sinh.

• The even functions form a commutative algebra over the reals. However, the odd functions do not form an

algebra over the reals, as they are not closed under multiplication.

Page 13: Question bank Engineering Mathematics- ii

Important questions

1. find the Fourier series for the function 2

( )4

xf x x xπ π= + − ≤ ≤ uptu 2009

2. find the Fourier series for the function ( ) sin 0 2f x x x x π= ≤ ≤ uptu 2001

3. find the Fourier series for the function 0

( )0

k xf x

k x

π

π

− −=

≺ ≺

≺ ≺

uptu 2010

hence show that 1 1 1

1 ........3 5 7 4

π− + − + =

4. find the Fourier series for the function 0

( )0

x xf x

x x

π

π

− < <=

− < < uptu 2002, 2008

Hence show that

2

2 2 2 2

1 1 1 1........

1 3 5 7 8

π− + − + =

5. find the Fourier series for the function 2

( )4

xf x x xπ π= + − ≤ ≤ uptu 2009

6. find the Fourier series for the function 2( )f x x x xπ π= + − ≤ ≤ uptu 2003

Deduce that

2

2 2 2 2

1 1 1 1........

6 1 2 3 4

π= + + + +

7. find the Fourier series for the function 0

( ) ( 2 ) ( )0

x xf x and f x f x

x x

π ππ

π π

+ − < <= + =

− − < <

uptu 2006

8. find the Fourier series for the function 1 0

( ) ( 2 ) ( )1 0

xf x and f x f x

x

ππ

π

− − < <= + =

< <

9. find the Fourier series for the function ( ) sinf x x x xπ π= − ≤ ≤ uptu 2008

Deduce that 1 1 1 1 2

........1.3 3.5 5.7 7.9 4

π −+ + + + =

10. find the Fourier series for the function ( ) cosf x x x xπ π= − ≤ ≤ uptu 2008

11. Obtain the half range sine series for the function 2( )f x x= in the interval 0 3x< < uptu 2008

12. Find the half range cosine series for the function 2 0 1

( )2(2 ) 1 2

t xf x

t x

< <=

− < < uptu 2001, 06, 07

13. Obtain the fourier cosine series expansion of the periodic function define by the function uptu 2001

( ) sin , 0 1t

f t tl

π = < <

14. Find the half range sine series for the function ( )f x given in the range (0,1) by the graph OPQ as shown in

the figure. Uptu 2009

Page 14: Question bank Engineering Mathematics- ii

The Laplace Transformation

Pierre-Simon Laplace (1749-1827)

Laplace was a French mathematician, astronomer, and physicist who applied the Newtonian theory of

gravitation to the solar system (an important problem of his day). He played a leading role in the

development of the metric system.

The Laplace Transform is widely used in engineering applications (mechanical and electronic),

especially where the driving force is discontinuous. It is also used in process control.

What Does the Laplace Transform Do?

The main idea behind the Laplace Transformation is that we can solve an equation (or system of

equations) containing differential and integral terms by transforming the equation in "t-space" to

one in "s-space". This makes the problem much easier to solve. The kinds of problems where the

Laplace Transform is invaluable occur in electronics. You can take a sneak preview in

the Applications of Laplace section.

If needed we can find the inverse Laplace transform, which gives us the solution back in "t-space".

The Unit Step Function (Heaviside Function)

In engineering applications, we frequently encounter functions whose values change abruptly at

specified values of time t. One common example is when a voltage is switched on or off in an

electrical circuit at a specified value of time t.

The value of t = 0 is usually taken as a convenient time to switch on or off the given voltage.

Page 15: Question bank Engineering Mathematics- ii

The switching process can be described mathematically by the function called the Unit Step

Function (otherwise known as the Heaviside function after Oliver Heaviside).

The Unit Step Function

Definition: The unit step function, u(t), is defined as

0 0{ }

1 0

tu t

t

<=

>

That is, u is a function of time t, and u has value zero when time is negative (before we flip the

switch); and value one when time is positive (from when we flip the switch).

Laplace Transformation

Question : find the Laplace transform of the function 1, 1 2

( )3 , 2 3

t tf t

t t

− < <=

− < < uptu 2009

Ans 2 3

2

12

s s se e e

s

− − − − +

Question: find the Laplace transform of the function

2 , 1 2

( ) 1, 2 3

7 3

t t

f t t t

t

< <

= − < < >

uptu 2007

Ans ( ) ( )2 3

2

3 3 2

22 3 3 5 1

s se e

s s ss s s

− −

− + + + −

Question: if ( )( )

22

2

2cos

4

sL t

s s

+=

+, find 2(cos )L at Ans

( )

2 2

2 2

2

4

s a

s s a

+

+ uptu 2006

Question: if { }( ) ( ),L f t F s= then prove that { } ( ) [ ]( ) 1 ( )n

nn

n

dL t f t F s

ds= − uptu 2005

Question: Obtain the laplace transform of 2. .sin 4tt e t Ans

( )( )

2

32

4 3 6 13

2 17

s s

s s

− −

− + uptu 2002

Page 16: Question bank Engineering Mathematics- ii

Question: find the laplace transform of the function ( ) . .sin 2tf t t e t−= Ans ( )

( )2

2

4 1

1 4

s

s

+

+ +

uptu 2002

Question: if { }( ) ( ),L f t F s= then prove that 1

( ) ( )s

L f t F s dst

=

∫ uptu 2005, 07

Question: find the laplace transform of the 0

sin( )

tat

f t dtt

= ∫ uptu 2004

Question: find the laplace transform of the cos cos

( )at bt

f tt

−= Ans

2 2

2 2

1log

2

s b

s a

+

+ uptu 2004

Question: if cos

( )at

e btf t

t

−= , find the laplace transform of ( )f t Ans

( )

2 2

2

1log

2

s b

s a

+

+

uptu 2003

Question: find { }L erf t and hence prove that { }( )

3/22

3 8. 2

4

sL t erf t

s s

+=

+ uptu 2001

Question : Express the following function in terms of unit step function : 1, 1 2( )

3 , 2 3

t tf t

t t

− < <=

− < <

uptu 2009

Ans

2 3

2 2 22

s s se e e

s s s

− − −

− +

Question : Express the following function in terms of unit step function :

0, 0 1

( ) 1, 1 2

1 2

t

f t t t

t

< <

= − < < >

uptu 2002

Ans

2

2 2

s se e

s s

− −

Question : find the laplace transform of the function , 0

( ), 2

t tf t

t t

π

π π π

< <=

− < <

uptu 2003

Page 17: Question bank Engineering Mathematics- ii

Ans ( )2

1

1

s s

s

se e

s e

π π

π

π − −

− + −

+

Question : find the laplace transform of the function

sin , 0

( )2

0,

t t

f t

t

πω

ωπ π

ω ω

< <

= < <

uptu 2010

Ans

( )2 2 1s

s e

π

ω

ω

ω−

+ −

Question : Draw the graph of the following periodic function and find the Laplace transform of the function

, 0( ) 2002

2 , 0 2

t for t af t uptu

a t for t a

< ≤=

− < <

Question : state and prove Convolution theorem.

OR

If [ ] [ ] [ ]1 1 1

1 1 2 2 1 2 1 2

0

( ) ( ) ( ) ( ) ( ). ( ) ( ). ( )

t

L F s f t and L F s f t then prove that L F s F s f x f t x dx− − −= = = −∫

Solution/proof : By using the definition we can write

1 2 1 2

0 0 0

( ). ( ) ( ). ( )

t t

stL f x f t x dx e f x f t x dx dt

∞−

− = −

∫ ∫ ∫

1 2 1 2

0 0 0

( ). ( ) ( ). ( )

t t

stL f x f t x dx e f x f t x dxdt

∞−

− = −

∫ ∫ ∫

By using the change of order the above equation reduced as

1 2 1 2

0 0

( ). ( ) ( ). ( )

t

st

s

L f x f t x dx e f x f t x dxdt

∞ ∞−

− = −

∫ ∫ ∫

On adding and subtracting x in the power of e we will get

Page 18: Question bank Engineering Mathematics- ii

( )

( )

( )

1 2 1 2

0 0

1 2 1 2

0 0

1 2 1 2

0 0

( ). ( ) ( ). ( )

( ). ( ) ( ). ( )

( ). ( ) ( ) . ( )

t

s t x x

s

ts t x x

s

ts t xsx

s

L f x f t x dx e f x f t x dxdt

L f x f t x dx e f x f t x dxdt

L f x f t x dx e f x dx e f t x dt

∞ ∞− − +

∞ ∞− − +

∞ ∞− −−

− = −

− = −

− = −

∫ ∫ ∫

∫ ∫ ∫

∫ ∫ ∫

Let t x z dt dz− = ⇒ =

Now the above equation reduced as

[ ]

1 2 1 2

0 0 0

1 2 1 2

0

1

1 2 1 2

0

( ). ( ) ( ) . ( )

( ). ( ) ( ). ( )

( ). ( ) ( ). ( )

t

sx sz

t

t

L f x f t x dx e f x dx e f z dt

L f x f t x dx F s F s

L F s F s f x f t x dx proved

∞ ∞− −

− =

− =

= −

∫ ∫ ∫

Question : Use the convolution theorem to evaluate

( )1

22 4

sL

s

+

uptu 2010 Ans 1

sin 24

t

Question : Use the convolution theorem to find the inverse of the function

( )2

2

s

s a+ uptu 2009

Ans [ ]3

1sin cos

2at at at

a−

Question : using the Convolution theorem find ( ) ( )

21

2,

2 2

sL a b

s s

≠ + −

uptu 2006,04

Ans2 2

sin sina at b bt

a b

Question : using the Convolution theorem find ( )( )

1

2 21 4

sL

s s

+ +

uptu 2006,04

Ans ( )1

cos cos 23

t t−

Page 19: Question bank Engineering Mathematics- ii

Question : using the convolution theorem, prove that ( )

21

3 2

1cos 1

21

tL t

s s

= + − +

uptu 2005

Question: Using Laplace transformation solve the differential equation uptu 2002

2

29 2 , (0) 1, 1

2

d xx t if x x

dt

π + = = = −

Ans.

[ ]1 4 12 sin 3 1

cos 2 cos3 cos 2 4cos3 4sin 35 5 5 3 5

tx t t t t t= + + = + +

Question : solve, using Laplace transform method ' " '(0) 2, (0) 8, 4 4 6 ty y y y y e−= − = + + =

uptu 2007

Ans.

2 26 8 2t t ty e e te− − −= − −

Question : solve 2

'

22 5 sin (0) 1, (0) 1xd x dy

y e x where y ydt dx

−+ + = = = uptu 2004

Ans.

( )1

sin sin 23

xy e x x

−= +

Question : apply Laplace transform to solve the equation2

2cos 2 , 0, 0 0

d x dxy t t t where y for t

dt dt+ = > = = =

Uptu 2005

Ans 5 4

sin sin 2 cos 29 9 3

ty t t t= − + −

Question: solve, by the method of laplace transform, the differential equation

( ) ( )2 2 sin , 0 0D n x a nt x Dx at tα+ = + = = = uptu 2002, 2010

Ans ( )2sin cos cos sin

2

ax nt nt nt

nα α = −

Question : using the laplace transform, solve the differential equation " 2 ' (0) 0 '(0) 1y ty y t when y and y+ − = = = Ans , ( )y t C y t c must be vanish= + = uptu 2003

Question : A particle moves in a line so that its displacement x from the fixed point O at any time t, is

given by " 4 ' 5 80sin 5x x x t+ + = using Laplace transform, find its displacement at any time t if initially

particle is at rest at 0x = . Uptu 2009

Ans. ( ) ( )22 cos 7sin 2 cos5 sin 5tx e t t t t

−= + − +

Page 20: Question bank Engineering Mathematics- ii

Question : Voltage atEe is applied at 0t = to a circuit of inductance L and R. show that the current (i) at time t is

( )/at Rt LEe e

R aL

− −−−

. uptu 2007

Question : Solve the simultaneous differential equations by using the laplace transform

3 2 , 0dx dx dy

y t ydt dt dt

− = + − = with the condition (0) (0) 0x y= = uptu 2008

Ans

2 22

3 33 3 3 3

,2 2 4 4 2 2

t tt t

x e y t e= + − + = + −

Question : Solve the simultaneous equation , sin , (0) 1, (0) 0tdx dy

y e x t x ydt dt

− = + = = = uptu 2006

Ans. 1 1

cos 2sin cos , sin cos sin2 2

t tx e t t t t y t t e t t = + + − = − + −

Question : The co-ordinates (x,y) of the particle moving along a plane curve at any time I are given by

2 cos 2 , 2 sin 2 , ( 0)dx dy

y t x t tdt dt

− = + = > it is given that at 0, 1 0.t x and y= = = show by using transforms that

the particle moves along the curve 2 24 4 5 4x xy y+ + = uptu 2003

Ans: ( )2 2 2 2 2 24 4 5 4sin 2 4cos 2 4 sin 2 cos 2 4x xy y t t t t+ + = + = + =

Question : Solve the simultaneous differential equations by Laplace transform 4 0, 2tdx dy dx

y y edt dt dt

+ − = + =

With the condition (0) (0) 0x y= = uptu 2008

Ans.

3

3 /4 41 5 8 1

,3 7 21 7

t

t t tx e e y e e

= − + = −

Question : solve the simultaneous equations ( ) ( )2 23 4 0, 1 0D x y x D y− − = + + = for 0,t > given that

0 2 0dy dx

x y and at tdt dt

= = = = = uptu 2004

Ans ( )2 cosh , 1 sinhx t t y t t= = −

Question : The co-ordinates (x,y) of the particle moving along a plane curve at any time I are given by

2 sin 2 , 2 cos 2 , ( 0)dx dy

y t y t tdt dt

+ = − = > it is given that at 0, 1 0.t x and y= = = show by using transforms that

the particle moves along the curve 2 24 4 5 4x xy y+ + = uptu 2003

Ans: ( )2 2 2 2 2 24 4 5 4sin 2 4cos 2 4 sin 2 cos 2 4x xy y t t t t+ + = + = + =

Page 21: Question bank Engineering Mathematics- ii

UNIT- V

APPLICATIONS OF PARTIAL DIFFERENTIAL EQUATIONS

Question: Explain the method of separation of variables. Uptu 2007

Question: Solve the following equation 2

22 0

z z z

x x y

∂ ∂ ∂− + =

∂ ∂ ∂ by the separation of variables method.

Ans. ( ) ( ){ }1 1 1 1

1 2 3

K x K x Kyz C e C e C e

+ + − + −= + uptu 2006

Question : Using the method of separation of variables, solve 32 ( ,0) 6 xu uu whereu x e

x t

−∂ ∂= + =

∂ ∂

Ans. 3 26 x tu e− −= uptu 2006

Question : solve the following equation by the method of separation of variables 2

costue x

x t

−∂=

∂ ∂

Given that 0u = when 0t = and 0 0u

when xt

∂= =

∂ Ans. ( )11 sinu e x

−= − uptu 2008

Question : Drive the one Dimensional wave equation 2 2

2

2 2

y ya

t x

∂ ∂=

∂ ∂ . uptu 2007

Question : Solve completely the equation 2 2

2

2 2

y ya

t x

∂ ∂=

∂ ∂,representing the vibrations of a string of length l,

fixed at both the ends, given that (0, ) 0, ( , ) 0, ( ,0) ( )y t y l t y x f x= = = and ( ,0) 0,0 .y x x lt

∂= < <

Ans. sin cosn

n x n cty b

l l

π π= uptu 2005

Question : A string is stretched and fastened to two points � apart. Motion is started by displacing the string in the

form sinx

y al

π =

from which it is released at a time 0t = show tat the displacement of any point at a

distance x from one end at time t is given by

( , ) sin cosn

x cty x t b

l l

π π =

uptu 2004, 09

Page 22: Question bank Engineering Mathematics- ii

Question: If a string of the length � is initially at rest in equilibrium position and each of its points is given the

velocity 3

0

sint

y xb

t l

π

=

∂ =

∂ , find the displacement ( , )y x t .

Ans: 3 3

( , ) 9sin sin sin sin12

bl x cl x cty x t

c l l l l

π π π π

π

= −

uptu 2001,06

Question : find the temperature in a bar of length 2 whose ends kept at zero and lateral surface insulated if the

initial temperature is 5

sin 3sin2 2

x xπ π+

Ans

2 2 2 225

4 45

sin 3sin2 2

t tc cx x

u e e

π ππ π− −

= + uptu 2007,09

Question : An insulated of of length � has its ends A and B maintained at 00 C and

0100 C respected until steady

conditions prevail. If B is suddenly reduced to 00 C and maintained at

00 C, find the temperature at a distance x

from A at time t.

Ans : ( )2 2 2

21 2001 . sin

tn cn

ln x

u en l

ππ

π

−+

= − uptu 2004,05

Question: Solve

2 2

2 20

u u

x y

∂ ∂+ =

∂ ∂which satisfies the conditions (0, ) (0, ) ( ,0) 0u y u y u x= = = and

( , ) sinn x

u x al

π =

Ans.

sinh

sin

sinh

n y

n x lu

n al

l

π

π

π

=

uptu 2004

Question : Solve

2 2

2 20,0 ,0

u ux y

x yπ π

∂ ∂+ = < < < <

∂ ∂, which satisfies the conditions

2(0, ) ( , ) ( , ) 0 ( ,0) sinu y u y u x and u x xπ π= = = =

Ans. 21,3,5,..

8 sin sinh ( )

( 4)sinhn

nx n yu

n n n

π

π π

=

− −=

−∑ (from AKTU Question Bank

Page 23: Question bank Engineering Mathematics- ii

SERIES SOLUTION OF SECOND ORDER DIFFERENTIAL EQUATIONS

Question : solve the series the legendre’s equation of order one

( )21 " 2 ' 2 0 ................ (1)x y xy y− − + =

Solution : here x=0 is an ordinary point. Let a trial solution of (1) in the form ogf series in x be

2 3

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

ny c c x c x c x c x= + + + + + + − − − − − −

Different equation (2) we get

2

1 2 3

2

2 3 4

' 2 3 ..............

" 2 6 12 ..........

y c c x c x

y c c x c x

= + + +

= + + +

On putting the values of y,y’ and y” in equation (1) we get

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

2 3 4 1 2 3 0 1 2 31 2 6 12 ..... 2 2 3 ..... 2 ..... 0x c c x c x x c c x c x c c x c x c x− + + + − + + + + + + + + =

( ) ( ) ( ) ( )2 3

2 0 3 1 1 4 2 2 2 5 3 3 32 2 6 2 2 12 2 4 2 20 6 6 2 .... 0c c c c c x c c c c x c c c c x+ + − + + − − + + − − + + = on equating

the powers of x to zero we get

2 0 3 4 2 0 5 7 9 6 0

1 1 1, 0, , 0

3 3 5c c c c c c c c c and c c= − = = = − = = = = −

On substitute these above values in equation (2) we get

2 4 6

0 1 0 0 0

2 4 6

0 1

1 1......

3 5

1 11 ....

3 5

y c c x c x c x c x

y c x x x c x

= + − − − −

= − − − − +

Question : Using the Frobenius method, obtain a series solution in powers of x for differential equation

( ) ( )2

22 1 1 3 0

d y dyx x x y

dx dx− + − + = UPTU 2010

Solution : we have ( ) ( )2

22 1 1 3 0 (1)

d y dyx x x y

dx dx− + − + = − − − − − − −

Here 2

1 2( ) ( )xP x and x P x are analytic at x=0. So x=0 is a regular singular point, as assume the solution in the form

Page 24: Question bank Engineering Mathematics- ii

( )

( )( )

0

1

0

22

20

1

m k

k

k

m k

k

k

m k

k

k

y a x

dya m k x

dx

d ya m k m k x

dx

∞+

=

∞+ −

=

∞+ −

=

=

= +

= + + −

On putting the values of

2

2, ,dy d y

ydx dx

in equation (1) we get

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

0 0 0

2 1 1 1 3 0m k m k m k

k k k

k k k

x x a m k m k x x a m k x a x∞ ∞ ∞

+ − + − +

= = =

− + + − + − + + =∑ ∑ ∑

( )( ) ( )( ) ( )( )1 1

0 0 0 0

0

2 1 2 1

3 0

m k

km k m k m k

k k k

k k k k

m k

k

k

a m k xa m k m k x a m k m k x a m k x

a x

+∞ ∞ ∞ ∞+ − + + −

= = = =

∞+

=

++ + − − + + − + + −

+ =

∑ ∑ ∑ ∑

We can rewrite as

2 2

0

2 2 1

0

2 2 2 2 2 2 3

2 2 2 2 2 2 0

m k

k

k

m k

k

k

a m mk m mk k k m k x

a m mk m mk k k m k x

∞+

=

∞+ −

=

− − + − − + − − +

+ + − + + − + + =

( )2 2 2 2 1

0 0

2 4 2 3 2 4 1 2 0m k m k

k k

k k

a m mk m k k x a m k m k k x∞ ∞

+ + −

= =

− − + − + + + + − + − = ∑ ∑

( )[ ] ( )[ ] 1

0 0

1 2 2 3 2 2 1 0 (2)m k m k

k k

k k

a m k m k x a m k m k x∞ ∞

+ + −

= =

+ − − − + + + + − = − − −∑ ∑

The coefficient of the lowest degree term 1m

x−

in equation (2) is obtain by substituting k=0 in second summation

of (2) only equating it by zero we get

( )0 02 1 0 0, 1/ 2, 0a m m m a− = ⇒ = = ≠

The coefficient of the next lowest degree termmx in (2) is obtained by substituting k=0 in first summation and k=1

in the second summation, we get

( )( ) ( )( )0 11 2 3 1 2 1 0a m m a m m+ − + + + + =

Page 25: Question bank Engineering Mathematics- ii

( )( )( )( )

( )( )

1

1

1 2 3

1 2 1

2 3

2 1

m ma

m m

ma

m

+ − += −

+ +

−= −

+

On equating the coefficient of 2

x

On putting 1k k→ + in second summation of (2) we get the coefficient of mx and equating to zero we get

( )( ) ( )( )

( )( )

( )

1

1

1

1 2 2 3 1 2 2 1 0

2 2 3

2 2 1

2 2 3

2 2 1

k k

k k

k k

a m k m k a m k m k

m ka a

m k

m ka a

m k

+

+

+

+ + − + + + + + + + =

− + += −

+ +

− −=

+ +

( )

( )

1 0

2 1

2 2 30

2 2 1

2 2 31

2 2 1

m kif k a a

m k

m kif k a a

m k

− −= =

+ +

− −= =

+ +

( )

( )

( )

3 2

4 3

5 4

2 2 32

2 2 1

2 2 33

2 2 1

2 2 34

2 2 1

m kif k a a

m k

m kif k a a

m k

m kif k a a

m k

− −= =

+ +

− −= =

+ +

− −= =

+ +

m=0 M=1/2

( )

1 0

2 1 0 0

3 2 0

4 3 0 0

5 4 0 0

2 3 4 5

1 0

23 4 5

1 0

3

1 13

3 3

1 1

5 5

3 3 1 3

7 7 5 35

5 5 3 1

9 9 35 21

1 3 11 3 ....

5 35 21

3 3 3 31 3 ....

1.3 3.5 5.7 7.9

a a

a a a a

a a a

a a a a

a a a a

y a x x x x x

xy a x x x x

= −

= − = − − =

= =

= = =

= = =

= − + + + + +

= − + + + + +

( )

1 0 0

2 1

3 4 5

1 3

2 22 0 0

2 0

12 3

2

12 1

2

12 1

20

12 3

2

...... 0

1

a a a

a a

a a a

y a x a x

y a x x

= = −

+

= =

+

= = = =

= −

= −

Page 26: Question bank Engineering Mathematics- ii

General solution

( )2

3 4 5

0

3 3 3 31 3 .... 1

1.3 3.5 5.7 7.9

xy A x x x x Ba x x

= − + + + + + + −

Ans

Question: using extended power series method find one solution of the differential equation 2" ' 0xy y x y+ + = .

Indicate the form of a second solution which is linearly independent of the first obtained above .

(UPTU 2007

Solution: we have 2( ) ( )xP x and x Q x are analytic therefore x=0 is a regular singular point of the equation

22

20

d y dyx x y

dx dx+ + =

Suppose

( )

1

22

2

( )

( ) 1

m k

k

m k

k

m k

k

y a x

dya m k x

dx

d ya m k m k x

dx

+

+ −

+ −

=

= +

= + + −

On putting all these above values of y , y’ and y” in (1) we get

( )

( )

( )

( ){ }

{ }

2 1 2

1 1 2

1 2

1 2

1

( ) 1 ( ) 0

( ) 1 ( ) 0

( ) 1 ( ) 0

( ) 1 1 0

( ) 1 1

m k m k m k

k k k

m k m k m k

k k k

m k m k

k k

m k m k

k k

m k

k k

x a m k m k x a m k x x a x

a m k m k x a m k x a x

a m k m k m k x a x

a m k m k x a x

a m k m k x a x

+ − + − +

+ − + − + +

+ − + +

+ − + +

+ −

+ + − + + + =

+ + − + + + =

+ + − + + + =

+ + − + + =

+ + − + +

∑ ∑ ∑∑ ∑ ∑∑ ∑

∑ ∑

∑( )

2

1 2

2 1 2

0

( ) 0

( ) 0 (2)

m k

m k m k

k k

m k m k

k k

a m k m k x a x

a m k x a x

+ +

+ − + +

+ − + +

=

+ + + =

+ + = − − − − − − − −

∑∑ ∑∑ ∑

The coefficient of the lowest degree term 1mx − in (2) is obtained by the putting k=0 in the first summation of (2)

only and equating it to zero. Then the indicial equation is 2 2

0 0 0 0,0a m m m= ⇒ = ⇒ =

Page 27: Question bank Engineering Mathematics- ii

Coefficient of next lowest degree term mx in (2) is obtained putting k=1 in the first summation only and equating it

to zero

( )2

1 11 0 0a m a+ = ⇒

Equating the cifficient of 1m

x+

for k =2 we get

( )2

2 22 0 0a m a+ = ⇒

Now on equating the coefficient of 2m kx + +

to zero we have

( )

( )

2

3

3 2

3 0

3

k k

kk

a m k a

aa

m k

+

+

+ + + =

= −+ +

I

( )

( )

( )

( ) ( ) ( )

( ) ( ) ( ) ( )

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

3 02

4 1 7 102

5 2 8 112

6 3 02 2 2

9 6 02 2 2 2

3 6 9

0 2 2 2 2 2 2

10

3

11 0, 0

4

12 0, 0

5

1 13

6 3 6

1 1

9 3 6 9

1 ............. (3)3 3 6 3 6 9

m

if k a am

if k a a a am

if k a a a am

if k a a am m m

a a am m m m

x x xy x a

m m m m m m

= = −+

= = − = = =+

= = − = = =+

= = − =+ + +

= − =+ + + +

= − + − + − − − − − −

+ + + + + +

For m=0

To find the first solution , put m=0 in (3) we get

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

3 6 9

1 0 2 2 2 2 2 21 ............. (4)

3 3 6 3 6 9

x x xy a

= − + − + − − − − −

To find the second independent solution, differentiate (3) w.r.t m then

Page 28: Question bank Engineering Mathematics- ii

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

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

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

3 6 9

0 2 2 2 2 2 2

3 6 6 9

3 3 2 2 3 3 2 2

0 9 9

2 3 2 2 3 3

log 1 .............3 3 6 3 6 9

2 2 2 2

3 3 6 3 6 3 6 9(5)

2 2.............

3 6 9 3 6 9

m

m

y x x xx x a

m m m m m m m

x x x x

m m m m m m m mx a

x x

m m m m m m

∂= − + − +

∂ + + + + + +

− − +

+ + + + + + + + + − − − − −

+ + + + + + + +

putting m=0 in equation (5) we get

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

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

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

3 6 9

2 0 2 2 2 2 2 2

3 6 6 9

3 3 2 2 3 3 2 2

0 9 9

2 3 2 2 3 3

log 1 .............3 3 6 3 6 9

2 2 2 2

3 3 6 3 6 3 6 9(6)

2 2.............

3 6 9 3 6 9

x x xy x a

x x x x

ax x

= − + − +

− − +

+ − − − − −

+ +

hence, the general solution is given by (4) and (6)

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

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

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

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

1 1 2 2

3 6 9 3 6 9

1 0 2 02 2 2 2 2 2 2 2 2 2 2 2

3 6 6 9

3 3 2 2 3 3 2 2

2 0 9 9

2 3 2 2 3 3

1 ............. log 1 .............3 3 6 3 6 9 3 3 6 3 6 9

2 2 2 2

3 3 6 3 6 3 6 9

2 2.............

3 6 9 3 6 9

y c y c y

x x x x x xy c a c x a

x x x x

c ax x

y c

= +

= − + − + + − + − +

− − +

+

+ +

= ( )( ) ( ) ( ) ( ) ( ) ( )

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

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

3 6 9

1 2 0 2 2 2 2 2 2

3 6 6 9

3 3 2 2 3 3 2 2

2 0 9 9

2 3 2 2 3 3

log 1 .............3 3 6 3 6 9

3 3 6 3 6 3 6 92.

.............3 6 9 3 6 9

x x xc x a

x x x x

c a Ansx x

+ − + − + +

− − +

+ +

Page 29: Question bank Engineering Mathematics- ii

Question : solve the differential equation

2

20

d y dyx y

dx dx+ − = (UPTU C.O, 2003

Solution : on comparing with the equation

2

2( ) ( ) 0

1 1( ) ( )

d y dyP x Q x y we have

dx dx

P x and Q xx x

+ + =

−= =

Because at x=0 both ( )P x and ( )Q x are not analytic

0x∴ = is a singular point

Also ( ) 1xP x = and 2 ( )x Q x x= −

Both ( )P x and ( )Q x are analytic at x=0 0x∴ = is a singular point

Now suppose that 1 2 3

0 1 2 3 ....... (1)m m m my a x a x a x a x

+ + += + + + + − − − − − − − −

Now ( ) ( ) ( )

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

1 1 2

0 1 2 3

2 1 1

0 1 2 3

' 1 2 3 .......

" 1 1 2 1 3 2 .......

m m m m

m m m m

y ma x m a x m a x m a x

y m m a x m m a x m m a x m m a x

− + +

− − +

= + + + + + + + − − − − − − − −

= − + + + + + + + + +

On substituting the all these above values in given equation we will get

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

( ) ( ) ( )( )

( )

2 1 1

0 1 2 3

1 1 2

0 1 2 3

1 2 3

0 1 2 3

1 1 2 1 3 2 .......

1 2 3 ......

....... 0

m m m m

m m m m

m m m m

x m m a x m m a x m m a x m m a x

ma x m a x m a x m a x

a x a x a x a x

− − +

− + +

+ + +

− + + + + + + + + +

+ + + + + + + +

− + + + + =

The coefficient of 1m

x−

=0 Then the indicial equation is

( ) 0 01 0 0,0m m a ma m− + = ⇒ = which are equal

Coefficient of next lowest degree term 0mx =

( ) ( ) ( )

( )

2

1 1 0 1 0

01 2

1 1 0 1

1

m ma m a a m a a

aa

m

+ + + − = ⇒ + =

=+

Equating the coefficient of 1 0mx + =

( )( ) ( ) ( )

( ) ( ) ( )

2

2 2 1 2 1

012 2 2 2

2 1 2 0 2

2 1 2

m m a m a a m a a

aaa

m m m

+ + + + − = ⇒ + =

= =+ + +

Page 30: Question bank Engineering Mathematics- ii

( ) ( ) ( )0

3 2 2 21 2 3

aa

m m m=

+ + +

Using equation (1)

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

2 3

0 2 2 2 2 2 21 .......... (2)

1 1 2 1 2 3

m x x xy a x

m m m m m m

= + + + + − − − −

+ + + + + +

( ) ( )

2 3

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

2! 3!m

x xy y a x x=

= = + + + + − − − −

To find the second independent solution, partially differentiate (1) w.r.t m then

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

( ) ( ) ( ) ( ) ( )

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

2 9

0 2 2 2 2 2 2

2

3 2 2

0

3

2 2 2

log 1 .............1 1 2 1 2 3

2 2 1 1

1 21 1 2

2 1 1 1.....

1 2 31 2 3

m

m

y x x xa x x

m m m m m m m

xx

m mm m mx a

xm m mm m m

∂= + + + +

∂ + + + + + +

− − +

+ ++ + + +

− + + − + + ++ + +

Now the second solution

( )( ) ( ) ( ) ( )

( ) ( )

2 32 3

2 0 02 2 2 2

0

2 3

1 0 2 2

1 1 1 1 1log 1 ............. 2 1 1 ......

2 2 32! 3! 2! 3!

1 1 1 1 1log 2 1 1 ......

2 2 32! 3!

m

y x xy a x x a x x x

m

y x a x x x

=

∂ = = + + + + − + + + + + −

= − + + + + + −

Hence the complete solution is

( )( ) ( ) ( ) ( )

( )( ) ( ) ( ) ( )

1 1 2 2

2 32 3

1 2 0 2 02 2 2 2

2 32

2 2 2 2

1 1 1 1 1log 1 .......... 2. 1 1 ......

2 2 32! 3! 2! 3!

1 1 1 1 1log 1 .......... 2. 1 1

2 2 32! 3! 2! 3!

y c y c y

x xy c c x a x c a x x x

x xy A B x x B x x

= +

= + + + + + − + + + + + −

= + + + + + − + + + + +

3 ......x Ans

Question: Solve in the series the differential equation

22 2

25 0

d yx x x y

dx+ + = UPTU 2002

Page 31: Question bank Engineering Mathematics- ii

Solution : we have

22 2

25 0

d yx x x y

dx+ + =

on comparing above given differential equation with

2

2( ) ( ) 0

5( ) ( ) 1

d y dyP x Q x y we have

dx dx

P x and Q xx

+ + =

= =

Because at x=0 both ( )P x and ( )Q x are not analytic

0x∴ = is a singular point

Also ( ) 5xP x = and 2 2( )x Q x x=

Both ( )P x and ( )Q x are analytic at x=0 0x∴ = is a singular point

Now suppose that 1 2 3

0 1 2 3 ....... (1)m m m my a x a x a x a x+ + += + + + + − − − − − − − −

Now ( ) ( ) ( )

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

1 1 2

0 1 2 3

2 1 1

0 1 2 3

' 1 2 3 ....... (2)

" 1 1 2 1 3 2 ....... (3)

m m m m

m m m m

y ma x m a x m a x m a x

y m m a x m m a x m m a x m m a x

− + +

− − +

= + + + + + + + − − − − − − − −

= − + + + + + + + + + − −

On substituting the all these above values in given equation we will get

( ) ( )( ) ( )( )

( ) ( )( )

( )

2 2 1

0 1 2

1 1

0 1 2

2 1 2 3

0 1 2 3

1 1 2 1 .......

5 1 2 ......

....... 0 (4)

m m m

m m m

m m m m

x m m a x m m a x m m a x

x ma x m a x m a x

x a x a x a x a x

− −

− +

+ + +

− + + + + + +

+ + + + + +

+ + + + + = − − − − − −

The coefficient of mx =0 Then the indicial equation is

( )

( )( )

0 0

2

0

0

1 5 0

4 0

4 0

0, 4

m m a ma

m m a

m m a

m

− + =

+ =

+ =

= −

which are distinct

Coefficient of next lowest degree term 1 0mx + =

( ) ( ) ( )( )1 1 1

1

1 5 1 0 5 1 0

0 (5) [ 5, 1

m ma m a m m a

a m

+ + + = ⇒ + + =

= − − − − − − ≠ − −

Equating the coefficient of 2 0m

x+ =

Page 32: Question bank Engineering Mathematics- ii

( )( ) ( )

( )( )

( )( )

2 2 0

2 0

02

2 1 5 2 0

2 6 0

(6)1 6

m m a m a a

m m a a

aa

m m

+ + + + + =

+ + + =

−= − − − − − −

+ +

Equating the coefficient of 2 0m

x+ =

( ) ( ) ( )

3 3 1

3 1

13 2 2 2

3

5 7 9

( 3)( 2) 5( 3) 0

( 3)( 7) 0

1 2 3

0

0 (7)

m m a m a a

m m a a

aa

m m m

a

similarly a a a

+ + + + + =

+ + + =

−=

+ + +

=

= = = − − − − − −

Coefficient of 4 0mx + =

4 4 2

4 2

024

( 4)( 3) 5( 3) 0

( 4)( 8)

. (8)( 4)( 8) ( 2)( 4)( 6)( 8)

m m a m a a

m m a a

aaa etc

m m m m m m

+ + + + + =

+ + = −

−= = − − − − −

+ + + + + +

These give

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

2 3

0 1 .......... (9)2 6 2 4 6 8

m x xy a x

m m m m m m

= − + + − − − −

+ + + + + +

On putting 0m = in eq (9) we get

2 4

1 0 0( ) 1 .......... (10)2.6 2.4.6.8

m

x xy y a=

= = − + + − − − −

On putting 4m = − in the series given by equation (9). The coefficients becomes infinite.

To skip this problem we will put 0 0 ( 4)a b m= + so that

( )( )

( )( ) ( )( )( )

2 3

0

44 .......... (11)

2 6 2 6 8

mm x x

y b x mm m m m m

+= + − + + − − − −

+ + + + +

( )( )

( )( )

222 4

0 2 22 3 2

3 32 768 20log 1 .............

8 12 16 76 96

mm my m m

b x x x xm m m m m m

+ +∂ + + = + − + ∂ + + + + +

Page 33: Question bank Engineering Mathematics- ii

Now the second solution

( )

( )

( )

2 44

2 0

4

4 6 2 44 4

2 0 0

4

4 64

2 0

4

log 1 .............4 4

log 0 0 ... 1 ....( 2)(2)(4) 16 ( 2)(2)(4) 16

log .....16 16

m

m

m

y x xy b x x

m

y x x x xy b x x b x

m

y x xy b x x

m

=−

− −

=−

=−

∂ = = + − +

∂ = = − + − + + + − +

∂ − −

∂ − = = − − +

2 44

0 1 ...(4) 4

x xb x−

+ − +

Hence the Hence

complete solution is

( )

( )

1 1 2 2

2 4 4 6 2 44 4

1 0 2 0 2 0

2 4 2 4 64

1 .......... log ..... 1 ...2.6 2.4.6.8 16 16 (4) 4

11 .......... 1 ... log .....

12 384 (4) 4 16 16

y c y c y

x x x x x xy c a c b x x c b x

x x x x xy A Bx B x An

− −

= +

= − + + + − − − + + − +

= − + + + + − + − + +

s

Question : solve the given differential equation " 2 ' 0xy y xy+ + = UPTUU 2003

Solution : we have

" 2 ' 0xy y xy+ + =

on comparing above given differential equation with

2

2( ) ( ) 0

2( ) ( ) 1

d y dyP x Q x y we have

dx dx

P x and Q xx

+ + =

= =

Because at x=0 ( )P x

0x∴ = is a singular point

Also ( ) 2xP x = and 2 2( )x Q x x=

Both at x=0 ( )P x and ( )Q x are analytic at x=0 0x∴ = is a singular point

Now suppose that 1 2 3

0 1 2 3 ....... (1)m m m my a x a x a x a x+ + += + + + + − − − − − − − −

Now ( ) ( ) ( )

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

1 1 2

0 1 2 3

2 1 1

0 1 2 3

' 1 2 3 ....... (2)

" 1 1 2 1 3 2 ....... (3)

m m m m

m m m m

y ma x m a x m a x m a x

y m m a x m m a x m m a x m m a x

− + +

− − +

= + + + + + + + − − − − − − − −

= − + + + + + + + + + − −

On substituting the all these above values in given equation we will get

Page 34: Question bank Engineering Mathematics- ii

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

( ) ( ) ( )( )

( )

2 1 1

0 1 2 3

1 1 2

0 1 2 3

1 2 3

0 1 2 3

1 1 2 1 3 2 .......

2 1 2 3 ......

....... 0

m m m m

m m m m

m m m m

x m m a x m m a x m m a x m m a x

ma x m a x m a x m a x

x a x a x a x a x

− − +

− + +

+ + +

− + + + + + + + + +

+ + + + + + + +

+ + + + + =

The coefficient of 1m

x−

=0 Then the indicial equation is

( )

( )0 0

2

0

2

1 2 0

0

0

0, 1

m m a ma

m m a

m m

m

− + =

+ =

+ =

= −

which are distinct

Coefficient of next lowest degree term 0mx =

( ) ( )

( )( )

( )

1 1

1

1

1 2 1 0

1 2 0

1 0

m ma m a

m m a

m a

+ + + =

+ + =

+ =

Since m+1 may be zero ,hence 1a is arbitrary (

1a becomes in determinant)

Hence the solution will contain 0a and 1a as arbitrary constants. The complete solution will be given by the

putting 1m = − in y

Equating the coefficient of 1 0mx + =

( )( ) ( )

( )( )

( )( )

2 2 0

2 0

02

2 1 2 0

2 3 0

3 2

m m a m a a

m m a a

aa

m m

+ + + + + =

⇒ + + + =

−=

+ +

Equating the coefficient of 2 0m

x+ =

( )( )

3 3 1

3 1

13

( 3)( 2) 2( 3) 0

( 3)( 4) 0

3 4

m m a m a a

m m a a

aa

m m

+ + + + + =

+ + + =

−=

+ +

Equating the coefficient of 3 0m

x+ =

Page 35: Question bank Engineering Mathematics- ii

4 4 2

4 2

24

04

( 4)( 3) 2( 4) 0

( 4)( 5) 0

( 4)( 5)

( 2)( 3)( 4)( 5)

m m a m a a

m m a a

aa

m m

aa

m m m m

+ + + + + =

+ + + =

−=

+ +

=+ + + +

Equating the coefficient of 4 0m

x+ =

5 5 3

5 3

5 3

35

15

( 5)( 4) 2( 5) 0

( 5)( 6) 0

( 5)( 6)

( 5)( 6)

( 3)( 4)( 5)( 6)

m m a m a a

m m a a

m m a a

aa

m m

aa so on

m m m m

+ + + + + =

+ + + =

+ + = −

−=

+ +

=+ + + +

On putting all above values in equation (1) we get

( )( ) ( )( )2 3 40 01

0 1

51

3 2 3 4 ( 2)( 3)( 4)( 5)

( 3)( 4)( 5)( 6)

m

a aaa a x x x x

m m m m m m m my x

ax

m m m m

+ − − + + + + + + + + +

=

+ + + + +

( )( )

( )( )

( )

2 4

0

3 5

1

2 4 3 51

0 11

1 ...3 2 ( 2)( 3)( 4)( 5)

....3 4 ( 3)( 4)( 5)( 6)

1 ... ....1.2 1.2.3.4. 2.3 2.3.4.5

m

m

x xa

m m m m m my x

x xa x

m m m m m m

Now

x x x xy x a a x−

=−

− + −

+ + + + + + =

− + − + + + + + +

= − + − + − + −

Hence the solution is

( )

( )

1

0 1

1cos sin

my y

y a x a xx

=−=

= +

Page 36: Question bank Engineering Mathematics- ii

( ) ( ) ( )0

3 2 2 21 2 3

aa

m m m=

+ + +

Using equation (1)

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

2 3

0 2 2 2 2 2 21 .......... (2)

1 1 2 1 2 3

m x x xy a x

m m m m m m

= + + + + − − − −

+ + + + + +

( ) ( )

2 3

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

2! 3!m

x xy y a x x=

= = + + + + − − − −

To find the second independent solution, partially differentiate (1) w.r.t m then

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

( ) ( ) ( ) ( ) ( )

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

2 9

0 2 2 2 2 2 2

2

3 2 2

0

3

2 2 2

log 1 .............1 1 2 1 2 3

2 2 1 1

1 21 1 2

2 1 1 1.....

1 2 31 2 3

m

m

y x x xa x x

m m m m m m m

xx

m mm m mx a

xm m mm m m

∂= + + + +

∂ + + + + + +

− − +

+ ++ + + +

− + + − + + ++ + +

Now the second solution

( )( ) ( ) ( ) ( )

( ) ( )

2 32 3

2 0 02 2 2 2

0

2 3

1 0 2 2

1 1 1 1 1log 1 ............. 2 1 1 ......

2 2 32! 3! 2! 3!

1 1 1 1 1log 2 1 1 ......

2 2 32! 3!

m

y x xy a x x a x x x

m

y x a x x x

=

∂ = = + + + + − + + + + + −

= − + + + + + −

Hence the complete solution is

( )( ) ( ) ( ) ( )

( )( ) ( ) ( ) ( )

1 1 2 2

2 32 3

1 2 0 2 02 2 2 2

2 32

2 2 2 2

1 1 1 1 1log 1 .......... 2. 1 1 ......

2 2 32! 3! 2! 3!

1 1 1 1 1log 1 .......... 2. 1 1

2 2 32! 3! 2! 3!

y c y c y

x xy c c x a x c a x x x

x xy A B x x B x x

= +

= + + + + + − + + + + + −

= + + + + + − + + + + +

3 ......x Ans