The document is a comprehensive question bank for Engineering Mathematics II at Jahangirabad Institute of Technology, detailing numerical problems and theories related to differential equations. It includes definitions, classifications (ordinary and partial), orders and degrees of differential equations, and various solution methodologies. Additionally, it contains numerous example problems with solutions, showcasing the application of the theories discussed.

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

2. 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 dy
a 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 : '
, ,
d
D y
d x
Second Derivative:
2
2 ''
2
, ,
d
D y
d x
Third Derivative :
3
3 '''
3
, ,
d
D y
d x
.
.
.
.
Nth
Derivative : , ,
n
n n
n
d
D 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
2
2 0
d y dy
y
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”.

3. For Example consider a differential equation
32
2
2 0
d y dy
y
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 dy
P 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 dy
P 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
2
3 4 0
d y dy
y
dx dx
− − = ----------------------------------------------------- (1)

4. 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 21
0
n n
n n
d y d y
a 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 sinx
y e c x c xα
β β= +
Case –V : when roots of Auxiliary Equation are imaginary and repeated twice than
{ } { }( )1 2 3 4cos sinx
y e c c x x c c x xα
β β= + + +
Case –VI : when roots of Auxiliary Equation are Irrational as α β± where β is positive than

5. ( )1 2cosh sinhx
y 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 ࢟ = ࢉ࢕࢓࢖࢒࢏࢓ࢋ࢔࢚࢘࢟ ࢌ࢛࢔ࢉ࢚࢏࢕࢔ (࡯. ࡲ)
Question : find the general solution of the differential equation
5 3
5 3
0
d y d y
dx dx
− = (2009)
2
1 2 3 4 5( ) x x
y 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
cos2
.
8
x
P I = Ans
Question : Find the particular integral of
2
2
2
2 1
d y
x x
dx
= + − (uptu 2013-14

6. 4 3 2
2
.
12 6 2
x x x
P I
 
= + − 
 
Ans
Question : Find the particular integral of
2
2
2
d y
y 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
2
5 6 sin3 cos2
d y dy
y x x
dx dx
+ − = +
Question : solve
2
2 3
2
2 xd y dy
y x e
dx dx
− + = UPTU 203-14
( )
3
2
1 2. 2 16 128
16
x
x e
C S C C x e x x x = + + + − +  Ans
Question : solve
3 2
2
3 2
6 1
d y d y dy
x
dx dx dx
− − = + UPTU 2001
3 2
3 2
1 2 3
1 25
6 18 3 6
x x x x
CS c c e c e x−  
= + + − + − 
 
Ans
Question : if ( )
2
2
0; ,
d x g
x a a b and g
dt b
+ − = are positive numbers and '
; 0 0
dx
x a whent
dt
= = = , show that
( )'
cos
g
x a a a t
b
= + −
(UPTU 2007
1 2cos sin
g g ag b
CS c x c x
b b b g
   
= + +       
Ans
Question : if
3 2
3 2
3 4 2 0
d y d y dy
y
dx dx dx
− + − =
(uptu 2001
Question : obtain the general solution of the differential equation '' '
2 2 cosx
y y y x e x− + = + uptu 2002
Ans ( ) [ ]1 2
1
. cos sin 1 sin
2 2
x
x 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 x
C S C e C e x x Ax x ns−
= + − + + −

7. Question : find the complete solution of
2
3
2
3 2 sin 2xd y dy
y xe x
dx dx
− + = + (uptu 2003
( )
3
2
1 2
3 1
. 3cos2 sin 2
2 2 20
x
x x e
C S c e c e x x x
 
 = + + + + −   
Question : solve the differential equation ( )2 2
1 cos ( 2007x
D y xe x uptu− = +
Question : find the complete solution of
2
2
3 2 sin 2xd y dy
y xe x
dx dx
− + = + (uptu 2001
Question : solve
2
2
2
sin
d y
a y ax
dx
+ = ( UPTU 2008
1 2
cos
. cos sin
2
x ax
C S c ax c ax
a
= + −
Question : solve
3 2
3 2
3 4 2 cosxd y d y dy
y e x
dx dx dx
− + − = + (UPTU 2001, 2006
Ans ( )1 2 3
3sin cos
. cos sin
10
x x x x
C S e c x c x c e
 
= + + +   
Question : solve ( )2
4 sin3 cos2D x x+ = + ( UPTU 2008
Ans ( ) ( )1 2
1 sin 2
. cos2 sin 2 sin 3
5 4
x x
C S c x c x x
 
= + − + 
 
Question : solve the differential equation ( )4 3 2 2
2 3 3 4sinx
D D D y e x+ − = + (UPTU2008
Ans ( ) ( ) [ ]0 3 2
1 2 3 3
3 2
. cos 2sin
20 5
x x x x
C S c c x e c e c e e x x−
= + + + + − −
Question : solve
3 2
3 2
3 2
3 log
d y d y dy
x x x y x x
dx dx dx
+ + + = + UPTU 2001
1
1 2
1 2 3
3 3 1
cos log sin log log
2 2 2
y c x x c x c x x x−
    
= + + + +       
      Ans
Question : solve ( )
2
2
2
log sin log
d y dy
x x y x x
dx dx
+ + = UPTU 2002

8. Ans ( ) ( ) ( ) ( ) ( ) ( )
2
1 2
1 1
. cos log sin log log sin log log cos log
4 4
C 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 y
dt dt dt
= − + + = − with condition (0) 1x = (0) 0y = Ans ( ) 2
1 2
t
x c c x e−
= + 2
( ) t
y t te−
=
Question: solve the simultaneous equations UPTU 2008
3 2 , 5 3
dx dy
x y x y
dt dt
= + = +
( ) ( )3 10 3 10
1 2
t t
Ans x c e c e
+ −
= +
( ) ( )
( )3 10 3 10
1 2
1
10 10
2
t t
y c e c e
+ +
= −
QUESTION: Question: solve the simultaneous equations UPTU 2008
5 2 , 2 0
dx dy
x y t x y
dt dt
+ − = + + =
Ans ( ) 3
1 2
1
9 27
t t
x c c t e−
= + + + 3 3 32
1 2
4 2
2 27 9
t t tc t
y c e c te e− − − 
= + + + −  
1
1
27
c = − 2
6 2
27 9
c = − = −
Question : The equations of the motions of a particle are given by UPTU 2009
0, 0
dx dy
wy wx
dt 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 2
x y r+ = Which is a equation of circle
Question : solve the following differential equation (uptu 2009
1, 1
dx dy
y x
dt dt
= + = +
Ans 1 2 1t t
x c e c e−
= + − , 1 2 1t t
y c e c e−
= − −
Question: Solve the simultaneous differential equations uptu 2005
2 2
2 2
3 4 0, 0
d x d y
x 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

9. 2 2
2 2
3 , 4 3 sin 2td x dx d y dx
x e y t
dt dt dt dt
+ + = − + =
1 2 3 4
1 2 1
2 3 4
2
cos sin cos3 sin3 cos2
5 15
1
cos 3 sin 3 sin 2 2 sin
15
1
2 cos 2 sin3 2 cos3
5
t
t
e
x 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 5 tdx dy
x y t x y e
dt 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
1
2sin cos
5
1
2cos 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 2
4 4 , 4 4 25 16 td x dx d y dy
x y y x e
dt 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
2
1
d y dy
dx dx
  
= −  
   
uptu 2002
Ans 1cos( )y x c c= − + +
Question : Solve
22
2
0
d y dy dy
x x
dx dx dx
 
+ − = 
 
uptu 2010
Ans 2
1 2log( 2 )y x c c= + +

10. Question : solve 2
" 4 ' (4 2) 0y xy x y− + − = given that
2
x
y e= is an integral included in the complementary
function . uptu 2004
Ans
2
1 2( )x
y e c x c= +

11. 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

12. 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.

13. Important questions
1. find the Fourier series for the function
2
( )
4
x
f 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 x
f 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 x
f 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
x
f 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 x
f x and f x f x
x x
π π
π
π π
+ − < <
= + =
− − < <
uptu 2006
8. find the Fourier series for the function
1 0
( ) ( 2 ) ( )
1 0
x
f x and f x f x
x
π
π
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 x
f 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 1
t
f t t
l
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

14. 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.

15. 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
t
u 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 t
f t
t t
− < <
= 
− < <
uptu 2009
Ans 2 3
2
1
2s s s
e 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
2
2 3 3 5 1
s s
e e
s s s
s s s
− −
− + + + −
Question: if ( )
( )
2
2
2
2
cos
4
s
L 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
d
L t f t F s
ds
= − uptu 2005
Question: Obtain the laplace transform of 2
. .sin 4t
t e t Ans
( )
( )
2
32
4 3 6 13
2 17
s s
s s
− −
− +
uptu 2002

16. Question: find the laplace transform of the function ( ) . .sin 2t
f t t e t−
= Ans
( )
( )
22
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 ds
t
∞
 
= 
 
∫ uptu 2005, 07
Question: find the laplace transform of the
0
sin
( )
t
at
f t dt
t
= ∫ uptu 2004
Question: find the laplace transform of the
cos cos
( )
at bt
f t
t
−
= Ans
2 2
2 2
1
log
2
s b
s a
+
+
uptu 2004
Question: if
cos
( )
at
e bt
f t
t
−
= , find the laplace transform of ( )f t Ans
( )
2 2
2
1
log
2
s b
s a
 +
 
+  
uptu 2003
Question: find { }L erf t and hence prove that { } ( )
3/22
3 8
. 2
4
s
L 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 t
f t
t t
− < <
= 
− < <
uptu 2009
Ans
2 3
2 2 2
2
s s s
e 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 s
e e
s s
− −
−
Question : find the laplace transform of the function
, 0
( )
, 2
t t
f t
t t
π
π π π
< <
= 
− < <
uptu 2003

17. 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
π
ω
ω
π π
ω ω

< <
= 
 < <

uptu 2010
Ans
( )2 2
1
s
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 a
f 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
st
L f x f t x dx e f x f t x dx dt
∞
−
   
− = −  
   
∫ ∫ ∫
1 2 1 2
0 0 0
( ). ( ) ( ). ( )
t t
st
L 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

18. ( )
( )
( )
1 2 1 2
0 0
1 2 1 2
0 0
1 2 1 2
0 0
( ). ( ) ( ). ( )
( ). ( ) ( ). ( )
( ). ( ) ( ) . ( )
t
s t x x
s
t
s t x x
s
t
s 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
2
2
4
s
L
s
−
 
 
 
+  
uptu 2010 Ans
1
sin 2
4
t
Question : Use the convolution theorem to find the inverse of the function
( )
22
s
s a+
uptu 2009
Ans [ ]3
1
sin cos
2
at at at
a
−
Question : using the Convolution theorem find
( ) ( )
2
1
2
,
2 2
s
L a b
s s
−
  
≠ 
+ −  
uptu 2006,04
Ans 2 2
sin sina at b bt
a b
−
−
Question : using the Convolution theorem find
( )( )
1
2 2
1 4
s
L
s s
−
  
 
+ +  
uptu 2006,04
Ans ( )
1
cos cos2
3
t t−

19. Question : using the convolution theorem, prove that
( )
2
1
3 2
1
cos 1
21
t
L t
s s
−
  
= + − 
+  
uptu 2005
Question: Using Laplace transformation solve the differential equation uptu 2002
2
2
9 2 , (0) 1, 1
2
d x
x t if x x
dt
π 
+ = = = − 
 
Ans. [ ]
1 4 12 sin3 1
cos2 cos3 cos2 4cos3 4sin3
5 5 5 3 5
t
x t t t t t= + + = + +
Question : solve, using Laplace transform method ' " '
(0) 2, (0) 8, 4 4 6 t
y y y y y e−
= − = + + = uptu 2007
Ans. 2 2
6 8 2t t t
y e e te− − −
= − −
Question : solve
2
'
2
2 5 sin (0) 1, (0) 1xd x dy
y e x where y y
dt dx
−
+ + = = = uptu 2004
Ans. ( )
1
sin sin 2
3
x
y e x x−
= +
Question : apply Laplace transform to solve the equation
2
2
cos2 , 0, 0 0
d x dx
y t t t where y for t
dt dt
+ = > = = =
Uptu 2005 Ans
5 4
sin sin 2 cos2
9 9 3
t
y 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 ( )2
sin cos cos sin
2
a
x nt nt nt
n
α 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 bevanish= + = 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 80sin5x x x t+ + = using Laplace transform, find its displacement at any time t if initially
particle is at rest at 0x = . Uptu 2009
Ans. ( ) ( )2
2 cos 7sin 2 cos5 sin5t
x e t t t t−
= + − +

20. Question : Voltage at
Ee is applied at 0t = to a circuit of inductance L and R. show that the current (i) at time t is
( )/at Rt LE
e e
R aL
− −
−
−
. uptu 2007
Question : Solve the simultaneous differential equations by using the laplace transform
3 2 , 0
dx dx dy
y t y
dt dt dt
− = + − = with the condition (0) (0) 0x y= = uptu 2008
Ans
2 22
3 3
3 3 3 3
,
2 2 4 4 2 2
t t
t t
x e y t e= + − + = + −
Question : Solve the simultaneous equation , sin , (0) 1, (0) 0tdx dy
y e x t x y
dt dt
− = + = = = uptu 2006
Ans.
1 1
cos 2sin cos , sin cos sin
2 2
t t
x 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 cos2 , 2 sin 2 ,( 0)
dx dy
y t x t t
dt 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 2
4 4 5 4x xy y+ + = uptu 2003
Ans: ( )2 2 2 2 2 2
4 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, 2 tdx dy dx
y y e
dt dt dt
+ − = + =
With the condition (0) (0) 0x y= = uptu 2008
Ans.
3
3 /4 4
1 5 8 1
,
3 7 21 7
t
t t t
x e e y e e− 
= − + = − 
 
Question : solve the simultaneous equations ( ) ( )2 2
3 4 0, 1 0D x y x D y− − = + + = for 0,t > given that
0 2 0
dy dx
x y and at t
dt 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 cos2 ,( 0)
dx dy
y t y t t
dt 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 2
4 4 5 4x xy y+ + = uptu 2003
Ans: ( )2 2 2 2 2 2
4 4 5 4sin 2 4cos 2 4 sin 2 cos 2 4x xy y t t t t+ + = + = + =

21. UNIT- V
APPLICATIONS OF PARTIAL DIFFERENTIAL EQUATIONS
Question: Explain the method of separation of variables. Uptu 2007
Question: Solve the following equation
2
2
2 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 Ky
z C e C e C e
+ + − + −
= + uptu 2006
Question : Using the method of separation of variables, solve 3
2 ( ,0) 6 xu u
u whereu x e
x t
−∂ ∂
= + =
∂ ∂
Ans. 3 2
6 x t
u e− −
= uptu 2006
Question : solve the following equation by the method of separation of variables
2
costu
e x
x t
−∂
=
∂ ∂
Given that 0u = when 0t = and 0 0
u
when x
t
∂
= =
∂
Ans. ( )1
1 sinu e x−
= − uptu 2008
Question : Drive the one Dimensional wave equation
2 2
2
2 2
y y
a
t x
∂ ∂
=
∂ ∂
. uptu 2007
Question : Solve completely the equation
2 2
2
2 2
y y
a
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 l
t
∂
= < <
∂
Ans. sin cosn
n x n ct
y 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 sin
x
y a
l
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 ct
y x t b
l l
π π   
=    
   
uptu 2004, 09

22. 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
sin
t
y x
b
t l
π
=
∂ 
= 
∂ 
, find the displacement ( , )y x t .
Ans:
3 3
( , ) 9sin sin sin sin
12
bl x cl x ct
y 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 3sin
2 2
x xπ π
+
Ans
2 2 2 2
25
4 4
5
sin 3sin
2 2
t t
c c
x x
u e e
π π
π π− −
= + uptu 2007,09
Question : An insulated of of length ݈ has its ends A and B maintained at 0
0 C and 0
100 C respected until steady
conditions prevail. If B is suddenly reduced to 0
0 C and maintained at 0
0 C, find the temperature at a distance x
from A at time t.
Ans : ( )
2 2 2
21 200
1 . sin
t
n c
n
l
n x
u e
n l
π
π
π
−
+
= − uptu 2004,05
Question: Solve
2 2
2 2
0
u u
x y
∂ ∂
+ =
∂ ∂
which satisfies the conditions (0, ) (0, ) ( ,0) 0u y u y u x= = = and
( , ) sin
n x
u x a
l
π 
=  
 
Ans.
sinh
sin
sinh
n y
n x l
u
n al
l
π
π
π
 
 
   =  
  
 
 
uptu 2004
Question : Solve
2 2
2 2
0,0 ,0
u u
x y
x y
π π
∂ ∂
+ = < < < <
∂ ∂
, which satisfies the conditions
2
(0, ) ( , ) ( , ) 0 ( ,0) sinu y u y u x and u x xπ π= = = =
Ans. 2
1,3,5,..
8 sin sinh ( )
( 4)sinhn
nx n y
u
n n n
π
π π
∞
=
− −
=
−
∑ (from AKTU Question Bank

23. SERIES SOLUTION OF SECOND ORDER DIFFERENTIAL EQUATIONS
Question : solve the series the legendre’s equation of order one
( )2
1 " 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 5
c 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 1
1 ....
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
2
2 1 1 3 0
d y dy
x x x y
dx dx
− + − + = UPTU 2010
Solution : we have ( ) ( )
2
2
2 1 1 3 0 (1)
d y dy
x 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

24. ( )
( )( )
0
1
0
2
2
2
0
1
m k
k
k
m k
k
k
m k
k
k
y a x
dy
a m k x
dx
d y
a m k m k x
dx
∞
+
=
∞
+ −
=
∞
+ −
=
=
= +
= + + −
∑
∑
∑
On putting the values of
2
2
, ,
dy d y
y
dx 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 x
a 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 term m
x 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+ − + + + + =

25. ( )( )
( )( )
( )
( )
1
1
1 2 3
1 2 1
2 3
2 1
m m
a
m m
m
a
m
+ − +
= −
+ +
−
= −
+
On equating the coefficient of 2
x
On putting 1k k→ + in second summation of (2) we get the coefficient of m
x 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 k
a a
m k
m k
a a
m k
+
+
+
+ + − + + + + + + + =
− + +
= −
+ +
− −
=
+ +
( )
( )
1 0
2 1
2 2 3
0
2 2 1
2 2 3
1
2 2 1
m k
if k a a
m k
m k
if k a a
m k
− −
= =
+ +
− −
= =
+ +
( )
( )
( )
3 2
4 3
5 4
2 2 3
2
2 2 1
2 2 3
3
2 2 1
2 2 3
4
2 2 1
m k
if k a a
m k
m k
if k a a
m k
m k
if 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
2
3 4 5
1 0
3
1 1
3
3 3
1 1
5 5
3 3 1 3
7 7 5 35
5 5 3 1
9 9 35 21
1 3 1
1 3 ....
5 35 21
3 3 3 3
1 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
x
y a x x x x
= −
= − = − − =
= =
 
= = = 
 
 
= = = 
 
 
= − + + + + + 
 
 
= − + + + + + 
 
( )
1 0 0
2 1
3 4 5
1 3
2 2
2 0 0
2 0
1
2 3
2
1
2 1
2
1
2 1
2
0
1
2 3
2
...... 0
1
a a a
a a
a a a
y a x a x
y a x x
 
− 
 = = −
 
+ 
 
 
− 
 = =
 
+ 
 
= = = =
= −
= −

26. General solution
( )
2
3 4 5
0
3 3 3 3
1 3 .... 1
1.3 3.5 5.7 7.9
x
y 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
2
2
2
0
d y dy
x x y
dx dx
+ + =
Suppose
( )
1
2
2
2
( )
( ) 1
m k
k
m k
k
m k
k
y a x
dy
a m k x
dx
d y
a 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 1m
x −
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= ⇒ = ⇒ =

27. Coefficient of next lowest degree term m
x 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 k
x + +
to zero we have
( )
( )
2
3
3 2
3 0
3
k k
k
k
a m k a
a
a
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
1
0
3
1
1 0, 0
4
1
2 0, 0
5
1 1
3
6 3 6
1 1
9 3 6 9
1 ............. (3)
3 3 6 3 6 9
m
if k a a
m
if k a a a a
m
if k a a a a
m
if k a a a
m m m
a a a
m m m m
x x x
y 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 2
1 ............. (4)
3 3 6 3 6 9
x x x
y a
 
= − + − + − − − − − 
  
To find the second independent solution, differentiate (3) w.r.t m then

28. ( )
( ) ( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( ) ( )
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 x
x x a
m m m m m m m
x x x x
m m m m m m m m
x 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 x
y x a
x x x x
a
x 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 x
y c a c x a
x x x x
c a
x 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 9
2.
.............
3 6 9 3 6 9
x x x
c x a
x x x x
c a Ans
x x
 
+ − + − + + 
  
 
− − + 
 
 
 + +
  

29. Question : solve the differential equation
2
2
0
d y dy
x y
dx dx
+ − = (UPTU C.O, 2003
Solution : on comparing with the equation
2
2
( ) ( ) 0
1 1( ) ( )
d y dy
P x Q x y wehave
dx dx
P x and Q x
x 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 m
y 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 0m
x =
( ) ( ) ( )
( )
2
1 1 0 1 0
0
1 2
1 1 0 1
1
m ma m a a m a a
a
a
m
+ + + − = ⇒ + =
=
+
Equating the coefficient of 1
0m
x +
=
( )( ) ( ) ( )
( ) ( ) ( )
2
2 2 1 2 1
01
2 2 2 2
2 1 2 0 2
2 1 2
m m a m a a m a a
aa
a
m m m
+ + + + − = ⇒ + =
= =
+ + +

30. ( ) ( ) ( )
0
3 2 2 2
1 2 3
a
a
m m m
=
+ + +
Using equation (1)
( ) ( ) ( ) ( ) ( ) ( )
2 3
0 2 2 2 2 2 2
1 .......... (2)
1 1 2 1 2 3
m x x x
y a x
m m m m m m
 
= + + + + − − − − 
+ + + + + +  
( ) ( )
2 3
1 0 0 2 2
( ) 1 .......... (3)
2! 3!
m
x x
y 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 x
a x x
m m m m m m m
x
x
m mm m m
x a
x
m m mm m m
 ∂
= + + + + 
∂ + + + + + +  
   
− − +  
+ ++ + +    
+  
   
− + + −  + + ++ + +    
Now the second solution
( )
( ) ( ) ( ) ( )
( ) ( )
2 3
2 3
2 0 02 2 2 2
0
2 3
1 0 2 2
1 1 1 1 1
log 1 ............. 2 1 1 ......
2 2 32! 3! 2! 3!
1 1 1 1 1
log 2 1 1 ......
2 2 32! 3!
m
y x x
y a x x a x x x
m
y x a x x x
=
   ∂     
= = + + + + − + + + + + −       
∂           
    
= − + + + + + −    
     
Hence the complete solution is
( )
( ) ( ) ( ) ( )
( )
( ) ( ) ( ) ( )
1 1 2 2
2 3
2 3
1 2 0 2 02 2 2 2
2 3
2
2 2 2 2
1 1 1 1 1
log 1 .......... 2. 1 1 ......
2 2 32! 3! 2! 3!
1 1 1 1 1
log 1 .......... 2. 1 1
2 2 32! 3! 2! 3!
y c y c y
x x
y c c x a x c a x x x
x x
y A B x x B x x
= +
      
= + + + + + − + + + + + −      
         
     
= + + + + + − + + + + +     
    
3
......x Ans
 
− 
  
Question: Solve in the series the differential equation
2
2 2
2
5 0
d y
x x x y
dx
+ + = UPTU 2002

31. Solution : we have
2
2 2
2
5 0
d y
x x x y
dx
+ + =
on comparing above given differential equation with
2
2
( ) ( ) 0
5( ) ( ) 1
d y dy
P x Q x y wehave
dx dx
P x and Q x
x
+ + =
= =
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 m
y 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 m
x =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
0m
x +
=
( ) ( ) ( )( )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 +
=

32. ( )( ) ( )
( )( )
( )( )
2 2 0
2 0
0
2
2 1 5 2 0
2 6 0
(6)
1 6
m m a m a a
m m a a
a
a
m m
+ + + + + =
+ + + =
−
= − − − − − −
+ +
Equating the coefficient of 2
0m
x +
=
( ) ( ) ( )
3 3 1
3 1
1
3 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
a
a
m m m
a
similarly a a a
+ + + + + =
+ + + =
−
=
+ + +
=
= = = − − − − − −
Coefficient of 4
0m
x +
=
4 4 2
4 2
02
4
( 4)( 3) 5( 3) 0
( 4)( 8)
. (8)
( 4)( 8) ( 2)( 4)( 6)( 8)
m m a m a a
m m a a
aa
a etc
m m m m m m
+ + + + + =
+ + = −
−
= = − − − − −
+ + + + + +
These give
( )( ) ( )( )( )( )
2 3
0 1 .......... (9)
2 6 2 4 6 8
m x x
y 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 x
y 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
4
4 .......... (11)
2 6 2 6 8
m m x x
y b x m
m m m m m
 +
= + − + + − − − − 
+ + + + +  
( )
( )
( )
( )
22
2 4
0 2 2
2 3 2
3 32 768 20
log 1 .............
8 12 16 76 96
m
m my m m
b x x x x
m m m m m m
 + +∂ + + = + − +
 ∂ + + + + + 

33. Now the second solution
( )
( )
( )
2 4
4
2 0
4
4 6 2 4
4 4
2 0 0
4
4 6
4
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 x
y b x x
m
y x x x x
y b x x b x
m
y x x
y b x x
m
−
=−
− −
=−
−
=−
 ∂ 
= = + − +   
∂   
   ∂ 
= = − + − + + + − +     
∂ − −     
 ∂ − 
= = − − +  
∂   
2 4
4
0 1 ...
(4) 4
x x
b x−  
+ − + 
 
Hence the Hence
complete solution is
( )
( )
1 1 2 2
2 4 4 6 2 4
4 4
1 0 2 0 2 0
2 4 2 4 6
4
1 .......... log ..... 1 ...
2.6 2.4.6.8 16 16 (4) 4
1
1 .......... 1 ... log .....
12 384 (4) 4 16 16
y c y c y
x x x x x x
y c a c b x x c b x
x x x x x
y 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 dy
P x Q x y wehave
dx dx
P x and Q x
x
+ + =
= =
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 m
y 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

34. ( ) ( )( ) ( )( ) ( )( )
( ) ( ) ( )( )
( )
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 0m
x =
( ) ( )
( )( )
( )
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
0m
x +
=
( )( ) ( )
( )( )
( )( )
2 2 0
2 0
0
2
2 1 2 0
2 3 0
3 2
m m a m a a
m m a a
a
a
m m
+ + + + + =
⇒ + + + =
−
=
+ +
Equating the coefficient of 2
0m
x +
=
( )( )
3 3 1
3 1
1
3
( 3)( 2) 2( 3) 0
( 3)( 4) 0
3 4
m m a m a a
m m a a
a
a
m m
+ + + + + =
+ + + =
−
=
+ +
Equating the coefficient of 3
0m
x +
=

35. 4 4 2
4 2
2
4
0
4
( 4)( 3) 2( 4) 0
( 4)( 5) 0
( 4)( 5)
( 2)( 3)( 4)( 5)
m m a m a a
m m a a
a
a
m m
a
a
m m m m
+ + + + + =
+ + + =
−
=
+ +
=
+ + + +
Equating the coefficient of 4
0m
x +
=
5 5 3
5 3
5 3
3
5
1
5
( 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
a
a
m m
a
a 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 aa
a a x x x x
m m m m m m m m
y x
a
x
m m m m
 
+ − − + + + + + + + + +
 =
 
+ 
+ + + + 
( )( )
( )( )
( )
2 4
0
3 5
1
2 4 3 5
1
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 x
a
m m m m m m
y x
x x
a x
m m m m m m
Now
x x x x
y x a a x−
=−
   
− + −  
+ + + + + +   
=  
   − + −  + + + + + +   
    
= − + − + − + −    
    
Hence the solution is
( )
( )
1
0 1
1
cos sin
m
y y
y a x a x
x
=−
=
= +

36. ( ) ( ) ( )
0
3 2 2 2
1 2 3
a
a
m m m
=
+ + +
Using equation (1)
( ) ( ) ( ) ( ) ( ) ( )
2 3
0 2 2 2 2 2 2
1 .......... (2)
1 1 2 1 2 3
m x x x
y a x
m m m m m m
 
= + + + + − − − − 
+ + + + + +  
( ) ( )
2 3
1 0 0 2 2
( ) 1 .......... (3)
2! 3!
m
x x
y 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 x
a x x
m m m m m m m
x
x
m mm m m
x a
x
m m mm m m
 ∂
= + + + + 
∂ + + + + + +  
   
− − +  
+ ++ + +    
+  
   
− + + −  + + ++ + +    
Now the second solution
( )
( ) ( ) ( ) ( )
( ) ( )
2 3
2 3
2 0 02 2 2 2
0
2 3
1 0 2 2
1 1 1 1 1
log 1 ............. 2 1 1 ......
2 2 32! 3! 2! 3!
1 1 1 1 1
log 2 1 1 ......
2 2 32! 3!
m
y x x
y a x x a x x x
m
y x a x x x
=
   ∂     
= = + + + + − + + + + + −       
∂           
    
= − + + + + + −    
     
Hence the complete solution is
( )
( ) ( ) ( ) ( )
( )
( ) ( ) ( ) ( )
1 1 2 2
2 3
2 3
1 2 0 2 02 2 2 2
2 3
2
2 2 2 2
1 1 1 1 1
log 1 .......... 2. 1 1 ......
2 2 32! 3! 2! 3!
1 1 1 1 1
log 1 .......... 2. 1 1
2 2 32! 3! 2! 3!
y c y c y
x x
y c c x a x c a x x x
x x
y A B x x B x x
= +
      
= + + + + + − + + + + + −      
         
     
= + + + + + − + + + + +     
    
3
......x Ans
 
− 
  