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Title: Ordinary differential equation
Description: Differential equation of first order and first degree,beenoulie equation
Description: Differential equation of first order and first degree,beenoulie equation
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Mathematics-II [BT-202]
Dr
...
Lecture Plan
Lecture
No
...
F
Method to find P
...
I and Rule-3-4 &
Problems
Homogenous LDE and Problems
Simultaneous Differential Equations
References
R1: 3-86
R1: 3-86
R1: 3-86
R1: 3-86
R1: 3-86
R1: 3-86
R1: 3-86
R1: 3-86
R1: 3-86
R1: 3-86
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R1: 3-86
R1: 3-86
Teaching
Methodology*
Reference
to Course
Outcomes
Chalk/Board
CO1
Chalk/Board
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CO1
CO1
CO1
CO1
Chalk/Board
CO1
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Reference Books:
R1: Dr, Sonendra Gupta, Engineering Mathematics-II, Dhanpat Rai Publishing Company(P), New Delhi
R2: B
...
Grewal, Higher Engineering Mathematics, Khanna Publishers, 36th Edition, 2010
R3: N
...
Bali and Manish Goyal, A text book of Engineering Mathematics, Laxmi Publications, Reprint,
2008
...
Dr
...
1
Concept of Differential equation:
A equation is said to be differential equation which is involves one dependent variable and one or more
than one independent variable and also involves the differential coefficient of dependent variable with respect to
their independent variable
...
2
Types of Differential equation:
There are two types of differential equation:
1
...
2
...
1
...
Example : suppose z = f (x, y) be any two variable function, then
1
...
x 2
2 z
x 2
y
2
2 z
2 y
2x y
1
...
e
...
1
...
d2y
dx
2
5
dy
8y 0
dx
dy 2
4
...
2
dy
dy
y0
3
...
5
Order of Differential equation:
The order of differential equation is the order of the highest ordered derivative occurring in the
differential equation
...
6
Degree of a Differential equation:
The degree of a differential equation is the power of the highest derivative of that differential equation
...
Sonendra Gupta [9893455006]
Page 3
Mathematics-II [BT-202]
Note: 1
2
...
Degree and order never be negative and zero
...
7
Types of Ordinary differential equation:
Differential equations in which the dependent variable and its derivative (or differential equation) is of
first degree are called the linear differential equation
...
8
Non-Linear ordinary Differential equation:
The differential equation which is not linear i
...
, the differential coefficient of higher degree are called
the non-linear differential equation
...
No
...
1
1
Linear
dy
cos x tan x
dx
2
2
...
dy 2
1
dx 2 dx
d2y
3/ 2
4
...
d2y
2
dx 2
6
...
9
General solution or complete solution
The General (Complete) solution of a differential equation is the solution in which the number of
arbitrary constants is equal to the order of the differential equation
...
1
...
S):
A Particular solution of a differential equation is the solution which is obtained from its general solution by
giving particular values to the arbitrary constants
...
Note : - The solution of a differential equation of nth order is its particular solution if it contains less than n
arbitrary constant
...
Sonendra Gupta [9893455006]
Page 4
Mathematics-II [BT-202]
Module-I
E-Notes
1
...
Equations solvable by separation of the variables
2
...
3
...
1
...
If the differential equation can be written in the form;
f ( x)
dy
1
f 2 ( y ) dy f1 ( x) dx
dx f 2 ( y )
We say those variables are separable, y on left hand side and x on right hand side
...
f2 ( y) dy f1( x) dx c , where c be integral constant (which we take any form as per the requirement
i
...
,
of solving the problems)
...
13
Homogeneous function:
A function f (x, y) in x and y is a homogeneous function of degree n, if the degree of each term is n
...
Suppose the function,
f ( x, y) a0 xn a1xn 1 y a2 x n 2 y 2 a3x n 3 y3
...
y
2
...
f ( x, y )
4
...
f ( x, y ) y tan -1
x
Test for Homogeneity: f (t x, t y ) t n f ( x, y )
1
...
Putting y = v x and solve it
...
Sonendra Gupta [9893455006]
Page 5
Mathematics-II [BT-202]
1
...
A B
, then putting x = u + h, y = v + k where h & k are constant
...
Case 2: When
A B
Lecture-02
1
...
e, x
...
Write the D
...
in the standard form,
dy
P yQ
dx
2
...
…
(1)
P dx
3
...
F
...
The solution of equation (1) is,
y
...
F C Q
...
F
...
e
...
Working Rule:
1
...
E
...
Find the value of P and Q
...
Find the Integrating factor such as I
...
e
4
...
Sonendra Gupta [9893455006]
Page 6
Mathematics-II [BT-202]
Module-I
E-Notes
x
...
F C Q
...
F
...
F
...
I
...
c I
...
Q dx
Problem 2:
x
dx
1cos
x2
y
...
1 x 2 c sin x
Solve: 1 y 2 dx tan 1 y x dy
y
...
2019]
dx tan 1 y x
dy
1 y2
Solution: Given:
dx
x
tan 1 y
dy 1 y 2 1 y 2
Answer
… (1)
This is Linear differential equation of first order
...
F
...
e
Putting,
tan 1 y
c
tan 1 y
tan 1 y t dt
x
...
e
dy
1 y2
dy
1 y2
c t
...
Sonendra Gupta [9893455006]
Page 7
Mathematics-II [BT-202]
1
Module-I
x
...
etan y c et t 1
x
...
et et
y
1
E-Notes
y
c e tan
1
y
tan1 y 1
Problem 3 Solve the differential equation
Answer 1
...
F
...
x dx
y
...
F
...
I
...
dx
x3
3
Thus the required solution is
x
...
y c
x3
3
Answer
dy y
sin x
cos x
dx x
x
Solution: Given differential equation is
dy y
sin x
cos x
dx x
x
This is linear differential equation of first order
...
Solve:
I
...
e
P dx
e
1/ x dx
[RGPV Dec
...
I
...
c Q
...
F
...
x dx
x
c x cos x dx sin x dx
Dr
...
Solve: 1 x 2 2 xy 4 x 2
dx
[RGPV Dec
...
1 x c
]
3
dy
y
2 sin x
2
...
2019]
[Answer: x 2 y c x 2 cos x 2 x sin x 2 cos x ]
3
...
Solve
dx
[Answer: y e x c e x ]
Dr
...
dx
dy y
y3
Problem 1
Solve:
dx x
Solution: Given differential equation is
dy y
y3
dx x
This is Bernoulli’s equation
...
17
y 3
dy y 2
1
dx
x
… (2)
y 2 v
Putting,
dy
1 dv
dx
2 dx
From equation (2) and (3), we get
1 dv v
1
2 dx x
dv 2
v 2
dx x
This is LDE in v
...
F
...
I
...
c Q
...
F
...
2
...
Sonendra Gupta [9893455006]
Q v y 1
Answer
Page 10
Mathematics-II [BT-202]
Module-I
E-Notes
dy
x sin 2 y x3
...
cos 2 y
dx
This is Bernoulli’s equation
...
v x3
dx
This is LDE in v
...
F
...
ex
...
I
...
c Q
...
F
...
x
...
dx
v ex
2
2
x2
x2
2
x 2 t x dx
1
dt
2
c 12 t
...
e 1 et
2
2
1 2
e x tan y c e x t 1
2
2
1 2
e x tan y c e x x 2 1
2
2
e x tan y c
Dr
...
dy
2 y tan x y 2 tan 2 x
dx
2
...
y y 2 e x / 2 sin x
dx
3
...
y 1 x
...
2
1
tan 3 x
[Answer: sec2 x c
]
y
3
[Answer: y 1 e x
2
/2
c cos x ]
y2
1
y 2 /2
2
1 ]
[Answer: c e
2
x
dy
1
dx
dy tan y
1 x e x sec y
dx 1 x
[Answer: sin y c e x 1 x ]
Lecture-04
1
...
2015]
… (1)
Here, M y cos x sin y y and N sin x x cos y x
N
M
cos x cos y
cos x cos y and
x
y
Clearly,
M N
y
x
Therefore, equation (1) is exact, and then the solution is
y is contant M dx Independent of x N dy c
y constant y cos x sin y y dx Independent of x sin x x cos y x dy c
y sin x x sin y x y No
...
Sonendra Gupta [9893455006]
Answer
Page 12
Mathematics-II [BT-202]
Problem 2:
Module-I
E-Notes
Solve: 1 4 xy 2 y 2 dx 1 4 xy 2 x 2 dy 0
[RGPV June 2017]
Solution: Given the differential equation is
1 4xy 2 y2 dx 1 4xy 2x2 dy 0
… (1)
Here M 1 4 xy 2 y 2 and N 1 4 xy 2 x 2
N
M
4 y 4x
4 x 4 y and
x
y
Clearly
M N
x
2
y
x
y
Equation (1) is exact
...
Therefore the solution of equation (1) is
y is contant M dx Independent of x N dy c
x/ y
y is con tan t 1 e dx c
x y e x/ y c
Dr
...
1
...
x y 2 dx x 2 y 3 dy 0
[Answer: x 2 2 xy 4 x 2 y 2 6 y 2c ]
3
...
20
Rules for finding the IF when differential equation is not exact:
M N
If differential M x, y dx N x, y dy 0 is not exact i
...
, then we will find the integrating
y
x
factor by the following methods
...
]
Method 1: If differential equation is homogeneous and Mx Ny 0 , then
I
...
1
Mx Ny
Method 2: If the given differential equation can be written in the form y f1( x y) dx x f 2 ( x y) dy 0 and
Mx Ny 0 , then
Problem 4
I
...
1
Mx Ny
Solve: x 2 y 2 xy 2 dx x3 3x 2 y dy 0
Solution: Given the differential equation is
x2 y 2xy2 dx x3 3x2 y dy 0
… (1)
Here M x 2 y 2 xy 2 and N x3 3 x 2 y
M
N
x 2 4 xy and
3x 2 6 xy
y
x
Clearly
M N
y
x
Equation (1) is not exact
...
Sonendra Gupta [9893455006]
Page 14
Mathematics-II [BT-202]
Module-I
Since Equation (1) is homogeneous then
Mx Ny x x 2 y 2 xy 2 y x3 3x 2 y
x3 y 2 x 2 y 2 x3 y 3 x 2 y 2
x2 y 2 0
The integrating Factor I
...
Multiplying equation (1) by
1
2 2
E-Notes
1
1
2 2
Mx Ny x y
, we get
x y
x 2 y 2 xy 2
x3 3 x 2 y
dx
dy 0
2 2
2 2
x y
x y
x 3
1 2
y x dx 2 y dy 0
y
1 2
x 3
Equation (2) is exact and M and N 2
y x
y
y
… (2)
Therefore the solution of equation (2) is
y is contant M dx Independent of x N dy c
1
2
3
y is contant y x dx Independent of x y dx c
x
2 log x 3log y c
y
x
2 log x 3log y c
y
Problem 5
Answer
Solve: y 1 x y dx x 1 x y dy 0
Solution: Given differential equation is
y 1 x y dx x 1 x y dy 0
… (1)
Here M y x y 2 and N x x 2 y
M
N
1 2 x y and
1 2x y
y
x
Clearly
M N
y
x
Equation (1) is not exact
...
Sonendra Gupta [9893455006]
Page 15
Mathematics-II [BT-202]
Module-I
Mx Ny x y x y 2 y x x 2 y
x y x2 y 2 x y x2 y 2
2 x2 y 2 0
The integrating Factor I
...
Multiplying equation (1) by
1
2x 2 y 2
E-Notes
1
1
2 2
Mx Ny 2 x y
, we get
y x y2
x x2 y
dx
2 2 dy 0
2 2
2 x y
2 x y
1 1
1
1 1
1
2 dx 2 dy 0
2 x y x
2 x y
y
1
1
1
1
2 dx 2 dy 0
y
x y x
x y
1
1
1
1
Equation (2) is exact and M
and N
2
2
y
x y x
xy
… (2)
Therefore the solution of equation (2) is
y is contant M dx Independent of x N dy c
1
1
1
2 dx
dx c
y is contant x y x
Independent of x y
1
log x log y log c
xy
x 1
log
cy xy
x
1/ x y
e
cy
1/ x y
x c y e
x c y e
Dr
...
1
...
x3
3 y3
log y log c ]
y x y2 dx x x2 y dy 0
[Answer: x c y e x y ]
3
...
Method 3: If
1 M N
x dx
x , then I
...
e
N y
x
Method 4: If
1 N M
y dy
y , then I
...
e
M x
y
Problem 6
Solve: x 2 y 2 2 x dx 2 y dy 0
Solution: Given differential equation is
x2 y2 2x dx 2 y dy 0
… (1)
Here M x 2 y 2 2 x and N 2 y
M
N
2 y and
0
y
x
Clearly
M N
y
x
Equation (1) is not exact
...
F
...
Sonendra Gupta [9893455006]
Page 17
Mathematics-II [BT-202]
Module-I
E-Notes
x2 y2 2x e xdx 2 y e xdy 0
Equation (2) is exact and M e x x 2 y 2 2 x and N 2 y e x
… (2)
Therefore the solution of equation (2) is
y is contant M dx Independent of x N dy c
x 2
2
y is contant e x y 2 x dx No term free from x c
x 2
2
x
e x 2 x dx y e dx c
x 2 e x y 2e x c
Problem 7
Solve: 3x 2 y 4 2 x y dx 2 x3 y 3 x 2 dy 0
Answer
[RGPV Dec
...
Since Equation (1) is not homogeneous and not in the form y f1 x y dx x f 2 x y dy 0 , then
1 N M
1
6 x 2 y 3 2 x 12 x 2 y 3 2 x
2 4
M x
y 3 x y 2 x y
1
2
2
y 3x y 2 x
3
2
y 3x y 2 x
3
6 x 2 y 3 4 x
3 6 x 2 y 3 2 x
2
y
y
1
y dy
2/ y dy
e
e 2log e y 2
The integrating Factor I
...
e
y
Multiplying equation (1) by
1
y2
, we get
3x 2 y 4 2 x y
2 x3 y 3 x 2
dx
dy 0
y2
y2
Dr
...
1
...
x y3 y dx 2 x 2 y 2 x y 4 dy 0
2
...
x
...
Sonendra Gupta [9893455006]
Page 19
Mathematics-II [BT-202]
Module-I
E-Notes
Lecture-07
Chapter – 2
LINEAR DIFFERENTIAL EQUATION FIRST ORDER AND HIGHER DEGREE
2
...
Here we
discuss the following methods
Method 1
Solvable by p
Method 2
Solvable by x
Method 3
solvable by y
Method 4
Clairaut’s Equation
Method 1 [Solvable by p]
Concept: If the differential equation in polynomial form of p, then it is solvable by p
...
2019]
Solution: Given: p 2 5 p 6 0
p 2 p 3 0
p 2 and p 3
Now,
p2
dy
2
dx
Integrating on both sides, we get
y 2 x c1
y 2 x c1 0
Now,
...
(2)
The required solution is,
y 2 x c y 3x c 0
Problem 2
where c1 c2 c
Solve: x 2 p 3 y 1 x 2 y p 2 y 3 p 0 , where p
Answer
dy
dx
[RGPV June 2019]
Solution: Given: x 2 p 3 y 1 x 2 y p 2 y 3 p 0
Dr
...
(1)
Integrating on both sides, we get
1
1
x c2 x c2 0
y
y
and
...
(3)
where c1 c2 c3 c
Answer
2
Problem 3
dy
dy
Solve: x 2 xy 6 y 2 0
dx
dx
[RGPV Dec
...
Sonendra Gupta [9893455006]
Page 21
Mathematics-II [BT-202]
Taking, p
Module-I
E-Notes
dy
dx
x 2 p 2 xyp 6 y 2 0
…(1)
x 2 p 2 3 xyp 2 xyp 6 y 2 0
xp px 3 y 2 y px 3 y 0
px 3 y px 2 y 0
px 3 y 0
dy
dx
3
y
x
x
log y 3log x log c1
dy
dx
2
y
x
log y 2log x log c2
log y log c2 x 2
y x2c2
px 2 y 0
c
log y log 13
x
c
y 13
x
x3 y c1 0
… (2)
dy
2y 0
dx
y x2 c2 0
… (3)
The required solution is,
x3 y c y x2 c 0
Answer
IMPROVE YOUR SELF
Solve the following differential Equations:
1
...
Solve: p 2 2 py cot x y 2 0
[RGPV Dec
...
Solve: x 2 xy 6 y 2 0
dx
dx
[Answer: x3 y c y x 2 c 0 ]
Dr
...
2019]
Solution: Given differential Equation is,
y 2 px y 2 p 3
… (1)
Clearly it is solvable for x, then
y
2x y2
...
r
...
y, we get
2
p
...
dp
dy
2
dp
2
y
2
p
p
2
y
dy
p2
2 1
y dp
dp
2
2 py 2
2 p2 y
p p p dy
dy
dx
dy
y
2
2 2 py
p
1
dp y
1 2 p3 y
1 2 p3 y
p
dy p 2
1
dp
2 p2 y
p
dy
dp y
dy p
1
dp
dy
p
y
Integrating both sides, we get
log p log y log c
c
log p log
y
c
p
y
Putting the value of p in equation (1), we get
c
c
y 2 x y2
y
y
3
y 2 2c c 3
Answer
Problem 2
Solve: y 2 log y x y p p 2
[RGPV June 2015]
Solution: Given differential equation is
Dr
...
r
...
y, we get
1
dp
dp
p y log y
...
1
y
dy
dx
dy
dy
p2
y2
1 1 log y y log y dp 1 dp p
p p
p
p 2 dy y dy y 2
y log y 1 dp log y p
2 2
2
dy
y
p
p
y
y 2 log y p 2 dp y 2 log y p 2
2
dy
p
y
p y2
dp 1 1
dy p y
dp dy
p
y
Integrating both sides, we get
log p log y log c
log p log y c
p yc
Putting in equation (1), we get
y 2 log y x y y c y c
2
log y c x c 2
Answer
IMPROVE YOUR SELF
Solve the following differential Equations:
1
...
y p 2 2 x p y 0
[Answer: y 2 c 2 2cx ]
p
3
...
Sonendra Gupta [9893455006]
Page 24
Mathematics-II [BT-202]
[Answer: y
1
p 1
2
Module-I
E-Notes
c]
Lecture-08
Method 3 [Solvable by y]
Concept: The differential equation is said to be solvable for y when y can be separate in term of x such
that
y f x, p
Problem 1
Solve: y p x x 4 p 2
Solution: Given differential equation is
y p x x4 p2
…(1)
Differentiate with respect to x, we get
dy
dp
dp
p
...
x 4 2 p 4 x3 p 2
dx
dx
dx
p p x
...
Dr
...
cos p
dx
dx
dx
dx
dp
p cos p p sin p cos p
dx
E-Notes
dy
Q p dx
dp
dx
p p sin p
dx sin p dp
Integrating both sides, we get
x cos p c
cos p c x
Since sin p 1 cos 2 p 1 c x
2
Putting in equation (1), we get
y 1 c x p c x
2
1 c x y
2
p
cx
Putting the value of p in equation (1) we get
2
1 c x y
cos
cx
cx
Answer
IMPROVE YOUR SELF
Solve the following differential Equations:
1
...
y 2 px tan 1 x p 2
c2
[Answer: y c x tan 1 ]
4
3
...
Sonendra Gupta [9893455006]
Page 26
Mathematics-II [BT-202]
Concept:
Module-I
E-Notes
Method 4 [Clairaut’s Equation]
If the differential equation can we written in the form
y p x f p , then it is called Clairaut’s equation and solution is y c x f c
Problem 1
Solve: y px p 1 p
Solution: Given differential equation is
y px p 1 p
y px
… (1)
p
p 1
y px
p
p 1
This is required Clairaut’s equation, and then required solution is
c
y cx
c 1
Problem 2
Answer
Solve: p 2 x 2 a 2 2 p x y y 2 a 2 0
Solution: Given differential equation is
p2 x2 a2 2 p x y y 2 a2 0
p2 x2 a2 p2 2 p x y y 2 a2 0
y px 2 a 2 p 2 1
y p x a p2 1
y p x a p2 1
… (1)
p2 x2 2 p x y y 2 a2 p2 1
This is required Clairaut’s equation, and then required solution is
y c x a c2 1
Answer
IMPROVE YOUR SELF
Solve the following differential Equations:
1
...
sin px cos y cos px sin y p
[Answer: y cx sin 1 c ]
3
...
Sonendra Gupta [9893455006]
Page 27
Mathematics-II [BT-202]
Module-I
E-Notes
[Answer: y c 1 cx c 1 c ]
S
...
Video Lectures Links
Topic
Lecture No
Link
1
Lecture 0
Differential Equation-Ordinary Differential Equation
https://youtu
...
be/nyKypjThv0c
3
Lecture 2
Ordinary Differential Equation-Variable Separable form
https://youtu
...
be/s-JdP_ohMmA
25
Lecture 2
Homogeneous Linear Differential Equation of Higher Order in Hindi
https://youtu
...
be/nyxipXkzCfo
Ordinary Differential Equation-Homogeneous Differential Equation
of first order
Ordinary Differential Equation-Non-Homogeneous Differential
Equation of First order
Ordinary Differential Equation-Linear Differential Equation of First
order
Ordinary Differential Equation-Bernoulli's Equation of First order
Ordinary Differential Equation-Exact differential Equation of First
order and First Deg
Ordinary Differential Equation-Non Exact differential Equation of
First order
Ordinary Differential Equation-Non Exact differential Equation of
First order
Ordinary Differential Equation-Non Exact differential Equation of
First order
Ordinary Differential Equation-Non Exact differential Equation of
First order
Ordinary Differential Equation-Solve by Inspection Method
Ordinary Differential Equation-First Order and Higher degree
Solvable for p
Ordinary Differential Equation-First Order and Higher degree
Solvable for y
Ordinary Differential Equation-First Order and Higher degree
Solvable for x
Ordinary Differential Equation-First Order and Higher degree
equation Clairaut's Equation
Linear Differential Equations of Higher order with constant
coefficients (Updated)
Linear Differential Equations of Higher order with constant
coefficients (updated)
Linear Differential Equations of Higher order with constant
coefficients (Updated)
Linear Differential Equations of Higher order with constant
coefficient (Updated)
Linear Differential Equations of Higher order with constant
coefficient (Updated)
Linear Differential Equations of Higher order with constant
coefficient (Updated)
Dr
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Lecture-09
CHAPTER – 3
LINEAR DIFFERENTIAL EQUATION OF HIGHER ORDER WITH CONSTANT COEFFICIENT
3
...
an y Q
…… (1)
Where a1, a2 ----- an-1, an are constant and Q is the function of independent variable x (only) is called the linear
differential equation of higher order with constant coefficient
...
an y Q
f D y Q
… (2)
The Solution of Equation is,
y C
...
I
Note: (i)
...
2
… (3)
If Q = 0, then P
...
F
Method to find complementary function when Q = 0
From equation (2), we have f D y 0
The Auxiliary equation (A
...
) is
f m 0
a0 mn a1 mn 1 a2 mn 2
...
Case 1: When the roots of A
...
are distinct and real
Say m m1, m2 , m3
C
...
c1 em1 x c2 em2 x c3 e m3 x
Case 2:When the roots of A
...
are real and repeated
Say
m m1, m1, m1
C
...
(c1 c2 x c3 x 2 ) e m1 x
Case 3:When the roots of A
...
are in pair of complex number
Say, m i
Dr
...
F
...
F
...
F
...
E
...
F
...
F
...
F
...
E
...
F
...
2019]
Solution: Given:
D2 1 y 0
The Auxiliary Equation is,
m2 1 0
m 1, 1
The C
...
is,
C
...
c1 e x c2 e x
The Complete Solution is,
y C
...
c1 e x c2 e x
Dr
...
2019]
Solution: Given, D 2 4 D 3 y 0
… (1)
The A
...
equation is,
m 2 4m 3 0
m 3 m 1 0
m 1, 3
The complete solution is
y C
...
c1e x c2e3x
Answer
IMPROVE YOUR SELF
Solve the following differential Equations:
1
...
3
...
I
...
an y Q
a0 D n a1 D n 1 a2 D n 2
...
I
...
I
...
D Q
dx
1
Q Q dx
2
...
Sonendra Gupta [9893455006]
Page 31
Mathematics-II [BT-202]
Module-I
E-Notes
Method 1:
When Q = eax
1
1 ax
(i)
...
1
1
ea x ea x
...
ea x
e , when f (a ) 0
f ( D)
f D
D
(iv)
...
E
...
F
...
I
...
F
...
I
...
I
...
E
...
Sonendra Gupta [9893455006]
Page 32
Mathematics-II [BT-202]
Module-I
E-Notes
m 2 3m 2 0
m 2 m 1 0
m 2, 1
C
...
c1e2 x c2e x
The C
...
is
P
...
1
D 3D 2
2
1
e2 x
e2 x
D 2 D 1
e2 x
1 1
e2 x
D 2 D 1
1 1 2x
1
e
e2 x
D 2 2 1
D
2
1
1
...
1
D 2 2
D
P
...
e 2 x x
The required solution is,
y c1e2 x c2e x e2 x x
Problem 3 Solve the differential equation
Answer
d2y
dx
2
6
dy
9 y 5 e3 x
dx
[RGPV June 2018]
Solution: Given differential equation is
d2y
dx
2
6
dy
9 y 5 e3 x
dx
… (1)
D 2 6 D 9 y 5 e3 x
The A
...
is
m 2 6m 9 0
m 32 0
Now,
m 3, 3
C
...
C1 x C2 e3 x
P
...
P
...
1
D 6D 9
1
2
3
2
5 e3 x
6 3 9
5 e3 x
Here f (3) 0
5 3x
e
64
Dr
...
F
...
I
...
E
...
F
...
F
...
I
...
1
3
3
0 1
D
1
1
3
1
xx
e 3
...
I
...
Sonendra Gupta [9893455006]
Answer
[RGPV Dec
...
E
...
F
...
I
...
I
...
F
...
I
...
d2y
dy
3 2 y e5 x
2
dx
dx
2
...
2019][Answer:
3
...
Sonendra Gupta [9893455006]
y c1e 2 x c2 xc3 x 2c4 e x
x3 e x
]
18
3
3 x
1/2 x
x c2 sin
x e ]
[Answer: y e c1 cos
2
2
Page 35
Mathematics-II [BT-202]
Module-I
E-Notes
Method 2: When Q = xm, where m I+
1
1
x m f ( D) x m
f ( D)
1
Where f ( D) is expand as per Binomial theorem
...
(1 D ) 1 1 D D 2 D 3 D 4
...
(1 D ) 1 1 D D 2 D3 D 4
...
(1 D) 2 1 2 D 3D 2 4 D3 5 D 4
...
(1 D) 2 1 2 D 3D 2 4 D3 5 D 4
...
E
...
Complementary function is
C
...
c1 c2e x c2e2 x
Now,
P
...
1
D 3D 2 D
3
1
2D
2
x2
D 2 3D
1
2
1
x2
2
1 D 2 3D D 2 3D
2
1
...
1
2D
1
1 2
7
2 1
x 2 2 6 x 4 0 18 0
...
I
...
Sonendra Gupta [9893455006]
Page 36
Mathematics-II [BT-202]
Module-I
y c1 c2e x c2e 2 x
1 x3 3 x 2 7 x
2 3
2
2
d2 y
dy
6 9 y x 2 2e2 x
2
dx
dx
Solution: Given differential equation is
Problem 2
E-Notes
Answer
[RGPV Dec
...
d
D
Suppose,
dx
D
2
6 D 9 y x 2 2e2 x
The A
...
is
m 2 6m 9 0
m 3
2
0
m 3, 3
The C
...
F
...
I
...
x 2 2
e
2
9
3
3
2 3
1 2 2
2 2
2
2x
x 3 D x 3 D x
...
I
...
F
...
I
...
Sonendra Gupta [9893455006]
Page 37
Mathematics-II [BT-202]
Module-I
y c1 xc2 e3 x
1 2 4x
x
9
3
2
2 e2 x
3
E-Notes
Answer
IMPROVE YOUR SELF
Solve the following differential Equations:
1
1
[Answer: y c1 cos 2 x c2 sin 2 x x 2 ]
4
2
1
...
d3y
dx
3
...
2
2
f (D )
f (a )
(ii)
...
1
D a
1
2
2
sin ax
cos ax
x
cos ax, When f (a 2 ) 0
2a
x
sin ax, When f (a 2 ) 0
2a
D a
Problem 1
Solve the differential equation
2
2
( D 2 D 1) y sin x
Solution: Given differential equation is
( D 2 D 1) y sin x , Here, D
d
dx
The A
...
is,
m2 m 1 0
m
1 1 4 1 3 i
1
3
i
2
2
2 2
3
3
x c2 sin
x
C
...
e x /2 c1 cos
2
2
Now,
P
...
1
sin x
1
D D 1
1 D 1
1
sin x cos x
D
2
Dr
...
E
...
F
...
F
...
I
...
1
5 D 1 D 1
4 D 1 1
1
1
3
1
D 1 2 cos 2 x e x
...
x
5
2 1
4
P
...
D3 D 2 D 1
cos 2 x
1
2 2 D 2 2 D 1
1
5
cos 2 x
3 1
x
e
D 1 1 12
3 1
1
x
D 1 cos 2 x 4 D 1 e
1
3
2sin 2 x cos 2 x x e x
25
4
The solution is,
Dr
...
2
...
d2y
dx 2
d2y
dx 2
d2y
dx 2
1
3sin 2 x cos 2 x 1 ]
10
dy
2 y 4 cos 2 x
dx
[Answer: y c1e x c2e 2 x
8
dy
9 y 40sin 5 x
dx
[Answer: y c1 cosh
4
dy
4 y x 2 e x cos 2 x
dx
1
3
1
[Answer: y c1 xc2 e 2 x x 2 2 x e x sin 2 x ]
4
2
8
3
7 x c2
5
5cos 5 x 2sin 5 x ]
29
Method 4: When Q = V
...
V eax
V
f ( D)
f ( D a)
Method 5: When Q = x
...
V x
V
...
This method is applicable when power of x is an unity
...
This method is fail when V operate on f (D), which is zero
...
2019]
Solution: Given differential equation is
d2y
dy
4 y e x cos x
dx
… (1)
D2 2D 4 y e x cos x
… (2)
dx
Suppose D
d
dx
2
2
The A
...
is
m 2 2m 4 0
Dr
...
F
...
I
...
F
...
I
...
I
...
E
...
F
...
I
...
Sonendra Gupta [9893455006]
Page 41
Mathematics-II [BT-202]
Module-I
E-Notes
1
1
2 x
e2 x 2
sin 2 x e
2
sin 2 x
D D
2 D
D 4
D4
1
e2 x
sin 2 x e2 x 2
sin 2 x
D 4 D 4
D 16
1
e 2 x
e 2 x D 4 2
sin 2 x
D sin 2 x 4sin 2 x
20
2 16
P
...
e 2 x
e 2 x
2cos 2 x 4sin 2 x
cos 2 x 2sin 2 x
20
10
The solution is,
y c1e
Problem 3
3 x
c2e
2 x
e2 x
cos 2 x 2sin 2 x
10
Answer
Solve: D 2 2 D 1 y x cos x
[RGPV Dec
...
E
...
F
...
I
...
I
...
Sonendra Gupta [9893455006]
Page 42
Mathematics-II [BT-202]
y c1 xc2 e x
Module-I
E-Notes
x
1
sin x sin x cos x
2
2
Answer
IMPROVE YOUR SELF
Solve the following differential Equations:
1
...
d2y
dx 2
d2y
dx
2
3
dy
2 y x ex
dx
x2
[Answer: y c1e x c2e 2 x e x x ]
2
dy
y x e x sin x
dx
[Answer: y c1 xc2 e x e x x sin x 2cos x ]
2
3
...
1 m
1 m ai x
1
x m sin ax I
...
x (cos ax i sin ax) I
...
x e
f ( D)
f ( D)
f ( D)
1 m
1 m a ix
1
x m cos ax R
...
x (cos ax i sin ax) R
...
x e
f ( D)
f ( D)
f ( D)
Where m ≥ 1 and m is positive integer
...
D
1
Q e x e x Q dx
(iv)
...
Dr
...
1 Concept:
The standard form of HLDE is,
a0 x n
dny
dx n
d n 1 y
a1 x n 1
dx n 1
a2 x n 2
d n2 y
dx n 2
...
To reduce HLDE in LDE of higher order,
x e z z log x
And x
d
d
d2
d3
D, x 2 2 D ( D 1), x3 3 D ( D 1) ( D 2) as D
dz
dx
dx
dx
Solve: x 2
Problem 1
d 2y
dx
2
2x
dy
4y x 2 2log x
dx
[RGPV June 2019, 2015]
Solution: Given differential equation is
x2
d 2y
dx 2
2x
dy
4y x 2 2log x
dx
… (1)
This is homogeneous linear differential equation
...
E
...
F
...
I
...
Sonendra Gupta [9893455006]
2
2z
Page 44
Mathematics-II [BT-202]
Module-I
1
22 3(2) 4
e
2z
1 D 2 3D
1
4
4
1
2z
1 2 z 1 D 2 3D
e 1
...
6
2
4
1 2z 1
1
1
z 3
e z (0 3) e2 z
6
2
4
6
2 8
P
...
E-Notes
x 2 log x 3
6
2
8
The required solution is,
y C
...
P
...
y c1 x 1 c2 x 4
Problem 2
Solve: x
2
x 2 log x 3
6
2
8
d2y
2
x
Answer
dy
4y 0
dx
[RGPV Dec
...
x ez
So put
z log x and
dz 1
dx x
d
d
d2
D, x 2 2 D ( D 1) as D
dz
dx
dx
then equation (1), becomes
and x
D( D 1) D 4 y 0
D2 4 y 0
The A
...
is,
m2 4 0
m 2i
C
...
c1 cos z c2 sin z c1 cos log x c2 sin log x
The complete solution is
y C
...
c1 cos log x c2 sin log x
Dr
...
x ez
So put
z log x and
dz 1
dx x
d
d
d2
D, x 2 2 D ( D 1) as D
dz
dx
dx
then equation (1), becomes
and x
D( D 1) D 1 y z
D 2 2 D 1 y z
The A
...
is,
m 2 2m 1 0
m 12 0
m 1, 1
C
...
c1 zc2 e z c1 log x c2 x
P
...
1
D 1
2
2
1 D
z
z 1 2 D 3D 2
...
I z 2
...
F
...
I
...
x
2
d2y
dx
2
5x
dy
4 y x log x
dx
Dr
...
x
2
d2y
dx
3
...
2019]
… (2)
…(3)
…(4)
Eliminate y from equation (3) and (4), we get
D 5 D 7 x D 5 y 0
2 x D 5 y 0
(Subtraction)
D 5 D 7 2 x 0
D2 12D 37 x 0
… (5)
The A
...
is,
m 2 12m 37 0
m
12 122 4 1 37 12 144 148
2 1
2
m
12 2 i
6 i
2
x e6t c1 cos t c2 sin t
Answer
Differentiate w
...
t
...
Sonendra Gupta [9893455006]
Page 47
Mathematics-II [BT-202]
Module-I
E-Notes
dx
e6t 6c1 c2 cos t 6c2 c1 sin t
dt
From equation (1), we get
dx
y 7x
dt
7 e6t c1 cos t c2 sin t e6t 6c1 c2 cos t 6c2 c1 sin t
e6t 7c1 6c1 c2 cos t 7c2 6c2 c1 sin t
y e6t c1 c2 cos t c1 c2 sin t
Answer
Problem 2
Solve the following simultaneous differential equations
...
E
...
F
...
I
...
I
...
Sonendra Gupta [9893455006]
Page 48
Mathematics-II [BT-202]
x c1 tc2 e
Module-I
4 t
4 et e 2t
25 36
E-Notes
… (6)
Answer
dx
4 e t 2e 2 t
e 4 t 4 c1 tc2 c2
dt
25
36
From equation (1), we have
dx
y
5 x et
dt
and
e 4 t 4 c1 tc2 c2
4 e t 2e 2 t
4 e t e 2t t
5 c1 tc2 e 4 t
e
25
36
25 36
y e 4t 4c1 4tc2 c2 5c1 5tc2
y c2 tc2 c1 e
et
7
e 2t
25 25
et
7
e 2t
25 25
4t
Answer
IMPROVE YOUR SELF
Solve the following differential Equations:
dy
dx
x cos t
y sin t and
1
...
dt
dt
[Answer: x c1et c2et and y c1et c2et sin t ]
[Answer: x c1e4t c2et and y
3
...
No
...
be/GtZnXWdTD5Q
2
Lecture 1
Ordinary Differential Equation-Formation of Differential Equation
https://youtu
...
be/C_2ScUovh8Y
4
Lecture 3
5
Lecture 4
6
Lecture 5
7
Lecture 6
Ordinary Differential Equation-Bernoulli's Equation of First order
https://youtu
...
be/DkztYqeADIE
Ordinary Differential Equation-Homogeneous Differential Equation
of first order
Ordinary Differential Equation-Non-Homogeneous Differential
Equation of First order
Ordinary Differential Equation-Linear Differential Equation of First
order
Dr
...
be/uJMDxnr2B8s
https://youtu
...
be/bJ1TqnXV_D8
Page 49
Mathematics-II [BT-202]
Module-I
Ordinary Differential Equation-Non Exact differential Equation of
First order
Ordinary Differential Equation-Non Exact differential Equation of
First order
Ordinary Differential Equation-Non Exact differential Equation of
First order
Ordinary Differential Equation-Non Exact differential Equation of
First order
E-Notes
9
Lecture 8
10
Lecture 9
11
Lecture 10
12
Lecture 11
13
Lecture 12
14
Lecture 13
15
Lecture 14
16
Lecture 15
17
Lecture 16
18
Lecture 1
19
Lecture 2
20
Lecture 3
21
Lecture 4
22
Lecture 5
23
Lecture 6
24
Lecture 1
Homogeneous Linear Differential Equation of Higher Order in Hindi
https://youtu
...
be/Abyz2W7aYQQ
26
Lecture 1
Simultaneous Ordinary Differential Equations in Hindi
https://youtu
...
be/JL9eJxw5V0g
Ordinary Differential Equation-Solve by Inspection Method
Ordinary Differential Equation-First Order and Higher degree
Solvable for p
Ordinary Differential Equation-First Order and Higher degree
Solvable for y
Ordinary Differential Equation-First Order and Higher degree
Solvable for x
Ordinary Differential Equation-First Order and Higher degree
equation Clairaut's Equation
Linear Differential Equations of Higher order with constant
coefficients (Updated)
Linear Differential Equations of Higher order with constant
coefficients (updated)
Linear Differential Equations of Higher order with constant
coefficients (Updated)
Linear Differential Equations of Higher order with constant
coefficient (Updated)
Linear Differential Equations of Higher order with constant
coefficient (Updated)
Linear Differential Equations of Higher order with constant
coefficient (Updated)
Dr
...
be/Gw3xu3NzlX4
https://youtu
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be/XCn6UXbyYtc
https://youtu
...
be/iOpkpv_cE68
https://youtu
...
be/EU93LOvQoCw
https://youtu
...
be/1tkpbmKnI2Q
https://youtu
...
be/me2Y7Rckj8c
https://youtu
...
be/VJ2_8ZfduoM
https://youtu
...
be/Yh-oNF2_qRs
Page 50
Title: Ordinary differential equation
Description: Differential equation of first order and first degree,beenoulie equation
Description: Differential equation of first order and first degree,beenoulie equation