This document discusses first order ordinary differential equations (ODEs) and their applications. It begins by defining differential equations and classifying them as ordinary or partial based on the number of independent variables. First order ODEs are then introduced, along with methods for forming and solving them. Examples are given of linear and non-linear first order ODEs. The document also outlines some common applications of first order ODEs, such as modeling Newton's law of cooling, natural growth and decay processes, and trajectories in physics problems.

1. By
Dr.K. LakshmiNarayan
Dr.K . KondalaRao
MATHEMATICSII
VIDY JYOTHI INSTITUTE OF TECHNLOGY
1

2. Ordinary Differential
Equations & Vector
Calculus
Unit I
Unit II
Unit III
Unit IV
Unit V
First order
ODE and their
Applications
Higher Order
Linear
Differential
Equations
Laplace
transforms
Vector
Integration
Multiple
Integrals &
Vector
Differentiation
2

3. UNIT-I :First order ODE and their Applications
Course Outcome:
Classify the various types of differential equations of first order and first
degree and apply the concepts of differential equations to the real world
problems. 3

4. UNIT-I I: Higher Order Linear Differential Equations
UNIT-II:
Higher Order Linear
Differential Equations
Linear differential equations of second and
higher order with Constant Coefficients
and Variable Coefficients
Method of variation of
parameters
Course Outcome:
Solve higher order differential equations and apply the concepts of
differential equations to the real world problems. 4
   
x
V
x
x
V
e
x
ax
ax
e
k
ax
k
ax
,
&
,
cos
,
sin
,
f(x)
type
the
of
term
RHS
ts
Coefficien
Constant

Variable
Coefficients
Cauchy- Eulers
and Legendre’s
Equations
Applications:
Electric Circuits

5. UNIT-I II: Laplace transforms
UNIT-II:
Laplace
transforms
Inverse
Transforms
Applications
Course Outcome:
Find the Laplace Transform of various functions and apply to find
the solutions of differential equations.
Laplace
Transforms
Introduction
First shifting Theorem,
Laplace Transforms of
derivatives and integrals
 Unit step function
second shifting theorem
 Dirac’s delta function
Convolution theorem
 Periodic function
Differentiation and
integration of transforms
Application of Laplace
transforms to ordinary
differential equations.
5

6. UNIT-I V: Multiple Integrals& Vector Differentiation
UNIT-IV:
Multiple Integrals &
Vector Differentiation
Course Outcome:
Evaluate the multiple integrals and identify the vector differential
operators physically in engineering problems.
Multiple
Integrals
Vector
differentiation
Double and triple
integrals
Change of order of
integration
Change of
variables.
Gradient
Divergence
Curl
Properties
Laplacian and
second order
operators.
.
6

7. UNIT-V: Vector Integration
UNIT-II:
Vector
Integration
Course Outcome:
Evaluate the line, surface and volume integrals and converting them
from one to another by using vector integral theorems.
Line integral,
work done,
Surface Integrals
Volume integrals.
Green’s Theorem
Stoke’s Theorem
Gauss Divergence
Theorems
(Only Verification)
Vector
Integration
Vector
integrals
theorems
7

8. COURSE OUTCOMES
After learning the contents of this course the students must be able to:
Classify the various types of differential equations of first order and first degree
and apply the concepts of differential equations to the real world problems.
Solve higher order differential equations and apply the concepts of differential
equations to the real world problems.
Find the Laplace Transform of various functions and apply to find the solutions
of differential equations.
Evaluate the multiple integrals and identify the vector differential operators
physically in engineering problems.
Evaluate the line, surface and volume integrals and converting them from one
to another by using vector integral theorems.
8

10. UNIT-I
First order ODE and their Applications
10

11. TOPICS
Introduction Formation Solutions Applications
Exact &
Non Exact
Linear Bernoulli’s
Orthogonal
Trajectories
Newton’s Law
of Cooling
Natural law of
Growth &
Decay
11

12.  Differential equations have wide applications in various engineering and science
disciplines.
 In general, modeling of the variation of a physical quantity, such as temperature,
pressure, displacement, velocity, stress, strain, current, voltage, or concentration of a
pollutant, with the change of time or location, or both would result in differential
equations.
 Similarly, studying the variation of some physical quantities on other physical
quantities would also lead to differential equations.
 In fact, many engineering subjects, such as mechanical vibration or structural
dynamics, heat transfer, or theory of electric circuits, are founded on the theory of
differential equations.
 It is practically important for engineers to be able to model physical problems using
mathematical equations, and then solve these equations so that the behavior of the
systems concerned can be studied.
 That is Many of general laws in all branches of Engineering, Physics, Biology,
Astronomy, Populations Studies, Economics, etc., can be expressed through
“Mathematical Modelling” in the form of an equation connecting the variables and
their rates of change, resulting in Differential Equations.
12

13. Basic Definition of Differential Equation:
An equation involving an independent variable, dependent variable and the
differential coefficients of dependent variable with respect to independent variable is
called a Differential Equation.
(OR)
An equation which involves the differential co-efficients is called a Differential
Equations
Classification of Differential Equation:
Generally Differential Equations are classified into two categories “Ordinary
and Partial” depending on the number of independent variables appearing in the
equation.
Ordinary Differential Equation:
Differential Equations in which the derivatives involved are with respect to a
single independent variables are called Ordinary Differential Equations.
That is an ordinary differential equation is differential equation which the
dependent variable ‘y’ depends only on one independent variable say ‘x’.
Example: and
are ODE’s
0
)
,.......,
,
,
,
( )
(



 n
y
y
y
y
x
F
y
x
dx
dy
Tanx
y
dx
dy
dx
y
d




 2
3
5 2
2
13

14. Partial Differential Equation:
If the derivatives are with partial derivatives, that is they are with respect to
two or more independent variables then the equation is called Partial Differential
Equations.
That is a Partial Differential Equation is one in which the dependent variable
‘y’ depends on two or more independent variable say ‘x, t, ……’.
Example : 0
,.......)
,
,
,
,
,
,
( 2
2
2
2









t
y
t
y
x
y
x
y
y
t
x
F
14

15. Order of A Differential Equation:
The order of a differential equation is the order of the highest order derivative
involved in the differential equation.
Degree of A Differential Equation:
The degree of a differential equation is the degree of the highest order
derivative, when the differential co-efficients are made free from radicals and
fractions so far derivatives are concerned.
Examples:
15

16. Some More Examples
16

17. Linear Differential Equation:
An nth order ordinary differential equations in the dependent variable ‘y’ is said
to be Linear in ‘y’ if
 ‘y’ and all its derivatives are of same degree ‘1’
 No product terms of ‘y’ and/or any of its derivatives are present
 No transcendental functions of ‘y’ and/or its derivatives occur.
Non-Linear Differential Equation:
An ordinary differential equation which is not linear is called as non-linear
 The General form of nth order Linear Ordinary Differential Equation in ‘y’ with
variable co-efficients is
(1)
 Where are given functions of and
 If all the coefficients are the constants then the above equation is
known as nth order Linear Ordinary Differential Equation with constant coefficients.

)
(
)
(
)
(
)
(
)
(
)
( 1
2
2
2
1
1
1
0 x
Q
x
a
dx
dy
x
a
dx
y
d
x
a
dx
y
d
x
a
dx
y
d
x
a n
n
n
n
n
n
n
n










 




)
(
...
),........
(
),
(
),
(
&
)
( 2
1
0 x
a
x
a
x
a
x
a
x
Q n
0
)
(
0 
x
a
'
'x
n
a
a
a
a ,......
,
, 2
1
0
17

18. Solutions of Differential Equation:
A Solution or Primitive of a differential equation is relation free from
derivatives between the variables involved, which satisfies the differential equation.
(OR)
Any relation connecting the variables of an equation and not involving their
derivatives, which satisfies the given differential equation is called Solution.
Types of Solutions
General Solution:
The solution of a differential equation that contains a number of arbitrary
constants equal to the order of the differential equation is called the General
Solution or The Complete Integral or Complete Primitive.
Particular Solution:
The solution obtained from the general solution by giving particular values to
the arbitrary constants is called Particular Solution.
Singular Solution:
Singular Solution of a differential equation are solutions of differential
equations which cannot be obtained from the general solution.
18

19. 2.Formation of Differential Equations:
Suppose the relation between the dependent variable ‘y’ and independent
variable ‘x’ with constants from which the differential equation
is to be formed is
(2.1)
and the Differential Equation is
(2.2)
So, the Differential Equation formed should be free from the constants
.
So, the Differential Equation can be formed by eliminating the constants
, from the relation
Differentiating the equation (2.1) n -times, we get n-equations and given
equation (2.1), from these ‘n+1’ equations , eliminating , we get
an equation of the form (2.2) which the required Differential Equation of nth order.
n
c
c
c
c ,.......,
,
, 3
2
1
0
)
,.......,
,
,
,
,
( 3
2
1 
n
c
c
c
c
y
x
f
0
)
,.......,
3
,
,
,
,
(
3
2
2

n
n
dx
y
d
dx
y
d
dx
y
d
dx
dy
y
x
f
n
c
c
c
c ,.......,
,
, 3
2
1
.
0
)
,.......,
,
,
,
,
( 3
2
1 
n
c
c
c
c
y
x
f
n
c
c
c
c ,.......,
,
, 3
2
1
n
c
c
c
c ,.......,
,
, 3
2
1
19

20. Examples:
1. Form the differential equation of simple harmonic motion given by
.
Solution: (1)
(2)
2. Form the differential equation by eliminating ‘c’ from
3. Form the differential equation of all circles of radius
4. Form the differential equation of all circles passing through the origin and having
their centre's on the x-axis.
)
( B
nt
Cos
A
X 

equation
al
differenti
required
the
is
Which
n
x
x
n
x
n
B
nt
Cos
A
dt
x
d
get
we
t
to
respect
with
ating
Differenti
n
B
nt
Sin
A
dt
dx
get
we
t
to
respect
with
ating
Differenti
B
nt
Cos
A
x
is
lation
Given
0
(1)]
equation
[From
)
(
,
'
'
)
2
(
)
(
,
'
'
)
1
(
)
(
Re
2
2
2
2















C
y
Sin
x
Sin
ii
c
x
y
i 


 
 1
1
2
2
)
(
)
(
)
(
.
)
(
)
( 2
2
2
r
k
y
h
x 



0
2
2
:
int 2
2




 c
fy
gx
y
x
circles
the
of
Equation
H
20

21. 5. Form the differential equation by eliminating a, b, c from the relation
Solution: (1)
(2)
(3)
(4)
x
x
x
e
c
e
b
e
a
y 

 3
2
0
27
8
9
4
3
2
)
4
(
&
)
3
(
),
2
(
),
1
(
,
,
lim
27
8
,
'
'
9
4
,
'
'
3
2
,
'
'
)
1
(
3
2
3
2
3
2
3
2
3
2
3
2
3
2
3
2



































x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
e
e
e
y
e
e
e
y
e
e
e
y
e
e
e
y
from
c
b
a
inating
E
e
c
e
b
e
a
y
get
we
x
to
respect
with
te
differntia
Again
e
c
e
b
e
a
y
get
we
x
to
respect
with
te
differntia
Again
e
c
e
b
e
a
y
x
to
respect
with
ating
Differenti
e
c
e
b
e
a
y
relation
Given
21

22. 6. Form the differential equation from the relation
.
)
(
0
6
7
0
28
7
8
3
4
1
0
0
28
7
0
8
3
0
4
1
1
1
1
,
,
0
1
27
8
1
9
4
1
3
2
1
1
1
1
4
4
1
3
3
1
2
2
3
2
equation
al
differenti
required
the
is
Which
tion
simplifica
After
y
y
y
y
y
y
y
y
y
y
y
y
y
y
y
y
R
R
R
R
R
R
R
R
R
y
y
y
y
e
e
e x
x
x

















































 
x
x
e
B
e
A
xy 2


22

23. Methods find the Solutions of a Differential Equations of the First Order
and First Degree:
There are different methods of solving the Differential Equations of first order
first degree depending upon the type of the equations.
 Variable Separable Method
 Homogeneous Differential Equations Method
 Equations reducible to Homogeneous Equations(Non-Homogeneous Equations)
 Linear Differential Equations
 Bernoulli’s Differential Equations (Non-Linear Differential Equations)
 Exact Differential equations
 Equations Reducible to Exact Differential Equations (No-Exact)
Out of theses we already discussed in the previous classes. Now we will
discuss Exact, Linear and Bernoulli’s Methods to solve the Ordinary Differential
Equations of fist order first degree.
23

24. Definition:
A differential equation which can be obtained from its solution or primitive by
direct differentiation is said to be an Exact without any further transformation such
as elimination etc.
Condition for Exactness:
If are two real valued functions which have continuous
partial derivatives, then a necessary and sufficient condition for differential equation
to be exact is
Procedure to solve the Exact Differential Equation:
Step1: Test for the Exactness of the differential equation by
Step2: If the differential equation is exact, integrate terms in M with respect to ‘x’
keeping ‘y’ as constant.
Step3: Integrate the terms in N which are free from ‘x’ with respect to ‘y’.
Step 4: The sum of Step2 and Step3 equate to an arbitrary constant will be the
required solution
That is Solution of exact differential equations is
)
,
(
)
,
( y
x
N
and
y
x
M
0

 Ndy
Mdx
x
N
y
M





0

 Ndy
Mdx
x
N
y
M





C
Ndy
Mdx
x
from
free
Terms
t
cons
as
y

 
 '
'
tan 24

25. Example :
equation
al
differenti
given
for
solution
required
the
is
Which
c
y
x
x
y
C
dy
xydx
y
C
Ndy
Mdx
is
equation
al
differenti
Exact
of
solution
the
Now
Exact
is
equation
al
differenti
Given
x
N
y
M
y
x
x
N
x
y
y
M
xy
x
N
xy
y
M
Where
Ndy
Mdx
of
form
the
in
is
Which
dy
xy
x
dx
xy
y
dy
xy
x
dx
xy
y
is
Equation
al
differenti
Given
Solution
dy
xy
x
dx
xy
y
Equation
l
diffeentia
the
Solve
x
from
free
Terms
t
cons
as
y
x
from
free
Terms
t
cons
as
y













































2
2
'
'
tan
2
'
'
tan
2
2
2
2
2
2
2
2
0
2
2
2
&
2
2
)
2
(
&
2
0
)
1
(
0
)
2
(
)
2
(
)
2
(
)
2
(
:
)
2
(
)
2
(
.
1
25

26. Example :
equation
al
differenti
given
for
solution
required
the
is
Which
c
Sinx
Siny
Cosx
C
Cotxdx
dx
Cosy
Cosx
C
dy
dx
Cotx
Cosy
Cosx
C
Ndy
Mdx
is
equation
al
differenti
Exact
of
solution
the
Now
Exact
is
equation
al
differenti
Given
x
N
y
M
Siny
Cosx
x
N
Siny
Cosx
y
M
Siny
Sinx
N
Cotx
Cosy
Cosx
M
Where
Ndy
Mdx
of
form
the
in
is
Which
dy
Siny
Sinx
dx
Cotx
Cosy
Cosx
is
Equation
al
differenti
Given
Solution
dy
Siny
Sinx
dx
Cotx
Cosy
Cosx
Equation
l
diffeentia
the
Solve
t
cons
as
y t
cons
as
y
x
from
free
Terms
t
cons
as
y
x
from
free
Terms
t
cons
as
y




































 




)
log(
0
)
(
&
&
0
)
1
(
0
)
(
)
(
:
0
)
(
)
(
.
2
tan tan
'
'
tan
'
'
tan
26

27. Example 2:
Exact
an
not
is
equation
al
differenti
Given
x
N
y
M
y
x
xy
x
N
y
x
xy
y
M
y
x
y
x
x
N
y
x
xy
y
M
Where
Ndy
Mdx
of
form
the
in
is
Which
dy
y
x
y
x
x
dx
y
x
xy
y
is
Equation
al
differenti
Given
Solution
dy
y
x
y
x
x
dx
y
x
xy
y
Equation
l
diffeentia
the
Solve




































2
2
2
2
2
3
2
3
2
2
2
3
2
3
2
2
2
3
2
3
2
2
3
1
&
3
1
&
0
)
1
(
0
)
(
)
(
:
0
)
(
)
(
.
3
27

28. Non-Exact Differential Equations:
(That is Equations Reducible to Exact Differential Equations)
There are some differential equations which are not directly exact but can be
converted to exact differential equations.
Integrating Factor:
A function, by multiplying the given differential equation can be converted to
exact differential equation is called an Integrating factor.
That is a function which converts non-exact differential equation to exact
differential equation is called Integrating Factor.
There are two methods to find an Integrating factor
(i) Integrating factor by inspection method.
(ii) Integrating factor by using some rules.
Integrating factor by Inspection Method (Group of Terms):
If a student knows the following differentials, some differential equations can
be reduced to exact using these differentials.
28

29. Some Basic Derivatives:
2
4
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
1
2
2
2
2
1
2
2
2
2
2
2
4
2
2
2
2
)
(
.
16
2
2
)
(
.
8
2
2
)
(log(
.
15
2
)
(
.
7
)
(log
.
14
2
)
(
.
6
)
(log
.
13
)
(tan
.
5
)
1
(
.
12
)
(tan
.
4
)
1
(
.
11
)
(
.
3
log
2
1
.
10
)
(
.
2
2
2
)
(
.
9
)
(
.
1
y
dy
e
dx
ye
y
e
d
y
ydy
x
dx
xy
y
x
d
y
x
ydy
xdx
y
x
d
x
dx
y
xydy
x
y
d
xy
ydx
xdy
x
y
d
y
dy
x
xydx
y
x
d
xy
xdy
ydx
y
x
d
y
x
xdy
ydx
y
x
d
y
x
ydx
xdy
xy
d
y
x
ydx
xdy
x
y
d
y
x
ydx
xdy
xy
d
y
xdy
ydx
y
x
d
y
x
ydx
xdy
y
x
y
x
d
x
ydx
xdy
x
y
d
x
dx
xy
ydy
x
x
y
d
ydx
xdy
xy
d
x
x
x

























































29

30. Examples:
equation
al
differenti
given
for the
solution
required
the
is
Which
0
)
(
2
0
y
2
0
2xy
as
written
be
can
equation
The
:
0
)
(
equation
al
differenti
the
Solve
1.
2
2
2
C
y
e
x
g
Integratin
y
e
d
xdx
dy
e
dx
ye
xdx
dy
e
dx
ye
dx
Solution
dy
e
dx
e
xy
y
x
x
x
x
x
x
x
x
















equation
al
differenti
given
for the
solution
required
the
is
Which
tan
)
(tan
,
'
'
by
equation
al
differenti
given
the
dividing
Therefor
)
(tan
that
know
We
)
(
is
equation
al
differenti
given
The
:
)
(
equation
al
differenti
the
Solve
2.
1
1
2
2
2
2
2
2
1
2
2
2
2
c
ax
y
x
g
Integratin
adx
y
x
d
adx
y
x
xdy
ydx
get
we
y
x
y
x
xdy
ydx
y
x
d
dx
y
x
a
xdy
ydx
Solution
dx
y
x
a
xdy
ydx






















30

31. Rules to find Integrating Factor:
Rule 1:
Example:
equation
al
differenti
s
homogeneou
is
and
equation
al
differenti
exact
an
not
is
equation
al
differenti
given
is
That
.
&
4
&
0
Ndy
Mdx
of
form
the
in
is
Which
)
1
(
0
)
(
is
equation
al
differenti
Given
:
.
0
)
(
equation
al
differenti
the
Solve
.
1
3
3
3
4
4
3
4
4
3
4
4
x
N
y
M
y
x
N
y
y
M
xy
N
y
x
M
Here
dy
xy
dx
y
x
Solution
dy
xy
dx
y
x

























factor.
g
integratin
an
is
Ny
Mx
1
then
0,
Ny
Mx
if
and
y'
'
&
x'
'
in
s
homogeneou
is
0
Ndy
Mdx
equation
al
differenti
given
the
If





31

32. Multiplying Equation (1) by integrating factor , we get
equation
al
differenti
exact
an
is
)
2
(
Equation
.
4
&
4
&
1
0
dy
N
dx
M
of
form
the
in
is
Which
)
2
(
0
1
0
)
(
1
)
1
(
5
3
5
3
5
3
5
4
5
3
5
4
5
3
5
4
4
5
x
N
y
M
x
y
x
N
x
y
y
M
x
y
N
x
y
x
M
Here
dy
x
y
dx
x
y
x
dy
x
xy
dx
x
y
x
x











































5
1
x
32

33. Sol:
equation
al
differenti
given
of
solution
required
the
is
Which
4
log
1
1
0
)
1
(
is
equation
al
differenti
exact
of
solution
the
Now
4
4
tan tan
5
4
'
'
tan
5
4
'
'
tan
C
x
y
x
C
dx
x
y
dx
x
C
dy
dx
x
y
x
C
dy
N
dx
M
t
cons
as
y t
cons
as
y
x
from
free
Terms
t
cons
as
y
x
from
free
Terms
t
cons
as
y














 




33

34. Rule 2:
Example:
.
product xy
argument
the
of
functions
the
are
)
(
&
)
(
Where
factor.
g
integratin
an
is
Ny
Mx
1
then
0,
Ny
Mx
if
and
0
)
(
)
(
form
the
of
is
0
Ndy
Mdx
equation
al
differenti
given
the
If
2
1
2
1
xy
f
xy
f
xdy
xy
f
ydx
xy
f







equation
al
differenti
exact
an
not
is
equation
al
differenti
given
is
That
.
3
2
&
6
2
&
2
0
Ndy
Mdx
of
form
in the
is
hich
)
1
(
0
)
(
)
2
(
is
equation
al
differenti
Given
:
.
0
)
(
)
2
(
equation
al
differenti
the
Solve
.
1
2
2
2
2
2
3
2
3
2
2
2
2
2
2
2
2
2
2
x
N
y
M
y
x
xy
x
N
y
x
xy
y
M
y
x
y
x
N
y
x
xy
M
Here
W
dy
y
x
xy
x
dx
y
x
xy
y
Solution
dy
y
x
xy
x
dx
y
x
xy
y




























34

35. 0
dy
N
dx
M
of
form
in the
is
hich
)
2
(
0
)
3
1
3
1
(
)
3
2
3
1
(
0
)
3
3
(
)
3
2
3
(
0
)
3
(
)
3
2
(
3
1
)
1
(
,
3
1
factor
g
integratin
by
(1)
Equation
g
Multyplyin
factor
g
integratin
an
is
3
1
1
0
3
2
)
(
)
2
(
Ny
Mx
Now
.
0
)
(
)
(
of
form
in the
is
equation
given
the
and
2
2
3
3
2
3
3
3
2
3
3
3
2
3
3
2
3
3
2
3
2
3
3
3
2
2
3
3
3
3
3
3
3
3
3
3
2
2
3
3
2
2
2
3
2
3
2
2
2
1




































W
dy
y
xy
dx
x
y
x
dy
y
x
y
x
y
x
y
x
dx
y
x
y
x
y
x
xy
dy
y
x
y
x
y
x
dx
y
x
y
x
xy
y
x
get
we
y
x
y
x
Ny
Mx
y
x
y
x
y
x
y
x
y
x
y
y
x
y
x
x
y
x
xy
xdy
xy
f
ydx
xy
f
35

36. Sol:
 
equation
al
differenti
given
of
solution
required
the
is
Which
3
log
1
log
3
1
log
3
2
3
1
3
1
1
3
2
1
3
1
3
1
)
3
2
3
1
(
is
equation
al
differenti
exact
of
solution
the
Now
equation
al
differenti
exact
an
is
)
2
(
Equation
.
3
1
&
3
1
3
1
3
1
&
3
2
3
1
1
1
2
'
'
tan tan
2
'
'
tan
2
'
'
tan
2
2
2
2
2
2
C
C
Where
C
x
y
xy
C
y
x
xy
C
dy
y
dx
x
dx
x
y
C
dy
y
dx
x
y
x
C
dy
N
dx
M
x
N
y
M
y
x
x
N
y
x
y
M
y
xy
N
x
y
x
M
Here
x
from
free
Terms
t
cons
as
y t
cons
as
y
x
from
free
Terms
t
cons
as
y
x
from
free
Terms
t
cons
as
y






















































 




36

37. Rule 3:
Example:
factor.
g
integratin
an
is
then
),
(
M
N
1
if
and
equation
al
differenti
exact
an
nt
is
0
Ndy
Mdx
equation
al
differenti
given
the
If
)
(

















dx
x
f
e
x
f
y
N
x
only
x
of
function
x
f
x
xy
y
y
x
N
y
M
x
N
y
M
y
x
N
y
y
M
xy
N
y
x
M
Here
dx
y
x
xydy
Solution
dx
y
x
xydy
'
'
)
(
2
2
2
2
N
1
Now
equation
al
differenti
exact
an
not
is
equation
al
differenti
given
is
That
.
2
&
2
2
&
)
1
(
0
Ndy
Mdx
of
form
the
in
is
Which
)
1
(
0
)
1
(
2
is
equation
al
differenti
Given
:
.
0
)
1
(
2
equation
al
differenti
the
Solve
.
1
2
2
2
2
2
2
















































37

38. Sol:
equation
al
differenti
exact
an
is
)
2
(
Equation
.
2
&
2
2
&
1
1
0
dy
N
dx
M
of
form
the
in
is
Which
)
2
(
0
2
1
1
0
)
1
(
2
1
)
1
(
,
1
factor
g
integratin
by
(1)
Equation
g
Multyplyin
factor
g
integratin
an
is
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
log
2
2
)
(
x
N
y
M
x
y
x
N
x
y
y
M
x
y
N
x
x
y
M
Here
dy
x
y
dx
x
x
y
dx
x
y
x
dy
x
xy
x
get
we
x
x
e
e
e x
dx
x
dx
x
f






























































 


38

39. Sol:
Rule 3:
equation
al
differenti
given
of
solution
required
the
is
Which
1
1
1
1
1
0
1
1
is
equation
al
differenti
exact
of
solution
the
Now
2
2
2
tan tan tan
2
2
2
'
'
tan
2
2
2
'
'
tan
xC
x
y
C
x
x
y
x
C
dx
x
dx
x
y
dx
C
dy
dx
x
x
y
C
dy
N
dx
M
t
cons
as
y t
cons
as
y t
cons
as
y
x
from
free
Terms
t
cons
as
y
x
from
free
Terms
t
cons
as
y
































  




factor.
g
integratin
an
is
then
),
(
N
M
1
if
and
equation
al
differenti
exact
an
nt
is
0
Ndy
Mdx
equation
al
differenti
given
the
If
)
(

















dy
y
g
e
y
g
x
M
y
39

40. Example:
)
2
(
0
)
1
(
2
0
0
)
(
2
1
)
1
(
,
1
factor
g
integratin
by
(1)
Equation
g
Multyplyin
factor
g
integratin
an
is
'
'
)
(
2
2
2
2
M
1
Now
equation
al
differenti
exact
an
not
is
equation
al
differenti
given
is
That
.
2
&
2
&
2
0
Ndy
Mdx
of
form
the
in
is
Which
)
1
(
0
)
(
2
is
equation
al
differenti
Given
:
.
0
)
(
2
equation
al
differenti
the
Solve
.
1
2
2
2
2
2
2
2
2
2
log
2
2
)
(
2
2
2
2
2
2































































dy
y
x
dx
y
x
dy
y
x
y
dx
y
xy
y
get
we
y
y
e
e
e
only
y
of
function
y
g
y
xy
x
x
y
M
x
N
x
N
y
M
x
x
N
x
y
M
x
y
N
xy
M
Here
dy
x
y
xydx
Solution
dy
x
y
xydx
y
dy
y
dy
y
g
40

41. Sol:
equation
al
differenti
given
of
solution
required
the
is
Which
1
2
is
equation
al
differenti
exact
of
solution
the
Now
equation
al
differenti
exact
an
is
)
2
(
Equation
.
2
&
2
1
&
2
0
dy
N
dx
M
of
form
the
in
is
Which
2
2
2
'
'
tan
'
'
tan
2
2
2
2
yC
y
x
C
y
y
x
C
dy
dx
y
x
C
dy
N
dx
M
x
N
y
M
y
x
x
N
y
x
y
M
y
x
N
y
x
M
Here
x
from
free
Terms
t
cons
as
y
x
from
free
Terms
t
cons
as
y












































41

42. Definition:
Procedure to Solve the Linear Differential equation in ‘y’.
 Find Integrating Factor
 Multiply by Integrating factor and write the equation of the form
 Integrate to get the general solution
.
'
'
&
.
form
in the
written
be
can
it
if
,
'
'
in
equation
al
differenti
Linear
be
to
said
is
0
equation
al
differenti
A
only
x
of
functions
are
Q
P
Where
Q
Py
dx
dy
y
Ndy
Mdx






dx
x
f
e
)
(
F
I.
 
  .)
.
(
)
. F
I
Q
F
I
y
dx
d

   
 c
dx
F
I
Q
F
I
y .)
.
(
)
.
42

43. Definition:
Procedure to Solve the Linear Differential equation in ’x’.
 Find Integrating Factor
 Multiply by Integrating factor and write the equation of the form
 Integrate to get the general solution
.
'
'
&
.
form
in the
written
be
can
it
if
,
'
'
in
equation
al
differenti
Linear
be
to
said
is
0
equation
al
differenti
A
only
y
of
functions
are
Q
P
Where
Q
Px
dy
dx
x
Ndy
Mdx






dy
y
f
e
)
(
F
I.
 
  .)
.
(
)
. F
I
Q
F
I
x
dy
d

   
 c
dy
F
I
Q
F
I
x .)
.
(
)
.
43

44. Example:
44
)
1
(
F
I.
factor
g
Integratin
Now
1
)
(
&
1
2
)
(
Where
),
(
)
(
of
form
in the
is
Which
)
2
(
1
1
2
1
1
2
)
1
(
2
)
1
(
is
equation
al
differenti
Given
:
2
)
1
(
equation
al
differenti
the
Solve
1.
2
)
1
log(
1
2
)
(
2
2
2
2
2
2
2
2
2
2
x
e
e
e
x
Cotx
x
Q
x
x
x
P
x
Q
y
x
P
dx
dy
x
Cotx
y
x
x
dx
dy
dx
x
Cotx
dx
x
xy
dy
dx
Cotx
xydx
dy
x
Sol
dx
Cotx
xydx
dy
x
x
dx
x
x
dx
x
P

































45. Sol:
45
 
 
 
 
 
 
 
equation
al
differenti
given
for the
solution
required
the
is
Which
1
)
log(sin
)
log(sin
1
1
)
1
(
1
1
get
we
sides,
both
on
g
Integratin
)
1
(
1
)
1
)
.
)(
(
)
.
as
equation
in the
write
and
factor
g
Integratin
with the
(2)
eqution
Multiply
2
2
2
2
2
2
2
2
2
x
C
x
y
C
x
x
y
C
Cotxdx
x
y
C
dx
x
x
Cotx
x
y
x
x
Cotx
x
y
dx
d
F
I
x
Q
F
I
y
dx
d


























46. Example:
46
 
    y
y
y
dy
y
dy
y
P
e
y
y
xe
dy
d
F
I
y
Q
F
I
x
dy
d
e
e
e
y
y
y
Q
y
y
P
y
Q
x
y
P
dy
dx
y
y
x
y
dy
dx
dy
x
y
dx
y
Sol
dy
x
y
dx
y
1
1
1
2
tan
2
1
tan
tan
1
1
)
(
2
1
2
2
1
2
1
2
1
2
1
tan
)
.
)(
(
)
.
as
equation
in the
write
and
factor
g
Integratin
with the
(1)
eqution
Multiply
F
I.
factor
g
Integratin
Now
1
tan
)
(
&
1
1
)
(
Where
),
(
)
(
of
form
in the
is
Which
)
1
(
1
tan
1
1
)
(tan
)
1
(
is
equation
al
differenti
Given
:
)
(tan
)
1
(
equation
al
differenti
the
Solve
2.





































47. Sol:
47
equation
al
differenti
given
for the
solution
required
the
is
Which
)
1
(tan
)
1
(
)
)
(
(
1
1
,
tan
1
tan
get
we
sides,
both
on
g
Integratin
1
1
1
1
1
1
tan
1
tan
tan
tan
2
1
tan
2
1
tan
y
t
y
t
t
y
t
y
y
y
Ce
y
x
C
e
t
xe
C
dt
dt
e
t
dt
d
dt
e
t
xe
C
dt
te
xe
dt
dy
y
then
t
y
Put
C
dy
e
y
y
xe































 



0
)
2
3
(
equation
al
differenti
the
Solve
4.
sec
2
equation
al
differenti
the
Solve
3.







dy
xy
x
ydx
x
Co
x
x
yCot
y

48. Definition:
Procedure to Solve the Bernoulli’s Differential equation in ‘y’:
 Divide given differential equation by
 Put
 Substitute these and write in the form of
 Solve as a linear differential equation in ‘t’ and apply the solution for the
given differential equation.
only
x
of
functions
are
Q
P
Where
y
x
Q
y
x
P
dx
dy
y
Ndy
Mdx
n
&
.
)
(
)
(
form
in the
written
be
can
it
if
,
'
'
in
equation
al
differenti
s
Bernoulli'
be
to
said
is
0
equation
al
differenti
A





)
(
)
( 1
1 x
Q
t
x
P
dx
dt



48
n
y
dx
dt
n
dx
dy
y
t
y n
n


 

1
1
that
So
,
1

49. Example:
49
2
6
6
5
2
5
6
6
6
3
6
6
6
6
3
6
3
1
5
1
get
we
(2),
in
values
these
Substitute
5
1
5
get
we
sides,
both
on
x'
'
respect to
with
ating
Differenti
,
Put
)
2
(
1
1
1
get
we
sides,
both
on
by
(1)
equation
Dividing
)
1
(
is
equation
al
differenti
Given
:
Solve
1.
x
t
x
dx
dt
dx
dt
dx
dy
y
dx
dt
dx
dy
y
t
y
x
y
x
dx
dy
y
y
y
x
x
y
y
x
dx
dy
y
y
y
x
y
dx
dy
x
Sol
y
x
y
dx
dy
x

























50. Sol:
50
 
   
 
equation
al
differenti
given
for the
solution
required
the
is
Which
2
5
2
5
2
5
5
get
we
sides,
both
on
g
Integratin
5
)
(
5
)
.
)(
(
)
.
as
equation
in the
write
and
factor
g
Integratin
with the
(2)
eqution
Multiply
F
I.
factor
g
Integratin
Now
5
)
(
&
5
)
(
Where
),
(
)
(
of
form
in the
is
Which
)
3
(
5
5
5
3
5
5
3
2
5
3
5
3
5
2
5
5
)
log(
5
1
5
)
(
2
2
Cx
x
y
Cx
x
t
C
x
x
t
C
dx
x
x
t
x
x
x
x
t
dx
d
F
I
x
Q
F
I
t
dx
d
x
e
e
e
x
x
Q
x
x
P
x
Q
t
x
P
dx
dt
x
t
x
dx
dt
x
dx
x
dx
x
P


















































51. Orthogonal Trajectories:
If two family of curves are such that each member of cuts each member of the
other family at right angles, then the members of one family are known as the
Orthogonal Trajectories of the members of other family.
(OR)
A Curve which cuts every member of a given family of curves at a right angles
is an orthogonal trajectories of the given family.
Orthogonal Trajectories in Cartesian Coordinates:
Procedure:
Step1: Suppose f( x, y, c)=0 is the given family of curves, where ‘c’ is the constant.
Form the differential equation by eliminating the arbitrary constant
Suppose the differential equation is (1)
Step2: Form the differential equation of the family, which is orthogonal to the given
family by replacing . That is (2)
Step3: Solving the differential equation (2), we get the Orthogonal trajectories of the
given family.
51
0
,
, 






dx
dy
y
x
F
dy
dx
by
dx
dy
 0
,
, 









dy
dx
y
x
F

52. Family of Curves:
A family of curves is a set of curves, each of which is given by a function or
parameterization in which one or more of the parameters is variable. In general, the
parameter(s) influence the shape of the curve in a way that is more complicated than
a simple Linear transformation. Sets of curves given by an implicit relation may also
represent families of curves.
Fig (1) Fig (2) Fig (3)
52

53. Example:
For instance, each member of the family y = mx of straight lines
through the origin is an orthogonal trajectory of the
family x2 + y2 = r2 of concentric circles with center the origin (see Figure 4).
We say that the two families are orthogonal trajectories of each other.
Fig (4)
53

54. Example:
54
)
2
(
replace
ies
trajector
orthogonal
the
ing
correspond
equatin
al
differenti
get the
To
family.
given
the
ing
correspond
equatin
al
differenti
the
is
Whch
)
2
(
0
0
3
2
3
2
get
we
,
x'
'
respect to
with
(1)
ate
Differenti
Given
)
1
(
curves
of
family
Given
:
curves
of
family
the
of
es
Trajectroi
orthogonal
the
Find
1.
3
1
3
1
3
1
3
1
1
3
2
1
3
2
3
2
3
2
3
2
3
2
3
2
3
2
in
dy
dx
by
dx
dy
y
x
dx
dy
dx
dy
y
x
dx
dy
y
x
a
y
x
Solution
a
y
x




















55. Sol:
55
curves
of
family
given
the
of
ies
trajector
orthogonal
Which the
3
4
4
3
4
3
4
3
get
we
(3),
equation
the
Integrate
ies
trajector
orthogonal
the
ing
correspond
equatin
al
differenti
Which
)
3
(
0
1
1
3
4
3
4
3
4
3
4
1
3
1
1
3
1
1
3
1
1
3
1
3
1
3
1
3
1
3
1
3
1
3
1
C
C
Where
C
y
x
C
y
x
C
y
x
C
dy
y
dx
x
dy
y
dx
x
y
x
dy
dx



























56. Self Orthogonal :
If the differential equation corresponding to the given family of curves and
the differential equation corresponding to the orthogonal trajectories are
same, then the given family curves are Self Orthogonal.
Example:
56
   
 
 
 
 
 
 
 
 
)
3
(
)
2
(
0
0
0
0
2
2
get
we
,
x'
'
respect to
with
(1)
ate
Differenti
Given
)
1
(
1
curves
of
family
Given
:
.
orthogonal
self
are
1
curves
of
family
that the
Show
2.
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
y
y
x
b
a
y
y
y
y
x
a
y
y
xb
b
b
and
y
y
x
b
a
x
y
y
x
a
y
y
xb
a
a
Now
y
y
x
a
y
y
xb
y
y
x
a
y
y
xb
y
y
a
y
y
x
xb
a
y
y
b
x
b
y
y
a
x
b
y
a
x
Solution
b
y
a
x














































































57. Sol:
57
 
 
 
 
 
 
 
 
 
 
 
 
     
    
 
    
.
orthogonal
self
are
curves
of
family
given
the
Therefore
same.
are
ies
trajector
orthogonl
the
of
equation
al
differenti
the
and
family
given
the
of
equation
al
differenti
the
Therefore
same.
are
(4)
&
(5)
Equation
ies
trajector
orthogonal
the
ing
correspond
equatin
al
differenti
Which
)
5
(
1
)
2
(
1
replace
ies
trajector
orthogonal
the
ing
correspond
equatin
al
differenti
get the
To
family.
given
the
to
ing
correspond
equatin
al
differenti
Which the
)
4
(
1
1
1
get.
we
(1)
in
(3)
and
(2)
Substitute
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
b
a
y
y
y
x
y
y
x
b
a
y
y
y
x
y
y
x
in
y
by
y
b
a
y
y
y
x
y
y
x
b
a
y
y
y
x
y
y
y
y
x
x
b
a
y
y
y
x
y
b
a
y
y
x
x
b
a
y
y
y
y
x
y
b
a
x
y
y
x
x
y
y
x
b
a
y
y
y
y
y
x
b
a
x
x






















































































58. Orthogonal Trajectories in Polar Coordinates:
Procedure:
Step1: Suppose f( r, θ, c)=0 is the given family of curves, where ‘c’ is the constant.
Form the differential equation by eliminating the arbitrary constant
Suppose the differential equation is (1)
Step2: Form the differential equation of the family, which is orthogonal to the given
family by replacing . That is (2)
Step3: Solving the differential equation (2), we get the Orthogonal trajectories of the
given family.
Example:
58
0
,
, 








d
dr
r
F
dr
d
r
by
d
dr 

2
 0
,
, 2








dr
d
r
r
F



59. Example:
59
get
we
(3),
equation
the
Integrate
ies
trajector
orthogonal
the
ing
correspond
equatin
al
differenti
Which
)
3
(
2
2
)
2
(
replace
es
trajectori
orthogonal
the
ing
correspond
equatin
al
differenti
get the
To
family.
given
the
ing
correspond
equatin
al
differenti
the
is
Whch
)
2
(
2
))
1
(
(
2
2
2
2
2
1
get
we
,
'
'
respect to
with
(1)
ate
Differenti
Given
)
1
(
)
1
(
curves
of
famil
Given
:
)
1
(
Cardioich
the
of
es
Trajectroi
orthogonal
the
Find
1.
2
2
2
r
dr
Cot
d
rCot
dr
d
r
in
dr
d
r
by
d
dr
rCot
d
dr
From
Cos
Cos
rSin
Sin
Cos
r
aSin
d
dr
Cos
a
r
Solution
Cos
a
r


































60. Sol:
Exercise:
60
curves
of
family
given
the
of
ies
trajector
orthogonal
Which the
2
log
log
)
2
log(
2
1
log
2
log
2
2
2
C
r
Cos
C
r
Cos
C
r
dr
d
Tan
C
r
dr
Cot
d











 









.
2
)
(
)
1
(
(i)
Cardioich
the
of
es
Trajectroi
orthogonal
the
Find
.
4
.
)
(
curves
of
family
the
of
es
Trajectroi
orthogonal
the
Find
3.
).
1
(
orthogonal
self
are
parabolas
of
system
that the
Show
.
2
.
)
(
)
(
curves
of
family
the
of
es
Trajectroi
orthogonal
the
Find
1.
2
2
3
2
2
2




Cos
a
r
ii
Cos
a
r
b
n
Sin
r
Cos
a
r
x
ay
ii
ax
y
x
i
n
n










61. Statement:
The rate of change of the temperature ‘T’ of a body at any time ‘t’ is
proportional to the difference between ‘T’ and the temperature of
surrounding medium. This is known as Newton’s law of cooling.
61
)
1
(
)
(
is
Colling
of
Law
s
Newton'
Solutionof
)
(
)
(
)
log(
)
(
get
we
sides,
both
on
g
Integratin
)
variables
of
Separation
Using
(
)
(
constant
ality
proportion
the
is
k
Where
)
(
)
(
0
0
0
0
0
kt
A
C
kt
A
C
kt
A
C
kt
A
A
A
A
A
A
Ce
T
t
T
e
C
Where
Ce
T
T
e
e
T
t
T
e
T
T
C
kt
T
T
C
kdt
T
T
dT
kdt
T
T
dT
T
T
k
dt
dT
T
T
dt
dT







































62. Method of Solving:
Step I: Identify TA , the temperature of the surrounding medium, so that the
general solution is given by (1)
Step II: Use Two conditions given to determine the constant of integration ‘C’,
and known proportionality constant ‘k’.
Step III: Substituting ‘C’ and ‘k’ obtained from step II in (1)
(a): The value of temperature ‘T’ for a given time ‘t’ (or)
(b): The value of time ‘t’ for a given temperature ‘T’ can be
obtained from (1)
62

63. Example:
63
0
0
0
)
(
)
(
)
log(
)
(
get
we
sides,
both
on
g
Integratin
)
variables
of
Separation
Using
(
)
(
constant
ality
proportion
the
is
k
Where
)
(
)
(
Cooling
of
Law
s
Newton'
by
Then
t'.
'
any time
at
water
of
re
temperatu
be
T"
"
Let
:
Solution
?
25
be
re
temperatu
the
When will
(b)
hour
2
1
after
and
min
20
after
re
temperatu
the
Find
(a)
.
20
e
temperatur
room
a
in
20
to
up
warm
5min to
takes
10
ture
at tempera
Water
1.
0
0
C
kt
A
C
kt
A
C
kt
A
A
A
A
A
A
e
C
Where
Ce
T
T
e
e
T
t
T
e
T
T
C
kt
T
T
C
kdt
T
T
dT
kdt
T
T
dT
T
T
k
dt
dT
T
T
dt
dT
c
c
c
c








































64. Sol:
64
)
4
(
30
40
)
(
becomes
(3)
eqution
is
That
08109
.
0
5
405465108
.
0
3
2
ln
5
3
2
30
40
20
30
40
)
5
(
get
we
(3),
From
20
T(5)
min,
5
at t
is
That
.
20
is
e
temperatur
min water
5
After
)
3
(
30
40
)
(
becomes
(2)
Eqution
30
40
10
40
)
0
(
get
we
(2),
From
10
T(0)
0,
at t
is
That
.
10
is
rature
ater tempe
Intially w
)
2
(
40
)
(
),
1
(
40
Substitute
)
1
(
)
(
is
Colling
of
Law
s
Newton'
Solutionof
)
08109
.
0
(
5
5
5
0
t
k
k
k
kt
k
kt
A
kt
A
e
t
T
k
k
e
e
e
T
c
c
e
t
T
C
C
Ce
T
c
c
Ce
t
T
get
we
in
c
T
Ce
T
t
T


























































65. Sol:
2. A murder victim is discovered and a lieutenant from the Forensic science laboratory
is summoned to estimate the time of death. The body is located in a room that is
kept at a constant temperature of 68°F. The lieutenant arrived at 9.40PM and
measured the body temperature as 94.4°F at that time. Another measurement of the
body temperature at 11PM is 89.2°F. find the estimated time of death.
3. If the temperature of the body is changing from 100°c to 70°c in 15 minutes, find
when the temperature will be 40°c, if the temperature of air is 30°c.
65
s
t
t
e
e
e
e
t
T
From
c
c
c
e
e
T
From
t
t
t
t
min
547
.
8
2
1
ln
)
08109
.
0
(
2
1
30
40
25
30
40
25
30
40
)
(
),
4
(
?
t
,
25
T(t)
at
is
That
?
25
be
re
temperatu
the
When will
(a)
34
30
40
30
40
)
20
(
),
4
(
?
T(20)
min,
20
at t
is
That
is?
e
temperatur
min water
20
After
(a)
)
08109
.
0
(
)
08109
.
0
(
)
08109
.
0
(
)
08109
.
0
(
6218
.
1
20
)
08109
.
0
(









































66. Law of Natural Growth and Decay:
Growth Statement:
Let B(t) be the population at any time ‘t’. Assume that population
growth at a rate directly proportional to the amount of Population
present at that time.
66
)
2
(
)
(
is
Growth
of
Law
Natural
Solutionof
)
(
)
ln(
get
we
sides,
both
on
g
Integratin
)
variables
of
Separation
Using
(
constant
ality
proportion
the
is
k
Where
Growth
of
equation
al
differenti
the
is
0
0
0
0
0
kt
C
kt
C
kt
C
kt
Ce
t
B
e
C
Where
Ce
B
e
e
t
B
e
B
C
kt
B
C
kdt
B
dB
kdt
B
dB
kB
dt
dB
B
dt
dB






















67. Law of Natural d Decay:
Rate of decay of Radio Active Materials:
Statement: Let M(t) be the radio active substance at any time ‘t’.
Assume that the radio active material M(t) at any time ‘t’ decays
which proportional to the amount of present at that time.
67
)
3
(
)
(
is
Decay
of
Law
Natural
Solutionof
)
(
)
ln(
get
we
sides,
both
on
g
Integratin
)
variables
of
Separation
Using
(
constant
ality
proportion
the
is
k
Where
Decay
of
equation
al
differenti
the
is
0
0
0
0
0
kt
C
kt
C
kt
C
kt
Ce
t
M
e
C
Where
Ce
M
e
e
t
M
e
M
C
kt
M
C
kdt
M
dM
kdt
M
dM
kM
dt
dM
M
dt
dM






























68. By
Dr. KONDALA RAO. KANAPARTI