Chapter 2
1
Chapter 2
2
Example
Chapter 2
3
Chapter 2
4
EXAMPLE
Chapter 2
5
Solution
Chapter 2
6
Chapter 2
7
Chapter 2
8
Method for solving First Order
Differential Equations
Chapter 2
9
Methods
Variable Separable
Reducible to variable separable
Exact Differential Equation
Integrating Factor
Separable Variable
x is independent variable and y is dependent variable
or
or
are separable forms of the differential equation
General solution can be solved by directly integrating both the sides
+c
Where c is constant of integration
DO YOU REMEMBER INTEGRATION
Chapter 2
FORMULA
11
Separation of Variables
Definition
A differential equation of the type y’ = f(x)g(y) is separable.
Example
y   xy and y  
but y  
x
are separable
y
xy
is not separable.
xy
Separable differential equations can often be solved with direct integration. This
may lead to an equation which defines the solution implicitly rather than directly.
Example
y 
x
y
  ydy   xdx
y 2 x2


 C1  y 2  x 2  C
2
2
EXAMPLE:
Chapter 2
13
EXAMPLE:
Chapter 2
14
To find the particular solution, we apply the given initial
condition, when x =1, y = 3
is solution of initial value problem
Chapter 2
15
Chapter 2
16
Chapter 2
17
Note1:
If we have
2 y  ln y dy  2xdx
 2 y  ln y dy   2 xdx
Integrating by parts
y 2   y ln y  y   x 2  c
y
y
y
2 y  ye dy  2 xdx
ye
dy

y
e

e

 2 y  ye dy    2 xdx
y  ye  e   x  c
y
Integrating by parts
Note.2.
If we have
y
2
Note.3.
 ln ydy  y ln y  y
y
y
2
2 y  sin y dy  2 xdx

 2 y  sin y dy   2 xdx
If we have
1
1

y  y sin
2
1
y  1 y
2
 x
Chapter 2
2
sin 1 ydy  y sin 1 y  1  y 2
c
18
Chapter 2
19
Chapter 2
20
Chapter 2
21
Method
Reducible
to
separable
Homogeneous Equations
Homogenous Differential Equations
A differential equation
dy
 f ( x, y ) is
dx
Homogenous differential equation if
every t, where t  R
f tx, ty   t n f  x, y  for
Chapter 2
23


xydy  x 2  y 2 dx
Example:1.
Show that differential equation
is homogenous differential equation.
dy x 2  y 2

dx
xy
Solution:
Differential equation is homogeneous
x2  y2
f x, y  
xy
t x t y
f tx, ty  
t 2 xy
2

2
2
2


t 2 x2  y 2
t 2 xy
x
2
 y2
 f  x, y 
xy
Differential equation is homogeneous
Chapter 2
24
METHOD for solving Homogenous differential equations
Substitute
OR
Substitute
y  ux
dy
du
ux
dx
dx
dy  udx  xdu
x  vy
dx
dv
v y
dy
dy
dx  vdy  ydv
Chapter 2
25
.
Using substitution the homogeneous differential equation
is reduce to separable variable form.
Example:2
Solve the homogenous differential equation
dy x 2  y 2

dx
xy
Solution:
Rewriting in the form :
x
substitute

M x, y dx  N x, y dy  0
 y dx  xydy  0
y  ux and dy  udx  xdu
2
2
Chapter 2
26
x
2

 x u dx  ux udx  xdu  0
2
2
2
x 2 dx  x 2 u 2 dx  u 2 x 2 dx  ux 3 du  0
x 2 dx  ux 3 du  0
x dx  ux du
2
3
2
x dx
 udu
3
x
dx
 udu
x
dx
 x   udx
is variable separable form
1  y2 
ln x   2   c
2  x Chapter 2
u2
ln x 
c
2
is general solution.
27
Note.
Selection of substitution Differential Equation depends on
M x, y  and Nx, y 
number of terms of coefficients
1.
If
1  2  3dx  1dy  0 ,
2.
If
1dx  1  2  3dy  0,
3.
If
then take
y  ux
then take
x  vx
1  2dx  1  2dy  0,
Chapter 2
then take
x = vy or y = ux
28
Example:.
Solve the Differential Equation by using appropriate substitution
y
2

 xy  x dx  x dy  0
2
2
(1 / 2)
Solution: Differential equation is homogeneous as degree of each term is same,
hence we can use either y = ux or x = vy as substitution
Let y  ux
dy  udx  xdu
Substituting y and dy in the given equation
u
2

x 2  ux 2  x 2 dx  x 2 udx  xdu  
u 2 x 2 dx  ux 2 dx  x 2 dx  x 2 udx  x 3 du  
u 2 x 2 dx  x 2 dx  x 3 du  


x 2 u 2  1 dx  x 3 du
Chapter 2
29
Separating variable u and x
(2 / 2)
dx
du

is Separable form
2
x 1 u
Integrating both the sides
dx
du
 x  1 u2
ln x  tan 1 u  c
 y
ln x  tan    c.
x
1
is general solution of the differential equation
Chapter 2
30
Example:
Show that differential equation
dy
3 xy
 4x 2  9 y 2
dx
Solution:

(1 / 2)
is homogeneous

3xydy  4 x 2  9 y 2 dx  
y  ux,
dy  udx  xdv


3 x.ux  udx  xdu   4 x 2  9u 2 x 2 dx 
3 x 2u 2 dx  3ux 3 du  4 x 2 dx  9u 2 x 2 dx 


3ux 3 du  4 x 2 dx  6u 2 x 2 dx  x 2 4  6u 2 dx
Chapter 2
31
3udu
dx

x
4  6u 2
is Separable form
(2 / 2)
Integrating both the sides
3udu
dx
 4  6u 2   x
Let
1 dz
dx


4 z
x
z  4  6u 2
dz  12udu
1
ln z  ln x  c
4
1 
y2 
ln 4  6 2   ln x  c.
4 
x 
is general solution of the differential equation
Chapter 2
32
Chapter 2
33
Homogeneous Differential Equation
Chapter 2
Homogeneous Differential Equation
Chapter 2
(1 / 3)
Chapter 2
34
Homogeneous Differential Equation
Chapter 2
(2 / 3)
Chapter 2
35
Homogeneous Differential Equation
Chapter 2
(3 / 3)
Chapter 2
36
Chapter 2
37
Homogeneous Differential Equation
Chapter 2
Homogeneous Differential Equation
Chapter 2
(1 / 2)
Chapter 2
38
Homogeneous Differential Equation
Chapter 2
(2 / 2)
is general solution Chapter
of differential
equation
2
39
Chapter 2
40
Differential Equation
Chapter 2
Chapter 2
41
Differential Equation
Chapter 2
Chapter 2
Differential Equation
is general solution of differential equation
Chapter 2
42