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LEIBNITZ'S LINEAR EQUATION
The standard form of linear equation of first order commonly known
as Leibnitz's linear equation
i
...


BERNOULLI’S EQUATION
A first order DE that can be written in the form

dy
 P ( x) y  Q ( x ) y n
...

Note:(i )

It is named after the Swiss Mathematician Jacob Bernoulli (1654
-1705) who is known for his basic work in probability distribution
theory
...

(iii ) It is clear that when n>1, the DE is non-linear
...


METHOD OF SOLUTION (WHEN n=0)
Consider the equation
i
...
where n=0
...


Problem:-01
Solve

dy
 y  cos x
...
----(1)
dx

It is of the form

dy
 P( x) y  Q ( x ) ----(2)
dx

Comparing the equations (1) and (2), we have
P  1,

& Q  cos x

Integration factor
e Pdx  e  dx  e  dx  e x

General solution
ye pdx   Qe pdx dx  c
ye  x   cos xe  x dx  c

  e ax cos bxdx 

e ax
[a cos bx  b sin bx]
a 2  b2

Here a=-1, b=1, substituting in the formula, we get
ye

x

ye

x

e x

[sin x  cos x ]  c
(1)2  (1) 2

e x

[sin x  cos x ]  c
2

Which is the required solution
...

x dx x

Solution:The given DE is

1 dy y
 tan x  cos x
...

dx

It is of the form

(Multiply by x)

dy
 P( x) y  Q ( x ) ----(2)
dx

Comparing the equations (1), (2), we get
P  tan x,

& Q  x cos x

Integration factor
1

e  Pdx  e  tan xdx  e log cos x  sec x

General solution
ye pdx   Qe pdx dx  c
y sec x   x cos x sec xdx  c
y sec x   xdx  c

x2
 y sec x 
c
2

Which is the required solution
...

dx

Solution :The given DE is

dy
 y cot x  sin 2 x
...


Problem:-04
Solve

dy

 y cot x  4 x cos ecx , given that y=0 when x=
dx
2

Solution:The given DE is

dy
 y cot x  4 x cos ecx -----(1)
dx

It is of the form

dy
 P( x) y  Q ( x ) ----(2)
dx


2

Given condition y(x)=0 when x= -----(3)
Comparing the equations (1), (2), we get
P  cot x,

&

Q  4 x cos ecx

Integration factor
e Pdx  e cot xdx  elog sin x  sin x

[ y  y( x)]

General solution
ye pdx   Qe pdx dx  c
y sin x   4 x cos ecx sin xdx  c
y sin x   4 xdx  c

y sin x  4

x2
 c  2 x 2  c ----(4)
2

To find the constant C
We use the given condition (3) in (4), we get
i
...
sin( )  2( )  c
2
4
0
...

2
2

Which is the required solution
...
e

dy
 P( x) y  Q( x) y n
...

dx

dy
 P( x) y  Q( x) y
...

i
...


Problem:-01
Solve

dy
 xy  4 y
dx

Solution:The given DE's is

dy
 xy  4 y -----(1)
dx

The equation (1) is of the form

dy
 P( x) y  Q( x) y
...


Problem:-02
Solve

dy
 ( x 2  2) y  (tan x) y
dx

Solution:The given DE's is

dy
 ( x 2  2) y  (tan x) y -----(1)
dx

The equation (1) is of the form

dy
 P( x) y  Q( x) y
...


METHOD OF SOLUTION (WHEN n=2,3,4,
...
where n=2,3
...
----(1)
dx

Divide the equation by yn
1 dy
1
1
 P ( x) n y  Q( x) n y n
...

n
y dx
y

y n

dy
 P( x) y1n  Q( x)
...

dx
1 dz
 P( x) z  Q( x)
...

dx
dz
 P1 ( x) z  Q1 ( x)
...

This equation can be solved using following procedure
Integration factor
I
...


Problem:-1
Solve

dy y
  x2 y6
dx x

Solution:The given DE is

dy y
  x2 y6
dx x

-----(1)

It is of the form

dy
 P( x) y  Q( x) y n
...
F= e

 P ( x ) dx

e



5
dx
x

General solution
ze  P ( x ) dx   Q ( x)e P ( x ) dx dx  c

zx 5   (5 x2 ) x 5 dx  c
zx5  5  x3 dx  c

5

 e 5 log x  e log x  x 5

zx5  5
zx 5 
z

x 31
x2
 c  5
c
3  1
2

5x 2
c
2

5x3
 cx5
2

(Multiply x 5 )

Which is the general solution of equation (4)
...

Sub z=y-5, we get
y 5 

5x3
 cx5
2

(Multiply x5 )

Which is the required solution
...
----(3)
dy

Multiply equation (2) by x-2, we get
x 2

dx
 x 2 yx  x 2 y 3 x 2
dy

x 2

dx
 yx 1  y 3 -----(4)
dy

Let z=x1-n=x1-2=x -1
Differentiate z with respect to x, we get
dz
  x 11   x2
dx

dz   x2 dx

Update the above values in equation (4), we get


dz
 yz  y 3
dy

dz
 yz   y 3
dy

(Multiply by -1)

It is of the form

dz
 P( y) z  Q( y) ----(6)
dy

Comparing the equations (5) and (6), we get
P(y)=y

and

Q(y)=-y3

Integrating factor
I
...


1 t
te  et  c
2









 tet  2et  c

y2
t=
Sub
2 in the above equation, we get
ze

ze



y2
2

y2
2

 (2  t )e t  c

2

 (2  y )e

y2
2

c

Which is the general solution of equation (5)
...

Sub z=x-1, we get
1

x e

y2
2

2

 (2  y )e

1
 (2  y 2 )  ce
x

y2
2

 y2
2

c

(Multiply on both sides e

 y2
2

)

Which is the required solution
...
F= e  P ( x ) dx  e  2 xdx  e x

2

General solution
ze  P ( x ) dx   Q ( x)e P ( x ) dx dx  c
2

2

2

2

ze x   x3 e x dx  c
ze x   x2 e x xdx  c ----(5)
Let t=x2 ,

dt
dt
 2 x   xdx
dx
2

Substitute the above values in the right side of equation (6), we get
2
 dt 
ze x   (t )et    c
2

2

ze x 
2

ze x 

1 t
 te dt  c
2

1   et
(t ) 
2   1


 et

(1)



1


  c


 chain rule of integration  udv  uv ' u ' v '' u '' v '''
...

The general solution of (1) is obtained by replacing z into y
...


EXERCISE
Solve the following DE's
dy

y

dz

z

2 6
1
...
(logz)1  1  cx

2
2
...
dx  y tan x  y sec x

3 2
4
...
2 xy '  10 x y  y

dy x 2  y 2
6
...
x( x  y )dy  y dy  0

dy

tan y

x
8
...
e  dx  1  e
dy

2
10
...
6 y ' 2 y  ty

3
12
...


1 1 t

 ce t
y3
2

Ans

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