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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

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