Page 155 - DMTH202_BASIC_MATHEMATICS_II

Basic Mathematics-II

Notes
x
 dx  1 log 1 (  x )  1
2
2
(
2 1 x ) 2

I.F.  e e  4 1 (   x ) 4 .
 The solution is
x
F
I
I
F
z . ( . .)   . ( . .) dx  c
2
1 1
 
2
or z . 1 (  x ) 4   x 1 (  x ) 4 dx  c
2
2
1

1
2
  1 (   x ) 4 . ( 2x) dx  c
4
3
2 4
1 1 x )
(

   c
4  3 
 
 4 
1
1 2
2 4
or z   1 (  x )  c 1 (  x )
3
1
1 2
2 4
or y   1 (  x )  c 1 (  x )
3
Which is the required solution.
Example:
2 dy
Solve xy ( 1 xy )  1.

dx
Solution:
The given equation can be written as
dx 2 3
 y  x y .
dy
2
Dividing by x , we have
3
x  2 dx  y x  1  y . …..(1)
dy
-1
Putting x = z
 x  2 dx  dz
dy dy

or x  2 dx  dz .
dy dy

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