Page 151 - DMTH202_BASIC_MATHEMATICS_II

Basic Mathematics-II

Notes Thus (2) becomes
-1 -1
-1
x e tan y  e tan y (tan y -1 )  c
-1
-1
or x  (tan y – 1)  ce tan y ,
which is the required solution.

Example:
2
Solve x (1–x ) dy + (2x y–y-ax ) dx = 0.
2
3
Solution:
The given equation is equivalent to
dy 2x 2  1 ax 2
 y  . ……(1)
dx x (1  x 2 ) 1  x 2

2x 2  1 ax 2
Here P = 2 , Q  2 .
x (1  x ) 1  x

 2  2
 2x  1  1  2x
  2 dx  dx
 pdx  (1x  x  ) ( x x 2  1)
F
 I . .  e e   e
1  2x 2
 dx
( x x  1) (x  1)
 e

  1 1 1 
      dx
 x 2 ( x  1) 2 ( x  1) 
e  
 1 
log    
  x x 2 1  1
 e   .
x x 2 1

Thus the solution of (1) is

1 ax 2 1
. y   2 . dx  c
x x 2  1 1  x x x 2 1

x
c
  a  3 dx
(1  x 2 ) 2
a dt 2
   3 , (t x  1)
c
2 t 2
a 1
c
t
  . 2( ) 2
2

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