Page 137 - DMTH202_BASIC_MATHEMATICS_II

Basic Mathematics-II

Notes Integrating, we get

 1 v 
log x   v 2v   1 dv
 1 1 
    v  2v    dv
1
1
 log v  log 2v   1 logc 1
2

or 2log x  2log v log 2v 1 log ,c  2log c 1 log  c

2 2
 x v 
or log    lgc
 2v  1 
2 2
x v
or  c
2v  1
 y 2 
2
x  2 
 x   c    y 
v
or     x  
y
2    1
x
 
xy 2
or
2y  x

or xy 2  c 2y x 
which is the required general solution.

2
Example: Solve xdy y dx  x 2  y dx .

Solution:

dy y  x 2  y 2
Here  ….(1)
dx x

dy y  x 2  y 2
Putting y = vx, so that 
dx x
 (1) reduces to
dv 2
v  x   1 v

v
dx
dv 2

or x  1 v
dx
dv dx
or 2  .
1  v Dx

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