Page 136 - DMTH202_BASIC_MATHEMATICS_II

Unit 10: Homogeneous Equations

Notes
dy dv
v
,
Putting y  vx so that   x
dx dx
dy 1  v 2
 (1) reduces to v x 
dx 2v
dv 1  v 2 1 3v 2
v
or x   
dx 2v 2v
2v dx
or 2 dv 
1  3v x
Integrating, we get

1 2
 log  1 3v   log x c

3
or 3log x  log  1 3v 2   3c

or log x 3  1 3v 2   3c

 3y 3  3c
3
c
or x  1  2   e  ' : y  vx 
 x 
 3y 2 
3
c
or x  1  2   '
 x 
or  x x 2  3y 2   'c

which is the required general solution.

2
Example: Solve y dx   xy  x 2  dy  0
Solution:

dy y 2
Here   2 …(1)
dx xy  x
dv dv
v
Putting y = vx, so that   x
dx dx
dv v 2
 (1) reduces to v x 
dx 1  v

dv v  2v 2
or x  
dx 1  v

dx  1 v dv
or   2
x v  2v

LOVELY PROFESSIONAL UNIVERSITY 131