# General solution for homogeneous equation

i am having trouble finding the general solution for the given homogeneous equation:

x2yy' = (2y2 - x2)
x2dy = (2y2 - x2) dx

i turned it into the following:
(2y2 - x2) dx - x2 dy = 0
then i used substitution of y = xv and got
(2(xv)2 - x2 - x2v) dx - x3 dv = 0
then i turned this into the following ... i think this may be the part i did wrong
(2v2 - 1 - v) dx - x dv = 0
then the integrating factor was
1/x(2v2 - 1 - v)
so...
dx/x - dv/(2v2 - 1 - v) = 0
i am stuck here. i know the first part is ln x, but i'm not sure if the second part is [ln (2v2 - 1 - v)]/(4v - 1)...if this part is right then i am stuck again here. i don't know how to get to the final solution which is

y(x) = x + Cx4/1 - 2Cx3

plz help. thx.

Last edited:

Char. Limit
Gold Member
i am having trouble finding the general solution for the given homogeneous equation:

x2dy = (2y2 - x2) dx

i turned it into the following:
(2y2 - x2) dx - x2 dy = 0
then i used substitution of y = xv and got
(2(xv)2 - x2 - x2v) dx - x3 dv = 0
then i turned this into the following ... i think this may be the part i did wrong
(2v2 - 1 - v) dx - x dv = 0
then the integrating factor was
1/x(2v2 - 1 - v)
so...
dx/x - dv/(2v2 - 1 - v) = 0
i am stuck here. i know the first part is ln x, but i'm not sure if the second part is [ln (2v2 - 1 - v)]/(4v - 1)...if this part is right then i am stuck again here. i don't know how to get to the final solution which is

y(x) = x + Cx4/1 - 2Cx3

plz help. thx.

Better if you write your final solution as

$$y(x) = \frac{x+Cx^4}{1-2Cx^3}$$

that doesn't help me :\