- #1
asteg123
- 12
- 0
I'm really stomped with this problem... i can't seem to get the answer...
anyway... here's the problem..
(2(x^3) - (y^3))y'=3(x^2)y
and i need to get the general solution...
SO, here's what i did...
I Let
u=x^3 and
du=3x^2dx
so what happens is
2udy-(y^3)dy= ydu
and that's where i got stuck...
i tried using
d(y/u)=(udy-ydu)/u^2
but i can't seem to get rid of the 2 in 2udy and it would be much of a problem if i did that...
could anyone help me??
anyway... here's the problem..
(2(x^3) - (y^3))y'=3(x^2)y
and i need to get the general solution...
SO, here's what i did...
I Let
u=x^3 and
du=3x^2dx
so what happens is
2udy-(y^3)dy= ydu
and that's where i got stuck...
i tried using
d(y/u)=(udy-ydu)/u^2
but i can't seem to get rid of the 2 in 2udy and it would be much of a problem if i did that...
could anyone help me??