- #1

asteg123

- 12

- 0

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??