Question

Solve (x^3)y'=2(y^2+x^2*y-x^4) explicitly.

Answer 1

@SithsAndGiggles

Answer 2

Do I multiply the right side?

Answer 3

\[\begin{align*}
x^3y'&=2y^2+2x^2y-2x^4\\\\
y'&=\frac{2y^2}{x^3}+\frac{2y}{x}-2x
\end{align*}\]
I'd try the homogeneous substitution next, \(y=vx\), then \(y'=xv'+v\).
\[\begin{align*}
xv'+v&=\frac{2v^2}{x}+2v-2x\\\\
v'&=\frac{2v^2}{x^2}+\frac{v}{x}-2
\end{align*}\]
Another homogeneous sub should do the trick.

Answer 4

After you solved for v', what would you do?

Answer 5

Substitute again, \(v=tx\) so \(v'=xt'+t\).
\[xt'+t=2t^2+t-2\]
which will be a separable equation.

Answer 6

I'll come back later due to something. Excuse me.

Answer 7

Thank you!