Question

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

Answer 1

@SithsAndGiggles

Answer 2

like this???

Answer 3

\[\begin{align*}3x(y^2)y'&=y^3+x\end{align*}\]
Let's try the Bernoulli substitution, \(v=y^3\), so \(v'=3y^2y'\).
\[\begin{align*}xv'&=v+x\\\\
v'-\frac{1}{x}v&=1\end{align*}\]
The eq. is linear. Find the integrating factor, etc.

Answer 4

So the integrating factor is 1/x, right?

Answer 5

Yes

Answer 6

Got it! Thanks!

Answer 7

You're welcome!