How to solve $(3y^3-x)y'=3y$?

Question

I am having trouble with this equation:

$$(3y^3-x) y'=3y$$

and it seems that none of the methods I know works. Maybe what I need is a hint or a push to a right direction. Thank you in advance.

Answer 1

You can revert the dependency, $x$ as a function of $y$ satisfies the linear ODE $$ 3yx'(y)+x(y)=3y^3 $$ which you can solve with standard methods. Then try to find an inverse function of the solution.

Answer 2

If you are familiar with exact/non-exact differential equations and finding integrating factors, you can also try this.

Your equation can be rewritten as $$ 3y dx + (x-3y^3)dy = 0 .$$

Then $\mu = y^{-\frac{2}{3}}$ is an integrating factor. Multiplying the DE with $\mu$ gives $$ 3 y^{\frac{1}{3}} dx + (xy^{-\frac{2}{3}} - 3 y^{\frac{4}{3}}) dy = 0 .$$

Then the answer is $\phi(x,y)=c$, with $c$ constant and $\phi(x,y)$ the solution of the system $\phi_x = 3 y^{\frac{1}{3}}$, $\phi_y = xy^{-\frac{2}{3}} - 3 y^{\frac{4}{3}}$.

(Since you asked for hints, I didn't fill in the details.)