Problems with an exact differential equation I'm trying to solve this exact differential equation:
$$y'(t) = \frac{t^3+y^3}{ty^2}$$
First of all I split the eq. into this one:
$$(t^3+y^3)-(ty^2)y'(t) = 0$$
Let now be  $\alpha(t,y)=(t^3+y^3)$ and $\beta=-(ty^2)$.
I have to find a primitive $F:\mathbb R x \mathbb R \longrightarrow \mathbb R$ such that $\partial_t F=\alpha$ and $\partial_y F = \beta$.
We can compute that:
$$\partial_y \alpha -\partial_t\beta \neq0 $$and in fact it's
$$\partial_y \alpha -\partial_t\beta = 4y^2 $$
Now I can compute a factor $\mu = e^{-\int y^2\,dy}$ but after that I'm not ale to continue.
In fact if I put the factor in the equation, I don't find that the relation
$$\partial_y \mu \alpha -\partial_t \mu\beta = 0$$
is true. How can I continue? Any hint?

$$y'(t) = \frac{t^3+y^3}{ty^2}$$
This DE is homogeneous so substitute $y=wt$:
$$ w't+w= \frac{t^3+(wt)^3}{t(wt)^2}$$
$$ w't+w= \frac{1+w^3}{w^2}$$
$$ w't= \frac{1}{w^2}$$
This DE is separable.
