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?
 A: $$y'(t) = \frac{t^3+y^3}{ty^2}$$
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.
A: The problem with your method is that your equation is not already exact. To make it exact I will first write it in the form:
\begin{align}
3y^2y’-\frac{3}{t}y^3=3t^2.
\end{align}
Note that the derivative of $y^3$ is $3y^2y’$, which is why I’ve multiplied by $3/t$. Now I will seek an integrating factor $E(t)$ so that it becomes exact,
\begin{align}
3Ey^2y’-\frac{3E}{t}y^3&=(Ey^3)’=3Ey^2y’+E’y^3,\\
-\frac{3E}{t}&=E’\longrightarrow E=t^{-3}.
\end{align}
So the equation is exact under the integrating factor $t^{-3}$. Integrating we arrive at
\begin{align}
\left(\frac{y}{t}\right)^3=t^3+C,\\ \\
y^3=t^3(t^3+C).
\end{align}
I’ve effectively skipped the partial differential equation problem that would have resulted from your method by the note that $(y^3)’=3y^2y’$. For the general problem, you can seek an integrating factor of the form $E(t,y)$, but this will lead to a PDE which is solved via solving the original ODE, i.e. it isn’t helpful unless you have a particular solution already.
