Need a hint to solve the ODE $(t^3+e^y)y'= 3t^2$. I'm practicing on ODE and I found this ODE, the problem is that i don't know what technique to use or where to start : it's 

$$(t^3+e^y)y'= 3t^2$$

Please give me a Hint , Thank You So Much For Your Help. 
 A: In another post its been asked how this equation is a Bernoulli Equation. That post was directed here because of repetition. Here is how it can be solved using Bernoulli form.
$$(t^3 + e^y) \frac{dy}{dt} = 3t^2$$
$$ t^3 + e^y = 3t^2 \frac{dt}{dy}$$
$$ \frac{dt}{dy} - \frac{1}{3}t = \frac{1}{3}e^y t^{-2}$$
Now it has the form of Bernoulli Eq:
$$ \frac{dt}{dy} + f(y)t = g(y) t^n$$
The substitution:  $\displaystyle u = t^{1-n}$  will transform the DE into first order linear and then can be solved using integrating factor.
A: Multiply with $e^{-y}$ and try to find a helpful substitution, $u=e^{-y}$ or $u=t^3e^{-y}$.

Or integrate then directly
$$
3t^2e^{-y}-t^3e^{-y}y'-y'=0\implies t^3e^{-y}-y=c
$$
which then indeed can be solved using the Lambert-W function as
$$
t^3e^c=(c+y)e^{c+y}\implies y=-c+W(t^3e^c)
$$
A: Let $y=\ln u$ then
$$(t^3+u)\dfrac{du}{u}= 3t^2dt$$
$$t^3du+udu= 3t^2udt$$
$$t^3du-3t^2udt=-udu$$
$$\dfrac{t^3du-udt^3}{t^6}=-\dfrac{udu}{t^6}$$
$$d\left(\dfrac{u}{t^3}\right)=-\left(\dfrac{u}{t^3}\right)^2\dfrac{du}{u}$$
$$\int\dfrac{d\left(\frac{u}{t^3}\right)}{-\left(\frac{u}{t^3}\right)^2}=\int\dfrac{du}{u}$$
$$\dfrac{1}{\left(\frac{u}{t^3}\right)}=\ln u+C$$
$$t^3=u\ln u+Cu$$
