Solve $(x^3+e^y)y'=3x^2$. Help me please I can't solve it. Solve  $(x^3+e^y)y'=3x^2$. Help me please I can't solve it.
 A: Multiply both sides by $e^{-y}$:
$$
 x^3 e^{-y} y' + y' = 3 x^2 e^{-y}
$$
Rearrange this to yield:
$$
y' = \frac{d}{dx}\left(x^3 e^{-y}\right)\, ,
$$
Or:
$$
y(x) = C + x^3 e^{-y(x)}
$$
As noted above, the solution to this for $y(x)$ involves the Lambert $W$ function:
$$
y(x) = C + W\left(e^{-C} x^3\right)\, .
$$
The Lambert function cannot be expressed in terms of more elementary functions.
A: Let $u=x^3$. Then, $\frac{du}{dx}=3x^2$. $\frac{dy}{du}=\frac{y'}{3x^2}$.
The expression is,
$$y'=\frac{3x^2}{x^3+e^y}$$
$$\frac{dy}{du}=\frac{1}{u+e^y}$$
Let $v=u+e^y$. Then, $\frac{dv}{du}=1+e^y\frac{dy}{du}$.
$$\frac{\frac{dv}{du}-1}{e^y}=\frac1v$$
$$\frac{\frac{dv}{du}-1}{v-u}=\frac1v$$
$$v\frac{dv}{du}-v=v-u$$
$$v\frac{dv}{du}=2v-u$$
$$\frac{dv}{du}=2-\frac uv$$
Let $v=ku$. Then, $\frac{dv}{du}=k+u\frac{dk}{du}$.
$$k+u\frac{dk}{du}=2-\frac1k$$
$$\int \frac{dk}{2-k-\frac1k}=\int\frac{du}{u}$$
A: suppose we make two change of variables $$t = x^3, u = e^y \to dt = 3x^2 dx,\  du = udy \tag 1$$  now the differential equation $$(x^3 + e^y) dy = 3x^2 dx $$ becomes $$(t+u)\frac{du}{u} = dt $$ which we will rewrite as forced linear equation in $t,$ that is, $$\frac{dt}{du} = \frac{t+u}u=1+\frac t u \tag 2$$ now, $(2)$ has the general solution $$ t = u \ \ln u+Cu\tag 3$$ going back to the original variables, we get $$x^3=ye^y+Ce^y. $$
