Bernoulli equation with $e^y$ term I'm having a hard time solving the following differential equation:
$$(x^3 + e^y)y' = 3x^2$$
I'm familiar with the approach of introducing $z=y^{1-2}=y^{-1}$, but that doesn't do the trick. Am I missing something?
 A: $$3x^2x'-x^3-e^y=0$$
$$(x^3)'-x^3-e^y=0$$
$$u'-u-e^y=0$$
Where $u(y)=(x(y))^3$. Can you continue?
A: It is :
$$(x^3 + e^y)y' = 3x^2 \Leftrightarrow (x^3+e^y)y' - 3x^2 = 0$$
Now, let $R(x,y) = -3x^2$ and $S(x,y) = e^y + x^3$. Then, this is not an exact equation, because :
$$\frac{\partial R(x,y)}{\partial y} = 0 \neq 3x^2 = \frac{\partial S(x,y)}{\partial x}$$
We will find an integrating factor $\mu(y)$ such that the differential equation
$$\mu(y)R(x,y) + \frac{\mathrm{d}y(x)}{\mathrm{d}x}\mu(y)S(x,y) = 0$$
is exact.
But, this means that :
$$\frac{\partial}{\partial y}\bigg(\mu(y)R(x,y)\bigg)= \frac{\partial}{\partial x}\bigg(\mu(y)S(x,y)\bigg) \Rightarrow -3\frac{\mathrm{d}\mu(y)}{\mathrm{d}y}x^2 =3\mu(y)x^2 $$
$$\Rightarrow \dots \Rightarrow \mu(y) = e^{-y}$$
Multiplying your differential equation by $\mu(y)$ now, will yield you an exact equation (check why) with :
$$P(x,y) = -3x^2e^{-y}, \quad Q(x,y) =e^{-y}x^3 + 1$$
Now, define a function $f(x,y)$ such that :
$$\frac{\partial f(x,y)}{\partial x} = P(x,y) \quad \text{and} \quad \frac{\partial f(x,y)}{\partial y} = Q(x,y)$$
Then, the solution will be given by $f(x,y) = c_1$ where $c_1$ is an arbitrary constant.
Now, it is :
$$\frac{\partial f(x,y)}{\partial x} = -3x^2e^{-y} \implies f(x,y) = -\int3x^2e^{-y}\mathrm{d}y =-e^{-y}x^3 +g(y)$$
Now, differentiate the $f(x,y)$ above with respect to $y$ in order to find $g(y)$, which yields :
$$\frac{\partial f(x,y)}{\partial y} = e^{-y}x^3 + \frac{\mathrm{d}g(y)}{\mathrm{d}y}$$
Substituting that $f(x,y)$ now into the equation 
$$\frac{\partial f(x,y)}{\partial y} = Q(x,y) \implies g(y) = y$$
Thus, $f(x,y) = -e^{-y}x^3 + y$ which means that the solution to the given differential equation is :
$$f(x,y) = c_1 \Rightarrow -e^{-y}x^3 + y = c_1 \implies y(x) = \mathrm{W}(e^{-c_1}x^3) + c_1$$
If you are not familiar with the product log function $\mathrm{W}$, the expression before it is as good as a solution and as rigorous as the final one.
Note : I have left some computations for you along the way.
A plot for the slope field of the given differential equation and solution :
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And some sample initial values plots :
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A: Diff. equation we write as
$$3x^2dx-(x^3+e^y)dy=0$$
Integrating factor is $\mu=e^{-y}$. Exact equation
$$3x^2e^{-y}dx-(x^3e^{-y}+1)dy=0$$
has solution
$$x^3e^{-y}-y=C$$
