How to find the interval of existence of this Bernoulli Equation? Question: $\frac{dy}{dx} - y + 2 = xe^x (y-2)^{1/3}$ with initial conditions $y(0) = 3$
My solution: $y(x) = (2xe^x - 6e^x +Ce^{\frac23 x})^{\frac 32} + 2$
Using the initial condition I found $C=7$. Now when I plot my answer I get a function that looks like it is asymptotically increasing as $\lim_{y \to \infty} $. But it seems like the interval of validity is $(-\infty, \infty)$.
How should I approach this?
 A: A differential equation is Bernoulli when can be written as $y'=a(x)y+b(x)y^{\alpha}$ and $\alpha\in \mathbf{R}$ so a priori I do not see how the title it relates with the differential equation.
On other hand, for the context: in the following, I will consider real value root. So the domain of $x\mapsto x^{1/3}:=\sqrt[3]{x}$ and $x\mapsto x^{2/3}:=\sqrt[3]{x^{2}}$ is all real set. $^{(1)}$
Setting $f(x,y)=xe^{x}(y-2)^{1/3}-2+y$ wich is continuous everywhere and $f_{y}(x,y)=\frac{e^{x}x}{3(y-2)^{2/3}}+1$ which is continuous everywhere except on the line $y=2$. Then a rectangle can be drawn about the initial point $(0,3)$ in which $f$ and $f_{y}$ are continuous. Therefore, by  Picard–Lindelöf theorem the initial value problem has a unique solution in some interval about $x=0$.
I have checked your solution and I think it works. So I will only answer your question about the existence interval of the solution.
Considering your definition of  $y(x)$ and using the definition $^{(1)}$, then $y(x)$ can be written as $$y(x)=\sqrt[3]{(2xe^{x}-6e^{x}+7e^{\frac{2}{3}x})^{2}}+2,$$
which is well-defined over all real set. Therefore, the interval of validy is $]-\infty,+\infty[$.
