Solving an exact IVP and solution domain

Question

$e^{2y}+x^2+2xe^{2y}\frac{dy}{dx}=0 \\ y(-1)=1$

This IVP is exact and:

\begin{align*} \frac{d(e^{2y}x + \frac{x^3}{3})}{dx}=0 \\ \iff e^{2y}x + \frac{x^3}{3}=k; \ k \in \mathbb{R} \\ \underbrace{\iff}_{y(-1)=1} e^{2y}x + \frac{x^3}{3}=-e^2-\frac{1}{3} \end{align*}

$$e^{2y}x + \frac{x^3}{3}=-e^2-\frac{1}{3} \\ \iff e^{2y}x = -\frac{x^3}{3}-e^2-\frac{1}{3} \\ \iff e^{2y} = \frac{-\frac{x^3}{3}-e^2-\frac{1}{3}}{x} \\ \iff y=\frac{1}{2}\log(\frac{-\frac{x^3}{3}-e^2-\frac{1}{3}}{x}); D_{y}=]{\sqrt[3]{-e^2-\frac{1}{3}}}, 0[ $$ Is the solution and its domain correct?

Answer 1

Your solution for the $y$ value is correct.

Maybe correct your working for the domain, because:

1. $-\dfrac{x^3}{3}-e^2-\dfrac13=0 \implies x^3=-3e^3-1$
2. If $\dfrac{-\dfrac{x^3}{3}-e^2-\dfrac13}{x}$ is $0$ or undefined then the natural log of that is undefined. This means open set rather than closed i.e $D_y=\left(\sqrt[3]{-3e^2-1},0\right)$