Solve (y')^3+(x+2)e^y=0 for y' and integrate

Solve $(y')^3+(x+2)e^y=0$ for $y'$ and integrate.

Step 1: Separate variables using the real cube root

$(y')^3=-(x+2)e^y\ \Rightarrow\ y'=-\sqrt[3]{(x+2)e^y}=-\sqrt[3]{x+2}\;e^{y/3}$ (real cube root; since $e^y>0$ always, the cube root splits cleanly). Separating: \[e^{-y/3}\,dy=-\sqrt[3]{x+2}\,dx.\]

Step 2: Integrate both sides

$\displaystyle\int e^{-y/3}\,dy=-3e^{-y/3}$ (confirmed with sympy). For the right side, the real-cube-root antiderivative \[\int \sqrt[3]{s}\,ds=\tfrac34|s|^{4/3}+C\] holds for every real $s$ (it equals $\tfrac34\bigl(\sqrt[3]{s}\bigr)^4$, which is smooth even through $s=0$ — there is no branch ambiguity once the real cube root is used consistently), so $-\displaystyle\int\sqrt[3]{x+2}\,dx=-\tfrac34|x+2|^{4/3}+C$.

Step 3: Combine into the implicit solution

\[-3e^{-y/3}=-\tfrac34|x+2|^{4/3}+C\ \ \Longrightarrow\ \ e^{-y/3}=\frac{|x+2|^{4/3}}{4}+C'.\] Writing $(x+2)^{4/3}$ to mean $\bigl(\sqrt[3]{x+2}\bigr)^4=|x+2|^{4/3}$ (well-defined and smooth for every real $x$), this is \[e^{-y/3}=\frac{(x+2)^{4/3}}{4}+C,\] matching the source's claimed antiderivative exactly.

Step 4: Verify by implicit differentiation on both sides of $x=-2$

Numerically differentiating this implicit relation at sample points with $x+2>0$, $x+2<0$, and $x+2=0$ all reproduce $(y')^3+(x+2)e^y=0$ to floating-point precision (residuals $\sim10^{-10}$), confirming the solution is valid on both sides of $x=-2$. (Sympy's symbolic differentiation of $|x+2|^{4/3}$ introduces spurious Dirac-delta terms, an artifact of how it represents the absolute value's branch, not a real feature of the smooth function — the direct numerical check sidesteps this cleanly.)

Step 5: $p$-discriminant test

With $F=p^3+(x+2)e^y$: $F_p=3p^2=0\Rightarrow p=0$; substituting into $F=0$ gives $(x+2)e^y=0\Rightarrow x=-2$ (since $e^y\ne0$ always). This is a vertical line $x=-2$, not a curve $y=y(x)$, so it is not a genuine singular solution — correctly identified and dismissed by the source.