This is a simple yet ugly ODE (arising from Euler-Lagrange equations): $$yy''-(y')^2=y^4$$
What method could I use to quickly solve it?
I began to notice that by dividing by $y^2$ I can write it as $$ (y'/y)'=y^2 $$ which implies that $y'=y^3t+cy$ for some constant $c$. But even there I don't know how to proceed.
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$\begingroup$introduce a variable $u$ by $$y' = yu, \quad u = \frac {y'} y \tag 1 .$$ then $$u' = \frac{y''} y - \frac{y'^2}{y^2} =\frac{yy''-y'^2}{y^2} = y^2 \tag 2$$ dividing $(1)$ by $(2),$ we have $$\frac{dy}{du} = \frac uy\implies u^2 = y^2 + c , y'^2 = u^2y^2 =y^2(y^2 + c) $$ that is $$y' = \pm y\sqrt{c+y^2} \to \int\frac{dy}{y\sqrt{c+y^2}} = \pm x + d.$$
$\endgroup$ $\begingroup$You can write: $$y'' = \frac{dy'}{dx}=\frac{dy'}{dy}\frac{dy}{dx}=y'\frac{dy'}{dy}=\frac12\frac{dy'^2}{dy}.$$ Now we introduce $p=y'^2$, and rewrite your equation: $$\frac y2 \frac{dp}{dy}-p=y^4.$$ Solve $$\frac y2 \frac{dp}{dy}-p=0;\quad \frac{dp}{p}=2\frac{dy}{y}\Longrightarrow p=Cy^2.$$
Let's try $p=Cy^n$: $$\frac y2\frac{dp}{dy}-p=\frac12 Cny^n - Cy^n=y^4,$$ and $n=4$, $C=1$. So we have general solution $p=y^4 +Cy^2$, or $y'^2=y^4 + Cy^2$. Hence $$\pm\int\frac{dy}{\sqrt{y^4+Cy^2}}=x + D$$
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