Euler Lagrange, weird Second Order DE

Question

I have found the Euler-Langrange equation of the functional: $$\int_{a}^{b} y'^2+y^4\,\mathrm dx$$ to be $y'' = 2y^3$

How do I solve this non-linear DE?

Answer 1

$$y''=2y^3$$ $$2y''y'=4y^3y'$$ $$y^4=y'^2+c_1$$ $$y'=\pm \sqrt{y^4-c_1}$$ $$x=\pm\int \frac{\mathrm dy}{\sqrt{y^4-c_1}}+c_2$$ This is an elliptic integral : http://mathworld.wolfram.com/EllipticIntegral.html

The inverse function $y(x)$ involves the sn Jacobi elliptic function : http://mathworld.wolfram.com/JacobiEllipticFunctions.html $$y(x)= C_1 \text{sn}\left(C_1(x+c_2)\:|\:-1 \right)$$ with $C_1=c_1^{1/4}$

Answer 2

$$y'' = 2y^3$$ Substitute $$p=\frac {\mathrm dy}{\mathrm dx} \implies y''=\frac {\mathrm dp}{\mathrm dx}=\frac {\mathrm dp}{\mathrm dy}\frac {\mathrm dy}{\mathrm dx}=p'p$$ Then the equation is separable $$pp'=2y^3$$ Integrate $$\int p\,\mathrm dp=2\int y^3\,\mathrm dy$$ $$p^2+K=\frac {y^4}2$$ $$\frac {\mathrm dy}{\mathrm dx}=\pm \sqrt {\frac {y^4}2+C}$$ $$.....$$