Solution of $y''+y^2=0$?

Question

I'm trying to find solution for $$ \frac{d^2y}{dx^2}+y^2=0 $$

Can this differential equation(it is nonlinear, isn't it?) be solved analytically?

Or do I have to resort to numerical method?

The solution of $$ \frac{d^2y}{dx^2}-y^2=0 $$ should be fine, too.

Answer 1

substitute $p=y'$ and conclude $y''=p\frac{dp}{dy}$ because $$y''=\frac{dy'}{dx}=\frac{dp}{dx}=\frac{dp}{dy}\frac{dy}{dx}=\frac{dp}{dy}y'=\frac{dp}{dy}p$$

then get $p\frac{dp}{dy}+y^2=0 \Rightarrow pdp=-y^2dy\Rightarrow p^2=-\frac{y^3}{3}+C \Rightarrow p= \pm \sqrt{C-\frac{y^3}{3}}$ then substitute for $p$ $$\frac{dy}{dx}=\pm\sqrt{C-\frac{y^3}{3}} \Rightarrow \frac{dy}{\pm\sqrt{C-\frac{y^3}{3}}} = dx$$ this last integral cant be evaluated.

also we can use this method for the second equation too: $$p\frac{dp}{dy}-y^2=0\Rightarrow p^2=\frac{y^3}{3}+C \Rightarrow p= \pm \sqrt{C+\frac{y^3}{3}}\Rightarrow \frac{dy}{\pm\sqrt{C+\frac{y^3}{3}}} = dx$$ again this integral can't be evaluated.

Answer 2

$$y''+y^2=0$$ $$2y''y'+2y^2y'=0 \quad\to\quad y'^2+\frac{2}{3}y^3=c_1$$ $$y'=\pm \sqrt{c_1-\frac{2}{3}y^3}\quad\to\quad dx=\pm\frac{dy}{\sqrt{c_1-\frac{2}{3}y^3}}$$ $$x=\pm\int\frac{dy}{\sqrt{c_1-\frac{2}{3}y^3}}+c_2$$ This integral can be expressed on closed form thanks to elliptic integrals.

On inverse, $y(x)$ is a Weierstrass elliptic function, from Eq.(35) with $g_2=0$ in : http://mathworld.wolfram.com/WeierstrassEllipticFunction.html