# Nonlinear Ordinary differential equation

I would like to solve:

$$\frac{d^2y}{dx^2} - \frac{2}{y^2}=0$$

with $y(0)=a$ and $y'(0)=0$

Where $a$ is a known constant.

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@Mike: Oops....(I had suggested integrating twice). – Arturo Magidin Oct 20 '10 at 20:03

Change variables so that the independent variable is $y$ and the dependent on $p=y'$. Then $p\dot p=y''$ (dots stand for derivatives w.r.t. $y$, primes w.r.t. $x$) and the equation becomes $$p\dot p-\frac2{y^2}=0$$ or $$\frac{d}{dy}(\frac{p^2}2)=\frac2{y^2}.$$ This can be solved by integrating into a first order equation for $y$ as a function of $x$.

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If one finds this change of variables mysterious, another way of phrasing it is that we just multiply the ODE by $2 dy/dx$ and note that it then becomes $\frac{d}{dx}[ (dy/dx)^2 + 4/y ] = 0$. (This is the standard trick in mechanics for 1-dim "particle in potential" systems $d^2 q/dt^2=-\nabla V(q)$.) Hence $(dy/dx)^2 + 4/y = C$, where $C=4/a$ according to initial conditions. Solving for $dy/dx$ gives a separable first order ODE for $y$ which can be integrated to give $x$ as a function of $y$. That's as far as I got; the expression seems too messy to be invertible by a simple closed formula. – Hans Lundmark Oct 20 '10 at 20:37
Messy is right... o_O – J. M. Oct 21 '10 at 1:26
@Hans, this substitution applies more generally to any equation in which the independent variable does not appear explicitly, and allows you to lower the order by one. One natural way to 'find' it is to look for symmetries of the equation, à la Petter J. Olver (See his extraordinarily beautiful book Applications of Lie Groups to Differential Equations for details) – Mariano Suárez-Alvarez Oct 21 '10 at 13:55
I know, I already have that book. :) – Hans Lundmark Oct 21 '10 at 14:47

Maybe this is the same technique, but what I have done before for $y''=f(y)$ is treated as $$\frac{{\rm d}y'}{{\rm d}x} = f(y)$$ $$\frac{{\rm d}y'}{{\rm d}y} \frac{{\rm d}y}{{\rm d}x} = f(y)$$ $$y'\,\frac{{\rm d}y'}{{\rm d}y} = f(y)$$ $$\int y'\,{\rm d}y' =\int f(y)\,{\rm d}y +K_0$$ $$\frac{1}{2} y'^2 = g(y)$$ with $g(y)=\int f(y)\,{\rm d} y + K_0$ $$y' = \sqrt{2\,g(y)}$$ $$\frac{{\rm d}y}{\sqrt{2\,g(y)}} = {\rm d}x$$ $$x = \int \frac{1}{\sqrt{2\,g(y)}}\,{\rm d}y+K_1$$

With your example $$f(y)=2/y^2$$ $$g(y)=\int 2/y^2\,{\rm d}y+K_0 = -2/y +K_0$$ $$x = \int \frac{1}{\sqrt{2\,\left(-2/y +K_0\right)}}\,{\rm d}y+K_1$$ and with the IC given $$x = \frac{\sqrt{a}}{4} \left( 2 a \ln{\left(\sqrt{y-a}+\sqrt{y}\right)}-a \ln{a}+2 \sqrt{y (y-a)} \right)$$

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