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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.

Thanks in advance.

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    @Mike: Oops....(I had suggested integrating twice).2010-10-20

3 Answers 3

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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.2010-10-20
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    [Messy is right](http://tinyurl.com/26uqp9n)... o_O2010-10-21
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    @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)2010-10-21
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    I know, I already have that book. :)2010-10-21
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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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Hint $$\frac{d^2y}{dx^2} - \frac{2}{y^2}=0$$ Multiply by $2y'$ $$2y'\frac{d^2y}{dx^2} - \frac{4y'}{y^2}=0$$ $$((y')^2)' +4 \left(\frac{1}{y}\right)'=0$$ Then integrate $$(y')^2 +\frac{4}{y}=K_1$$ $$.......$$