# A secondary nonlinear ODE

How to solve this particular ODE:

$$\frac{d^2y}{dx^2}=\frac{a^2}{y^2}-\frac{b^2}{y^3}$$

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The best questions on this site include not only a statement of the problem, but also the context in which you encountered the problem and a description of what you have tried already. That information makes it possible for the answers to be written in a more personalized manner. – Carl Mummert Feb 5 '13 at 13:07

Your ODE is of the form $y''=-V'(y)$, which is typical in mechanics. In your case, the potential is $V(y)=a^2 y^{-1}-\frac{1}{2} b^2 y^{-2}$. Multiply the equation by $y'$ and integrate; this gives $\frac{1}{2} (y')^2 = -V(y) + E$, where $E$ is a constant of integration. Thus $dy/dx = \pm \sqrt{2(E-V(y))}$, which is a separable ODE that can (in principle) be integrated to give $x$ as a function of $y$, and then you can (in principle) invert that relation to get $y$ as a function of $x$.

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Multiply both sides of ODE by $\frac{dy}{dx}$ and rearrange terms, we get: $$\frac{d}{dx}\left[ \frac12 \left(\frac{dy}{dx}\right)^2 + \frac{a^2}{y} - \frac{b^2}{2y^2}\right] = 0 \implies \left(\frac{dy}{dx}\right)^2 = 2 E - \frac{2a^2}{y} + \frac{b^2}{y^2}$$ for some constant $E$. Just for illustration, let's look at case when $E > 0$. Introduce a parametrization $t \mapsto ( x(t), y(t) )$ by: $$\frac{dx}{dt} = \frac{1}{\sqrt{2E}} y(t)$$ then $y$ satisfies: $$( \frac{dy}{dt} )^2 = y^2 - \frac{a^2}{E} y + \frac{b^2}{2E} = ( y - \frac{a^2}{2E} )^2 + \Omega^2$$ where $\Omega^2 = \frac{b^2}{2E} - \frac{a^4}{4E^2}$.

For $E \in ( \frac{a^4}{2b^2}, \infty )$, $\Omega$ is real and the ODE has a solution: \begin{align} & y(t) = \frac{a^2}{2E} \pm \Omega \sinh t\\ \implies & x(t) = \frac{1}{\sqrt{2E}}\left( \frac{a^2}{2E} t \pm \Omega \cosh t\right) \end{align} For $E \in ( 0, \frac{a^4}{2b^2} )$, $\Omega = i|\Omega|$ is imaginary and the ODE has a solution: \begin{align} & y(t) = \frac{a^2}{2E} \pm |\Omega| \cosh t\\ \implies & x(t) = \frac{1}{\sqrt{2E}}\left( \frac{a^2}{2E} t \pm |\Omega| \sinh t\right) \end{align} The case for $E < 0$ is similar except you get solutions involving $\sin(t)$ and $\cos(t)$.

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Maple comes up with the implicit solutions $$-C_{{1}}\sqrt {-2\,{a}^{2}y \left( x \right) {C_{{1}}}^{2}+{b}^{2}{C_{ {1}}}^{2}+ y \left( x \right) ^{2}}+{a}^{2}{C_{{1}}}^{ 3}\ln \left( C_{{1}} \right) -{a}^{2}{C_{{1}}}^{3}\ln \left( -{a}^{2 }{C_{{1}}}^{2}+y \left( x \right) +\sqrt {-2\,{a}^{2}y \left( x \right) {C_{{1}}}^{2}+{b}^{2}{C_{{1}}}^{2}+ y \left( x \right) ^{2}} \right) -x-C_{{2}}=0$$ and $$C_{{1}}\sqrt {-2\,{a}^{2}y \left( x \right) {C_{{1}}}^{2}+{b}^{2}{C_{{ 1}}}^{2}+ y \left( x \right) ^{2}}-{a}^{2}{C_{{1}}}^{3 }\ln \left( C_{{1}} \right) +{a}^{2}{C_{{1}}}^{3}\ln \left( -{a}^{2} {C_{{1}}}^{2}+y \left( x \right) +\sqrt {-2\,{a}^{2}y \left( x \right) {C_{{1}}}^{2}+{b}^{2}{C_{{1}}}^{2}+ y \left( x \right) ^{2}} \right) -x-C_{{2}}=0$$ where $C_1$ and $C_2$ are arbitrary constants.

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Thx, but how we solve this by hand? – Ryan Feb 5 '13 at 8:21
Don't except Robert to post you the entire solution – LoveFood Mar 12 '14 at 22:05