Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

How to solve this particular ODE:

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

share|improve this question
1  
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
add comment

3 Answers 3

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 $x$.

share|improve this answer
add comment

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 $\cos(t)$.

share|improve this answer
add comment

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.

share|improve this answer
    
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 at 22:05
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.