# Solving the differential equation $x x''(t)=\frac{1}{t^3-t}$

Solve the following differential equation $$x x''(t)=\frac{1}{t^3-t}$$ I tried to integrate both members: $$x(t) x'(t)-\int [x'(t)]^2 dt=\frac{1}{t}+\log\left|\frac{1}{t}-1\right|$$ but the situation does not improve. At this point, I think there is a mistake in the text of exercise.

I would appreciate some help with this problem.

Thank you very much.

• Does this problem have a physical background? Has it any initial condition? – Dmoreno Jun 13 '15 at 18:53
• No, it appeared in an old exam. I have a question: either there is a mistake in the text, or I can not seem to solve the exercise. – Mark Jun 13 '15 at 18:56
• Neither do I. Mathematica fails to find a closed-form solution as well. Maybe you're right and there is a typo in the statement of the problem. – Dmoreno Jun 13 '15 at 19:19

As stated in the comments, Mathematica doesn't give a closed form for the solution. Still, there is some information we can find anyway.

For $t\gg1$ we have $$\frac{1}{t^3-t}\approx\frac{1}{t^3},$$ so that asymptotically a solution $x(t)$ will satisfy $$x(t)x''(t) = \frac{1}{t^3}.$$ We can find a solution of this by assuming $x_{t\gg1}(t)=\beta t^\alpha$ for some $\alpha,\beta\in\mathbb{R}$, then $$x(t)x''(t) = \beta^2\alpha(\alpha-1)t^\alpha t^{\alpha-2}.$$ It follows that $2\alpha - 2 = -3$, and thus $\alpha = -\tfrac{1}{2}$, and $\beta^2\frac{3}{4} = 1$, so that $$x(t) = \pm\frac{\sqrt{3t}}{2}$$ for $t\gg1$. Maybe something more can be derived by working on perturbations of this asymptotic solution, but I'm not sure.

It would also be interesting to get some solutions for $0<t<1$.

Hint:

$$xx''(t)=\dfrac{1}{t^3-t}$$

$$x\dfrac{d^2x}{dt^2}=\dfrac{1}{t^3-t}$$

$$x\dfrac{d}{dt}\left(\dfrac{dx}{dt}\right)=\dfrac{1}{t^3-t}$$

$$x\dfrac{d}{dx}\left(\dfrac{1}{\dfrac{dt}{dx}}\right)\dfrac{dx}{dt}=\dfrac{1}{t^3-t}$$

$$-x\dfrac{\dfrac{d^2t}{dx^2}}{\left(\dfrac{dt}{dx}\right)^3}=\dfrac{1}{t^3-t}$$

$$t(1-t^2)x\dfrac{d^2t}{dx^2}=\left(\dfrac{dt}{dx}\right)^3$$