How to solve the equation${d^2x}\over {dt^2}$-t${{d^3x}\over {dt^3}}$+$({{d^3x}\over {dt^3}})^3$=0?

Question

How to solve the equation $$\frac{d^2 x}{dt^2} -t \frac{d^3 x} {dt^3} + \left(\frac{d^3x}{dt^3}\right)^3 = 0.$$

I supposed $\frac{d^2 x}{dt^2} = p$ , and then I can know that $p-t \frac{dp}{dt}+\left(\frac{dp}{dt}\right)^2=0$, however, I don't know how to continue. Can somebody tell me how to solve the differential equation?

Answer 1

That is a good approach.

We have:

$$x'' = p \implies x''' = p'$$

Rewriting, we have:

• $p + (p')^3 - t p' = 0$
• $p = t p'-(p')^3$
• $p' = t p'' + p' - 3(p')^2 p''$
• $p' = p' + p''(t-3(p')^2)$
• $p''(t - 3(p')^2) = 0$

This gives us two items to solve:

$$p'' = 0, t - 3(p')^2 = 0$$

Can you take it from here?