How can I solve the second order nonlinear ODE $y'' = 1/y^2$ or alternatively $y''(y^2) =$ constant?

Question

How can I solve the above ODE with respects to time? It is a nonlinear, second order differential equation which can be expressed in two ways:

(1) $y'' = 1/y^2 $

(2) $y''(y^2) =$ constant.

$y'' = d^2y / dt^2$

Answer 1

First multiply through by $y'$ giving $y'y''=y'y^{-2}$.

Note that $y'y^{-2}=-\frac{d}{dt}(y^{-1})$, and $y'y''=\frac{1}{2}\frac{d}{dt}((y')^{2})$, so the ODE becomes;

$$\frac{d}{dt}((y')^{2})=-2\frac{d}{dt}(y^{-1})$$

Integrate with respect to $t$ to find;

$$(y')^{2}=-2y^{-1}+C$$

Where $C$ is a constant of integration. Solving, we have $y'=\sqrt{C-2y^{-1}}$, where we take the positive root because;

$$y''=y^{-1}\Rightarrow y'=\int\frac{1}{y^{2}}dt\geq0$$

This is now a separable first order ODE. Separating and integrating with respect to time yields;

$$t(y)=\int\frac{\sqrt{y}}{\sqrt{Cy-2}}dy$$

This integral can be solved with the substitution $u=\sqrt{Cy-2}$, (although it is kind of messy), then you have your solution in the form of $t(y)$. Im not sure if it can be inverted to give $y(t)$ anlytically, so hopefully $t(y)$ is good enough!

Answer 2

You can substitute $$p=\frac {dy}{dx} \implies y''=\frac {dp}{dx}=p\frac {dp}{dy}$$ $$\implies y''=\frac 1 {y^2} \implies pp'=\frac 1{y^2}$$ This is separable $$\int pdp=\int \frac {dy}{y^2}$$