Derivation step in ODE solution in textbook

[D_Tr]

This is not homework but is part of the solution process of an ODE and I cannot understand how the author made a derivation step. After a change of variable in the original ODE, the ODE in the new independent variable has a standard method of solution. But instead of using this method, the author takes a shortcut but the exact steps are skipped and I cannot reproduce the transition on my own.

Homework Statement

How do we go from: y'' + 2y^{-3} = 0 to: y' = \sqrt{2(c_1 + 1/y^2)} ?

The only hint given by the author is: "...but in this case we can multiply through by y' and integrate directly to obtain..."

Homework Equations

N/A

The Attempt at a Solution

I tried myltiplying through by y' and integrating both sides by parts but that gave me nothing I could use to go further.. I cannot even get back the first eq. by differentiating the second..

 

[vanhees71]

Think about the chain and product rule of integration. If I say more, it's more or less the solution ;-)).

BTW: The textbook is incomplete! In front of the square root should be a \pm sign!

 

[D_Tr]

Thanks vanhees71! Your comment helped
So the intermediate steps are:

<br /> y&#039;&#039;y&#039; = -2y^{-3}y&#039; \implies y&#039;&#039;y&#039; = \frac{d(y^-2)}{dt} \implies \int{y&#039;&#039;y&#039;}dt = y^{-2} + c1<br />
<br /> \implies \frac{y&#039;^{2}}{2} = y^{-2} + c1 + c2 \implies y&#039; = \pm \sqrt{2(c + y^{-2})}<br />