TITLE: Solving the DE: $y''+1/(2y^3)=0$.
QUESTION [1 upvotes]: \begin{align}
y''+\frac{1}{2y^3}=0\\
\end{align}
This is a second order differential equation which doesn't contain $x$ explicitly.
Let $y'=p(y)$, so that $y''=p(y)*p'(y)$.
\begin{align}
pp'+\frac{1}{2y^3}=0\\
\int{pdp}=\int{-\frac{1}{2y^3}dy}\\
\frac{p^2}{2}=-\frac{1}{2}*\frac{-1y^{-2}}{2}+C_1\\
p^2=\frac{1}{2y^2}+C_1\\
p=\pm\sqrt{\frac{1+C_1y^2}{2y^2}}\\
\end{align}
$p$ was equal to $y'$
\begin{align}
y'=\pm\sqrt{\frac{1+C_1y^2}{2y^2}}\\
\end{align}
I've tried to continue with trigonometric substitution but that didn't work for me.
Can you please help me out?
Thank you in advance!
Edit: adding final solution
Because of the answers below, I found the solution:
\begin{align}
\frac{\sqrt2ydy}{\sqrt{1+C_1y^2}}=\pm{x}dx\\
\int{\frac{\sqrt2ydy}{\sqrt{1+C_1y^2}}}=\int{\pm{x}dx}\\
\frac{1}{2C_1}\int{\frac{d({1+C_1y^2})}{\sqrt{1+C_1y^2}}}=\pm\frac{1}{\sqrt2}x\\
2\sqrt{1+C_1y^2}=\pm{\frac{2C_1}{\sqrt{2}}}(x+C_2)\\
\sqrt{1+C_1y^2}=\pm{\frac{C_1}{\sqrt{2}}}(x+C_2)\\
2(1+C_1y^2)=C_1^2(x+C_2)\\
\end{align}

REPLY [1 votes]: Multiplying both sides of the ODE by $2y'$ concludes that:$$2y'y''+y'/(y^3)=0$$by integrating$$(y')^2-\dfrac{1}{2}\dfrac{1}{y^2}=C_1$$or$$y'=\pm \dfrac{\sqrt 2}{2}\dfrac{\sqrt{C_2y^2+1}}{y}$$which yields to$$\dfrac{yy'}{\sqrt{C_2y^2+1}}=\pm \dfrac{\sqrt 2}{2}$$which results in$$\dfrac{\sqrt{C_2y^2+1}}{C_2}=\pm \dfrac{\sqrt 2}{2}x +C_3$$
{"set_name": "stack_exchange", "score": 1, "question_id": 2801129}