# Solving the DE: $y''+1/(2y^3)=0$.

\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.

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}

• – Shaun May 29 '18 at 21:28
• Please edit the question using MathJax instead of a picture. This makes the question easier to read, edit, and search. – Shaun May 29 '18 at 21:33
• Okay! I'll try to edit it. – bravdwal May 29 '18 at 21:36
• After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark $\checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?. – Shaun May 29 '18 at 22:56
• I've edited everything, I also marked the answer of Doug M – bravdwal May 30 '18 at 7:24

$y' = \frac {\sqrt {Cy^2 + 1}}{\sqrt 2 y}\\ \int \frac {y}{\sqrt {Cy^2 + 1}} \ dy =\int \frac {1}{\sqrt 2} \ dt \\ \frac 1C \sqrt {Cy^2 + 1} = \frac {1}{\sqrt 2} t + D$

• Thank you for your help! – bravdwal May 29 '18 at 22:10

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$$\dfrac{\sqrt{C_2y^2+1}}{C_2}=\pm \dfrac{\sqrt 2}{2}x +C_3$$

• Thank you for helping me out. After reading the other tips, I found this solution. – bravdwal May 29 '18 at 22:12

Try to find $x$ as a function of $y^2$ instead, for taking it next from there, in terms of $y^2$:

$$dx= \frac{\sqrt2y dy}{\sqrt{1+2 C y^2}} = \frac{d y^2}{\sqrt{2(1+2 C y^2})}$$

leading to inverse circular/hyperbolic function solutions.

• Thank you for the tip! – bravdwal May 29 '18 at 22:10