Solving a differential equation $F-y'F_{y'}=C$, with $F(y,y')= \frac{1+2y'^2}{3y^3\sqrt{1+y'^2}}$ If $$F= F(y,y')= \frac{1+2y'^2}{3y^3\sqrt{1+y'^2}},$$
where $y=y(x)$ and $y'= y'(x)=\frac{dy}{dx}$, then how to solve the differential equation: 
$$F-y'F_{y'}=C, $$ 
that is: 
$$F(y,y')-\frac{dy}{dx}\frac{\partial F}{\partial y'}=C,$$
where $C$ is some constant? What are the keys steps? I got a very hairy differential equation when I tried solving this directly by doing the calculations. Is there some substitution trick etc. that I should apply here? 
 A: We have that
$$\begin{align} \\
F(y,y') &= \frac{1 + 2y'^{2}}{3y^{3} \sqrt {1+y'^{2}}} \\
&= \frac{1}{3y^{3}}\bigg[(1 + 2y'^{2})(1 + y'^{2})^{\frac{-1}{2}}\bigg]
\end{align} $$
As there is no explicit $x$ dependence, we can find a first integral of the form
$$\begin{align} \\
y'\frac{\partial F}{\partial y'} - F &= y'\cdot\frac{1}{3y^{3}}\bigg[4y'(1+y'^{2})^{\frac{-1}{2}} + (1 + 2y'^{2})\cdot\frac{-1}{2}\cdot2y'(1+y'^{2})^{\frac{-3}{2}}\bigg] - \frac{1}{3y^{3}}\bigg[(1 + 2y'^{2})(1 + y'^{2})^{\frac{-1}{2}}\bigg] \\
&= \frac{1}{3y^{3}}\bigg[(3y'^{2} + 2y'^{4})(1 + y'^{2})^{\frac{-3}{2}}\bigg] - \frac{1}{3y^{3}}\bigg[(1 + 2y'^{2})(1 + y'^{2})^{\frac{-1}{2}}\bigg] \\
&= \frac{1}{3y^{3}}\bigg[\frac{3y'^{2} + 2y'^{4} -(1 + 3y'^{2} + 2y'^{4})}{(1+y'^{2})^{\frac{3}{2}}}\bigg] \\
&= \frac{1}{3y^{3}}\bigg[\frac{-1}{(1+y'^{2})^{\frac{3}{2}}}\bigg] \\
&= C
\end{align} $$
(I omitted a few steps of algebra, if you would like me to put them in just write a comment below).
Hence we get
$$\begin{align} \\
(1+y'^{2})^{\frac{3}{2}} &= \frac{-1}{3y^{3}C} \\
\implies (1+y'^{2})^{3} &= \bigg(\frac{1}{9y^{6}C^{2}}\bigg) \\
\end{align} $$
Where in the last step we squared both sides.
Therefore,
$$\begin{align} \\
1 + y'^{2} &= \bigg(\frac{1}{9y^{6}C^{2}}\bigg)^{\frac{1}{3}} \\
\implies y'^{2} &= \bigg(\frac{1}{9y^{6}C^{2}}\bigg)^{\frac{1}{3}} - 1 \\
&= \alpha^{2}\cdot\frac{1}{y^{2}} - 1 \\
\implies y' &= \pm \bigg(\alpha^{2}\cdot\frac{1}{y^{2}} - 1\bigg)^{\frac{1}{2}} \\
&= \pm \bigg(\frac{\alpha^{2} - y^2}{y^{2}}\bigg)^{\frac{1}{2}} \\
&= \pm \frac{\sqrt{\alpha^{2} - y^2}}{y} \\
\end{align} $$
With 
$$\alpha^{2} = \bigg(\frac{1}{9C^{2}}\bigg)^{\frac{1}{3}}$$
Separating and integrating we get
$$ \int \frac{y}{\sqrt{\alpha^{2} - y^2}}  dy = \pm \int dx $$
where the LHS can be solved using a trig substitution.
For the future, a general method for solving maximising/minimising problems goes like this (this solution recipe is a grave oversimplification, but it may help you get used to these types of problems):
1) If there is no independent variable dependence ($x$ or $t$ or whatever other variable it might be), find a first integral using the formula
$$ y'\frac{\partial F}{\partial y'} - F = C $$
2) Substitute your functions $\frac{\partial F}{\partial y'}$ and $F$ into your equation.
3) Solve for $y'$
4) Separate and integrate
If you have any questions or you see a mistake in my algebra, just let me know in the comments.
