Can you solve $y=\frac{a}{2}x^2\left(y'-\frac{1}{y'}\right)^2+x\left(y'-\frac{1}{y'}\right)+ax^2+c$? I've recently come across this differential equation, but I am having trouble proceeding toward a solution.
$y=\frac{a}{2}x^2\left(y'-\frac{1}{y'}\right)^2+x\left(y'-\frac{1}{y'}\right)+ax^2+c$
where $y'=\frac{dy}{dx}$ and $a$ and $c$ are constants
There appears to be a nice symmetry and I assume some sort of substitution may be in order, but so far, have had no luck.
Originally, it took the form 
$y=\frac{a}{2}x^2\left(\frac{(y')^2-1}{y'}\right)^2+x\left(\frac{(y')^2-1}{y'}\right)+ax^2+c$
and I thought that perhaps a trig substitution might help.
Any thoughts?

I don't know if an implicit solution fits your needs, anyway with this solution you could at least find $x$ as a function of $y$.
Put $y'=\tan z$, $y=-\log k|\cos z|$, $z= \pm \arccos {e^{-y} \over k}$. So: $$
-\log k|\cos z|{\sin^2 z \over \cos^2 z}={a \over 2}{x^2 \over \cos^2 z}+x {\tan z \over \cos^2 z}+ax^2{\sin^2 z \over \cos^2 z}+c {\sin^2 z \over \cos^2 z}
$$
Simplify the denominators and notice that $\sin z=\sqrt{1-{e^{-2y} \over k^2}}$ and $\tan z={\sqrt{1-{e^{-2y} \over k^2}} \over {e^{-y} \over k}}$.
We can write: $$ ax^2 \left( {3 \over 2}-{e^{-2y} \over k^2} \right)+x{\sqrt{1-{e^{-2y} \over k^2}} \over {e^{-y} \over k}}+(c+y)\left(1-{e^{-2y} \over k^2} \right)=0.$$
You can now solve the quadratic to find values for $x$ in function of every value of $y$.
