I'm trying to take
$z\frac{d^2z}{dw^2}+\left(\frac{dz}{dw}\right)^2+\frac{\left(2w^2-1\right)}{w^3}z\frac{dz}{dw}+\frac{z^2}{2w^4}=0$
to a tractable DE through substitution of variables.
I'll type a summary of what I did so far, and would be very thankful if anyone could point out my mistakes.
First, I did $z=\sqrt y$, such that $\frac{dz}{dw}=\frac{1}{2\sqrt y}\frac{dy}{dw}$, and $\frac{d^2z}{dw^2}=\frac{1}{2\sqrt y}\frac{d^2y}{dw^2}-\frac{1}{4y^{3/2}}\left(\frac{dy}{dw}\right)^2$. When substituted in the equation above, I got
$\frac{d^2y}{dw^2}+\frac{(2w^2-1)}{w^3}\frac{dy}{dw}+\frac{1}{w^4}y=0$.
Then I tried $t=\frac{1}{w}$, such that $\frac{dy}{dw}=-t^2\frac{dy}{dt}$, and $\frac{d^2y}{dw^2}=t^4\frac{d^2y}{dt^2}+2t^3\frac{dy}{dt}$.
From here, I ended at
$\frac{d^2y}{dt^2}+t\frac{dy}{dt}+y=0$.
Now, I was hoping that last one would have regular singularities, possibly a Fuchsian equation, as that is what I'm studying. It didn't happen to be so, though I arrived at a solvable DE. Did I do something wrong, miss something, or may I proceed taking the last one to
$\frac{d}{dt}\left(\frac{dy}{dt}+ty\right)=0 \Rightarrow y(x)=ke^{(-x^2/2)}$
and take this as a solution?