Non-linear second-order ODE to Fuchsian equation

Homework Statement

$$z\frac{d^2z}{dw^2}+\left(\frac{dz}{dw}\right)^2+\frac{\left(2w^2-1\right)}{w^3}z\frac{dz}{dw}+\frac{z^2}{2w^4}=0$$

(a) Use $z=\sqrt y$ to linearize the equation.
(b) Use $t=\frac{1}{w}$ to make singularities regular.
(c) Solve the equation.
(d) Is the last equation obtained a Fuchsian equation?
(e) Discuss what you learn about $z(w)$ with this strategy.

Homework Equations

Source equation:
$$z\frac{d^2z}{dw^2}+\left(\frac{dz}{dw}\right)^2+\frac{\left(2w^2-1\right)}{w^3}z\frac{dz}{dw}+\frac{z^2}{2w^4}=0$$

Substitutions:
$$z=\sqrt y$$, $$t=\frac{1}{w}$$

The Attempt at a Solution

[/B]
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

$$\frac{d^2y}{dt^2}+t\frac{dy}{dt}+y=0$$.

This doesn't look right. For it could be rewritten as $\frac{d}{dt}\left(\frac{dy}{dt}+ty\right)=0 \Rightarrow y(x)=ke^{(-x^2/2)}$, and going back the substitutions I'd have $z(w)=\sqrt{ke^{-(1/w)^2/2}}$, which when substituted in the original equation yields only the trivial solution $k=0$.

Where did I go wrong?
I didn't type all my calculations to not clutter. Please ask if necessary.