Non-linear second-order ODE to Fuchsian equation

[carlosbgois]

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.

 

[lurflurf]

Your solution is not general it should have two constants.
It does work for all k check it again, or better yet check the general solution.
One possible method to add a constant is variation of a constant.
Allow k to be a function of t or w.