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## Homework Statement

[tex]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[/tex]

(a) Use [itex]z=\sqrt y[/itex] to linearize the equation.

(b) Use [itex]t=\frac{1}{w}[/itex] to make singularities regular.

(c) Solve the equation.

(d) Is the last equation obtained a Fuchsian equation?

(e) Discuss what you learn about [itex]z(w)[/itex] with this strategy.

## Homework Equations

Source equation:

[tex]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[/tex]

Substitutions:

[tex]z=\sqrt y[/tex], [tex]t=\frac{1}{w}[/tex]

## The Attempt at a Solution

[/B]

First, I did [itex]z=\sqrt y[/itex], such that [itex]\frac{dz}{dw}=\frac{1}{2\sqrt y}\frac{dy}{dw}[/itex], and [itex]\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[/itex]. When substituted in the equation above, I got

[tex]\frac{d^2y}{dw^2}+\frac{(2w^2-1)}{w^3}\frac{dy}{dw}+\frac{1}{w^4}y=0[/tex].

Then I tried [itex]t=\frac{1}{w}[/itex], such that [itex]\frac{dy}{dw}=-t^2\frac{dy}{dt}[/itex], and [itex]\frac{d^2y}{dw^2}=t^4\frac{d^2y}{dt^2}+2t^3\frac{dy}{dt}[/itex].

From here, I ended at

[tex]\frac{d^2y}{dt^2}+t\frac{dy}{dt}+y=0[/tex].

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

Where did I go wrong?

I didn't type all my calculations to not clutter. Please ask if necessary.