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I have a basic ODE:

[tex]y''(x)+\frac{1}{4}y'(x)=f(x)[/tex]

on 0<x<L

With Boundary conditions:

[tex]y(0)=y(L)=0[/tex]

For which I would like to construct a Green Function.

Rather than just plain ask for help, I'll show you what I've been thinking and maybe someone wiser can help/correct me:

We must solve:

[tex]G''(x,x')+\frac{1}{4}G'(x,x'')=\delta(x-x')[/tex]

Since G vanishes at the boundaries we can expand as a fourier sine series:

[tex]G(x,x')=\sum_{n=1}^\infty \gamma_n sin\frac{n\pi x}{L}[/tex]

and:

[tex]\delta(x-x')=\sum_{n=1}^\infty A_n(x')sin\frac{n\pi x}{L}[/tex]

Integrating the delta function I get:

[tex]A_n(x')=\frac{2}{L}sin\frac{n\pi x}{L}[/tex]

I take the first and second derivative of the equation for G

and plug everything into the differential equation for G:

[tex]\sum_{n=1}^\infty \gamma_n (\frac{-n^2 \pi^2}{L^2})sin\frac{n\pi x}{L}+\frac{1}{4} \sum_{n=1}^\infty \gamma_n (\frac{n \pi}{L})cos\frac{n\pi x}{L}=\sum_{n=1}^\infty \frac{2}{L}sin\frac{n\pi x'}{L}sin\frac{n\pi x}{L}[/tex]

Now I am not sure what to do from here to get the required Green function...can someone guide me please?

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# Need to find Green Function to solve ODE

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