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I am stuck on a possibly trivial question. I have the Greens function for the equation

p_0(x)y''+p_1(x)y'+p_2(x)y=0

with the boundary value $y(\alpha)=y(\beta)=0$ and I need to solve the equation

p_0(x)y''+p_1(x)y'+p_2(x)y=r(x)

with the boundary conditions $y(\alpha)=0, y(\beta) = A \not =0$

I know that if the boundary conditions of the second was the same as the first, then I could do

$y(x)=\int_{\alpha}^{\beta} G(x,t) r(t) dt$

But what does one do with the different boundary conditions?

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    The FAQ explicitly says you can answer your own question. I think you should do that so that the question doesn't appear unanswered.2011-04-21

1 Answers 1

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To those interested, I got the solution. It uses the following trivial theorem:

Given that $y_1$ is a solution to the problem

p_0(x)y''+p_1(x)y'+p_2(x)y=0

with BC $y(\alpha)=A, y(\beta) = B$

and that $y_2$ is a solution to the problem

p_0(x)y''+p_1(x)y'+p_2(x)y=r(x)

with BC $y(\alpha)=0, y(\beta) = 0$

Then,

$y=y_1+y_2$ is a solution to

p_0(x)y''+p_1(x)y'+p_2(x)y=r(x)

with boundary conditions $y(\alpha)=A, y(\beta) = B$

So the original problem is easily solved.