- #1

BiGyElLoWhAt

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## Homework Statement

Find the green's function for y'' +2y' +2y = 0 with boundary conditions y(0)=y'(0)=0

and use it to solve y'' + 2y' +2y = e^(-2x)

## Homework Equations

##y = \int_a^b G(x,z)f(z)dz##

## The Attempt at a Solution

I'm going to rush through the first bit. If you need a specific step, I can include it later, but I think a lot of the early steps are pretty self-explanatory.

##\lambda^2 +2\lambda + 2 = 0 \to \lambda = -1 \pm i##

##G' = 1##

##G(x,z) = \left\{ \begin{array}{ll}

Ae^{-(1-i)x} + Be^{-(1+i)x} & z<x \\

Ce^{-(1-i)x} + De^{-(1+i)x} & x<z

\end{array} \right. ##

##y(0)= 0 \to A=-B##

##y'(0)=0 \to -(1-i)A = B(1+i) \to A=B=0##

Continuity:

##Ce^{-(1-i)z} + De^{-(1+i)z} = 0 \to C = -De^{-1+i)z}e^{(1-i)z} = -De^{2iz}##

##-(1-i)Ce^{-(1-i)z} -(1+i)De^{-(1+i)z} = 1 \to (1-i)De^{2iz}e^{-(1-i)z} -(1+i)De^{-(1+i)z} = 1##

##1= D[(1-i)e^{(-1+i)z} - (1+i)e^{-(1+i)z}]=1##

##D = \frac{1}{(1-i)e^{(-1+i)z} - (1+i)e^{-(1+i)z}} ##& ##C=\frac{-e^{-2iz}}{(1-i)e^{(-1+i)z} - (1+i)e^{-(1+i)z}}##

Which leaves us with

##G(x,z) = \left\{ \begin{array}{ll}

0 & z<x \\

\frac{-e^{-(2iz)}e^{-(1-i)x} + e^{-(1+i)x}}{(1-i)e^{(-1+i)z} - (1+i)e^{-(1+i)z}} & x<z

\end{array} \right. ##

Assuming this is correct thus far, I am not sure how to get y.

I'm thinking that the limits of my integral should be from 0 to infinity, but G is zero from 0 to x, so really we need to integrate from x to infinity.

I let wolfram do some simplification for me and we have:

##(\frac{1}{2} + \frac{i}{2} )\int_x^\infty \frac{e^{-(1+i)z} +e^{2iz}}{e^{2iz}-i}##

Which I am not quite sure how to evaluate. It looks like it should converge to a value, since the numerator is "smaller" than the denominator (break it up into 2 terms, and you have [e^-stuff/ e^+things] + [e^stuff/e^stuff + i]) Is there a simple way to evaluate this? Wolfram exceeds computation time.

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