Solving 2nd-Order ODEs: y'' + 2y' + y = f(t); y0=y0'=0

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Homework Statement


y'' + 2y' + y = f(t); y0=y0'=0
f(t) is piecewise -- 1 for 0 < t < a; 0 for t > a

Use
y(t) = ∫G(t,t') f(t') dt' with bounds 0 to infinity


2. The attempt at a solution

I don't really have any logical attempt. My highest math is diffy q 1, Calc 3 and LA 1, I can't find a single resource (besides the 2 pages in my textbook) that addresses this topic in language that I understand.
 
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Lanza52 said:

Homework Statement


y'' + 2y' + y = f(t); y0=y0'=0
f(t) is piecewise -- 1 for 0 < t < a; 0 for t > a

Use
y(t) = ∫G(t,t') f(t') dt' with bounds 0 to infinity


2. The attempt at a solution

I don't really have any logical attempt. My highest math is diffy q 1, Calc 3 and LA 1, I can't find a single resource (besides the 2 pages in my textbook) that addresses this topic in language that I understand.
See the Example section on this page - http://en.wikipedia.org/wiki/Green's_function.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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