Laplace Transform: Solving y''-2y'+2y=0

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Homework Statement



y''-2y'+2y = 0

Homework Equations



y(0)=0
y'(0)=1

The Attempt at a Solution



e^at sin(bt)

I did this problem earlier with some help, but if someone could post the steps along the way so I could do a few more similar to it, I would appreciate it.
 
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Where's your problem? Do you have problems finding the inverse Laplace transform once you are done with substituting the initial values?

Can you get this expression:

L(y) \ = \frac{1}{s^2-2s+2}
 
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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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