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

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SUMMARY

The forum discussion focuses on solving the second-order linear differential equation y'' - 2y' + 2y = 0 using the Laplace Transform method. The initial conditions provided are y(0) = 0 and y'(0) = 1. The Laplace Transform of the function is expressed as L(y) = 1/(s^2 - 2s + 2). The discussion emphasizes the need for step-by-step guidance to understand the process of applying the Laplace Transform and finding the inverse transform.

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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:

[tex]L(y) \ = \frac{1}{s^2-2s+2}[/tex]
 
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