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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Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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