# How to solve this problem? Use the Laplace transform to solve the given initial value problem. y'' - 2y' +2y = 0; y(0)=1, y'(0)=0

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### 1 Answer

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

`=> L(y''-2y'+2y) = L (0) `

`==> L{y''}(s) - 2L{y'}(s) + 2{y}(s) = 0`

Now let `y(s)= L{y}(s) `

`==gt L{y'}(s)= sy(s) - y(0)= sy(s) - 1`

` ==gt L{y''}(s)= s^2 Y(s) - sy(0) - y'(0) = s^2 y(s) -s`

Now we will substitute into (2)

`==gt s^2 y(s)-s - 2 - 2s( y(s) -1) + 2 y(s) =0 `

`==gt s^2 y(s) - 2s y(s) + 2 y(s) = 4+s `

`==gt y(s) ( s^2 -2s + 2) = s+4 `

`==gt y(s)= (s+4)/(s^2 -2s+2) `

`==gt y(s)= (s+4)/((s-1)^2 +1) `

`==gt y(s)= ((s-1)+5)/((s-1)^2+1) `

`==gt y(s)= (s-1)/((s-1)^2 + 1) + 5(1/((s-1)^2+1) )`

`==gt y(t)= e^t cost + 5(e^t sint) `

`==gt y(t)= e^t cost + 5e^t sint`