Use the method of the Laplace transform to solve y\'\'-2y\'+2y=0 ; y(0)=0, y\'(0

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similar example Use laplace transform to solve: y''+2y'+y= e^(-t), y(0)= -1, y'(0)= 1? The Laplace transform of an exponential e^(at) is L{e^(at)} = 1/(s-a) which can be obtained from the definition or any table of transforms. So the transform of the right hand side is L{e^(-t)} = 1/(s+1). Taking the transform of the left, using the results for the Laplace transform of derivatives gives L{y''+2y'+y} = s²Y(s) - sy(0) - y'(0) + 2[sY(s) - y(0)] + Y(s) =(s² + 2s +1)Y(s) + s - 1 + 2 =(s+1)²Y(s) + (s+1) where Y(s) = L{y(t)}. So taking the transform of both sides of the equation gives the equation for Y: (s+1)²Y(s) + (s+1) = 1/(s+1) ==> Y(s) = 1/(s+1)^3 - 1/(s+1). To get y(t), we have to take the inverse Laplace transform. We can use the shifting theorem that say that if L{f(t)} = F(s), then L{e^(at)f(t)} = F(s-a). Note that L{t²/2} = 1/s^3. The function 1/(s+1)^3 is just 1/s^3 with s replaced by s+1. So L^(-1){1/(s+1)^3) = t²/2 e^(-t). We know from before that L^(-1){1/(s+1)} = e^(-t). So taking the inverse transfrom of Y(s) gives the solution y(t) = t²/2 e^(-t) - e^(-t) which can be written as y(t) = e^(-t) (t²/2 - 1).

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