Question
Using the method of Laplace transforms, solve the following initial value problem:
Answer :
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Taking the Laplace transform of both sides of the differential equation (y'' - 3y' + 2y = 6e^{-t}), we use the formulas (\mathcal{L}{y'} = sY(s) - y(0)) and (\mathcal{L}{y''} = s^2 Y(s) - s y(0) - y'(0)). Let (Y(s) = \mathcal{L}{y(t)}). Then: [ \mathcal{L}{y'' - 3y' + 2y} = \mathcal{L}{6e^{-t}} ] [ (s^2 Y(s) - 3s - 3) - 3(s Y(s) - 3) + 2 Y(s) = \frac{6}{s+1} ] [ (s^2 Y(s) - 3s - 3) - 3s Y(s) + 9 + 2 Y(s) = \frac{6}{s+1} ] [ (s^2 - 3s + 2) Y(s) + (-3s - 3 + 9) = \frac{6}{s+1} ] [ (s^2 - 3s + 2) Y(s) + (-3s + 6)
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