Answer to Question #239732 in Differential Equations for sam

Question

Use the linearity property of Laplace transform to find L[5e^-2t + t + 2e^2t]

Answer 1

$\mathcal{L}(5e^{-2t} + t + 2e^{2t})\\ =\mathcal{L}(5e^{-2t}) + \mathcal{L}(t) + \mathcal{L}(2e^{2t})\\ =5\mathcal{L}(e^{-2t})+\mathcal{L}(t)+2\mathcal{L}(e^{2t})\\ =\frac{5}{s+2}+\frac{1}{s^2}+\frac{2}{s-2}\\ \frac{5(s^3-2s^2)+s^2-4+2(s^3+2s^2)}{(s^2-4)s^2}\\ =\frac{7s^3-5s^2-4}{(s^2-4)s^2}$