Solving Differential Equations
the inverse Laplace transform: y(t) = L−1{Y(s)} = L−1{1 s+1}−L−1{s+1 (s+1)2 +1} = (e−t −e−t cost)u(t) which is the solution to the initial value problem. Exercises Use Laplace transforms to solve: 1. dx dt +x = 9e2t x(0) = 3 2. d2x dt2 +x = 2t x(0) = 0 x0(0) = 5 Answers 1. x(t) = 3e2t 2. x(t) = 3sint+2t 38 HELM (2008): Workbook 20 ...
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