'Use Laplace transformation to solve the initial value problem <" +r' = e-t, =(0) = 0, *(0) ='
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Step 1: Take the Laplace transform of both sides of the differential equation: L{y''(t)} + L{y'(t)} = L{e^{-t}} Using the properties of Laplace transform, we get: s^2 Y(s) - s y(0) - y'(0) + s Y(s) - y(0) = 1/(s+1) Substituting the initial conditions y(0) = 0 Show more…
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