Solve the initial value problem below using the method of Laplace transforms. Table of Laplace Transforms y" - 4y = 24t – 4 e – 2t, y(0) = 0, y'(0) = 7 Click here to view the table of Laplace transforms. Click here to view the table of properties of Laplace transforms. f(t) F(s) = L{f}(s) 1 ,s>0 y(t) = 1 (Type an exact answer in terms of e.) 1 e at s>0 s-a O Properties of Laplace Transforms n! tn, n= 1,2,.. sin bt s? + b2 s>0 L{f+ g} = L{f} + L{g} L{cf} = cL{f} for any constant c 2{e atf(t)} (s) = L{}(s - a) L(f'> (s) = sL{f}(s) – f(0) L(f'} (s) = s?L{}{(s) – sf(0) – f'(0) L{f{n)} (s) = s"£{f}(s) – s^ - 1 f(0) – sn - 2f'(0) – ... - f(n - 1) (0) cos bt s>0 n! e aten, n= 1,2,.. (s- a)n+1, s>a b e at sin bt ,s>a (s - a)2 + b2 dn L(1"f(1)} (s) = ( – 1)n - ds" L-1F, +Fa) = L- -(L{f}(s)) S-a at oor ht

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