MATH 0290 Midterm: Math 0290 Exam 2 (0290) 2016 Fall Solution -189
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S o l u t i o n s: by using laplace transform solve the initial-value problem y + 4y = 4 cos 2t, y(0) = 0, y (0) = 0. After applying laplace transform to the equation we get (s2 + 4)y (s) = 4s (s2 + 4)2 = (cid:18) 2 s2 + 4(cid:19) . = (l[sin 2t]) = l[t sin 2t] Therefore y(t) = t sin 2t: consider the initial-value problem y + y = g(t), y(0) = 0, where g(t) = . 0, t, for 0 t < 3 for t 3 (a) describe the function g(t) in terms of the heaviside function. Solution: g(t) = th(t 3) = (t 3)h(t 3) + 3h(t 3) (b) by using laplace transform solve the initial-value problem. L[y + y] = (s + 1)y (s) L[g(t)] = l [(t 3)h(t 3) + 3h(t 3)] = e 3s .
