9780070109100, Chapter 15 1, Problem 1E

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Chapter 15.1, Problem 1E Step-by-step solution Step 1 of 4 1. (a) With a = 4 and b = 12, we have yc = Ae-4t, yp = 12 = 3. The general solution is y(t) = Ae-4t + 3. Setting t = 0, we get y(0) = A + 3, thus A = Y (0) -3 = 2 -3 = -1. The definite solution is y(t) = -e-4t + 3. Step 2 of 4 (b) yc = Ae-(-2)t, yp = = 0. The general solution is y(t) = Ae2t. Setting t=0, we have y(0) = A; i.e., A = 9. Thus the definite solution is y(t) = 9e2t. Step 3 of 4 (c) = Ae = = Thus y(t) = Ae + Setting we get y(0) = A- i.e., A = y(0) = 0 = The definite solution is y(t) = (1 Step 4 of 4 (d) Upon dividing through by 2, we get the equation + 2y = 3. Hence = Ae Yp = and y(t) = + Setting t = 0, we get y(0) = A + implying t = = 0. The definite solution is y(t) =