9780070109100, Chapter 16 6, Problem 1E

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Chapter 16.6, Problem 1E Step-by-step solution Step 1 of 4 1. Given y"(t) + ay'(t) + by = t-1, the variable term t-1 has successive derivatives involving t-2, t-3, and giving an infinite number of forms. If we let y(t) = B1t-1 + B2t-2 + B3t-3 + B4t-4 + There is no end to the y(t) expression. Thus we cannot use it as the particular integral. (a) Try yp in the form of y = B1t + B2. Then y'(t) = B1 and y"(t) = 0. Substitution yields B1t + (2B1 + B2) = t, thus B1 = 1; moreover, 2B1 + B2 = 0, thus B2 = -2. Hence, yp = 2. Step 2 of 4 (b) Try yp in the form of y = B1t2 + B2t + B3. Then we have y'(t) = 2B1t + B2, and y"(t) = 2B1. Substitution now yields B1t2 + (8B1 + B2) t+ (2B1 + 4B2 + B3) = 2t2. Thus B1 = 2, B2 = -16, and B3 = 60. Hence, yp = 2t2 - 16t + 60. Step 3 of 4 (c) Try yp in the form of y = Bet. Then y'(t) = y"(t) = Bet. Substitution yields 4Bet = et; thus B = Hence, Yp = Step 4 of 4 (d) Try yp in the form of y = B1 sint + B2 t. Then we have y' (t) = B1 COS t -B2 sint, and y" (t) = -B1 sint - B2 cost. Substitution yields (2B1 - B2) sint+(B1 + 2B2) cost = sint. Thus B1 = and = Hence, Yp = sin t t.