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# sol5 - SOLUTION Consider y = A cos 3 t B sin 3 t Then y =-3...

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MAP 2302 Fall-11 NAME: Section 0100 Quiz 5 1. Solve the initial value problem y 00 + 2 y 0 + 2 y = 0 y (0) = 2 y 0 (0) = 1 . SOLUTION: The characteristic equation r 2 + 2 r + 2 = 0 has complex roots - 1 ± i . Hence the general solution is y = C 1 e - t cos t + C 2 e - t sin t . Its derivative is y 0 = C 1 e - t ( - sin t - cos t ) + C 2 e - t (cos t - sin t ). The condidtion y (0) = 2 implies that C 1 = 2. The condition y 0 (0) = 1 implies that 1 = - 2 + C 2 . Thus, C 2 = 3. The answer is y = 2 e - t cos t + 3 e - t sin t. 2. Find a particular solution to the differential equation y 00 - y 0 + 9 y = 3 sin 3
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Unformatted text preview: SOLUTION: Consider y = A cos 3 t + B sin 3 t . Then y =-3 A sin 3 t + 3 B cos 3 t . and y 00 =-9 A cos 3 t-9 B sin 3 t . Plug y,y , and y 00 to the equation to obtain-9 A cos 3 t-9 B sin 3 t +3 A sin 3 t-3 B cos 3 t +9 A cos 3 t +9 B sin 3 t = 3 sin 3 t. After cancellation we have 3 A sin 3 t-3 B cos 3 t = 3 sin 3 t. Therefore, A = 1 and B = 0. The answer is y = cos 3 t. 1...
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