m3354_f05_q3_sol

m3354_f05_q3_sol - We have n = 3 so (1-n ) =-2. Let y = u-2...

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Math 3354 Sec. 2, Quiz # 3 Name KEY 1. ( Linear Equation ) Solve the initial value problem u 0 - u = e 2 t , u (0) = 2. The integrating factor is I = e R ( - 1) dt = e - t . Multiplying by I we get ± ue - t ² 0 = e 2 t e - t = e t Integrate both sides to get ue - t = e t + C Now use u (0) = 2 to get 2 = u (0) = 1 + C or C = 1. Substitute C = 1 and multiply by e t to get u = e 2 t + e t 2. ( Bernoulli Equation ) Find the general solution of u 0 = u - u 3 (you must solve as Bernoulli not separable. Implicit ).
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Unformatted text preview: We have n = 3 so (1-n ) =-2. Let y = u-2 and the equation for y becomes y =-2 y + 2 . This is rst order linear and we write it as y + 2 y = 2. This I = e R (2) dt = e 2 t . Multiply by I to get ye 2 t = 2 e 2 t Integrate both sides to get ye 2 t = e 2 t + C Finally u-2 = 1 + Ce-2 t ....
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This note was uploaded on 10/19/2011 for the course MATH 2254 taught by Professor Gilliam during the Fall '05 term at Texas Tech.

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