2065f09ex1a

2065f09ex1a - Name: Exam 1 Instructions. Answer each of the...

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Name: Exam 1 Instructions. Answer each of the questions on your own paper. Be sure to show your work so that partial credit can be adequately assessed. Credit will not be given for answers (even correct ones) without supporting work. Put your name on each page of your paper. A table of Laplace transforms and a short table of integrals are appended to the exam. 1. [16 Points] Solve the initial value problem: y 0 = p ty , y (1) = 4. I Solution. This equation is separable. After separating variables, it becomes y ¡ 1 = 2 y 0 = p t , which in difierential form is y ¡ 1 = 2 dy = p tdt and integration then gives 2 y 1 = 2 = 2 = 3 t 3 = 2 + C . The initial condition y (1) = 4 means that y = 4 when t = 1. This gives 2 ¢ 2 = 2 = 3 + C so C = 10 = 3. Dividing by 2 gives p y = t 3 = 2 = 3 + 5 = 3. Hence, the solution of the initial value problem is y = ± 1 3 t 3 = 2 + 5 3 2 : J 2. [16 Points] Find the general solution of: y 0 ¡ (2 =t ) y = t 2 cos t I Solution. This is linear with coe–cient function p ( t ) = ¡ 2 =t , so that an integrating factor is ( t ) = e R ( ¡ 2 =t ) dt = e ¡ 2ln t = t ¡ 2 . Multiplication of the difierential equation by the integrating factor gives t ¡ 2 y 0 ¡ 2 t ¡ 3 t 3 y = t ¡ 2 t 2 cos t = cos t; and the left hand side is recognized (by choice of ( t )) as a perfect derivative: d dt ( t ¡ 2 y ) = cos t: Integration then gives t ¡ 2 y = sin t + C; where C is an integration constant. Multiplying through by t 2 gives y = t 2 sin t + Ct 2 : J 3. [16 Points] Solve the initial value problem: y 0 + 5 y = y 2 , y (0) = 1 I Solution. This equation is a Bernoulli equation. Dividing by y 2 gives the equation y ¡ 2 y 0 + 5 y ¡ 1 = 1 : Letting z = y ¡ 1 and noting that z 0 = ¡ y ¡ 2 y 0 we see that the equation is transformed into the equation ¡ z 0 + 5 z = 1, or after multiplying by ¡ 1: z 0 ¡ 5 z = ¡ 1. This is Math 2065 Section 1 September 23, 2009 1
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Name: Exam 1 a linear equation with an integrating factor ( t ) = e ¡ 5 t so multiplying by e ¡ 5 t gives ( e ¡ 5 t z ) 0 = ¡ e ¡ 5 t , and integrating gives
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2065f09ex1a - Name: Exam 1 Instructions. Answer each of the...

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