2065s10exfa

# 2065s10exfa - Name: Solutions Final Exam Instructions....

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Name: Solutions Final Exam Instructions. Answer each of the questions on your own paper. Put your name on each page of your 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. A table of Laplace transforms and the statement of the main partial fraction decomposition theorem have been appended to the exam. In Exercises 1 { 8, solve the given diﬁerential equation. If initial values are given, solve the initial value problem. Otherwise, give the general solution. Some problems may be solvable by more than one technique. You are free to choose whatever technique that you deem to be most appropriate. 1. [ 12 Points ] 2 tyy 0 = 1 + y 2 , y (2) = 3. I Solution. This equation is separable. Separate the variables to get 2 y 1 + y 2 y 0 = 1 t . Write in diﬁerential form and integrate to get: Z 2 y 1 + y 2 dy = Z dt t : Integrating gives ln(1 + y 2 ) = ln t + C , and taking the exponential of both sides gives 1 + y 2 = Bt where B = e C is a constant. Thus, y 2 = Bt ¡ 1 and using the initial condition y = 3 when t = 2 gives 9 = 2 B ¡ 1 so that B = 5. Hence y ( t ) = p 5 t ¡ 1 : J 2. [ 12 Points ] y 0 + 2 ty = t , y (0) = ¡ 3. I Solution. This equation is linear with p ( t ) = 2 t so that an integrating factor is ( t ) = e R p ( t ) dt = e t 2 . Multiplying the equation by e t 2 gives e t 2 y 0 + 2 te t 2 y = te t 2 : Thus, d dt e t 2 y · = te t 2 ; and integrating gives e t 2 y = Z te t 2 dt = 1 2 e t 2 + C: Hence, solving for y gives y = 1 2 + Ce ¡ t 2 . Now apply the initial conditions y (0) = ¡ 3 to get ¡ 3 = y (0) = 1 2 + C , which gives C = ¡ 7 2 . Therefore, y ( t ) = 1 2 ¡ 7 2 e ¡ t 2 : J Math 2065 Section 1 May 12, 2010 1

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Name: Solutions Final Exam 3. [ 10 Points ] y 00 + 10 y 0 + 29 y = 0. I Solution. The characteristic polynomial is q ( s ) = s 2 + 10 s + 29 = ( s + 5) 2 + 4, which has roots ¡ 5 § 2 i . Thus, the solutions are given by y ( t ) = c 1 e ¡ 5 t cos 2 t + c 2 e ¡ 5 t sin2 t: J 4. [ 10 Points ] 4 y 00 +12 y 0 +9 y = 0. The characteristic polynomial is q ( s ) = 4 s 2 +12 s +9 = (2 s + 3) 2 , which has a single root ¡ 3 = 2 of multiplicity 2. Thus, the solutions are given by y ( t ) = c 1 e ¡ 3 t= 2 + c 2 te ¡ 3 s= 2 : 5. [ 12 Points ] y 00 ¡ 4 y 0 + 3 y = 0 = 0, y (0) = 1, y 0 (0) = ¡ 1. The characteristic polynomial is q ( s ) = s 2 ¡ 4 s + 3 = ( s ¡ 3)( s ¡ 1), which has roots 1 and 3. Thus, the general solution of the diﬁerential equation is y ( t ) = c 1 e t + c 2 e 3 t . The initial conditions give the equations for c 1 , c 2 : 1 = y (0) = c 1 + c 2 ¡ 1 = y 0 (0) = c 1 + 3 c 2 : Solve these equations to get c 1 = 2, c 2 = ¡ 1. Thus, the solution of the initial value problem is y ( t ) = 2 e t ¡ e 3 t : 6. [ 12 Points ] y 00 + 2 y 0 ¡ 8 y = 4 e 2 t .
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2065s10exfa - Name: Solutions Final Exam Instructions....

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