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Winter 2006 - Reynold's Class - Final Exam (Version B)

Winter 2006 - Reynold's Class - Final Exam (Version B) -...

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Final Exam, Version B Math 20D, Lecture C, Winter 2006 22 March 2006 Name: ID #: Section Time: Read all of the following information before starting the exam: Show all work, clearly and in order, if you want to get full credit. I reserve the right to take off points if I cannot see how you arrived at your answer (even if your final answer is correct). Justify your answers algebraically whenever possible to en- sure full credit. When you do use your calculator, sketch all relevant graphs and explain all relevant mathematics. Circle or otherwise indicate your final answers. Please keep your written answers brief; be clear and to the point. I will take points off for rambling and for incorrect or irrelevant statements. This test has 11 problems and is worth 100 points. It is your responsibility to make sure that you have all of the pages! Good luck! # Score 1a 1b 2a 2b 3 4 5 6 7 8 9 10 11 Total
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1. [ 8 points ] Classify each of the following equations as (i) either ordinary or partial differential equations, (ii) either first, second or third order, (iii) either linear or nonlinear: a. [ 4 pts ] d 2 y dt 2 + y cos( t ) = ln( t 2 ) Solution: ordinary (1 pt), 2nd-order (1 pt), linear (2 pts). b. [ 4 pts ] dy dt + e 2 t cosh( y ) = 4 Solution: ordinary (1 pt), 1st-order (1 pt), nonlinear (2 pts).
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2. [ 8 points ] Determine whether each of the two functions is a solution to the given differential equation: a. [ 4 pts ] y + 4 y + 3 y = t ; y 1 ( t ) = t 3 , y 2 ( t ) = e - t + t 3 Solution: y 1 is a solution to the differential equation (2 pts). y 2 is a solution to the differential equation (2 pts). b. [ 4 pts ] 4 2 y ∂x 2 = ∂y ∂t ; y 1 ( x, t ) = e - 4 t sin( x ) , y 2 ( x, t ) = e - t sin x 2 Solution: y 1 is a solution to the differential equation (2 pts). y 2 is a solution to the differential equation (2 pts).
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3. [ 9 points ] Solve the following linear first-order initial value problem y + 2 y = e - 2 t , y (1) = 0 Solution: We first compute the integrating factor (2 pts setup, 1 pt correct μ ), μ ( t ) = exp 2 dt = e 2 t . We then compute the general solution y as (2 pts setup, 1 pt correct general solution) y ( t ) = 1 μ μe - 2 t dt + c = e - 2 t e 2 t e - 2 t dt + c = te - 2 t + ce - 2 t . We insert the initial condition to solve for the constant c , 0 = y (1) = e - 2 + ce - 2 obtaining c = - 1 (2 pts for correct c), so the particular solution to the differential equation is (1 pt for final solution) y ( t ) = te - 2 t - e - 2 t .
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4.
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