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Unformatted text preview: 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 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), 2ndorder (1 pt), linear (2 pts). b. [ 4 pts ] dy dt + e 2 t cosh( y ) = 4 Solution: ordinary (1 pt), 1storder (1 pt), nonlinear (2 pts). 2. [ 8 points ] Determine whether each of the two functions is a solution to the given differential equation: a. [ 4 pts ] y 0000 + 4 y 000 + 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). 3. [ 9 points ] Solve the following linear firstorder 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 Z 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 Z μe 2 t dt + c = e 2 t Z 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 . 4. [ 9 points ] Solve the following separable firstorder initial value problem y = 3 x 2 e x 2 y 5 , y (0) = 1 Solution: We first convert this to the appropriate form (1 pt), (2 y...
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This note was uploaded on 06/12/2008 for the course MATH 20D taught by Professor Mohanty during the Spring '06 term at UCSD.
 Spring '06
 Mohanty
 Math

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