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Unformatted text preview: Math 251 March 31, 2005 Second Exam ANS KEY 1. In Parts a and b determine the form of a particular solution y p = y p ( t ) having the least number of unknown constants. DO NOT DETERMINE the unknown constants appearing in your answers in Parts a and b . a. 2pt y 00- 14 y + 49 y = 2 t 2 e 7 t ANS. t 2 ( At 2 + Bt + C ) e 7 t 7 is a double root of the characteristic polynomial. Take off 1pt more missing factor of t 2 . Take off 2pts for any other error. b. 2pt y 00- 50 y + 49 y = 3 te t ANS. t ( At + B ) e t 1 is a simple root of the characteristic polynomial. Take off 1pt more missing factor of t . Take off 2pts for any other error. c. 7pt Without using Laplace transforms, find a particular solution to the following ODE: y 00 + 3 y = e t sin 2 t (In this part you need to determine the unknown constant(s) in the solution. ANS. Solve the complexified equation and then take imaginary part y 00 + 3 y = e (1+2 i ) t Plug in y C = Ae (1+2 i ) t since 1 + 2 i is not a root of the characteristic polynomial. (2pts) y C = A (1 + 2 i ) e (1+2 i ) t and y C = A (- 3 + 4 i ) e (1+2 i ) t .(1pt) Therefore, L [ y C ] = A (3 + (- 3 + 4 i )) e (1+2 i ) t . (2pt) We conclude that A = 1 / (4 i ) =- i/ 4 and hence the imaginary part of y C is y p = (- 1 / 4) cos(2 t ) (2pt) Alternatively, plug in y p = e t ( A cos(2 t ) + B sin(2 t )). (2pt) Taking the derivative: y p = e t (( A + 2 B ) cos(2 t ) + ( B- 2 A ) sin(2 t )). (1pt) and again: y 00 p = e t ((- 3 A + 4 B ) cos(2 t ) + (- 3 B- 4 A ) sin(2 t )). (2pt) Therefore, L [ y p ] = e t (4 B cos(2 t )- 4 A sin(2 t )). (1pt) We conclude that 4 B = 0 and- 4 A = 1, ie, y p = (- 1 / 4) cos(2 t ). (1pt) 2. Assume that acceleration due to gravity g is equal to 10 meter/sec 2 . An object with mass 2 kg stretches a spring 2 . 5 meters to the equilibrium position. Assume that there is no damping device attached and also assume that at time t = 0 the object is released 1 meter below its equilibrium position with a upward velocity of 4 meter/sec....
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