sol4 - ACM 95b/100b Solutions for Problem Set 4 Sean Mauch...

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Unformatted text preview: ACM 95b/100b Solutions for Problem Set 4 Sean Mauch sean@caltech.edu February 8, 2002 1 Problem 1 Problem. Use the shifting theorems to assist in solving the following initial value problems. 1. 4 y 00- 4 y + 37 y = 0 , y (0) = 3 , y (0) = 3 2 2. y 00 + y = r ( t ) , y (0) = 0 , y (0) = 0 , r ( t ) = ( t, < t < 1 , t > 1 Solution. 1. We consider the problem 4 y 00- 4 y + 37 y = 0 , y (0) = 3 , y (0) = 3 2 . We take the Laplace transform of the differential equation and solve for y . 4( s 2 y- sy (0)- y (0))- 4( s y- y (0)) + 37 y = 0 (4 s 2- 4 s + 37) y = 12 s- 6 y = 6 2 s- 1 4 s 2- 4 s + 37 y = 3 s- 1 / 2 s 2- s + 37 / 4 y = 3 s- 1 / 2 ( s- 1 / 2) 2 + 9 We take the inverse Laplace transform to obtain the solution. y = 3e t/ 2 cos(3 t ) Alternate Method. We could also solve this problem without the Laplace transform. We substitute y = e t into the differential equation. 4 2- 4 + 37 = 0 = 1 2 3 1 The general solution of the differential equation is y = c 1 e t/ 2 cos(3 t ) + c 2 e t/ 2 sin(3 t ) . We substitute this solution into the initial conditions to determine the constants. y (0) = 3 c 1 = 3 y (0) = 3 2 3 2 + 3 c 2 = 3 2 c 2 = 0 y = 3e t/ 2 cos(3 t ) 2. We consider the problem y 00 + y = r ( t ) , y (0) = 0 , y (0) = 0 , r ( t ) = ( t, < t < 1 , t > 1 We take the Laplace transform of the differential equation. s 2 y- sy (0)- y (0) + y = Z 1 t e- st d t ( s 2 + 1 ) y = 1 s 2- 1 s + 1 s 2 e- s y = 1 s 2 ( s 2 + 1)- 1 s ( s 2 + 1) + 1 s 2 ( s 2 + 1) e- s y = 1 s 2- 1 s 2 + 1 + s s 2 + 1- 1 s + 1 s 2 + 1- 1 s 2 e- s We take the inverse Laplace transform to obtain the solution. y = t- sin t + (cos( t- 1)- 1 + sin( t- 1)- ( t- 1)) H ( t- 1) y = t- sin t + (cos( t- 1) + sin( t- 1)- t ) H ( t- 1) Alternate Method. We could also solve this problem without the Laplace transform. First we solve the problem on the interval [0 ... 1]. The homogeneous solution is y h = c 1 cos t + c 2 sin t. We determine a particular solution y p = t by inspection. The general solution is y = c 1 cos t + c 2 sin t + t. We substitute this solution into the initial conditions to determine the constants. y (0) = 0 c 1 = 0 y (0) = 0 c 2 + 1 = 0 y = t- sin t, for t [0 ... 1] 2 Now we solve problem on the interval [1 ... ). The general solution is y = c 1 cos t + c 2 sin t. We determine initial conditions at t = 1 from the solution on the interval [0 ... 1]. y (1) = 1- sin(1) , y (1) = 1- cos(1) We substitute the solution into the initial conditions to determine the constants. y (1) = 1- sin(1) , y (1) = 1- cos(1) c 1 cos(1) + c 2 sin(1) = 1- sin(1) ,- c 1 sin(1) + c 2 cos(1) = 1- cos(1) c 1 = cos(1)- sin(1) , c 2 = cos(1) + sin(1)- 1 y = (cos(1)- sin(1))cos t + (cos(1) + sin(1)- 1)sin t Now we have the solution of the problem....
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This note was uploaded on 01/08/2011 for the course ACM 95b taught by Professor Nilesa.pierce during the Winter '09 term at Caltech.

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sol4 - ACM 95b/100b Solutions for Problem Set 4 Sean Mauch...

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