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Unformatted text preview: Laplace Transform PS April 9, 2011 Problem 1 Solve the following initial value problem y 00 + 2 y + y = 4 e t , y (0) = 2 ,y (0) = 1 . (1) Solution : Take the Laplace transform of both sides of the differential equation: s 2 Y sy (0) y (0)  {z } L { y 00 } +2 ( sY y (0))  {z } L { y } + Y {z} L { y } = 4 s + 1  {z } L { 4 e t } . (2) Using the intial conditions, we have s 2 Y 2 s + 1 + 2( sY 2) + Y = 4 s + 1 , Solving for Y , we have Y ( s 2 + 2 s + 1) = 4 s + 1 + 2 s + 3 . Y = 4 ( s + 1)( s 2 + 2 s + 1) + 2 s + 3 s 2 + 2 s + 1 = 4 ( s + 1) 3 + 2 s + 3 ( s + 1) 2 . Take the inverse Laplace of both sides of the equation: L 1 { Y } = y ( t ) = L 1 4 ( s + 1) 3 + L 1 2 s + 3 ( s + 1) 2 . For the first inverse Laplace, recall the sshifting theorem: L t 2 = 2 s 3 L e at t 2 = 2 ( s a ) 3 . Therefore, L 1 4 ( s + 1) 3 = 2 L 1 2 ( s + 1) 3 = 2 t 2 e t . For the second inverse Laplace transform, write L 1 2 s + 3 ( s + 1) 2 = L 1 2( s + 1) + 1 ( s + 1) 2 = 2 L...
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This note was uploaded on 05/02/2011 for the course GE 207K taught by Professor None during the Spring '10 term at University of Texas at Austin.
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