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2 - laplace_ps

# 2 - laplace_ps - Laplace Transform PS Problem 1 Solve the...

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Laplace Transform – PS April 9, 2011 Problem 1 Solve the following initial value problem y 00 + 2 y 0 + y = 4 e - t , y (0) = 2 , y 0 (0) = - 1 . (1) Solution : Take the Laplace transform of both sides of the differential equation: s 2 Y - sy (0) - y 0 (0) | {z } L { y 00 } +2 ( sY - y (0)) | {z } L { y 0 } + 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 s-shifting 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 - 1 1 s + 1 + L - 1 1 ( s + 1) 2 = 2 e - t + te - t . (3) 1 c HF, 2011

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Laplace Transform – PS April 9, 2011 Combining the results of the two inverse Laplaces, we have y ( t ) = 2 t 2 e - t + 2 e - t + te - t .
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