S 3 s s 2 2 invert the laplace transform y l 1 3

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Unformatted text preview: Y − s y (0) − y ￿ (0) − 4 sY − y (0) + 4Y = = = Use the given initial conditions and solve for Y . ￿ ￿ s 2 − 4s + 4 Y = ( s − 2 )2 Y = Y = 1 2 ( s − 2 )3 2 ( s − 2 )3 2 ( s − 2 )5 ￿ ￿ L t 2 e 2t ￿￿ L t 2 s ￿→s −2 ￿￿ 2! s 3 s ￿→s −2 2 Invert the Laplace transform. y L −1 = = ￿ (3) “Evaluate L f (t ) , where f (t ) = ￿ 2 t <3 −2 3 ≤ t ￿ 2 ( s − 2 )5 ￿ 4! −1 1 L 3·4 1 4 2t te 12 = ￿ ￿ ( s − 2 )5 ￿ .” Rewrite in terms of U . f ( t ) = 2 − 4U ( t − 3) Apply the Laplace transform. F (s ) = L { 2 − 4 U ( t − 3 )} = 2 · L {1} − 4e −3s L {1} ￿ ￿ 2 − 4e −3s L {1} = = = ￿ ￿ (4) “Evaluate L f (t ) , where f (t ) = 2 L { 1 } − 4 L { U ( t − 3 )} 2 − 4e −3s s ￿ 0 0≤t < 3π 2 sin (t ) Rewrite in terms of U . f (t ) = = ≤t 3π 2 .” ￿ ￿ 3π sin (t ) U t − 2 ￿ ￿￿ ￿ 3π 3π − cos t − U t− 2 2 Apply the Laplace transform. F (s ) = ￿ ￿ ￿￿ ￿￿ 3π 3π L − cos t − U t− 2 2 3π 2s = −e...
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