Y 2 y y 2 t 3 u t 3 y 0 2 y 0 1 4 take

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Unformatted text preview: invert the Laplace transform. ￿￿ ￿￿ ￿s￿ 11 1 ￿ s ￿ 1 −s 1 1 1 Y= − −e + e −s 2 −2 4s 4 s2 + 4 4 s 4 s +4 s +4 ￿￿ ￿￿ ￿s￿ 11 1 1 ￿ s ￿ 1 −s 1 1 = −2 − −e + e −s 2 4s s + 4 4 s2 + 4 4 s 4 s +4 ￿￿￿ ￿￿ ￿ 1 1 ￿ s ￿ 1 −s 1 1 −s ￿ s ￿ 11 −1 y=L −2 − −e +e 4s s + 4 4 s2 + 4 4 s 4 s2 + 4 ￿￿ ￿ ￿ ￿ ￿ ￿s￿1 ￿ 1 −1 1 1 −1 2 1 −1 1 1 s￿ = L −L −L − L −1 e −s · + L −1 e −s · 2 4 s 2 s2 + 4 4 s2 + 4 4 s 4 s +4 11 1 1 1 = − sin (2t ) − cos (2t ) − U (t − 1) + cos (2 (t − 1)) U (t − 1) 42 4 4 4 1 − 2 sin (2t ) − cos (2t ) (cos (2 (t − 1)) − 1) = + U ( t − 1) 4 4 ￿￿ ￿ (6) “Use Laplace transforms to solve for y, where y + 2 y + y = ￿ 0 0≤t <3 2 ( t − 3) 3 ≤ t Rewrite in terms of U . y ￿￿ + 2 y ￿ + y = 2 (t − 3) U (t − 3) ; y (0) = 2; y ￿ (0) = 1.” 4 Take the Laplace transform. ￿ ￿ L y ￿￿ + 2 y ￿ + y ￿￿ ￿￿ ￿￿ L y ￿￿ + 2L y ￿ + L y ￿2 ￿ ￿ ￿ s Y − s...
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This note was uploaded on 01/26/2014 for the course MATH 527 taught by Professor Boucher during the Spring '07 term at New Hampshire.

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