Decompose 1 s s 2 4 s 2 y 1 4y s2 4 y

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Unformatted text preview: = se − 2 s −2 s +1 = −e π − 32 s L {cos (t )} s ·2 s +1 3π ￿￿ (5) “Use Laplace transforms to solve for y, where y + 4 y = ￿ 1 0≤t <1 0 1≤t Rewrite in terms of U . y ￿￿ + 4 y = 1 − U (t − 1) . ; y (0) = 0; y ￿ (0) = −1.” 3 Apply the Laplace transform. ￿ ￿ L y ￿￿ + 4 y ￿￿ ￿￿ L y ￿￿ + 4L y ￿2 ￿ s Y − s y (0 ) − y ￿ (0 ) + 4 Y = L { 1 − U ( t − 1 )} = 1 e −s − s s = L { 1 } − L { U ( t − 1 )} Use the given initial values and solve for Y . ￿ Decompose 1 s (s 2 +4) ￿ s 2 Y + 1 + 4Y ￿ = ￿ s2 + 4 Y + 1 = Y = ￿ ￿ s2 + 4 Y = into partial fractions. 1 e −s − s s 1 e −s − s s 1 e −s − −1 s s 1 e −s 1 ￿ ￿− ￿ ￿− 2 s +4 s s2 + 4 s s2 + 4 A Bs +C +2 s s +4 = = ￿ ￿ A s 2 + 4 + (B s + C ) s 2 (A + B ) s + C s + 4A s 1 = ￿ 1 s2 + 4 ￿ 1 Break this into three equations. A +B = 0 1 4 +B = 0 B = −1 4 Therefore ￿ 1 s s2 + 4 ￿= C =0 ←− 4A = 1 A= 1 4 ￿￿ 11 1￿ s ￿ − . 4s 4 s2 + 4 Substitute into the expression for Y , then...
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