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Chem Differential Eq HW Solutions Fall 2011 141

# Chem Differential Eq HW Solutions Fall 2011 141 - Section...

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Section 8.2 Further Properties of The Laplace Transform 141 y = ( U 0 ( t - 1) - U 0 ( t - 4)) + (5 - t ) ( U 0 ( t - 4) - U 0 ( t - 5)) = U 0 ( t - 1) - U 0 ( t - 4) + ( t - 5) U 0 ( t - 5) - ( t - 5) U 0 ( t - 4) = U 0 ( t - 1) - U 0 ( t - 4) + ( t - 5) U 0 ( t - 5) - ( t - 4) U 0 ( t - 4) + U 0 ( t - 4); Y = e - s s - e - 4 s s + e - 5 s s 2 - e - 4 s s 2 + e - 4 s s = e - s s + 1 s 2 ( e - 5 s - e - 4 s ) 17. Let y ( t ) = sin t , then Y ( s ) = 1 s 2 +1 ; so if f ( t ) = U 0 ( t - 1) sin( t - 1) , then F ( s ) = e - s s 2 + 1 . 21. Let y ( t ) = t , then Y ( s ) = Γ(3 / 2) s 3 / 2 , where Γ(3 / 2) = 1 2 Γ(1 / 2) = π 2 . If φ ( t ) = 2 π t Φ( s ) = 1 s 3 / 2 ; and so if f ( t ) = 2 π t - 1 U 0 ( t - 1) then F ( s ) = e - s s 3 / 2 . 25. We will compute t * t in two different ways. First method: We have t L -→ 1 s 2 ; t
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