Chem Differential Eq HW Solutions Fall 2011 141

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

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Unformatted text preview: Section 8.2 y Further Properties of The Laplace Transform 141 ( 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); = e−s e−4s e−4s e−5s e−4s − + 2− 2+ s s s s s = Y = e−s 1 + 2 e−5s − e−4s s s 1 s2 +1 ; 17. Let y(t) = sin t, then Y (s) = so if f (t) = U 0(t − 1) sin(t − 1), e−s . s2 + 1 √ 21. Let y(t) = t, then Y (s) = then F (s) = Γ(3/2) , s3/2 φ( t ) = and so if f (t) = where Γ(3/2) = 1 Γ(1/2) = 2 π 2. If 2√ 1 t ⇒ Φ(s) = 3/2 ; π s 2√ e−s t − 1 U 0 (t − 1) then F (s) = 3/2 . π s 25. We will compute t ∗ t in two different ways. First method: We have L 1 ; s2 L 1 ; s2 L 11 1 · =4 s2 s2 s t −→ t −→ t∗t −→ 1 s4 −→ t3 = t ∗ t. 6 L−1 Alternatively, t t∗ t = (t − τ )τ dτ 0 = 29. We have F (s) = 1 s(s2 +1) 1 1 t τ2 − τ3 2 3 = 1 s · 1 , s2 +1 1 s 1 s2 + 1 t = 0 13 13 13 t − t = t. 2 3 6 and L−1 −→ 1; L−1 −→ sin t. So t f (t) = 1 ∗ sin t = sin τ dτ = 1 − cos t. 0 ...
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This note was uploaded on 12/22/2011 for the course MAP 3305 taught by Professor Stuartchalk during the Fall '11 term at UNF.

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