Chem Differential Eq HW Solutions Fall 2011 142

# Chem Differential Eq HW Solutions Fall 2011 142 - 142...

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142 Chapter 8 The Laplace and Hankel Transforms with Applications 33. Passing the equations y ±± + y = δ 0 ( t - 1) ,y (0) = 0 ± (0) = 0, through the Laplace transform, we get: s 2 Y - sy (0) - y ± (0) + Y = e - s ( s 2 +1) Y = e - s Y = e - s s 2 +1 . Thus the solution is y = L - 1 ± e - s s 2 ² . To compute this inverse transform, we observe sin t L -→ 1 s 2 ; U 0 ( t - 1) sin( t - 1) L e - s s 2 ; so y = U 0 ( t - 1) sin( t - 1). 37. Take the Laplace transform on both sides of y ±± +4 y = U 0 ( t - 1) e t - 1 and use the initial conditions y (0) = 1 ± (0) = 0, and you will get L ( y ±± y )= L ( U 0 ( t - 1) e t - 1 ) L ( y ±± )+ L (4 y L ( U 0 ( t - 1) e t - 1 ) ; s 2 Y - sy (0) - y ± (0) + 4 Y = e - s 1 s - 1 ( s> 1) s 2 Y Y = e - s s - 1 Y (4 + s 2 e - s s - 1 ; Y = e - s ( s - 1)( s 2 +4) , where we have used Theorem 1, Sec. 8.2. Thus the solution is the inverse Laplace
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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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