Chem Differential Eq HW Solutions Fall 2011 148

# Chem Differential Eq HW Solutions Fall 2011 148 - 148...

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148 Chapter 8 The Laplace and Hankel Transforms with Applications Solutions to Exercises 8.4 1. (a) Let z 2 = x u ( z, t )= u ( z 2 ,t u ( x, t ). Then 2 z dz dx =1 or dz dx = 1 2 z . So ∂x u ( x, t ˜ u ( ˜ u ∂z dz dx = ˜ u 1 2 z Similarly, 2 u 2 = 2 ± ˜ u 1 2 z ² = d dx ³ 1 2 z ´ ˜ u + 1 2 z ³ ˜ u ´ = - 1 2 z 2 dz dx ˜ u + 1 2 z 2 ˜ u 2 dz dx = - 1 4 z 3 ˜ u + 1 4 z 2 2 ˜ u 2 (b) Substituting what we found in (a) into (6) and using u in place of ˜ u to simplify notation, we get u tt = g ± z 2 ± - 1 4 z 3 u z + 1 4 z 2 u zz ² + 1 2 z u z ² = g 4 ± u zz + 1 z u z ² 5. Using Exercise 9 of Section 4.3, we Fnd H 0 ( x 2 N U 0 ( a - x ))( s µ 0 x 2 N U 0 ( a - x ) J 0 ( sx ) xdx = µ a 0 J 0 ( sx ) x 2 N +1 dx (change variables sx x ) = 1 s 2 N
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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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