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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 ) = ∂x ˜ u ( z,t ) = ˜ u ∂z dz dx = ˜ u ∂z 1 2 z Similarly, 2 u ∂x 2 = 2 ∂x ˜ u ∂z 1 2 z = d dx 1 2 z ˜ u ∂z + 1 2 z ∂x ˜ u ∂z = - 1 2 z 2 dz dx ˜ u ∂z + 1 2 z 2 ˜ u ∂z 2 dz dx = - 1 4 z 3 ˜ u ∂z + 1 4 z 2 2 ˜ u ∂z 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 + 1 2 z u z = g 4 u zz + 1 z u z 5. Using Exercise 9 of Section 4.3, we find H 0 ( x 2 N U 0 ( a -
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