Chem Differential Eq HW Solutions Fall 2011 70

# Chem Differential Eq HW Solutions Fall 2011 70 - 70 Chapter...

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Unformatted text preview: 70 Chapter 4 Partial Differential Equations in Polar and Cylindrical Coordinates This equation determines the coefficients b m ( m ≥ 1) in terms of the coefficients of y 1 . Furthermore, it will become apparent that k cannot be 0. Also, b is arbitrary but by assumption b = 0. Let’s take b = 1 and determine the the first five b m ’s. Recall from Exercise 1 y 1 = 1 1 · 6 x 3 8- 1 1 · 24 x 5 32 + 1 2 · 120 x 7 128 + ··· . So y 1 = 3 1 · 6 x 2 8- 5 1 · 24 x 4 32 + 7 2 · 120 x 6 128 + ··· and hence (taking k = 1) 2 kxy 1 = 6 k 1 · 6 x 3 8- 10 k 1 · 24 x 5 32 + 14 k 2 · 120 x 7 128 + ··· . The lowest exponent of x in 2 kxy 1- 5 b 1 x- 2 + ∞ m =2 ( m- 6) mb m + b m- 2 x m- 3 is x- 2 . Since its coefficient is- 5 b 1 , we get b 1 = 0 and the equation becomes 2 xy 1 + ∞ m =2 ( m- 6) mb m + b m- 2 x m- 3 . Next, we consider the coefficient of x- 1 . It is (- 4)2 b 2 + b . Setting it equal to 0, we find b 2 = b 8 = 1 8 ....
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