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Chem Differential Eq HW Solutions Fall 2011 72

# Chem Differential Eq HW Solutions Fall 2011 72 - 72 Chapter...

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72 Chapter 4 Partial Differential Equations in Polar and Cylindrical Coordinates 22. (a) Using (7), J 3 2 ( x ) = k =0 ( - 1) k k !Γ( k + 3 2 + 1) x 2 2 k + 3 2 = 2 x k =0 ( - 1) k k !Γ( k + 2 + 1 2 ) x 2 k +2 2 2 k +2 = 2 πx k =0 ( - 1) k k ! 2 2 2 k +1 k ! (2 k + 3)(2 k + 1)! x 2 k +2 2 2 k +2 (Γ( k + 2 + 1 2 ) = Γ( k + 1 + 1 2 )Γ( k + 1 + 1 2 ) then use Exercise 44(b)) = 2 πx k =0 ( - 1) k (2 k + 2) (2 k + 3)! x 2 k +2 (multiply and divide by (2 k + 2)) = 2 πx k =1 ( - 1) k - 1 (2 k ) (2 k + 1)! x 2 k (change k to k - 1) = 2 πx k =1 ( - 1) k - 1 [(2 k + 1) - 1] (2 k + 1)! x 2 k = 2 πx k =1 ( - 1) k - 1 (2 k )! x 2 k - 2 πx k =1 ( - 1) k - 1 (2 k + 1)! x 2 k = 2 πx - cos x + sin x x . 25. (a) Let u = 2 a e - 1 2 ( at - b ) , Y ( u ) = y ( t ) ,e - at + b = a 2 4 u 2 ; then dy dt = dY du du dt = Y ( - e - 1 2 ( at - b ) ); d 2 y dt 2 = d du Y ( - e - 1 2 ( at - b ) ) = Y e - at + b + Y a 2 e - 1 2 ( at - b ) . So Y e - at + b + Y a 2 e - 1 2 ( at - b ) + Ye - at + b = 0 Y + a 2 Y e - 1 2 ( at - b ) + Y = 0 , upon multiplying by e at - b . Using u = 2 a e - 1 2 ( at - b ) , we get Y + 1 u Y + Y = 0 u 2 Y + uY + u 2 Y = 0 , which is Bessel’s equation of order 0. (b) The general solution of u 2 Y + uY + u 2 Y = 0 is Y ( u ) = c 1 J 0 ( u ) + c 2 Y 0 ( u ). But Y ( u ) =
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