Chem Differential Eq HW Solutions Fall 2011 72

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

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Unformatted text preview: 72 Chapter 4 Partial Differential Equations in Polar and Cylindrical Coordinates 22. (a) Using (7), ∞ J 3 ( x) = 2 (−1)k k!Γ(k + 3 + 1) 2 k =0 2 x = ∞ 2 πx ∞ k =0 2 22k+1k! x2k+2 (−1)k k! (2k + 3)(2k + 1)! 22k+2 1 1 1 (Γ(k + 2 + ) = Γ(k + 1 + )Γ(k + 1 + ) then use Exercise 44(b)) 2 2 2 ∞ 2 (−1)k (2k + 2) 2k+2 (multiply and divide by (2k + 2)) x πx (2k + 3)! = = 2 πx 2 πx = 2 πx = 25. 2k + 3 2 (−1)k x2k+2 k!Γ(k + 2 + 1 ) 22k+2 2 k =0 = = x 2 k =0 ∞ k =1 ∞ k =1 ∞ 2 πx (a) Let u = k =1 2 a (−1)k−1(2k) 2k x (2k + 1)! (change k to k − 1) (−1)k−1[(2k + 1) − 1] 2k x (2k + 1)! (−1)k−1 2k x− (2k)! sin x − cos x + x 2 πx ∞ k =1 (−1)k−1 2k x (2k + 1)! . 1 e− 2 (at−b), Y (u) = y(t), e−at+b = a2 4 u2 ; then dy 1 1 a1 d2y d dY du = = Y (−e− 2 (at−b)); 2 = Y (−e− 2 (at−b)) = Y e−at+b +Y e− 2 (at−b). dt du dt dt du 2 So Y e−at+b + Y a − 1 (at−b) + Y e−at+b = 0 e2 2 ⇒ 1 a Y + Y e− 2 (at−b) + Y = 0, 2 1 2 upon multiplying by eat−b. Using u = a e− 2 (at−b), we get Y+ 1 Y +Y =0 u ⇒ u2Y + uY + u2Y = 0, which is Bessel’s equation of order 0. (b) The general solution of u2Y + uY + u2Y = 0 is Y (u) = c1 J0 (u) + c2Y0 (u). 1 2 But Y (u) = y(t) and u = a e− 2 (at−b), so 21 21 y(t) = c1J0 ( e− 2 (at−b)) + c2 Y0 ( e− 2 (at−b)). a a (c) (i) If c1 = 0 and c2 = 0, then 21 y(t) = c2Y0 ( e− 2 (at−b) ). a As t → ∞, u → 0, and Y0(u) → −∞. In this case, y(t) could approach either +∞ or −∞ depending on the sign of c2 . y(t) would approach infinity linearly as near 1 2 0, Y0 (x) ≈ ln x so y(t) ≈ ln a e− 2 (at−b) ≈ At. (ii) If c1 = 0 and c2 = 0, then 21 y(t) = c1 J0( e− 2 (at−b)). a ...
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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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