Chem Differential Eq HW Solutions Fall 2011 84

Chem Differential Eq HW Solutions Fall 2011 84 - 84 Chapter...

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84 Chapter 5 Partial Differential Equations in Spherical Coordinates Finally, Y 2 , 1 =( - 1) 1 Y 2 , - 1 = - Y 2 , - 1 = - 3 2 ± 5 6 π cos θ sin e - = - 3 2 ± 5 6 π cos θ sin e . ± 5. (a) If m = 0, the integral becomes ² 2 π 0 φdφ = 1 2 φ 2 ³ ³ ³ 2 π 0 =2 π 2 . Now suppose that m ± = 0. Using integration by parts, with u = φ, du = dφ, dv = e - imφ ,v = 1 - im e - imφ , we obtain: ² 2 π 0 = u ´µ¶· φ = dv ´ µ¶ · e - imφ = ¸ φ 1 - im e - imφ ³ ³ ³ ³ 2 π 0 + 1 im ² 2 π 0 e - imφ We have e - imφ ³ ³ φ =2 π = ¹ cos( ) - i sin( ) ³ ³ ³ φ =2 π =1 , and ² 2 π 0 e - imφ = ² 2 π 0 ( cos - i sin ) = º 0i f m ± =0 , 2 π if m So if m ± , ² 2 π 0 φe - imφ = 2 π - im = 2 π m i. Putting both results together, we obtain ² 2 π 0 - imφ = º 2 π m i if m ± , 2 π 2 if m . (b) Using n = 0 and m = 0 in (9), we get
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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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