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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 π 0 φdφ = 1 2 φ 2 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 π 0 = u φ = dv e - imφ = φ 1 - im e - imφ 2 π 0 + 1 im 2 π 0 e - imφ We have e - imφ φ =2 π = cos( ) - i sin( ) φ =2 π = 1 , and 2 π 0 e - imφ = 2 π 0 ( cos - i sin ) = 0 if m = 0 , 2 π if m = 0 So if m = 0, 2 π 0 φe - imφ = 2 π - im = 2 π m i. Putting both results together, we obtain 2 π 0 φe - imφ = 2 π m i if m = 0 , 2 π
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