Chem Differential Eq HW Solutions Fall 2011 73

# Chem Differential Eq HW Solutions Fall 2011 73 - Section...

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Section 4.7 Bessel’s Equation and Bessel Functions 73 As t →∞ , u ( t ) 0, J 0 ( u ) 1, and y ( t ) c 1 . In this case the solution is bounded. (ii) If c 1 ± =0and c 2 ± = 0, as t , u ( t ) 0, J 0 ( u ) 1, Y 0 ( u ) →-∞ . Since Y 0 will dominate, the solution will behave like case (i). It makes sense to have unbounded solutions because eventually the spring wears out and does not aﬀect the motion. Newton’s laws tell us the mass will continue with unperturbed momentum, i.e., as t , y ±± = 0 and so y ( t )= c 1 t + c 2 ,a linear function, which is unbounded if c 1 ± =0 . 33. (a) In (13), let u 2 = t ,2 udu = dt , then Γ( x ± 0 t x - 1 e - t dt = ± 0 u 2( x - 1) e - u 2 (2 u ) du =2 ± 0 u 2 x - 1 e - u 2 du. (b) Using (a) Γ( x )Γ( y )=2 ± 0 u 2 x - 1 e - u 2 du 2 ± 0 v 2 y - 1 e - v 2 dv =4 ± 0 ± 0 e - ( u 2 + v 2 ) u 2 x - 1 v 2 y - 1 dudv. (c) Switching to polar coordinates: u = r cos θ , v = r sin θ , u 2 + v 2 = r 2 , dudv = rdrdθ ; for ( u, v ) varying in the Frst quadrant (0 u< and 0 v< ), we have 0 θ π 2 , and 0 r< , and the double integral in (b) becomes Γ( x )Γ(
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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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