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 = 0 and 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 first quadrant (0 u< and 0 v < ), we have 0 θ π 2 , and 0 r< , and the double integral in (b) becomes
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• Fall '11
• StuartChalk
• Coordinate system, Spherical coordinate system, Polar coordinate system, Bessel function, Bessel Functions

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