Differential Equations Lecture Work Solutions 118

Differential Equations Lecture Work Solutions 118 - 1 c. +...

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1c . Θ 0 + µ Θ=0 Z 0 λZ =0 r ( rR 0 ) 0 +( λr 2 µ ) R =0 Z ( H )=0 | R (0) | < R ( a )=0 Solution as in 1a exactly ! But u z ( r, θ, 0) = α ( r, θ )= X m =0 X n =1 ( a nm cos + b nm sin ) J m ( ± λ nm r ) ± λ nm cosh ± λ nm ( H ) a nm ± λ nm cosh ± λ nm ( H )= R 2 π 0 R a 0 α ( r, θ )cos mθJ m ( λ nm r ) rd rdθ R 2 π 0 R a 0 cos 2 mθJ 2 m ( λ nm r ) rd rdθ a nm = R 2 π 0 R a 0 α ( r, θ )cos mθJ m ( λ nm r ) rd rdθ λ nm cosh λ nm ( H ) R 2 π 0 R a 0 cos 2
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This note was uploaded on 12/22/2011 for the course MAP 2302 taught by Professor Bell,d during the Fall '08 term at UNF.

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