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Unformatted text preview: ( u 1 ) rr + 1 r ( u 1 ) r + 1 r 2 ( u 1 ) = 0 , < r < 1 , < 2 , u 1 (1 , ) = sin 3 , < 2 . Subproblem #2 (to be solved after Fnding u 1 ( r, ) from Subproblem #1) ( u 2 ) t = ( u 2 ) rr + 1 r ( u 2 ) r + 1 r 2 ( u 2 ) , < r < 1 , < 2 , t > , u 2 (1 , , t ) = 0 , < 2 , t > , u 2 ( r, , 0) =u 1 ( r, ) , < r < 1 , < 2 . You can check, using linearity (or superposition), that u ( r, , t ) = u 1 ( r, ) + u 2 ( r, , t ) is a solution of the given problem. The solution of subproblem #1 follows immediately from the method of Section 4.5. We have u 2 ( r, ) = r 3 sin 3 ....
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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.
 Fall '11
 StuartChalk

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