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Unformatted text preview: MATH 450 Section 002 Homework 7 Solutions Winter 2009 Section 20.2, p. 1067, #15. (a) Let u ( x,y ) = f 2 x 2 + U ( x,y ), then U ( x,y ) satisfies U xx + U yy = 0 , U (0 ,y ) = 0 , U ( a,y ) = f 2 a 2 , U ( x, 0) = U ( x,b ) = f 2 x 2 . Seeking U ( x,y ) = X ( x ) Y ( y ), then X 00 ( x ) Y ( y ) + X ( x ) Y 00 ( y ) = 0 . Therefore X 00 ( x ) X ( x ) = Y 00 ( y ) Y ( y ) = 2 . This implies X ( x ) = A + Bx, if = 0 , C cos x + D sin x, if 6 = 0 , and Y ( y ) = E + Fy, if = 0 , Ge y + He y , if 6 = 0 . so u ( x,y ) = ( A + Bx )( E + Fy ) + ( C cos x + D sin x )( Ge y + He y ) . Now lets apply the boundary conditions. First, u (0 ,y ) = 0 = A ( E + Fy ) + C ( Ge y + He y ) , 1 so A = C = 0. Thus u ( x,y ) = Ix + Jxy + ( Pe y + Qe y )sin x. Second, u ( a,y ) = fa 2 2 = Ia + Jay + ( Pe y + Qe y )sin a, therefore, I = fa 2 , J = 0 , sin a = 0 . It follows from the above equations that = n a , n = 1 , 2 , 3 , . Thus u ( x,y ) = fa 2 x + X n =1 ( P n e n a y + Q n e n a y )sin n a x. Furthermore, one has (1) u ( x, 0) = fx 2 2 = fa 2 x + X n =1 ( P n + Q n )sin n a x, so P n + Q n = 2 a Z a ( fa 2 x f 2 x 2 )sin n a xdx. Finally, u ( x,b ) = fx 2 2 = fa 2 x + X n =1 ( P n e nb a + Q n e nb a )sin nx a dx, 2 so (2) P n e nb a + Q n e nb a = 2 a Z a ( fa 2 x f 2 x 2 )sin nx a dx. Solve the system (1) and (2) for P n and Q n , one has P n = e nb a 1 e nb a e nb a 2 a Z a ( fax 2 f 2 x 2 )sin nx a dx, Q n = e nb a 1 e nb a e nb a 2 a Z a ( fax 2 f 2 x 2 )sin nx a dx....
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This note was uploaded on 04/24/2010 for the course MATH 425 taught by Professor K during the Spring '10 term at University of MichiganDearborn.
 Spring '10
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 Math

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