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Differential Equations Lecture Work Solutions 61

# Differential Equations Lecture Work Solutions 61 - 5 b uxx...

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5b . u xx + u yy =0 u (0 ,y )= g ( y ) u ( L, y )=0 u y ( x, 0) = 0 u ( x, H )=0 X 0 λX =0 Y 0 + λY =0 X ( L )=0 Y 0 (0) = 0 Y ( H )=0 Using the summary of Chapter 4 we have Y n ( y )=co s ( n + 1 2 ) π H y, n =0 , 1 ,... λ n = n + 1 2 π H 2 n =0 , 1 ,... Now use these eigenvalues in the x equation: X 0 n ± ( n + 1 2 ) π H 2 X n =0 n =0 , 1 , 2 ,... Solve: X n = c n sinh  n + 1 2 π H x + D n Use the boundary condition: X n ( L )=0 X n ( L )= c n sinh  n + 1 2 π L H + D n =0 ( n + 1 2 ) πL H + D n =0 X n = c n sinh ( n + 1 2 ) π H ( x L ) ² u ( x, y )= X n =0 a n sinh " ( n + 1 2 ) π H ( x L ) # cos n + 1 2 ² π H y To Fnd the coeﬃcients a n , we use the inhomogeneous boundary condition:
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