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Differential Equations Lecture Work Solutions 89

# Differential Equations Lecture Work Solutions 89 - 19 2 u =...

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19. 2 u = 0 , 0 x π, 0 y π u ( x, 0) = π 2 x 2 + 2 πx 3 x 4 u ( x, π ) = 0 u (0 , y ) = u ( π, y ) = 0 The only difference is in the inhomogeneous condition, thus the general solution is the same u ( x, y ) = n =1 a n sinh n ( y π ) sin nx Now use the only inhomogeneous boundary condition π 2 x 2 + 2 πx 3 x 4 = u ( x, 0) = n =1 a n sinh sin nx The coeﬃcients a n sinh are those of the Fourier sine expansion of π 2 x 2 + 2 πx 3 x 4 , i.e. a n sinh = 2 π π 0 π 2 x 2 + 2 πx 3 x 4 sin nxdx Now we integrate (using integration by parts to reduce the powers of
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