Solve 2u = 2y sin 1/3x for the grid in Fig. 461 and uy (1, 3)

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Solve ∆2u = – π2y sin 1/3πx for the grid in Fig. 461 and uy (1, 3) = uy (2, 3) = 1/2√243, u = 0 on the other three sides of the square.
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