Differential Equations Lecture Work Solutions 90

Differential Equations Lecture Work Solutions 90 - Now to...

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Unformatted text preview: Now to our integral π 0 −π 2 x2 + 2πx3 − x4 sin nxdx π −π 2 = 0 + 2π − 6 n2 − 12 n2 − π 3x2 π x3 cos nx + 2 sin nx n n 0 0 π 0 −− x2 sin nxdx x sin nxdx π 4x3 π x4 cos nx + 2 sin nx n n 0 0 π 0 x2 sin nxdx The sine function vanishes at the end points and cos 0 = 1 and cos nπ = (−1)n so π 0 −π 2 x2 + 2πx3 − x4 sin nxdx −π 2 + = − 12π n2 π 0 −π 2 + = 12 n2 π x sin nxdx − 12 n2 + − 12π n2 − 2 12 −π 2 + 2 n n π 0 x2 sin nxdx 0 2π 3 π4 π (−1)n + (−1)n n n π2 (−1)n n π 0 x cos nxdx x sin nxdx After evaluating the integrals, we get −an sinh nπ = 24 2π 2 −2 n5 n 2 π 0 −2 8 sinh nπ 12 − π2 3 n for n even for n odd Thus for n odd, we have an = n2 π The solution is then 8 u(x, y ) = n=1,3,... n2 π sinh nπ 12 − π 2 sinh n(y − π ) sin nx n3 90 ...
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This note was uploaded on 12/22/2011 for the course MAP 2302 taught by Professor Bell,d during the Fall '08 term at UNF.

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