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Differential Equations Lecture Work Solutions 90

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

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Now to our integral π 0 π 2 x 2 + 2 πx 3 x 4 sin nxdx = π 2 π 0 x 2 sin nxdx + 2 π x 3 n cos nx π 0 + 3 x 2 n 2 sin nx π 0 6 n 2 π 0 x sin nxdx − − x 4 n cos nx π 0 + 4 x 3 n 2 sin nx π 0 12 n 2 π 0 x 2 sin nxdx The sine function vanishes at the end points and cos 0 = 1 and cos = ( 1) n so π 0 π 2 x 2 + 2 πx 3 x 4 sin nxdx = π 2 + 12 n 2 π 0 x 2 sin nxdx 12 π n 2 π 0 x sin nxdx 2 π n π 3 ( 1) n + π 4 n ( 1) n = π 2 + 12 n 2 π 2 n ( 1) n + 2 n π 2 + 12 n 2 π 0 x cos nxdx 12 π n 2 π 0 x sin nxdx After evaluating the integrals, we get a
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