Differential Equations Lecture Work Solutions 10

Differential Equations Lecture Work Solutions 10 - x as in...

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d. ku xx + Q =0 u xx = Q k = x 2 Integrate with respect to x we get u x ( x )= 1 3 x 3 + A Use the boundary condition u x ( L ) = 0 implies 1 3 L 3 + A =0 tha ti s A = 1 3 L 3 Integrating again with respect to x u = x 4 12 + 1 3 L 3 x + B Use the second boundary condition u (0) = 1 implies B =1 Therefore u ( x )= x 4 12 + L 3 3 x +1 e. ku xx
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Unformatted text preview: x as in the previous case u ( x ) = Ax + B Use the boundary conditions u (0) = 1 implies B = 1 u x ( L ) + u ( L ) = 0 implies A + ( AL + 1) = 0 that is A = 1 L + 1 Therefore u ( x ) = 1 L + 1 x + 1 10...
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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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