Differential Equations Lecture Work Solutions 9

# Differential Equations Lecture Work Solutions 9 - B = 1...

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5. ku xx + Q =0 a. ku xx =0 Integrate twice with respect to x u ( x )= Ax + B Use the boundary conditions u (0) = 1 implies B =1 u ( L ) = 0 implies AL + B =0 tha ti s A = 1 L Therefore u ( x )= x L +1 b. ku xx =0 Integrate twice with respect to x as in the previous case u ( x )= Ax + B Use the boundary conditions u x (0) = 0 implies A =0 u ( L ) = 1 implies AL + B
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Unformatted text preview: B = 1 Therefore u ( x ) = 1 c. k u xx = 0 Integrate twice with respect to x as in the previous case u ( x ) = Ax + B Use the boundary conditions u (0) = 1 implies B = 1 u x ( L ) = ϕ implies A = ϕ Therefore u ( x ) = ϕ x + 1 9...
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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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