Differential Equations Lecture Work Solutions 6

# Differential - u(0 = 0 we get B = 0 The other boundary condition will yield − Q k 1 2 A = 1 ⇒ A = Q 2 k 1 ⇒ u x = 1 Q 2 k ± x − Q 2 k x 2

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1. Since the temperature at both ends is zero (boundary conditions), the temperature of the rod will drop until it is zero everywhere. 2. ku xx + Q =0 u (0 .t )=0 u (1 ,t )=1 u xx = Q k Integrate with respect to x u x = Q k x + A Integrate again u = Q k x 2 2 + Ax + B Using the Frst boundary condition
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Unformatted text preview: u (0) = 0 we get B = 0. The other boundary condition will yield − Q k 1 2 + A = 1 ⇒ A = Q 2 k + 1 ⇒ u ( x ) = 1 + Q 2 k ± x − Q 2 k x 2 3. ±ollow class notes. 6...
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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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