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wik_chick88
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Homework Statement
[itex]T(x,t)[/itex] is the temperature distribution for [itex]t > 0[/itex] in a semi-infinite slab occupying [itex]x > 0[/itex]
[itex]T(x, 0) = T_0 e^{-ax}[/itex] for [itex]x > 0[/itex] (with [itex]a[/itex] positive constant)
[itex]T(0, t) = T_1[/itex] for [itex]t > 0[/itex]
[itex]\tau(x, s)[/itex] is the Laplace transform of [itex]T(x, t)[/itex]
show that [itex]\tau(x, s) = \frac{T_0}{s - Ka^2}e^{-ax} + [\frac{T_1}{s} - \frac{T_0}{s - Ka^2}] e^{-x\sqrt{\frac{s}{K}}}[/itex]
Homework Equations
The Attempt at a Solution
the non-homogeneous solution to the Laplace transform of the heat equation is [itex]\tau(x, s) = A(s) e^{-x\sqrt{\frac{s}{K}}}[/itex]
so i assumed that because [itex]T(x, 0) = T_0 e^{-ax}[/itex] then our general solution is
[itex]\tau(x, s) = A(s) e^{-x\sqrt{\frac{s}{K}}} + \frac{T_0 e^{-ax}}{s}[/itex]
is this correct?
we have [itex]T(0, t) = T_1[/itex], so [itex]\tau(0, s) = T_1[/itex]
thus [itex]T_1 = A(s) + \frac{T_0}{s}[/itex]
so [itex]A(s) = T_1 - \frac{T_0}{s}[/itex]
but this gives me a final solution of
[itex]\tau(x, s) = (T_1 - \frac{T_0}{s}) e^{-x\sqrt{\frac{s}{K}}} + \frac{T_0 e^{-ax}}{s}[/itex]
WHICH IS NOT RIGHT!
please somebody help me! I'm confused as to where the [itex]s - Ka^2[/itex] comes from...