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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...