Recent content by wik_chick88

excellent! so that gave me \frac{T_1}{s} = \tau_H (0, s) + \tau_N (0, s) = A(s) + \frac{T_0}{s - Ka^2} so A(s) = \frac{T_1}{s} - \frac{T_0}{s - Ka^2} so \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}}} which is the solution I am...

i found the solution to the homogenous ODE...i think! K \tau^{''} - s \tau = 0 try \tau_H = e^{rx}(where r is a constant) thus \tau_H^{''} = r^2 e^{rx} sub into ODE: Kr^2e^{rx} - se^{rx} = 0 e^{rx}(Kr^2 - s) = 0 Kr^2 - s = 0 Kr^2 = s r^2 = \frac{s}{K} r = ±\sqrt{\frac{s}{K}} thus...

assuming you just forgot to put the K before the \frac{d^2 \tau}{dx^2} this is what i got: K \frac{d^2 \tau}{dx^2} - s \tau = -T_0 e^{-ax} try \tau = Me^{-ax}, so \frac{d^2 \tau}{dx^2} = Ma^2 e^{-ax} so KMa^2 e^{-ax} - sMe^{-ax} = -T_0 e^{-ax} e^{-ax} (KMa^2 - sM) = -T_0 e^{-ax} M(Ka^2 -...

Homework Statement T(x,t) is the temperature distribution for t > 0 in a semi-infinite slab occupying x > 0 T(x, 0) = T_0 e^{-ax} for x > 0 (with a positive constant) T(0, t) = T_1 for t > 0 \tau(x, s) is the Laplace transform of T(x, t) show that \tau(x, s) = \frac{T_0}{s - Ka^2}e^{-ax} +...