I have the following equation
But if I let $T_t(x,t)=constant$ then my equation becomes at steady state,
since partial differentiation of a constant=0, right?
here $\omega_bp_bc_b=M, $a constant and $T_a,Q(m),Q_r$ are also constants as well
Now I take the Laplace transform of the second equation,
and I come up with $L(T(x,s))={MT_a+Q(m)+Q_r \over sM}+{k\over M}{d^2 L(T(x,s))\over dx^2}$. Is this correct?
Now I want to take the inverse Laplace transform. The boundary conditions and initial conditions are given by
$T_t(x,0)=T_c$
here $T_c$ is the body core temperature and since biologically the body core temperature tends to be stable $T_c$ is a constant.
How can I take the inverse Laplace of this?
I think it should be similar to
as is solved for a 2-D case in the page 38, equation 43 in the article here
I took the equation, letting $Q_r$ to be just similar to $Q_m$ for my purpose,
$k{d^2T(x,t)\over dx^2}+M(T_a-T_t(x,t))+Q_m+Q_r$. Then took the Laplace instead of taking it for t I took the transformation on x as you suggested.
Then I get,
$k(s^2L(T(s,t))-sT(0,t)-T'(0,t))+M {T_a\over S}-M L(T(s,t))+{Q_m+Q_r \over S}$.
If I let t=0, then I need to know the temperature at x=0 and assume that since at steady state this temperature doesn't change with time. Suppose I took it ass $T_t(0,0)=T_0$
then using boundary conditions I get,
$L(T(s,0))={ksT_0+h_0[T_s-T_0]-{MT_a \over s}-{(Q_m-Q_r)\over s}\over (ks^2 -M)}$ .
Is this correct. Can you check this please. And how am I going to get the inverse of this?
In the article here they say through simple calculations they can obtain the solution 
I have no idea how they have got this. I think this solution is similar to my problem