As mentioned in comment, neither Maple nor Mathematica can solve the heat pde analytically with the $K$ coefficient (thermal conductivity of the material) as function of space. Mathematica can solve it for scalar $k$ which I show below so you can use if you want to try for yourself. 

Heat PDE depends on time and space. You have not said what the initial conditions are. Assuming $u(x,0)=f(x)$ then

    ClearAll[u, x, t, Tm, Tk, h, f, n, z, k];
    pde = D[u[z, t], t] == k*D[u[z, t], {z, 2}]
    bc = {u[-h/2, t] == Tm, u[h/2, t] == Tc}
    ic = u[z, 0] == f[z]
    sol = DSolve[{pde, bc, ic}, u[z, t], {z, t}, Assumptions -> {h > 0, t > 0, k > 0}];


![Mathematica graphics](https://i.sstatic.net/a5JJe.png)

In the steady state (which you showed), i.e. as $t \to \infty$ then we see due to the term $e^{-\frac{\pi ^2 t K[1]^2}{h^2}}$ it will go to zero. Hence the sum term vanish and the solution at steady state becomes

![Mathematica graphics](https://i.sstatic.net/2q3PU.png)

This does not match what you showed, but $k$ used here is not function of space. 

May be Mathematica in next version could solve the heat pde with $k$ function of space.