Taking the inverse Laplace under boundary conditions I want to solve the problem as described here in this article

I am trying to solve it using Laplace Transformation.  
This is what I did:
Taking Laplace transformation of the equation I get
$$K[s^2\hat T_0(s)-sT_0(0)-T'_0(0)]+ {MT_a\over s}-M\hat T_0(s)+{Q_m \over s}=0$$ where $M$ is the constant $\omega_b\rho_b\mathcal C_b$. 
Using the boundary condition since $-K{dT_0(x)\over dx}=h_0[T_f-T_0(x)]$ and we have to convert the boundary condition from $x$ domain to Laplace domain and since at $x=0$, $h_0[T_f-T_0(0)]$ is a constant and  $L(a)=a/s$ where $a$ is a constant,
I get,
$$(Ks^2-M) \hat T_0(s)=KsT_0(0)-{h_0[T_f-T_0(0)] \over s}-{(Q_M+MT_a)\over s}.$$ 
Then taking the inverse Laplace I get,
$$T_0(x)=T_0(0)\cosh\left(\sqrt {M\over K}x\right)-h_0[T_f-T_0(0)]\left[{-x\over M}+{1\over M}\cosh\left(\sqrt {M\over K}x\right)\right]-(Q_m+MT_a)\left[{-x\over M}+{1\over M}\cosh\left(\sqrt {M\over K}x\right)\right].$$   
Is this correct. I am not sure if I did right when adjusting the boundary condition from $x$ domain to Laplace domain.  
Also I didn't need to use the condition $T_0(x)=T_c$ where $x=L$.
I can't use this to find $T_0(0)$,  since $T_0(x)=T_c $ is at $x=L$, right?
 A: Using the Laplace transform is useless and ill-advised here, it creates a lot of confusion.
Basically the first line is a second order differential equation with constant coefficients: 
$$K\frac{\mathrm d^2T_0}{\mathrm dx^2}-MT_0=-MT_a-Q_m.\tag{1}$$ 
The solutions of the homogeneous equation ($ K\frac{\mathrm d^2T_0}{\mathrm dx^2}-MT_0=0$) are of the form $x\mapsto \exp[\alpha x]$ and $x\mapsto\exp[-\alpha x]$, where $\alpha$  is
here simply given by $$\alpha=\sqrt{\frac{M}{K}}.$$
To solve the full equation (1), it remains to find one single solution called special solution.
In that case it is particularly simple because the second member is a constant: a constant solution $T_0(x)=\text{constant}$ is solution when the constant is chosen to be $T_a+Q_m/M$. 
So all solutions of (1) are of the form
$$ T(x)=T_a+\frac{Q_m}M+A\exp[\alpha x]+B\exp[-\alpha x],$$
where $A$ and $B$ are constants.
Applying the boundary conditions, you get the system for $A$ and $B$:
$$\left\{\begin{array}{rcl}T_a+\frac{Q_m}M+A\exp[\alpha L]+B\exp[-\alpha L]&=&T_c\\-K\alpha(A-B)&=&h_0\left[T_f-T_a-\frac{Q_m}M-A-B\right]\end{array}\right.$$
when you have found $A$ and $B$, you find $T_0(0)=T_a+\frac{Q_m}M+A+B$. 
