Usbrth

I am trying to solve the conduction problem for the temperatures $theta(x,t)$.
beginalign
fracpartialthetapartial t&=kfracpartial^2thetapartial x^2 \
theta(0,t)&=T_0e^-bt, t>0, b>0 tag1\
theta(x,0)&=0, x>0.
endalign

My attempt:

I took the Laplace transform with respect to t of the PDE.
beginalign
mathcalL_t(theta_t(x,t))&=kmathcalL_t(theta_xx(x,t)) \
smathcalL_t(theta(x,t))&=kfracd^2dx^2mathcalL_t(theta(x,t)) \
sbartheta&=kfracd^2dx^2bartheta.
endalign

Solving this ODE, I get
$$bartheta(x,t)=Ae^xsqrtfracsk+Be^-xsqrtfracsk, A,BinmathbbR.$$
To ensure $bartheta$ is finite, take $A=0$ as $|bartheta|rightarrowinfty$ as $|s|rightarrowinfty.$ Taking the Laplace transform of $(1)$ and imposing this boundary condition, I get $$bartheta(x,t)=fracT_0s+be^-xsqrtfracsk.$$ Assuming this is correct, how can I invert? A hint would be appreciated in (I have tried convolution theorem). I expect the result to be in terms of error functions.

Update:

$$mathcalL^-1left(frac1s+btimes e^-xsqrtfracskright)=e^-btastfrackxe^fracx^24tk2sqrtpi (kt)^3.$$ I have used the property $$mathcalL(f(ct))=frac1cFleft(fracscright).$$

Let $theta(x,t)=X(x)T(t)$ ,

Then $X(x)T'(t)=kX''(x)T(t)$

$dfracT'(t)kT(t)=dfracX''(x)X(x)=-s^2$

$begincasesdfracT'(t)kT(t)=-s^2\X''(x)+s^2X(x)=0endcases$

$begincasesT(t)=c_3(s)e^-kts^2\X(x)=begincasesc_1(s)sin xs+c_2(s)cos xs&textwhen~sneq0\c_1x+c_2&textwhen~s=0endcasesendcases$

$thereforetheta(x,t)=int_0^infty C_1(s)e^-kts^2sin xs~ds+int_0^infty C_2(s)e^-kts^2cos xs~ds$

$theta(0,t)=T_0e^-bt$ :

$int_0^infty C_2(s)e^-kts^2~ds=T_0e^-bt$

$C_2(s)=T_0deltaleft(s-sqrtdfracbkright)$

$thereforetheta(x,t)=int_0^infty C_1(s)e^-kts^2sin xs~ds+int_0^infty T_0deltaleft(s-sqrtdfracbkright)e^-kts^2cos xs~ds$

$theta(x,t)=int_0^infty C_1(s)e^-kts^2sin xs~ds+T_0e^-btcossqrtdfracbkx$