Hi i was wondering if anyone could help me show that
$$u(x,t)=\int_{0}^{t}\frac{e^{\sqrt{\tau}x}(e^{2\sqrt{\tau}(l-x)}-1)}{(a+t-\tau)(e^{2\sqrt{\tau}l}-1)}d\tau$$ is a solution to the heat equation problem given by $$\frac{\partial u}{\partial t}=\frac{\partial^{2} u}{\partial x^{2}},0<x<l,t>0$$
subject to $u(0,t)=e^{-at},t>0$, $u(l,t)=0,t>0$, $u(x,0)=0,0<x<l$
Only from the use of Laplace transformations.
I will now show my working. I get stuck at the last step.
Solution: Taking the Laplace transform of the pde in the $t$-direction gives
$$\mathcal{L}(\frac{\partial u}{\partial t})=\mathcal{L}(\frac{\partial^{2} u}{\partial x^{2}}).$$
From this we get that $$0=\frac{d^{2}}{dx^{2}}U(s,x)-SU(x,s).$$
Now note that this is a 2nd order ode, so the characteristic equation is $$m^{2}-s=0\implies m=\sqrt{s},$$
therefore $$U(x,s)=a(s)e^{\sqrt{s}x}+b(s)e^{-\sqrt{s}x}.$$
Now taking the Laplace transform of the first condition gives $$\mathcal{L}(u(0,t))=\mathcal{L}(e^{-at})=\frac{1}{a+s}.$$
Also, the Laplace transforms of the other conditions give: $$ \mathcal{L}(u(l,t))=\mathcal{L}(0)\Rightarrow U(l,s)=0, $$ and $$\mathcal{L}(u(x,0))=\mathcal{L}(0) \Rightarrow U(x,0)=0. $$ Applying these conditions to the characteristic equation is where I stopped as I'm confused where to go next.