# Solve using Laplace transforms

I want to solve the PDE
$$x\frac{\partial u}{\partial t} +\frac{\partial u}{\partial x}=x$$ $$u(x,0)=0, x>0$$ $$u(0,t)=0, t>0$$ for $$x>0,t>0$$ using a Laplace transform in t.
I Laplace transform the equation and get $$xs\hat{u}(x,s)+\frac{\partial}{\partial x}\hat{u}(x,s)=\frac{x}{s}$$ where $$\hat{u}(x,s)=L(u(x,t))=\int_{t=0}^{\infty} e^{-st}u(x,t)dt$$.
Solving this ODE, I obtain $$\hat{u}(x,s)=\frac{1}{s^2} + A(s)e^{\frac{-sx^2}{2}}.$$ I then apply the second boundary condition to get $$\hat{u}(x,s)=\frac{1}{s^2}-\frac{1}{s^2}e^{\frac{-sx^2}{2}}.$$ Now I need to inverse transform to get $$u(x,t)$$. I know that $$L^{-1}(\frac{1}{s^2})=t$$ but I don’t know how to find $$L^{-1}(\frac{1}{s^2}e^{\frac{-sx^2}{2}}).$$