I have a first order PDE that I want to solve using Laplace transform in order to practice that approach: $$\partial _x u=x \partial _t u$$
then, defining $v(x,s)=\mathcal L_t(u(x,t))(s)$, I use the laplace transform to turn this into: $$\partial _xv(x,s)-xs\cdot v(x,s)=-xu(x,0)$$ by substitution: $$\partial_x(e^{-\frac{1}{2}sx^2}v(x,s))=-xe^{-\frac{1}{2}sx^2}u(x,0)$$ integrating w.r.t. $x$, and solving for the integration constant gives: $$v(x,s)=e^{\frac{1}{2}sx^2}\left(v(0,s)-\int_0^xe^{-\frac{1}{2}sx^2}xu(x,0)dx\right)$$
Which means: $$\mathcal L_t(u(x,t))=e^{\frac{1}{2}sx^2}\mathcal L_t(u(0,t))-e^{\frac{1}{2}sx^2}\int_0^xe^{-\frac{1}{2}sx^2}xu(x,0)dx$$
Now, using the fact that $\mathcal L_t\left(\mu(t-c)f(t-c)\right)=e^{-sc}\mathcal L(f(t))$, where $\mu$ is the heaviside step function, we can rewrite this as:
$$\mathcal L_t(u(x,t))=\mathcal L_t\left(\mu(t-\frac 1 2 x^2)\cdot u(0,t-\frac{1}{2}x^2)\right)-e^{\frac{1}{2}sx^2}\int_0^xe^{-\frac{1}{2}sx^2}xu(x,0)dx$$
This is as far as I got. I don't know how to simplify it further, unless we assume $u(x,0)=0$, in which case, the solution is there for any boundary condition $u(0, t)$. However, I'd like to find the general solution for arbitrary initial conditions $u(x,0)$.
How do I simplify this further for a general solution in terms of arbitrary boundary and initial conditions?
note: I'd like to solve this using the Laplace transform, in order to practice using Laplace transform to solve PDE's. I know it can be solved more easily with simpler methods.
Solutions can be written as $u(x,t) = F(\sqrt{x^2 + 2 t})$ where $F$ is an arbitrary (differentiable) function. However, this requires the initial condition $u(x,0) = F(|x|)$ to be an even function.