I'm trying to solve this problem via Laplace transform, but something sounds me wrong.
\begin{cases} u_x-4xu_t=0\\ u(x,0)=0\\ u(0,t)=t+3 \end{cases}
My passages:
\begin{cases} \mathcal{L}[u_x]-\mathcal{L}[4xu_t]=\mathcal{L}[0]\\ u(x,0)=0\\ \mathcal{L}[u(0,t)]=\mathcal{L}[t+3] \end{cases}
\begin{cases} W_x(x,s)-4x(sW(x,s) - u(x,0)) =0\\ u(x,0)=0\\ \displaystyle W(0,s)=\frac{1}{s^2}+\frac{3}{s} \end{cases}
\begin{cases} W_x(x,s)-4xsW(x,s)=0\\ \displaystyle W(0,s)=\frac{1}{s^2}+\frac{3}{s} \end{cases}
Solving the first ODE, using $\Phi(s)$
\begin{cases} \displaystyle W(x,s) =\Phi(s)e^{2sx^2} \\ \displaystyle W(0,s)=\frac{1}{s^2}+\frac{3}{s} \end{cases}
So, obtaining $\Phi(s)$
$$\displaystyle \frac{1}{s^2}+\frac{3}{s}=W(0,s)=\Phi(s)$$
$$\displaystyle \Phi(s)=\frac{1}{s^2}+\frac{3}{s}$$
Results
$$\displaystyle W(x,s) =\left(\frac{1}{s^2}+\frac{3}{s}\right)e^{2sx^2}$$
With $\mathcal{L}^{-1}$, we obtain
$$u(x,t)=t+3+2x^2$$
that satisfies the PDE, the last condition, but not the second one. Where am I wrong? Thanks to anyone who can help me.