I am trying to solve the system $$\begin{cases} u_t+\omega\cdot\nabla_x u=f(t,x) \\ u(0,x)=u_0(x)\\ \end{cases}$$ for some $\omega\in\mathbb{R}^n$.
I have the system of ODEs $$\frac{dt}{ds}=1$$ $$\frac{dx_i}{ds}=\omega_i$$ $$\frac{du}{ds}=f(t,x)$$ Thus, using the initial conditions, I obtain $t=s$, $x=x_0+\omega s=x_0+\omega t$, and $u(t,x)=sf(t,x)+u_0(x_0)$. Thus, my solution is $u(t,x)=tf(t,x)+u_0(x-\omega t)$. This appears to satisfy the condition $u(0,x)=u_0(x)$. Is my solution correct?