QUESTION:
Solve PDE $$ u_t + t^3 u_x = u$$ Conditions: $u(t,0) = t$, $u(0,x) = 1 - e^{-x}$
ANSWER:
For $x \le\frac{t^4}{4}$ we have
$$\frac{dx}{dt} = t^3; x(0) = x_0 \to x(t) = \frac{t^4}{4} + x_0$$ $$\frac{du}{dt} = u; u(0) = 1 - e^{-x_0} \to u(t) = (1 - e^{-x_0})e^t$$
There's a little more to the answer, but I don't understand what method this is, or what steps have been skipped. Can someone explain?
Note I switched $(t,x)$ to $(x,t)$. Now what is really happening behind the scenes is merely a change of coordinate systems to one better suited to the PDE. Basically the PDE is saying that $\nabla_{(t^3,1)}u=u$ (directional derivative), so $u(x,t)=u(x_0,0)e^t$, where $x_0$ is the point on the x-axis you arrive at if you start at $(x,t)$ and travel backwards along the curve with tangent vector $(t^3,1)$ until reaches $t=0$. This curve is described by $\displaystyle\frac{d}{dt}(x(t),t)=(t^3,1)\implies (x(t),t)=(\frac{t^4}{4}+x_0,t)$, so we get that $u(x,t)=u(x-\frac{t^4}{4},0)e^t=(1-e^{-(x-\frac{t^4}{4})})e^t\,.$