Solve the lst order pde using the method of characteristics Solve  the following lst order $pde$ using the method of characteristics:
$ u_t + 3t^2 u_x = -u$ 
With $  u(x,0) =
  \begin{cases}
    e^{-1}     & \quad x<0\\
       e^{x-1}     & \quad 0<x<1\\
   e^{-(x-1)}     & \quad x>1\\
  \end{cases}
$
I found $u(x,t)= f(x_•)e^{-t}$
But i get stuck with inatial condition .
Thanks for help.

Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:
$\dfrac{dt}{ds}=1$ , letting $t(0)=0$ , we have $t=s$
$\dfrac{dx}{ds}=3t^2=3s^2$ , letting $x(0)=x_0$ , we have $x=x_0+s^3=x_0+t^3$
$\dfrac{du}{ds}=-u$ , letting $u(0)=f(x_0)$ , we have $u(x,t)=f(x_0)e^{-s}=f(x-t^3)e^{-t}$
$u(x,0)=\begin{cases}e^{-1}&x<0\\e^{x-1}&0<x<1\\e^{-(x-1)}&x>1\end{cases}$ :
$f(x)=\begin{cases}e^{-1}&x<0\\e^{x-1}&0<x<1\\e^{-(x-1)}&x>1\end{cases}$
$\therefore u(x,t)=\begin{cases}e^{-t}&x-t^3<0\\e^{x-t^3-1-t}&0<x-t^3<1\\e^{-(x-t^3-1)-t}&x-t^3>1\end{cases}=\begin{cases}e^{-t}&x<t^3\\e^{x-t^3-t-1}&t^3<x<t^3+1\\e^{-x+t^3-t+1}&x>t^3+1\end{cases}$
