Q:

How to solve this first order PDE using characteristics

Suppose $u = u(x,t)$. Is it possible to solve
$$u_t -u_x  = x^3u$$
using the method of characteristics?  I am struggling to do so.  I am being thrown off by the non-linearity on the RHS. 

A:

$${{u}_{t}}-{{u}_{x}}={{x}^{3}}u$$ 
Let $$u=u\left( x\left( s \right),t\left( s \right) \right)\Rightarrow \frac{du}{ds}=\frac{\partial u}{\partial t}\frac{dt}{ds}+\frac{\partial u}{\partial x}\frac{dx}{ds}$$
Hence let  $s=t$ thus 
$$\frac{du}{ds}=\frac{\partial u}{\partial t}+\frac{\partial u}{\partial x}\frac{dx}{dt}\Rightarrow \frac{dx}{dt}=-1\Rightarrow x\left( t \right)=x\left( 0 \right)-t$$
From the PDE we have therefore 
$$\frac{du}{dt}={{x}^{3}}u\Rightarrow \ln \left( \frac{u\left( x\left( t \right),t \right)}{u\left( x\left( 0 \right),0 \right)} \right)=\int\limits_{0}^{t}{{{x}^{3}}\left( s \right)ds}=\int\limits_{0}^{t}{{{\left( x\left( 0 \right)-s \right)}^{3}}ds}=\frac{1}{4}\left( x{{\left( 0 \right)}^{4}}-{{\left( x\left( 0 \right)-t \right)}^{4}} \right)$$
Therefore 
$$u\left( x\left( t \right),t \right)=u\left( x\left( 0 \right),0 \right)\exp \left( \frac{1}{4}\left( x{{\left( 0 \right)}^{4}}-{{\left( x\left( 0 \right)-t \right)}^{4}} \right) \right)$$
Let $$u\left( x,0 \right)=f\left( x \right)$$ then we find
$$u\left( x,t \right)=f\left( x+t \right)\exp \left( \frac{1}{4}\left( {{\left( x+t \right)}^{4}}-{{x}^{4}} \right) \right)$$

