PDE method of characteristics $u_t+u^2u_x=0$ with $u(x,0)=x$

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I'm confused on how to include the $u^2$ expression in the solution process

$$u_t +u^2u_x=0,\quad u(x,0)=x$$

where $u_t$ and $u_x$ denote the partial of u with respect to those variables

I'm actually unsure how to go about this as in our PDE class we haven't mentioned functions that contain the function $u$

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Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:

$\dfrac{dt}{ds}=1$ , letting $t(0)=0$ , we have $t=s$

$\dfrac{du}{ds}=0$ , letting $u(0)=u_0$ , we have $u=u_0$

$\dfrac{dx}{ds}=u^2=u_0^2$ , letting $x(0)=f(u_0)$ , we have $x=u_0^2s+f(u_0)=u^2t+f(u)$ , i.e. $u=F(x-u^2t)$

$u(x,0)=x$ :

$F(x)=x$

$\therefore u=x-u^2t$

$tu^2+u-x=0$

$u(x,t)=\begin{cases}x&\text{when}~t=0\\\dfrac{-1\pm\sqrt{4xt+1}}{2t}&\text{when}~t\neq0\end{cases}$