Solve the Cauchy problem: $u_t + u^2 u_x = 0$, $ u(0,x) = 2+x$ Solve the Cauchy problem: For $t>0$, $$u_y + u^2 u_x = 0$$ $$u(x,0) = 2+x$$
So by the characteristic method:$\frac{dx}{dt} = z^2$, $\frac{dy}{dt}=1$, $\frac{dz}{dt}=0$, parametrized by $\Gamma: (s,0,2+s)$. Then we have $z=2+s$, $x=t(2+s)^2 +s $ and $y=t$, which gives us $u=x-yu^2 + 2$. I think it's correct but when I plugged in to check, it only satisfies the initial condition. I appreciate if anyone could tell me where I did wrong.

$$u(x,y) \quad \begin{cases} u_y + u^2 u_x = 0 \\u(x,0) = 2+x \end{cases}$$
Your calculus is correct. The solution expressed on the form of implicit equation is :
$$u=x-yu^2 + 2$$
This agrees with the condition $u(x,0)=2+x$.
Probably you made a mistake in checking the agreement to the PDE.
$$u_x=1-2yuu_x \quad\to\quad u_x=\frac{1}{1+2yu}$$
$$u_y=-u^2-2yuu_y \quad\to\quad u_y=\frac{-u^2}{1+2yu}$$
$$u_y+u^2u_x=\frac{-u^2}{1+2yu}+u^2\frac{1}{1+2yu}=0$$ 
All agrees.
