Solving a certain nonlinear transport equation We want to solve
\begin{align}
(t+u)u_x +tu_t &= x-t \\
u(1,x) &= 1+x
\end{align}
Normally, I'd rewrite the equation as $\frac{t+u}{t}u_x + u_t = \frac{x-t}{t}$, then as per the method of characteristics, set $\frac{dx}{dt} = \frac{t+u}{t}$. Setting $v(t) = u(t, x(t)),$ we see $\dot{v}(t) = x(t)-t$. Now here, I'd normally solve for $v$, plug it back into the characteristic equation and proceed, but I'm not sure how to proceed. Might be something basic from ODE that I'm forgetting. Thanks. 

$$(t+u)u_x+tu_t=x-t$$
Characteristic differential equations : $\quad \frac{dx}{t+u}=\frac{dt}{t}=\frac{du}{x-t}$
A first family of characteristic curves comes from :
$\frac{dx}{t+u}=\frac{du}{x-t}=\frac{dx+du}{(t+u)+(x-t)}=\frac{dx+du}{x+u}=\frac{d(x+u)}{(x+u)}=\frac{dt}{t} \quad\to\quad \frac{x+u}{t}=c_1$
A second family of characteristic curves comes from :
$\frac{dx}{t+u}=\frac{dt}{t}=\frac{dx-dt}{(t+u)-t}=\frac{dx-dt}{u}=\frac{du}{x-t} \quad\to\quad u^2-(x-t)^2=c_2$
With any independent $c_1,c_2$ , all above is valid only on the characteristic curves. Outside, $c_1$ and $c_2$ are not independent. The relationship can be expressed on the form of an implicit equation $\Phi(c_1,c_2)=0$ :
$$\Phi\left(\frac{x+u}{t}\:,\:u^2-(x-t)^2\right)=0$$
or alternatively $c_1=F(c_2)$ where $F$ is any differentiable function :
$$\frac{x+u}{t}=F\left(u^2-(x-t)^2\right)$$
This is an implicit form for the general solution of the PDE.
Then, with the condition : $u(1,x)=1+x$
$$\frac{x+(1+x)}{1}=F\left((1+x)^2-(x-1)^2\right) \quad\to\quad F(4x)=2x+1$$
This determines the function $F(X)=\frac{X}{2}+1$ 
With $X=u^2-(x-t)^2$ put into the above general solution:
$$\frac{x+u}{t}=\frac{u^2-(x-t)^2}{2}+1 $$
This quadratic equation can be solved for $u$. One root $u=t-x$ doesn't satisfy the condition. The other root is the solution of the PDE consistent with the condition $u(1,x)=1+x$ :
$$u(x,t)=x-t+\frac{2}{t}$$
