burgers equation with source term how to solve $u_t + uu_x =f(x,t)$ with initial condition $u(x,0)= \phi(x)$
For the simple case $f(x,t)=1 $ and $\phi(x)=x$, I've tried to proceed like for the homogeneous one.
I find the characteristics are parabolae $$X(t)=\frac {t^2}{2} +X(0)(1+t)$$  
but the solution won't be constant along these characteristics because of the source term f!
I find that if I define $v(t)=u(X(t),t) \ \ $,  $v$ would be  solution to $v'=1$ hence $$v(t)=t+k=t+v(0)=t+u(X(0),0)=t+X(0)$$
plugging this into u I  get $$ u(X(t),t)=u(\frac {t^2}{2} +X(0)(1+t),t)$$
and then, solving for $X(0)$ in terms of $x$ $$u(x,t)=\frac{1}{1+t}(x-\frac {t^2}{2} )+t$$
Am I missing something here?
And how should I procced for a generic f and $\phi$?
 A: Your solution is correct. A slightly different way to obtain it is: write $u(x,t)=w(x,t)+t$, so that $w$ satisfies the equation $w_t+(w+t)w_x=0$. 
The function $w$ is constant along curves $X(t)$ that satisfy 
$X'(t)=\phi(X(0))+t$. This leads to 
$$X(t)=X(0)+\phi(X(0))t+\frac{t^2}{2}$$
which is easy to put in implicit form if $\phi$ is a nice function like $\phi(x)=x$. 
In this case,
$$X(t)=X(0)(1+t)+\frac{t^2}{2}$$
which in implicit form is 
$$\frac{x-t^2/2}{1+t}=C$$
Thus, $w$ is of the form 
$$w(x,t) = \Psi\left(\frac{x-t^2/2}{1+t}\right)$$
The initial condition is satisfied with $\Psi(z)=z$, i.e., 
$$w(x,t) =  \frac{x-t^2/2}{1+t} $$
Add $t$  to get $u(x,t)$. 
More generally, if $f$ depends only on $t$, you can write it as $f(x,t)=F'(t)$ for some $F$, and let $u(x,t)=w(x,t)+F(t)$ similar to the above. Then $w_t+(w+F(t))w_x=0$. The function $w$ is constant along curves $X(t)$ that satisfy 
$X'(t)=\phi(X(0))+F(t)$. This leads to 
$$X(t)=X(0)+\phi(X(0))t+\int_0^t F(s)\,ds$$
... and so on.  
I don't know how to handle $f$ that depends on $x$. 
Another version of this problem, where $f$ is a function of $u$, is considered in the paper Exact solution of generalized inviscid Burgers' equation where some more-or-less explicit formulas are derived for the solution. (One must be able to integrate, invert, and integrate again...) 
