Here is a problem:
$\left\{\begin{array}{l} 2 u_{t}-u_{x}=f(t, x), \quad t>0 \\ \left.u\right|_{t=0}=\varphi(x) \end{array}\right.$
My solution
$\frac{d t}{2}=\frac{d x}{-1}=\frac{d y}{f(x, c)}$
Find the first characteristic:
$\frac{d t}{2}=\frac{d x}{-1} \Rightarrow C_{1} = x+\frac{1}{2} t$
And the second one:
$\frac{d t}{2}=\frac{d u}{f(x, y)} \Rightarrow \frac{d u}{d t}=f\left(t, C_{1}-\frac{1}{2} t\right)$
And solution is a linear combination of them:
$u(x, t)=\varphi\left(C_{1}-\frac{1}{2} t\right)+\int_{0}^{t} f\left(s, C_{1}-\frac{1}{2} s\right) d s$
Am I right? If not, how to solve it correctly?
I agree with your result :
$$u(x, t)=\varphi\left(C_{1}-\frac{1}{2} t\right)+\int_{0}^{t} f\left(s, C_{1}-\frac{1}{2} s\right) d s$$ But I think that is not completely finished because in the final form of result $C_1$ depends on $x$ and $t$ : $$u(x, t)=\varphi\left(x\right)+\int_{0}^{t} f\left(s, x+\frac{t-s}{2} \right) d s$$