How to solve coupled linear 1st order PDE It is fairly straight forward to solve linear 1st order PDEs by the method of characteristics. For example, if
$\partial_tf+a\partial_xf=bf$ , 
we have that $\dfrac{df}{dt}=bf$ on the characteristic curve of $\dfrac{dx}{dt}=a$ . From this we deduce that 
$f(t,x)=g(C)e^{bt}$ where $x=at+C$ .
Now, how does this work when $f$ is multidimensional.
Can I solve equations on the following form by characteristics, or by any other means?
$\partial_tf_i(t,x)+\sum_jA_{ij}\partial_xf_j(t,x)=\sum_jB_{ij}f_j(t,x)$
where the components of $A$ and $B$ might be dependent on $x$ and $t$.
In particular, I am trying to solve the following,
$\begin{cases}\partial_tf+\dfrac{c}{t}\partial_xg=-\left(a+\dfrac{1}{t}\right)f\\\partial_tg+\dfrac{c}{t}\partial_xf=-\left(b+\dfrac{1}{t}\right)g\end{cases}$
where $f$ and $g$ are functions of $x$ and $t$ , where $t>t_0>0$ , $c\neq0$ .
Any help is highly appreciated.
Edit: Never mind the specific equation. It tured out to be the result of a erroneous derivation. But I still wonder if there is some procedure to solve the general equation above.
 A: $\begin{cases}\partial_tf+\dfrac{c}{t}\partial_xg=-\left(a+\dfrac{1}{t}\right)f~......(1)\\\partial_tg+\dfrac{c}{t}\partial_xf=-\left(b+\dfrac{1}{t}\right)g~......(2)\end{cases}$ , $c\neq0$
From $(1)$ ,
$\partial_tf+\dfrac{c}{t}\partial_xg=-\left(a+\dfrac{1}{t}\right)f$
$\dfrac{c}{t}\partial_xg=-\partial_tf-\left(a+\dfrac{1}{t}\right)f$
$\partial_xg=-\dfrac{t}{c}\partial_tf-\left(\dfrac{at}{c}+\dfrac{1}{c}\right)f~......(3)$
$\partial_{xt}g=-\dfrac{t}{c}\partial_{tt}f-\dfrac{1}{c}\partial_tf-\left(\dfrac{at}{c}+\dfrac{1}{c}\right)\partial_tf-\dfrac{a}{c}f$
$\partial_{xt}g=-\dfrac{t}{c}\partial_{tt}f-\left(\dfrac{at}{c}+\dfrac{2}{c}\right)\partial_tf-\dfrac{a}{c}f~......(4)$
From $(2)$ ,
$\partial_tg+\dfrac{c}{t}\partial_xf=-\left(b+\dfrac{1}{t}\right)g$
$\partial_{xt}g+\dfrac{c}{t}\partial_{xx}f=-\left(b+\dfrac{1}{t}\right)\partial_xg~......(5)$
Put $(3)$ and $(4)$ into $(5)$ ,
$-\dfrac{t}{c}\partial_{tt}f-\left(\dfrac{at}{c}+\dfrac{2}{c}\right)\partial_tf-\dfrac{a}{c}f+\dfrac{c}{t}\partial_{xx}f=-\left(b+\dfrac{1}{t}\right)\left(-\dfrac{t}{c}\partial_tf-\left(\dfrac{at}{c}+\dfrac{1}{c}\right)f\right)$
$-\dfrac{t}{c}\partial_{tt}f-\left(\dfrac{at}{c}+\dfrac{2}{c}\right)\partial_tf-\dfrac{a}{c}f+\dfrac{c}{t}\partial_{xx}f=\left(\dfrac{bt}{c}+\dfrac{1}{c}\right)\partial_tf+\left(\dfrac{abt}{c}+\dfrac{a+b}{c}+\dfrac{1}{ct}\right)f$
$\dfrac{c}{t}\partial_{xx}f=\dfrac{t}{c}\partial_{tt}f+\biggl(\dfrac{(a+b)t}{c}+\dfrac{3}{c}\biggr)\partial_tf+\left(\dfrac{abt}{c}+\dfrac{2a+b}{c}+\dfrac{1}{ct}\right)f$
$\partial_{xx}f=\dfrac{t^2}{c^2}\partial_{tt}f+\dfrac{(a+b)t^2+3t}{c^2}\partial_tf+\dfrac{abt^2+(2a+b)t+1}{c^2}f$
