How to solve vector-valued first order linear pde? Is there an analytical solution to the pde system?
$$\frac{\partial f}{\partial x} + \frac{\partial g}{\partial y} = 0$$
$$\frac{\partial f}{\partial y} - \frac{\partial g}{\partial x} = 0$$
More generally, how about
$$a\frac{\partial f}{\partial x} + b\frac{\partial g}{\partial y} = 0$$
$$c\frac{\partial f}{\partial y} - d\frac{\partial g}{\partial x} = 0$$
where $a,b,c,d$ are all constants.
Maybe now analytical solution depends on these coefficients and thus be hard or unclear, in which case a proof of existence is wanted. (Does here Frobenius theorem help?)
 A: These are the Cauchy-Riemann equations in disguise. They say that $f-ig$ is a holomorphic function of $z=x+iy$.
A: I assume $f, g$ are scalar fields.  Let $A(x,y) = f(x,y) \hat x + g(x,y) \hat y$.  This system is equivalent to
$$\nabla \cdot A = 0, \quad \nabla \times A = 0$$
The vector field $A$ is entirely determined by boundary values.  You can attack such a problem using the generalized Stokes theorem, but the expressions are a bit involved.
However, since this is a 2d problem, you can instead use complex analysis and the Cauchy integral formula.  The condition $\nabla \cdot A = 0, \nabla \times A = 0$ for a vector field is entirely equivalent to saying there is a corresponding complex function $a$ that is holomorphic.
Thus, at some complex point $z$, you can find the value of $a$ by
$$a(z) = \frac{1}{2\pi i} \oint_C \frac{a(z')}{z - z'} \, dz'$$
For some suitably chosen closed path with nice, known values on the boundary, this could have an analytic solution.
A: Here is an analytical solution for the first system

$$  f \left( x,y \right)={ F_1} \left( y-ix \right) +{ 
F_2} \left( y+ix \right),$$ 
$$ \,g \left( x,y \right) = i{ F_1} \left( y-i x\right)-i{ F_2} \left( y+ix \right) + {\it C} ,$$

where $F_1,F_2$ are arbitrary functions and $C$ is a constant. 
