I cannot solve the following partial derivative equation
$$\begin{cases} -\frac{\partial\phi(x,y)}{\partial x}+B_1(x,y) = y \\ ~\\ -\frac{\partial\phi(x,y)}{\partial y}+B_2(x,y) = -w^2 \sin(x)-ay\\ ~\\ \frac{\partial B_1(x,y)}{\partial x}+\frac{\partial B_2(x,y)}{\partial y}=0 \end{cases}$$
where $\omega$ and $a$ are parameters.
(If you are curious where I got this equation from, it comes from a Hemlholtz decomposition in a two-dimentional space).
I have no idea how to solve this equation to obtain $\phi, B_1$ and $B_2$.
I even could not solve it using mathematica:
DSolve[{
-D[v[x, y], x] + Ba[x, y] == y,
-D[v[x, y], y] + Bb[x, y] == -w^2 Sin[x] - a y,
D[Ba[x, y], x] + D[Bb[x, y], y] == 0
}, {v[x, y], Ba[x, y],Bb[x, y]},
{x, y}]
$$\begin{cases} -\frac{\partial\phi(x,y)}{\partial x}+B_1(x,y) = y \\ ~\\ -\frac{\partial\phi(x,y)}{\partial y}+B_2(x,y) = -w^2 \sin(x)-ay\\ ~\\ \frac{\partial B_1(x,y)}{\partial x}+\frac{\partial B_2(x,y)}{\partial y}=0 \end{cases}$$
$$\begin{cases} B_1(x,y) = y +\frac{\partial\phi(x,y)}{\partial x} \quad\implies\quad \frac{\partial B_1(x,y)}{\partial x} = \frac{\partial^2\phi(x,y)}{\partial x^2}\\ ~\\ B_2(x,y) = -w^2 \sin(x)-ay+\frac{\partial\phi(x,y)}{\partial y}\quad\implies\quad \frac{\partial B_2(x,y)}{\partial y}=-a+\frac{\partial^2 \phi(x,y)}{\partial y^2}\\ \end{cases}$$ $$\frac{\partial B_1(x,y)}{\partial x}+\frac{\partial B_2(x,y)}{\partial y}=0\quad\implies\quad \frac{\partial^2\phi(x,y)}{\partial x^2}+\frac{\partial^2 \phi(x,y)}{\partial y^2}=a \tag 1$$ The general solution of the associated homogeneous PDE is $\quad\phi_h=f(x+iy)+g(x-iy)\quad$ with arbitrary functions $f$ and $g$.
An obvious particular solution of the inhomogeneous PDE is $\quad \phi_p=\frac{a}{4}(x^2+y^2)\quad$ As a consequence, the solution of Eq.$(1)$ can be expressed on the form $\phi=\phi_h+\phi_p$ : $$\phi(x,y)=f(x+iy)+g(x-iy)+\frac{a}{4}(x^2+y^2)$$
And then: $\quad\begin{cases} B_1(x,y) = y +f'(x+iy)+g'(x-iy)+\frac{a}{2}x\\ ~\\ B_2(x,y) = -w^2 \sin(x)-ay+i\,f'(x+iy)-i\,g'(x-iy)+\frac{a}{2}y\\ \end{cases}$
where $f(X)$ and $g(X)$ are arbitrary differentiable functions.
$f'(X)=\frac{df(X)}{dX}$ and $g'(X)=\frac{dg(X)}{dX}\quad$ , with respectively $X=x+i\,y$ and $X=x-i\,y$ .