# How to solve PDE system

I have a system of PDEs

$\frac{\partial f}{\partial x}=1-f$

$\frac{\partial f}{\partial y}=-f$

$f(x,0)=1-e^{-x}$

$f(0,y)=0$

$x>0, y>0$

I've tried to integrate it manually, but the solution I've received is wrong $(1-e^{-x})\cdot e^{-y}$

How to solve this may be with some computer algebra system?

By a bootstrapping argument, you get that, if a solution $f$ exists, then $f \in \mathcal{C}^\infty(\mathbb{R}^2,\mathbb{R})$. Therefore, Schwarz's theorem holds, and you get $\frac {\partial^2 f} {\partial x \partial y} = \frac {\partial^2 f} {\partial y \partial x}$.
In your example, you get $$\frac {\partial} {\partial y} \left( \frac {\partial f} {\partial x} \right) = \frac {\partial} {\partial y} (1-f) = -\frac {\partial f} {\partial y} = f$$ and $$\frac {\partial} {\partial x} \left( \frac {\partial f} {\partial y} \right) = \frac {\partial} {\partial x} (-f) = -\frac {\partial f} {\partial x} = f-1.$$ Therefore, you get $f(x,y) = f(x,y) - 1$ for all $(x,y)$, which means that your system unfortunately has no solution.
The equation $\frac {\partial f} {\partial x} + f = 1$ yields $f(x,y) = 1 + g(y) e^{-x}$. Evaluating at $x=0$ yields $g(y) = -1$. Therefore, $f(x,y) = 1 + e^{-x}$, which is a contradiction since $\frac {\partial f} {\partial y} + f$ should vanish.