A function that is equal to the sum of its partial derivatives I recently learned about partial derivatives and I'm trying to solve the equation $$f_x(x,y)+f_y(x,y)=f(x,y)$$ I tried the assumptions $f=f(x)$ and $f=f(y)$ which led to the equations $f'(x)=f(x)$ and $f'(y)=f(y)$ respectively. This gives two solutions, $f=Ae^x$ and $f=Be^y$, and by using the fact that the sum of two solutions to the equation is also a solution, you get $f=Ae^x+Be^y$. Then I used the assumption $f=constant$ which is just $0+0=c$, so $f=0$ (which you get by setting $A=B=0$) is also a solution. This exhausts all solutions where $f_x$ or $f_y$ is $0$. I believe there are more solutions, but how do you find them? Is there a systematic way to go straight to the general solution?
 A: You can try applying the Fourier transform. Assuming that $f$ is a tempered distribution, it yields
$$i\xi_1\hat{f} + i\xi_2\hat{f} = \hat{f},$$
$$(1 - i\xi_1 - i\xi_2)\hat{f} = 0,$$
$$\hat{f} = 0,$$
$$f = 0.$$
Hence the only tempered solution is $0$. You can interpret this as saying that any nonzero solution grows very fast.
To try to find nonzero solutions, notice that the equation reads
$$Df(x, y) \cdot (1, 1) = f(x, y).$$
Let $(x_0, y_0) \in \mathbb{R}^2$ be arbitrary, and let $(x(t), y(t))$ be the integral curve of the vector field $(1, 1)$ starting at $(x_0, y_0)$, that is,
$$(x'(t), y'(t)) = (1, 1),$$
$$(x(0), y(0)) = (x_0, y_0).$$
Solving this trivial ODE yields
$$(x(t), y(t)) = (x_0 + t, y_0 + t).$$
Now let
$$z(t) = f(x(t), y(t)).$$
We have
$$z'(t) = Df(x(t), y(t)) \cdot (x'(t), y'(t)) = Df(x(t), y(t)) \cdot (1, 1) = f(x(t), y(t)) = z(t).$$
Hence
$$z(t) = e^tz(0).$$
This says
$$f(x_0 + t, y_0 + t) = e^tf(x_0, y_0).$$
Note that this holds for every $(x_0, y_0) \in \mathbb{R}^2$ and every $t \in \mathbb{R}$. Conversely, you can check that if $f$ satisfies the above equation, then by differentiating wrt $t$ and setting $t = 0$, you see that $f$ solves your PDE. Hence this is the "general solution" you were seeking.
