TITLE: Solving a simple partial differential equation.
QUESTION [0 upvotes]: I'm trying to find every continuously differentiable functions $f:\mathbb{R}^2\to \mathbb{R}$ such that $$\alpha \frac{\partial f}{\partial x}(x,y)+\beta\frac{\partial f}{\partial y}(x,y)=0 $$ where $\alpha,\beta$ are constants.
By inspection I can find a family of solution namely $f_c(x,y) = c(\beta x -\alpha y)$ for $c\in \mathbb{R}$ but I don't know that this is all of them, nor how I'd go about proving that it is (if it is) or finding all of them if it isn't. This is part of some practice sheet I have for a topology class, and this is in the section on the local inversion theorem and global inversion theorem, so I'm guessing it somehow involves this theorem but I don't know how. Any help or hint is appreciated.

REPLY [1 votes]: Along a line written parametrically as $x = x_0 + \alpha t$, $y = y_0 + \beta t$, by the chain rule $$ \dfrac{df}{dt} = \frac{\partial f}{\partial x} \dfrac{dx}{dt} + \frac{\partial f}{\partial y} \dfrac{dy}{dt} = \alpha \frac{\partial f}{\partial x} + \beta \frac{\partial f}{\partial y} = 0$$
which means $f$ is constant along such a line.  Such a line can also be written implicitly
as $$\beta x - \alpha y = \text{constant}$$
where the "constant" will be different for each line.  So the general solution is $$f(x,y) = g(\beta x - \alpha y)$$
where $g$ is an arbitrary differentiable function.
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