I'm currently taking an ODE course at my school and one of the problems given follows:
Suppose that a trajectory of $$(3x^2 - y)dx + (3y^2 - x)dy = 0$$ contains the point $(1,1)$. Show that it also contains the points $(1, -1), (-1, 1), (0, 1), (1, 0)$.
Note that the only methods for solving DEs we have discussed so far just involve guessing the form and plugging it in (basically undetermined coefficients). This class is heavily self-taught, so maybe I'm misunderstanding some part of it. I understand that we are examining the solution to the initial value problem where $y(1) = 1$.
Now, if we divide through by $dx$ and plug in the given point, we find that $y'(1) = -1$. However, $(x, y, y') = (1, -1, -1)$ is clearly not a solution. Furthermore, I don't know how to check the other points. Any help would be greatly appreciated!
Hint; $$\frac {\partial}{\partial y} (3x^2-y) = -1 = \frac {\partial}{\partial x} (3y^2-x)$$
Solve as an exact equation and plug in the values in the results. I suspect there's an easier way though as the result isn't pretty.