$\frac{dx}{d\tau}=xy-x\;(1), \quad \frac{dy}{d\tau}=-xy-\alpha-\beta\;(2)$
I'm preparing for my final in non-linear dynamics and I want to find the fixed points for this system i.e. when both equation (1) and (2) are equal to 0.
I see that I can factor equation (1) into $\frac{dx}{d\tau}=x(y-1) \rightarrow x^*=0, y^*=1$
And the I understand that that using y=1 in (2) gives the fixed point $(\beta-\alpha,1)$.
But here comes my confusion when using x=0 in equation (2) $0=-\alpha+\beta$ which is supposed to yield the second fixed point $(0,\frac{\beta}{\alpha})$
I don't understand how the y can exist explicitly when x is equal to 0, maybe I am just lacking som basic mathematical skills.
You can write $$xy-x=0$$ and $$-xy-\alpha-\beta=0$$ Since we have $$xy=x$$ from the first equation we get $$-x-\alpha-\beta=0$$ so $$x=-\alpha-\beta$$