# For this linear system:dx/(dt)=x+y−2 dy/(dt)=x−y−4 Classify (1,1)

For this linear system:
$\frac{dx}{dt}=x+y-2$
$\frac{dy}{dt}=x-y-4$
Classify $\left(1,1\right)$
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cydostwng6c
$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{x-y-4}{x+y-2}$
At $\left(1,1\right),\phantom{\rule{thickmathspace}{0ex}}\frac{dy}{dx}\ne 0$,$\left(1,1\right),\phantom{\rule{thickmathspace}{0ex}}\frac{dy}{dx}\ne 0$. Indeed, at $\left(1,1\right)$, $\frac{dy}{dx}$ is not defined.
$\frac{dy}{dx}=0$ when $x-y-4=0$
But this doesn't tell us too much: only that $x-y=4$. (Two variables and one equation. Infinitely many solutions.)
On the other hand, if you find the Jacobian and evaluate the eigenvalues, you'll find that your critical point is $\left(3,-1\right)$, and is a saddle point.
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