For this linear system:

$\frac{dx}{dt}}=x+y-2$

$\frac{dy}{dt}}=x-y-4$

Classify $(1,1)$

$\frac{dx}{dt}}=x+y-2$

$\frac{dy}{dt}}=x-y-4$

Classify $(1,1)$

Trystan Castaneda
2022-08-13
Answered

For this linear system:

$\frac{dx}{dt}}=x+y-2$

$\frac{dy}{dt}}=x-y-4$

Classify $(1,1)$

$\frac{dx}{dt}}=x+y-2$

$\frac{dy}{dt}}=x-y-4$

Classify $(1,1)$

You can still ask an expert for help

cydostwng6c

Answered 2022-08-14
Author has **13** answers

$\frac{dy}{dx}={\displaystyle \frac{\frac{dy}{dt}}{\frac{dx}{dt}}}={\displaystyle \frac{x-y-4}{x+y-2}}$

At $(1,1),\phantom{\rule{thickmathspace}{0ex}}{\displaystyle \frac{dy}{dx}}\ne 0$,$(1,1),\phantom{\rule{thickmathspace}{0ex}}{\displaystyle \frac{dy}{dx}}\ne 0$. Indeed, at $(1,1)$, $\frac{dy}{dx}$ 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 $(3,-1)$, and is a saddle point.

At $(1,1),\phantom{\rule{thickmathspace}{0ex}}{\displaystyle \frac{dy}{dx}}\ne 0$,$(1,1),\phantom{\rule{thickmathspace}{0ex}}{\displaystyle \frac{dy}{dx}}\ne 0$. Indeed, at $(1,1)$, $\frac{dy}{dx}$ 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 $(3,-1)$, and is a saddle point.

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