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Suppose $y_1(t)$, $y_2(t)$ and $y_3(t)$ are all solutions of the differential equation with different initial conditions as indicated below:

$y_1(t)$ solves the differential equation with initial condition $y(0) = −2$.

$y_2(t)$ solves the differential equation with initial condition $y(0) = 1.5$.

$y_3(t)$ solves the differential equation with initial condition $y(0) = −2.1$.

I am asked to find the limits of $y_1(t)$, $y_2(t)$ and $y_3(t)$ as $t$ approaches infinity.

I already have equilibrium solutions $y=-2, 0, 2$ where $y=0$ is stable and the rest are unstable. I am having trouble with the limit of $y_1(t)$ since the initial condition is equal to its equilibrium solution. Does the limit exist for $y_1(t)$? I also got $0$ and $-\infty$ as limits for $y_2(t)$ and $y_3(t)$ respectively. Am I doing it right?

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  • $\begingroup$ You are right for $y_2(t)$ and $y_3(t)$. When $y(0)$ is an equilibrium point, then $y(t)$ will be constant at the $y(0)$ value. $\endgroup$ – Nicholas Nov 15 '15 at 7:32
  • $\begingroup$ By constant, do you mean the limit of $y_1(t)$ as $t$ approaches infinity is just -2? $\endgroup$ – marg_ocruz Nov 15 '15 at 7:35
  • $\begingroup$ The equilibrium point gives the value of $y$ such that $\frac{dy}{dx}=0$. In other words, if $y(0)=-2$, then $y(t)=-2$ regardless of $t$. $\endgroup$ – Nicholas Nov 15 '15 at 7:37
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    $\begingroup$ Oh I see now. Thank you! $\endgroup$ – marg_ocruz Nov 15 '15 at 7:38

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