Table of Contents
Differential Equations
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Topic PlaylistFirst order ordinary differential equations that are autonomous can have equilibria points where a constant value is a solution to the differential equation. The long term behaviour of solutions to the ODE can be determined by drawing a phase line and analysing the stability of the equilibrium points. Equilibrium points can be stable, unstable, or semistable.
Explanation VideoAn autonomous first order ODE is when $y'(t)$ is only dependent on $y(t)$ and $t$ does not explicitly appear in the ODE.
Examples of autonomous ODEs are,
\begin{equation} y'=y+3 \qquad y' = y^{2} \qquad y'=y(4-y) \qquad y'=f(y) \end{equation}An equilibrium point is a value of $y$ that makes $y' = 0$. That constant value of $y$ will be a solution of of the ODE. The value of $y$ will neither increase nor decrease, hence the name equilibrium point.
| ODE | Equilibria |
|---|---|
| $y'=y+3$ | $y=-3$ |
| $y' = y^{2}$ | $y=0$ |
| $y'=y(4-y)$ | $y=0,4$ |
| $y'=f(y)$ | All $y$ values where $f(y)=0$ |
Explanation VideoEquilibria of an autonomous first order ODE also have a stability type. An equilibrium point can be stable, unstable, or semistable.
In this example, the equilibrium point $y=3$ is unstable because all nearby solutions diverge away from $y=3$. The equilibrium point $y=0$ is semistable because all nearby solutions above $y=0$ converge to $y=0$ and all solution below $y=0$ diverge from $y=0$. The equilibrium point $y=-2$ is stable because all nearby solutions converge towards $y=-2$.
Another way to determine stability is to plot $y'=f(y)$ versus $y$.
Solution VideoFind the equilibria and determine the stability of the autonomous ODE.
\begin{equation} y' = 4-y \end{equation} Solution VideoFind the equilibria and determine the stability of the autonomous ODE.
\begin{equation} y' = (y-2)(y+5) \end{equation} Solution VideoFind the equilibria and determine the stability of the autonomous ODE.
\begin{equation} y' = (y+1)^{2}(y-6) \end{equation} Solution VideoFind the equilibria and determine the stability of the autonomous ODE.
\begin{equation} y' = (y+1)(y+2)(y+3) \end{equation} Solution VideoFind the equilibria and determine the stability of the autonomous ODE.
\begin{equation} y' = y^{3}(y-2)^{2}(y-3)^{3} \end{equation}