4.2 Direction fields and numerical methods  (Page 3/14)

To determine the equilibrium solutions to the differential equation $y\prime =f(x,y),$ set the right-hand side equal to zero. An equilibrium solution of the differential equation is any function of the form $y=k$ such that $f\left(x,k\right)=0$ for all values of $x$ in the domain of $f.$

An important characteristic of equilibrium solutions concerns whether or not they approach the line $y=k$ as an asymptote for large values of $x.$

Now we return to the differential equation $y\prime =(x-3)(y^2-4),$ with the initial condition $y(0)=0.5.$ The direction field for this initial-value problem, along with the corresponding solution, is shown in [link] .

The values of the solution to this initial-value problem stay between $y=\mathrm{-2}$ and $y=2,$ which are the equilibrium solutions to the differential equation. Furthermore, as $x$ approaches infinity, $y$ approaches $2.$ The behavior of solutions is similar if the initial value is higher than $2,$ for example, $y\left(0\right)=2.3.$ In this case, the solutions decrease and approach $y=2$ as $x$ approaches infinity. Therefore $y=2$ is an asymptotically stable solution to the differential equation.

What happens when the initial value is below $y=\mathrm{-2}?$ This scenario is illustrated in [link] , with the initial value $y(0)=\mathrm{-3}.$

The solution decreases rapidly toward negative infinity as $x$ approaches infinity. Furthermore, if the initial value is slightly higher than $\mathrm{-2},$ then the solution approaches $2,$ which is the other equilibrium solution. Therefore in neither case does the solution approach $y=\mathrm{-2},$ so $y=\mathrm{-2}$ is called an asymptotically unstable, or unstable, equilibrium solution.

Stability of an equilibrium solution

Create a direction field for the differential equation $y\prime ={(y-3)}^2(y^2+y-2)$ and identify any equilibrium solutions. Classify each of the equilibrium solutions as stable, unstable, or semi-stable.

The direction field is shown in [link] .

The equilibrium solutions are $y=\mathrm{-2},y=1,$ and $y=3.$ To classify each of the solutions, look at an arrow directly above or below each of these values. For example, at $y=\mathrm{-2}$ the arrows directly below this solution point up, and the arrows directly above the solution point down. Therefore all initial conditions close to $y=\mathrm{-2}$ approach $y=\mathrm{-2},$ and the solution is stable. For the solution $y=1,$ all initial conditions above and below $y=1$ are repelled (pushed away) from $y=1,$ so this solution is unstable. The solution $y=3$ is semi-stable, because for initial conditions slightly greater than $3,$ the solution approaches infinity, and for initial conditions slightly less than $3,$ the solution approaches $y=1.$

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