Equilibrium Solutions to Differential Equations

Suppose that we have a differential equation $\frac{dy}{dt} = f(t, y)$. Sometimes it is easy to find some solutions immediately just by investigating the differential equation.

For example, consider the differential equation $\frac{dy}{dt} = 2y^2 + y$. Once such obvious solution to this differential equation is $y = 0$, since if $y = 0$ then the differential equation above reduces to $\frac{d}{dt} (0) = 0$ which is true. Furthermore, if we factor the righthand side of the differential equation above, we obtain that:

(1)

\begin{align} \quad \frac{dy}{dt} = 2y \left ( y + \frac{1}{2} \right ) \end{align}

Hence $y = -\frac{1}{2}$ is also a solution as you should verify. These solutions are special are we will give them a name in the definition below.

Definition: If $\frac{dy}{dt} = f(t, y)$ is a differential equation, then the Equilibrium Solutions are obtained by setting $\frac{dy}{dt} = 0$ and solving for $y$.

Let's look at some examples.

Example 1

Find the equilibrium solutions of the differential equation $\frac{dy}{dt} = y^2 - 9$.

The differential equation above can be rewritten as $\frac{dy}{dt} = (y + 3)(y - 3)$. Therefore $\frac{dy}{dt} = 0$ if $y = -3$ or $y = 3$ which are the equilibrium solutions.

Example 2

Find the equilibrium solutions of the differential equation $\frac{dy}{dt} = yt^2 - yt + y^2$.

The differential equation above can be rewritten as $\frac{dy}{dt} = y(t^2 - t + y)$. Therefore $y = 0$ is one of the equilibrium solutions. Another equilibrium solution is obtained when $t^2 - t + y = 0$, that is $y = t - t^2$.