# Stable, Semi-Stable and Unstable Equilibrium Solutions Examples 1

One the Stable, Semi-Stable and Unstable Equilibrium Solutions page we classified equilibrium solutions to a differential equation as either (asymptotically) stable, semi-stable, or unstable equilibrium solutions depending on whether or not there exists solutions to the differential equation that converge or diverge to these equilibrium solutions.

We will now look at some examples of stable, semi-stable, and unstable equilibrium solutions to differential equations.

## Example 1

**Find and classify the equilibrium solutions to the differential equation $\frac{dy}{dt} = e^y – 1$.**

Setting $\frac{dy}{dt} = 0$ implies that $y = 0$. Therefore, $y = 0$ is our only equilibrium solution.

If $-\infty then $\frac{dy}{dt} and so solutions below $y = 0$ diverge from $y = 0$.

If $0 then $\frac{dy}{dt} > 0$ and so solutions above $y = 0$ diverge from $y = 0$.

Therefore $y = 0$ is an unstable equilibrium solution.

## Example 2

**Find and classify the equilibrium solutions to the differential equation $\frac{dy}{dt} = y^2(y^2 – 1)$.**

The equilibrium solutions can be obtained by setting $\frac{dy}{dt} = 0$, that is, $y^2(y^2 – 1) = 0$. Therefore, $y = 0$, $y = 1$, and $y = -1$ are all equilibrium solutions.

If $-\infty then $\frac{dy}{dt} > 0$ since $y^2 > 0$ and $y^2 – 1 > 0$, and solutions converge to the equilibrium solution $y = -1$.

If $-1 , then $\frac{dy}{dt} since $y^2 > 0$ and $y^2 – 1 , and solutions converge to the equilibrium solution $y = -1$.

If $0 , then $\frac{dy}{dt} since $y^2 > 0$ and $y^2 – 1 , and solutions converge to the equilibrium solution $y = 0$.

If $1 , then $\frac{dy}{dt} > 0$ since $y^2 > 0$ and $y^2 – 1 > 0$ and solutions do not converge to any equilibrium solution.

From the information above, we note that $y = -1$ is an asymptotically stable equilibrium solution, $y = 0$ is a semi-stable equilibrium solution, and $y = 1$ is an unstable equilibrium solution.