What is the equilibrium solution of the equation y'(t) = 3y − 9? Is it stable or unstable?

Verified step by step guidance

1. Identify the equilibrium solution by setting the derivative equal to zero: solve the equation \$0 = 3y - 9\( for \)y$.
2. Rearrange the equation to isolate \(y\): \$3y = 9\(, then \)y = \frac{9}{3}$.
3. Determine the equilibrium solution value from the above step.
4. To analyze stability, consider the sign of the coefficient in the differential equation \(y'(t) = 3y - 9\) when rewritten as \(y'(t) = 3(y - 3)\).
5. Since the coefficient 3 is positive, conclude whether the equilibrium solution is stable or unstable based on whether perturbations grow or decay over time.

Key Concepts

Equilibrium Solution of a Differential Equation

An equilibrium solution occurs when the derivative y'(t) equals zero, meaning the function y(t) does not change over time. For the equation y'(t) = 3y − 9, setting y'(t) = 0 allows us to find constant solutions where the system is at rest.

Stability of Equilibrium Solutions

Stability refers to whether solutions near the equilibrium tend to move towards it (stable) or away from it (unstable) over time. This is often determined by analyzing the sign of the derivative of the right-hand side with respect to y at the equilibrium point.

Linear First-Order Differential Equations

The given equation y'(t) = 3y − 9 is a linear first-order differential equation. Understanding its structure helps in solving it and analyzing behavior near equilibrium points by using methods like separation of variables or integrating factors.