Question

Find the equilibrium solution of the following equations, make a sketch of the direction field, for \(t \geq 0,\) and determine whether the equilibrium solution is stable. The direction field needs to indicate only whether solutions are increasing or decreasing on either side of the equilibrium solution. $$y^{\prime}(t)=-6 y+12$$

Short answer

Answer: The equilibrium solution is y = 2. It is stable as solutions on either side of the equilibrium solution approach y = 2; the solution is decreasing when y > 2 and increasing when y < 2.

Step-by-step solution

Find the equilibrium solution

To find the equilibrium solution(s), we set the first derivative of y(t), \(y^{\prime}(t)\) to zero and solve for y. $$0=-6y + 12$$ Solve for y: $$6y = 12$$ $$y = 2$$ The equilibrium solution is y = 2.

Analyze stability

To analyze the stability of the equilibrium solution, we can look at the expression for the derivative of y(t), namely: $$y^{\prime}(t) = -6y + 12$$ Let's consider the equilibrium solution y = 2: 1. If y > 2, then \(y^{\prime}(t) = -6y + 12 < 0\), which means the solution is decreasing. 2. If y < 2, then \(y^{\prime}(t) = -6y + 12 > 0\), which means the solution is increasing. Since solutions on either side of the equilibrium solution are approaching y = 2, this equilibrium solution is stable.

Sketch the direction field

We need to sketch the direction field by indicating whether solutions are increasing or decreasing on either side of the equilibrium solution. 1. If y > 2, the derivative is negative, which means the solution is decreasing. We can draw arrows pointing downwards in the region above y = 2. 2. If y < 2, the derivative is positive, which means the solution is increasing. We can draw arrows pointing upwards in the region below y = 2. 3. At the equilibrium solution y = 2, the derivative is 0, which indicates arrows pointing horizontally to the right. Remember, this direction field only needs to show the behavior of the solutions around the equilibrium solution. If you sketch the direction field as described, you will see that solutions in the region above the equilibrium solution decrease, while solutions in the region below the equilibrium solution increase. This confirms our previous analysis that the equilibrium solution y = 2 is stable.