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. $$y^{\prime}(t)=-\frac{y}{3}-1$$

Short answer

Answer: The equilibrium solution for the given first-order differential equation is \(y = -3\), and it is stable.

Step-by-step solution

Find the equilibrium solution

To find the equilibrium solution, we need to set the derivative equal to zero: $$y^{\prime}(t) = 0$$ Substitute the given equation: $$0 = -\frac{y}{3} - 1$$ Solve for \(y\): $$y = -3$$ The equilibrium solution is \(y = -3\).

Determine the stability of the equilibrium solution

To determine the stability, we will analyze the sign of the derivative. Analyzing the given equation, we can notice that when \(y > -3\), the derivative is negative, and when \(y < -3\), the derivative is positive. This implies that the solutions will approach the equilibrium solution as time goes on. Therefore, the equilibrium solution \(y = -3\) is stable.

Sketch the direction field

To sketch the direction field, we will evaluate some points around the equilibrium solution to find out whether the solutions are increasing or decreasing on either side of the equilibrium solution. - When \(y > -3\), the derivative is negative, meaning the solutions are decreasing. Therefore, the arrow will point down on the right side of the equilibrium solution in the direction field. - When \(y < -3\), the derivative is positive, meaning the solutions are increasing. Therefore, the arrow will point up on the left side of the equilibrium solution in the direction field. In summary, the equilibrium solution is \(y = -3\) and is stable, with solutions decreasing when \(y > -3\) and increasing when \(y < -3\).