91Ó°ÊÓ

To find the equilibrium solution, we need to set \(y^{\prime}(t)\) to 0 and solve for \(y\). The given equation is: $$y^{\prime}(t)=12 y-18$$ Setting \(y^{\prime}(t)=0\), we get: $$12 y-18=0$$ Now, solve for \(y\): $$12 y=18 \Rightarrow y=\frac{18}{12}=\frac{3}{2}$$ Thus, the equilibrium solution is \(y=\frac{3}{2}\).

To sketch the direction field, we need to determine the behavior of the solutions on either side of the equilibrium solution (\(y=\frac{3}{2}\)). We only need to indicate if solutions are increasing or decreasing. For \(y<\frac{3}{2}\), let's choose \(y=1\) as an example: $$y^{\prime}(t)=12(1)-18=-6$$ Since \(y^{\prime}(t)=-6<0\), the solution at \(y=1\) is decreasing, and we can conclude that solutions are decreasing for all \(y<\frac{3}{2}\). For \(y>\frac{3}{2}\), let's choose \(y=2\) as an example: $$y^{\prime}(t)=12(2)-18=6$$ Since \(y^{\prime}(t)=6>0\), the solution at \(y=2\) is increasing, and we can conclude that solutions are increasing for all \(y>\frac{3}{2}\). Now, we can sketch the direction field: downward arrows for \(y<\frac{3}{2}\) and upward arrows for \(y>\frac{3}{2}\).

We need to determine if \(y=\frac{3}{2}\) is a stable equilibrium solution. We can deduce this from the direction field: - For \(y<\frac{3}{2}\), the solutions are decreasing, which means they will approach the equilibrium solution \(y=\frac{3}{2}\), as \(t\) increases. - For \(y>\frac{3}{2}\), the solutions are increasing, which means they will move away from the equilibrium solution, as \(t\) increases. Hence, the equilibrium solution \(y=\frac{3}{2}\) is unstable.