8. Consider the differential equation \[ \frac{d y}{d t}=f(y) \text { where } f(2)=0, f^{\prime}(2)=0, f^{\prime \prime}(2)<0 \] Which type of equilibrium point is \( y=2 \) ? (a) sink (b) source (c) node

To classify the equilibrium point \( y = 2 \) for the differential equation \[ \frac{dy}{dt} = f(y), \] we analyze the behavior of \( f(y) \) near \( y = 2 \). ### Given: 1. \( f(2) = 0 \): This means \( y = 2 \) is an equilibrium point. 2. \( f'(2) = 0 \): The first derivative at the equilibrium is zero, indicating that linearization (using only the first derivative) is inconclusive for determining stability. 3. \( f''(2) < 0 \): The second derivative is negative, providing more insight into the behavior near \( y = 2 \). ### Analysis: Let's consider a small perturbation around the equilibrium point by setting \( y = 2 + x \), where \( x \) is small. Expanding \( f(y) \) around \( y = 2 \) using a Taylor series: \[ f(2 + x) \approx f(2) + f'(2)x + \frac{1}{2}f''(2)x^2 = \frac{1}{2}f''(2)x^2 \] Since \( f''(2) < 0 \), we have: \[ \frac{dy}{dt} \approx -c x^2 \] where \( c > 0 \). #### Behavior: - **For \( y > 2 \) (\( x > 0 \))**: \( \frac{dy}{dt} < 0 \), so \( y \) decreases toward 2. - **For \( y < 2 \) (\( x < 0 \))**: \( \frac{dy}{dt} < 0 \), so \( y \) continues to decrease away from 2. This indicates that \( y = 2 \) is **attracting** from one side (\( y > 2 \)) and **repelling** from the other side (\( y < 2 \)). Such an equilibrium is termed **semi-stable**. ### Conclusion: Among the given options: - **Sink** typically refers to an equilibrium that is attracting on all sides. - **Source** refers to an equilibrium that is repelling on all sides. - **Node** is generally used in higher-dimensional systems. However, since the equilibrium at \( y = 2 \) is attracting from one side and repelling from the other (semi-stable), the best available choice is: **Answer:** (a) sink