Determine the critical points of the autonomous differential equation \( \frac{d y}{d x}=\left(\frac{1}{\pi} y+1\right)-\cos \frac{y}{2} \) \( -\frac{\pi}{2}, 0, \frac{\pi}{2} \) \( -\pi, 0, \pi \) \( 0,-\pi \) \( 0, \pi \) \( -\frac{\pi}{2}, \frac{\pi}{2} \)

To determine the critical points of the given autonomous differential equation \( \frac{d y}{d x}=\left(\frac{1}{\pi} y+1\right)-\cos \frac{y}{2} \), we need to find the values of \( y \) such that \( \frac{d y}{d x} = 0 \). Setting the equation to zero, we have: \[ \left(\frac{1}{\pi} y+1\right) - \cos \frac{y}{2} = 0 \] Multiply both sides by \( \pi \): \[ y + \pi = \pi \cos \frac{y}{2} \] Rearranging gives us: \[ y - \pi \cos \frac{y}{2} + \pi = 0 \] This equation can be challenging to solve analytically, and typically numerical methods or graphical methods might be used here. However, we can explore approximate values. Substituting potential values for \( y \) from the given options: 1. For \( y = 0 \): \[ 0 - \pi \cos(0) + \pi = 0 - \pi + \pi = 0 \] So, \( y = 0 \) is a critical point. 2. For \( y = \pi \): \[ \pi - \pi \cos \frac{\pi}{2} + \pi = \pi - 0 + \pi = 2\pi \neq 0 \] Not a critical point. 3. For \( y = -\pi \): \[ -\pi - \pi \cos \left(-\frac{\pi}{2}\right) + \pi = -\pi - 0 + \pi = 0 \] So, \( y = -\pi \) is also a critical point. 4. For \( y = \pm \frac{\pi}{2} \): \[ \frac{\pi}{2} - \pi \cos(\frac{\pi}{4}) + \pi = \frac{\pi}{2} - \frac{\pi}{\sqrt{2}} + \pi \neq 0 \] Similarly for \( -\frac{\pi}{2} \). Hence, the critical points determined from the options are: - \( y = 0 \) - \( y = -\pi \) The correct critical points are \( 0, -\pi \).