91Ó°ÊÓ

In a differential equation, an **equilibrium point** represents a value where the system is at rest, meaning the rate of change is zero. For the equation \( \frac{d y}{d x} = y(2-y)(y-3) \), we determine equilibrium by setting \( \frac{d y}{d x} = 0 \). This involves solving \( y(2-y)(y-3) = 0 \), which breaks down into simpler equations:

• \( y = 0 \)
• \( 2-y = 0 \rightarrow y = 2 \)
• \( y-3 = 0 \rightarrow y = 3 \)

These solutions \( y = 0, 2, 3 \) are the equilibrium points, where the function is not changing, thus providing stability analysis indicators later on.

**Stability analysis** checks if small changes in the system revert back or diverge further from equilibrium points. Once the equilibria for \( \frac{d y}{d x} = y(2-y)(y-3) \) are known as \( y = 0, 2, 3 \), we explore how the function behaves around these points.Near \( y = 0 \), the function switches from negative \( - \) to positive \( + \), suggesting stability. If disturbances occur, the system tends to return to equilibrium. For \( y = 2 \), the function transitions from positive \( + \) to negative \( - \), indicating instability, as disturbances drive the system away. Lastly, at \( y = 3 \), it again shifts from negative \( - \) to positive \( + \), reinforcing stability.

A **graphical representation** illuminates the nature of equilibrium points by visually depicting \( \frac{d y}{d x} = y(2-y)(y-3) \) against \( y \). You'll observe the graph crossing the \( y \)-axis at the equilibria: \( y = 0, 2, 3 \).- At \( y = 0 \), the graph's curve slides from below to above the axis.- At \( y = 2 \), it dips from above to below.- At \( y = 3 \), it resurfaces from below to above.
These graphical crossovers convey stability: equilibrium is stable if the curve reverts, and unstable if it departs.

**Eigenvalues** in differential equations provide more nuanced stability information for equilibrium points. Using the derivative of our function, known as the Jacobian, evaluated at each equilibrium yields these values.- For \( y = 0 \), \( f'(y) = 6y - 5y^2 \) evaluates to zero. A zero eigenvalue suggests additional checks, like higher-order derivatives, are necessary to determine stability.- At \( y = 2 \), \( f'(2) = -8 \) is negative, suggesting strong instability.- At \( y = 3 \), \( f'(3) = -27 \) is also negative, indicating stability.
By analyzing eigenvalues, we confirm our graphical findings and predict how small disturbances will affect the system's balance around these critical points.