91Ó°ÊÓ

In the study of differential equations, equilibrium values represent points where the system remains unchanged over time. For our exercise, we start by setting the derivative equation to zero: \( y' = \sqrt{y} = 0 \). This calculation helps us find the equilibrium value. By solving this equation, we discover that the equilibrium value is \( y = 0 \). It's crucial to recognize that this equilibrium is actually a boundary case due to the constraint \( y > 0 \). Therefore, we must consider it carefully when analyzing the system's behavior. Equilibrium points like \( y = 0 \) represent where the system might potentially settle, although, in this circumstance, it is considered more of a threshold or beginning point, since any positive \( y \) means the system moves away from this value.

Stability analysis involves determining whether small deviations from an equilibrium point will decay or amplify over time. For the given differential equation \( y' = \sqrt{y} \), we can assess stability by observing the behavior of \( \sqrt{y} \) as \( y \) approaches 0. Given that \( \sqrt{y} \) is always positive for \( y > 0 \), any initial positive value of \( y \) will lead to \( y \) increasing. This behavior indicates instability at \( y = 0 \), because the system moves away when slightly perturbed. In layman's terms, if you start even a tiny bit away from zero, \( y \) will grow, confirming that \( y = 0 \) is indeed an unstable point.

A phase line offers a simple method to visually represent the behavior of a differential equation's solutions across different values of \( y \). To create a phase line, mark the equilibrium point \( y = 0 \) on a number line. Then, analyze the sign of the derivative \( y' \) for regions around this point. For our example, since \( y' = \sqrt{y} > 0 \) for \( y > 0 \), we place arrows pointing upwards along the phase line to the right of zero, indicating that solutions are increasing. This graphic representation highlights that as soon as \( y \) is positive, the solution trends upward along the phase line, visually confirming the instability at the equilibrium point.

Solution curves graphically represent how solutions of a differential equation change over time for various initial conditions. To sketch them for the equation \( y' = \sqrt{y} \), start by considering initial conditions where \( y(0) = c > 0 \). Each initial condition corresponds to a different solution curve that starts at \( c \) and moves upwards. Since we've already established that both \( y' \) and \( y'' \) are positive when \( y > 0 \), we know that the solution curves not only rise but curve upward as well. This concave upward trajectory reflects the fact that the rate of increase accelerates. These curves depict how the solutions spread out from the initial condition and keep growing indefinitely, moving further away from the degenerate equilibrium at \( y = 0 \). Remember, each curve represents the "path" through which the system evolves with time, based on its starting point.