Question

Involve equations of the form \(d y / d t=f(y) .\) In each problem sketch the graph of \(f(y)\) versus \(y\), determine the critical (equilibrium) points, and classify each one as asymptotically stable, unstable, or semistable (see Problem 7 ). $$ d y / d t=a y-b \sqrt{y}, \quad a>0, \quad b>0, \quad y_{0} \geq 0 $$

Short answer

Question: Classify the equilibrium points of the differential equation \(dy/dt = ay - b\sqrt{y}\) and sketch the graph of \(f(y)\) versus \(y\). Answer: The equilibrium points are \(y_1=0\) (unstable) and \(y_2=\frac{b^2}{a^2}\) (semistable). The graph of \(f(y)\) has a hill-like shape with its peak at the semistable equilibrium point \(y=\frac{b^2}{a^2}\).

Step-by-step solution

Identify the differential equation

Here, we have a first-order differential equation given by: $$ \frac{dy}{dt} = f(y) = ay - b\sqrt{y} $$ where \(a > 0\) and \(b > 0\).

Find the equilibrium points

To finding the equilibrium points, we need to solve for \(y\) when \(\frac{dy}{dt} = 0\). In other words, $$ 0 = ay - b\sqrt{y} $$

Solve for equilibrium points (\(y_1\) and \(y_2\))

Rearrange the equation to solve for \(y\): $$ b\sqrt{y} = ay $$ Divide by \(y\) to isolate \(\sqrt{y}\) on one side: $$ \sqrt{y} = \frac{a}{b}y $$ Square both sides of the equation: $$ y=\frac{a^{2}}{b^{2}}y^{2} $$ Factor out \(y\): $$ y\left(1-\frac{a^{2}}{b^{2}}y\right)=0 $$ This gives us two equilibrium points: \(y_1=0\) and \(y_2=\frac{b^2}{a^2}\).

Determine the stability of each equilibrium point by analyzing the first derivative of \(f(y)\)

Calculate the first derivative of \(f(y)\) with respect to \(y\): $$ \frac{d}{dy}(ay - b \sqrt{y}) = a - \frac{b}{2\sqrt{y}} $$ Now, let's analyze the stability of equilibrium points. - For \(y_1=0\), the first derivative is: $$ \frac{d}{dy}f(y) \Big|_{y=0} = a - \frac{b}{2\sqrt{0}} = a $$ Since \(a > 0\), the first derivative is positive, and therefore \(y_1=0\) is unstable. - For \(y_2=\frac{b^2}{a^2}\), the first derivative is: $$ \frac{d}{dy}f(y) \Big|_{y=\frac{b^2}{a^2}} = a - \frac{b}{2\sqrt{\frac{b^2}{a^2}}} = 0 $$ The first derivative is zero, and we cannot determine the stability from it. Thus, \(y_2=\frac{b^2}{a^2}\) is semistable.

Sketch the graph of \(f(y)\) versus \(y\)

To sketch the graph of \(f(y)\) versus \(y\), first note the following properties of \(f(y)\): - The equilibrium points are \(y_1=0\) (unstable) and \(y_2=\frac{b^2}{a^2}\) (semistable). - \(f(y)\) is increasing for \(0\frac{b^2}{a^2}\). - As \(y\) approaches infinity, \(f(y)\) approaches negative infinity. Based on these properties, the graph of \(f(y)\) will have an unstable equilibrium point at \(y=0\) and a semistable equilibrium point at \(y=\frac{b^2}{a^2}\). The graph starts at positive infinity as \(y\) approaches \(0\) from the right, and as \(y\) approaches infinity, the graph approaches negative infinity. The graph will have a hill-like shape, with its peak at the semistable equilibrium point \(y=\frac{b^2}{a^2}\).

Key concepts

Equilibrium Points

Understanding equilibrium points is crucial for analyzing differential equations. Equilibrium points, also known as critical points, occur where the rate of change, represented by the derivative, equals zero. This means there is no net change at these points, and the system is in a state of balance.

To find the equilibrium points of a differential equation like \( \frac{dy}{dt} = ay - b\sqrt{y} \), we set \( \frac{dy}{dt} = 0 \) and solve for \( y \). Doing this yields the equilibrium points \( y_1=0 \) and \( y_2=\frac{b^2}{a^2} \), which are the values of \( y \) where the function's rate of change ceases.

Identifying these points is the first step in understanding the system's behavior. They can indicate positions of rest, constant solutions, or potential transitions in the system’s dynamics. For students struggling with finding equilibrium points, remember to look for where the change over time becomes zero, which can often be found by setting the right-hand side of the differential equation equal to zero and solving.

Stability Analysis

Stability analysis concerns itself with determining whether small disturbances to equilibrium points diminish or amplify over time. It informs us whether an equilibrium point is stable—meaning the system will return to equilibrium after a disturbance—or unstable—whereby the system will depart further from the equilibrium.

For the given equation, we assess stability by taking the derivative of \( f(y) \) and evaluating it at each equilibrium point. An equilibrium point where the derivative is negative is asymptotically stable since perturbations decay over time, leading the system back to equilibrium. Conversely, if the derivative is positive, the point is unstable as perturbations grow, pushing the system away from equilibrium.

Using \( y_1=0 \) and \( y_2=\frac{b^2}{a^2} \) for the equation \( \frac{dy}{dt} = ay - b\sqrt{y} \), we find \( y_1=0 \) to be unstable due to a positive derivative, and \( y_2=\frac{b^2}{a^2} \) to be semistable due to a zero derivative, meaning behavior around this point needs further analysis. For students, remember to closely examine the sign of the derivative at equilibrium points when conducting stability analysis.

Differential Equation Graphing

Graphing differential equations like \( \frac{dy}{dt} = ay - b\sqrt{y} \) provides a visual interpretation of the system's dynamics. This graph plots the rate of change of \( y \) with respect to \( y \) itself, revealing the function's behavior across different values.

Key components of the graph are the equilibrium points: \( y_1=0 \) and \( y_2=\frac{b^2}{a^2} \). At these points, the graph will show no slope. Between these points, the graph will rise, indicating positive growth, and it declines as \( y \) moves away from \( y_2 \). As \( y \) approaches infinity, the graph slopes downward, showing a negative change rate.

Visual learners might find it helpful to draw the graph to understand the system better. The graphical approach can complement analytical methods by providing insights into the overall structure and predicting system behavior under different conditions. Always plot the equilibrium points, note the increasing and decreasing intervals, and observe the end behavior for comprehensive understanding.