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 or unstable. $$ d y / d t=e^{-y}-1, \quad-\infty

Short answer

The equilibrium point for the differential equation is \(y = 0\), and it is asymptotically stable.

Step-by-step solution

Calculate critical points

To find the critical points, set \(dy/dt\) equal to 0: $$ 0 = e^{-y} - 1 $$ Solve for \(y\): $$ 1 = e^{-y} $$ Take the natural logarithm of both sides: $$ ln(1) = -y $$ Since \(ln(1) = 0\): $$ y = 0 $$ So the critical point is y = 0.

Determine stability of critical points

To determine the stability of the critical point, we will consider the sign of \(f(y)\) around the critical point. For \(y>0\): $$ f(y) = e^{-y} - 1 < 0 $$ Since \(e^{-y}\) is always positive and less than 1 for positive \(y\), \(f(y) < 0\) for \(y>0\). This means the function \(f(y)\) is decreasing for \(y > 0\). For \(y<0\): $$ f(y) = e^{-y} - 1 > 0 $$ Since \(e^{-y}\) is always positive and greater than 1 for negative \(y\), \(f(y) > 0\) for \(y < 0\). This means the function \(f(y)\) is increasing for \(y < 0\).

Classify the critical point

Since \(f(y)\) is increasing for \(y<0\) and decreasing for \(y>0\), the critical point at \(y=0\) is asymptotically stable. When \(y\) is smaller than the equilibrium point, the function \(f(y)\) is positive, and \(y\) increases towards the equilibrium. When \(y\) is larger than the equilibrium point, the function \(f(y)\) is negative, and \(y\) decreases towards the equilibrium. The equilibrium point for the given differential equation is: $$ y = 0, $$ which is asymptotically stable.

Key concepts

Critical Points

When studying differential equations, understanding the role of critical points (or equilibrium points) is pivotal. These points represent stationary states of the system where the change with time, or rate of the system's function, is zero. In mathematical terms, a critical point is found where the derivative of the function equals zero, given by the equation \(\frac{dy}{dt} = 0\).
In the provided exercise example, to find the critical point for the equation \(dy/dt = e^{-y} - 1\), the student is instructed to set this equal to zero and solve for \(y\). This process yields \(y = 0\) as the critical point.

Asymptotically Stable Equilibrium

A concept often paired with critical points is the notion of stability. Specifically, when a system's equilibrium point is tagged as 'asymptotically stable,' any small deviation from this point will result in the system eventually returning to its equilibrium state over time.
Understanding how to classify these points is crucial for predicting the long-term behavior of dynamic systems. To determine stability, observe the signs of the function around the critical point. If, for any initial point near the equilibrium, the system moves back toward it, then the equilibrium is asymptotically stable. The exercise presents such an analysis, showcasing that the critical point at \(y = 0\) is indeed asymptotically stable because function \(f(y)\) is negative when \(y > 0\) and positive when \(y < 0\), indicating that \(y\) will adjust back to zero over time.

Sketching Graphs of Functions

Sketching the graph of a function involved in a differential equation can be an insightful visual aid that assists with understanding the system's behavior. It reflects how the system evolves for different values of \(y\) over time. A graph illustrates the trends, such as increasing or decreasing behavior of the function, and clarifies points of equilibrium and their stability.
In the practice of sketching a graph for the given differential equation \(f(y) = e^{-y} - 1\), the student can plot points above and below the critical point to visualize how the function behaves. For example, values of \(y\) greater and smaller than 0 would demonstrate the decreasing and increasing nature of \(f(y)\), respectively. This graphical analysis seamlessly ties into how one can classify the equilibrium point via a visual medium, making the concept more tangible.