Question

In Exercises \(1-4\) find the equations of the trajectories of the given undamped equation. Identify the equilibrium solutions, determine whether they are stable or unstable, and plot some trajectories. $$ \mathrm{C} / \mathrm{G} \quad y^{\prime \prime}+y e^{-y}=0 $$

Short answer

**Answer:** The stability of the equilibrium solution at \(y = 0\) is stable because the derivative of the second equation with respect to \(y\) at the equilibrium point is negative, indicating that the trajectories will approach the equilibrium solution as time goes to infinity.

Step-by-step solution

Write down the given undamped equation

First, let's write down the given undamped equation: $$ y^{\prime\prime} + ye^{-y} = 0 $$

Transform into a first-order differential equation system

Let \(v = y'\) be the velocity of the system. Then the second-order differential equation can be transformed into the following first-order system: $$ \begin{cases} y' = v \\ v' = -ye^{-y} \end{cases} $$

Identify equilibrium solutions

To find the equilibrium solutions, we need to solve the system for \(y'\) and \(v'\) equal to 0: $$ \begin{cases} 0 = v \\ 0 = -ye^{-y} \end{cases} $$ From the first equation, we find \(v = 0\). Plugging this into the second equation, we find that the equilibrium solution occurs at \(y = 0\).

Determine the stability of the equilibrium solutions

To determine the stability of the equilibrium solutions, we'll analyze the signs of the derivative of the second equation with respect to \(y\). $$ \frac{d}{dy}(-ye^{-y}) = -e^{-y}+ye^{-y} $$ At the equilibrium point \(y = 0\), the derivative becomes: $$ -e^{0}+0\cdot e^{0} = -1 $$ Since the derivative at the equilibrium point is negative, the equilibrium solution at \(y = 0\) is stable.

Plot some trajectories

To plot trajectories, we need to analyze the phase plane of the system. It's difficult to find explicit trajectories analytically due to the complexity of the system. However, we could plot some trajectories using numerical methods or specialized software like Desmos or Geogebra. The system with equilibrium solution at \(y = 0\) and its stability will be represented by the phase portrait. The phase portrait will show the behavior of the system, and from the analysis, we know that the trajectories will approach the equilibrium solution as time goes to infinity. In conclusion, the undamped equation is given by the first-order differential equation system: $$ \begin{cases} y' = v \\ v' = -ye^{-y} \end{cases} $$ The equilibrium solution occurs at \(y = 0\) and is stable. To visualize the trajectories, you can use numerical methods or specialized software to plot the phase portrait.

Key concepts

Equilibrium Solutions

Exploring the equilibrium solutions of a differential equation is akin to finding the 'resting points' of a system. In the context of our undamped equation, the equilibrium solutions manifest where the system's change ceases, meaning both the velocity (\f\(v\f\)) and the acceleration (\f\(y''\f\)) are zero. Beginning with the equation \f\(y'' + ye^{-y} = 0\f\), we set up a system of first-order differential equations, translating the equation into a more analyzable form:\f[\begin{cases}y' = v \ v' = -ye^{-y}\f]Identifying the equilibrium solutions involves solving this system when \f\(y' = 0\f\) and \f\(v' = 0\f\) concurrently. In our case, this occurs when \f\(v = 0\f\) and \f\(y\f\) satisfies the equation \f\(-ye^{-y} = 0\f\). The latter indicates the solution \f\(y = 0\f\) fulfills the condition, signifying the occurrence of an equilibrium solution at this value.

Stability of Equilibrium

Determining whether an equilibrium solution is a haven of stability or a precipice of instability is imperative in understanding the system's long-term behavior. To deduce this characteristic, we examine the sign of the derivative of the acceleration expression with respect to \f\(y\f\). Thus, for our unamped equation's equilibrium at \f\(y = 0\f\), we delve into the derivative:\f[\frac{d}{dy}(-ye^{-y}) = -e^{-y} + ye^{-y}\f]At \f\(y = 0\f\), this becomes \f\(-e^0 = -1\f\), a negative value indicating a restoration force towards equilibrium—an emblem of stability. This negative derivative implies perturbations from equilibrium would attenuate over time, reassuring us that the equilibrium solution \f\(y = 0\f\) is indeed stable and will attract trajectories towards itself.

Phase Plane Analysis

The phase plane is the stage on which the system's dynamics are choreographed. Delving into phase plane analysis, we not only predict but also visualize the motion of the system over time. The phase plane, composed of the state variables \f\(y\f\) and \f\(v\f\), is where all possible states of the system are illustrated.The phase portrait—deriving from this analysis—captures all trajectories including their directions and stability. Unfortunately, the trajectories of our undamped equation \f\(y'' + ye^{-y} = 0\f\) are not straightforward to depict analytically due to the nonlinearity introduced by the \f\(e^{-y}\f\) term. Numerical methods or graphing software can serve as our allies here, offering a visual feast of how trajectories eventually convene or diverge around the equilibrium points identified. By scrutinizing the phase portrait, we validate the stability analysis, with trajectories spiraling towards our stable equilibrium at \f\(y = 0\f\) as time progresses.