Problem 28 - Calculus: Early Transcendental Functions

Question

Find the equilibrium solutions and determine which are stable and which are unstable. $$y^{\prime}=e^{-y}-1$$

Step-by-Step Solution

Answer

The equilibrium solution is \(y = 0\), and it is stable as \(y'\) changes from positive to negative as \(y\) increases through the equilibrium point and the second derivative at equilibrium is negative.

Step 1: Finding Equilibrium Solutions

Set \(y^{\prime}=0\) and solve the resulting equation for \(y\).\n So, we get, \(0 = e^{-y} - 1 \Rightarrow e^{-y} = 1\). This equation can be solved by taking the natural logarithm of both sides, giving \(y = 0\).\n Therefore, the equilibrium solutions are \(y = 0\).

Step 2: Checking for Stability

Test nearby points in the derivative equation to check if they result in positive or negative values.\n For \(y > 0\), \(y' = e^{-y} - 1\) is negative because \(e^{-y} < 1\).\n For \(y < 0\), \(y' = e^{-y} - 1\) is positive because \(e^{-y} > 1\).\n Since \(y'\) changes value from positive to negative as \(y\) increases through the equilibrium solutions, the equilibrium at \(y = 0\) is stable.

Step 3: Stability Analysis using the Second Derivative

Let's find the second derivative. The derivative of \(y' = e^{-y} - 1\) with respect to \(y\) will be \(y'' = -e^{-y}\).\n Now, evaluate this at the equilibrium point \(y = 0\):\n \(y''(0) = -e^{-0} = -1\).\n Since the second derivative is negative, by the Second Derivative Test, this confirms that the equilibrium at \(y = 0\) is stable, because when the second derivative of a function is negative at a particular point, the function attains a maximum value there.

Key Concepts

Stability Analysis

Stability analysis helps us determine the behavior of solutions near equilibrium points of differential equations. It tells us if these solutions will stay close or move away when slightly disturbed.

The process begins by setting the derivative of the function to zero. This gives us the equilibrium solutions, where the system does not change.

• If a slight disturbance results in a small positive value for the derivative, the solution is moving away, indicating an unstable equilibrium.
• If the derivative becomes negative, the solution is drawn back toward the equilibrium, signaling a stable state.

For example, in our exercise, the equilibrium of the function when solved was found at 0. Checking values slightly above 0 resulted in a negative derivative, pulling it back to equilibrium. Conversely, values below the equilibrium showed a positive derivative, also guiding the solution back to domestic territory. This behavior suggests a stable equilibrium.

Differential Equations

Differential equations involve equations with derivatives and are fundamental in mathematical modeling. They describe how a quantity changes over time.

In our context, the equation given is \(y^{\prime} = e^{-y} - 1\). To find equilibrium, we set this equal to zero and solve for \(y\). This step is crucial as equilibrium solutions denote the point where changes momentarily halt.

The relationship between variables determines the configuration we solve for. Here, by isolating \(y\), we reveal how changes in \(y\) impact its rate. Differential equations can be linear or nonlinear, with our particular case being nonlinear because of the exponential term \(e^{-y}\). Each type has specific solution techniques, and understanding them is crucial for solving real-world problems involving rates of change.

Second Derivative Test

The Second Derivative Test is a powerful tool in calculus, helping to confirm stability by analyzing the concavity of the function at particular points.

For the equation given, the second derivative is derived as \(y'' = -e^{-y}\). At the equilibrium point \(y = 0\), we calculate \(y''(0) = -1\). A negative second derivative indicates that the function is concave down at that point.

• Concave down functions make stable equilibria, akin to being at the 'top' of a hill where small movements lead to returning back to equilibrium.
• If it were positive, indicating concavity upward, the point would be unstable, like being at the 'bottom' of a bowl where small shifts might take you away from the equilibria.

By confirming that there is indeed a maximum at \(y = 0\), we reinforce the decision that this equilibrium is stable. The test provides an essential layer of verification alongside observing directional tendencies in the first derivative.