Problem 26 - Calculus: Early Transcendental Functions

Question

Find the equilibrium solutions and determine which are stable and which are unstable. $$y^{\prime}=y^{3}-1$$

Step-by-Step Solution

Answer

The equilibrium solutions are y= -1, y= \(\frac{1}{2}+\frac{\sqrt{3}}{2}i\) and y= \(\frac{1}{2}-\frac{\sqrt{3}}{2}i\). The first solution y=-1 is unstable. The solutions y=\(\frac{1}{2}+\frac{\sqrt{3}}{2}i\) and y=\(\frac{1}{2}-\frac{\sqrt{3}}{2}i\) cannot be determined for stability since y' is not defined there in the real number line.

Step 1: Find Equilibrium Solutions

The equilibrium solutions can be found by setting \(y^{3}-1=0\). Therefore, by solving this equation, the equilibrium solutions are y= -1, y= \(\frac{1}{2}+\frac{\sqrt{3}}{2}i\) and y= \(\frac{1}{2}-\frac{\sqrt{3}}{2}i\).

Step 2: Determine Stability

Now determine the stability of these solutions. This can be done by looking at the sign of the derivative. However, in our case the derivative is \(y^{3}-1\) which is not defined for \(\frac{1}{2}+\frac{\sqrt{3}}{2}i\) and \(\frac{1}{2}-\frac{\sqrt{3}}{2}i\), so we ignore these two solutions for this step. When y is less than -1, the derivative is positive, indicating an unstable equilibrium. When y is more than -1, the derivative is also positive, which means the equilibrium point at y = -1 is also unstable.

Key Concepts

Equilibrium Solutions

Equilibrium solutions in differential equations play an important role in understanding the long-term behavior of dynamic systems. An equilibrium solution occurs when the rate of change of a variable, like velocity or population, becomes zero. Mathematically, we find them by setting the differential equation equal to zero and solving for the variable. In our example with the equation \( y^{\prime} = y^{3} - 1 \), we set \( y^{3} - 1 = 0 \) to find the equilibrium solutions. This simplifies to \( y^{3} = 1 \).

• Solving \( y^{3} = 1 \), we find the real equilibrium solution is \( y = 1 \).
• Besides, there are complex equilibrium solutions due to cubic roots.
• These are \( y = \frac{1}{2} + \frac{\sqrt{3}}{2}i \) and \( y = \frac{1}{2} - \frac{\sqrt{3}}{2}i \).

Stability Analysis

Once we have equilibrium solutions, we must determine their stability. Stability analysis helps us understand whether small disturbances will die out over time or amplify.In our example, the derivative of the equation informs us about stability. If a small change in the variable results in a small change in the derivative, it suggests stability.

• The derivative is \( y^{\prime} = y^{3} - 1 \), which depends on the value of \( y \).
• For \( y = -1 \), examine if the derivative sign changes around this point.
• Since the derivative is positive on both sides of \( y = -1 \), the equilibrium is unstable. This means disturbances will not return to the equilibrium point.

For the complex solutions, stability analysis typically involves more advanced methods not shown here, but they often relate to behaviors like oscillations or complex system dynamics.

Cubic Functions

Cubic functions are polynomial equations of the form \( ax^3 + bx^2 + cx + d \). They are essential in this exercise because the equilibrium solutions come from a cubic equation.Understanding how cubic functions behave is crucial in solving them.

• The standard form allows us to predict the function's shape, having a characteristic "S" or backward "S" curve.
• There are up to three real roots and the possibility of complex roots, as seen in the example \( y^{3} - 1 = 0 \).

When you solve \( y^{3} - 1 = 0 \), it translates to \( y^{3} = 1 \), leading to one real root \( y = 1 \) and two complex roots. Understanding these roots helps in applications, from solving engineering problems to modeling natural phenomena.