Problem 2 - Calculus for Biology and Medicine

Question

Find the equilibria of the following differential equations. $$ \frac{d y}{d t}=y^{3}+y $$

Step-by-Step Solution

Answer

The equilibria is at \( y = 0 \).

Step 1: Set the differential equation to zero

To find the equilibria, we start by setting the given differential equation to zero. The equilibria are values of \( y \) such that \( \frac{d y}{d t} = 0 \). Thus, we set up the equation: \[ y^3 + y = 0 \]

Step 2: Factor the equation

We factor the equation \( y^3 + y = 0 \) by taking \( y \) as a common factor:\[ y(y^2 + 1) = 0 \]

Step 3: Solve for y

Now, we solve the factored equation for \( y \). The equation \( y(y^2 + 1) = 0 \) implies:1. \( y = 0 \)2. \( y^2 + 1 = 0 \)Since \( y^2 + 1 = 0 \) leads to \( y^2 = -1 \), which has no real solutions, the only equilibrium point in the real number system is \( y = 0 \).

Key Concepts

Equilibrium Points

In the realm of differential equations, equilibrium points are crucial because they indicate where the system stays constant over time; in other words, they don't change as time progresses. For a differential equation such as \( \frac{d y}{d t}=y^{3}+y \), an equilibrium point is a value of \( y \) where the rate of change \( \frac{d y}{d t} \) equates to zero. Setting this equation to zero helps identify these constant solutions.
Identifying equilibrium points requires solving \( y^{3} + y = 0 \). These points are critical when predicting the behavior of dynamical systems. Predominantly in models dependent on time, understanding these points allows us to determine conditions causing change or stability.

• In our example, the only equilibrium in the real number system turns out to be \( y = 0 \).

Recognizing these points lays the groundwork for examining more complex behaviors and stability in systems, making them integral to analyzing systems governed by differential equations.

Factoring Equations

Factoring equations is a method used to break down complex equations into simpler components, often making them easier to solve. For the equation \( y^3 + y = 0 \), factoring involves identifying and extracting common factors.
This approach can untangle intricate polynomials into manageable linear or quadratic components. In our given equation, we notice \( y \) is a common factor, allowing us to rewrite the equation as \( y(y^2 + 1) = 0 \).
Factoring transforms the problem into finding solutions for each factor:

• For \( y(y^2 + 1) = 0 \), we're assessing \( y = 0 \) and \( y^2 + 1 = 0 \).

Such factoring showcases the elegance of simplifying polynomial expressions and highlights potential solutions easily overlooked in their non-factored form. Factoring is thus indispensable in solving polynomial equations, offering a path to clearer resolution.

Solving Polynomial Equations

Solving polynomial equations means determining the values of the variable that satisfy the equation. Such solutions can inform important applications, from physics to engineering. In our case, solving \( y(y^2 + 1) = 0 \) guides us in finding where the original differential equation is at equilibrium.
The strategy involves setting each product component to zero. For \( y(y^2 + 1) = 0 \):

• \( y = 0 \) gives an immediate solution.
• \( y^2 + 1 = 0 \) implies looking for real or complex roots.

This latter scenario, \( y^2 = -1 \), does not yield real solutions because squaring a real number results in a non-negative product. Therefore, in real-number terms, \( y = 0 \) is the singular solution.
Recognizing when solutions exist or not helps strategize about next problem-solving steps and understand the nature of polynomial equations even in complex scenarios.