Question

Consider the differential equation \(y^{\prime}=f(x, y) .\) An equilibrium solution is any solution to the differential equation of the form \(y=c\), where \(c\) is a constant.

Short answer

The equilibrium solution is \( y = c \) where \( f(x, c) = 0 \) for all \( x \).

Step-by-step solution

Define the Equilibrium Solution

An equilibrium solution of a differential equation is a constant solution, which means that the derivative of the solution is zero. In this case, we are looking for solutions of the form \( y = c \), where \( c \) is a constant and \( y' = 0 \).

Substitute into the Differential Equation

Substitute \( y = c \) into the differential equation \( y' = f(x, y) \). This leads to \( 0 = f(x, c) \). The right-hand side should be evaluated at any \( x \), where the choice of \( x \) does not affect the solution, given that \( y \) is constant.

Solve for the Constant c

The equation \( f(x, c) = 0 \) implies that the function \( f \) must be zero for the constant \( c \). Solve for \( c \) by finding all values which make \( f(x, c) = 0 \) true for all \( x \).

Verify the Equilibrium Condition

Verify that the solution \( y = c \), with the determined \( c \), satisfies the condition \( y' = 0 \). Ensure \( f(x, c) = 0 \) for all \( x \) as part of confirming an equilibrium solution.

Key concepts

Constant Solutions

In the world of differential equations, a constant solution is an incredibly intuitive concept. When we talk about a constant solution, we mean a solution of the form \( y = c \), where \( c \) represents a constant value. The importance of this kind of solution lies in its simple nature: the function is flat, meaning it does not change over time. The derivative of a constant function is zero. This is the key characteristic of an equilibrium solution in differential equations. With the derivative simplified as zero, it indicates that there are no changes or movements, adhering to the steady-state behavior where the solution remains unchanged. This is fundamental when digging into the concept of equilibrium solutions.

Differential Equation Analysis

Analyzing differential equations involves exploring the relationship between functions and their derivatives. For an equation in the form \( y' = f(x, y) \), we typically seek functions \( y(x) \) such that the relationship holds. However, when we focus on equilibrium solutions, our analysis is simplified by considering scenarios where \( y' = 0 \). Substituting \( y = c \) into the differential equation transforms it into \( 0 = f(x, c) \). This essentially checks if the derivative remains zero across all points \( x \), verifying if \( y = c \) can indeed be considered as an equilibrium solution.

Equilibrium Condition Verification

Verification of the equilibrium condition is crucial in confirming constant solutions. Once we have \( y = c \), the next step is to substitute this back into the original differential equation. If \( 0 = f(x, c) \) holds true for all \( x \), then \( y = c \) maintains a flat state and is truly an equilibrium solution.

• Ensure the derived function satisfies the equation at any chosen point \( x \).
• A successful verification confirms that \( c \) is appropriately chosen, meaning there is no dependency on \( x \) in changing this flat state.
• This process is essential in differential equation analysis to isolate and validate true equilibrium scenarios.

By following these steps, we can confidently say that \( y = c \) is a valid solution, encapsulating the concept of an equilibrium in a neat mathematical form.