Problem 14
Question
In Problems 13 and 14, determine by inspection at least one solution of the given differential equation. $$ y^{\prime}=y(y-3) $$
Step-by-Step Solution
Verified Answer
The solutions found by inspection are \(y = 0\) and \(y = 3\).
1Step 1: Understand the Differential Equation
The given differential equation is \(y' = y(y - 3)\). This equation implies that the rate of change of \(y\) with respect to \(x\) is equal to the product \(y(y - 3)\).
2Step 2: Set y' to Zero for Equilibrium Solutions
One method to find solutions by inspection is to find constant solutions, which occur when \(y' = 0\). Setting \(y(y - 3) = 0\) gives potential solutions where either \(y = 0\) or \(y - 3 = 0\).
3Step 3: Solve for Equilibrium Solutions
From step 2, we have two potential constant solutions: \(y = 0\) and \(y = 3\). These are solutions to the differential equation where the derivative is zero, meaning \(y\) does not change with \(x\).
4Step 4: Verify the Solutions
Substituting \(y = 0\) into the original equation gives \(y' = 0(0 - 3) = 0\). Similarly, substituting \(y = 3\) gives \(y' = 3(3 - 3) = 0\). Both satisfy the equation, confirming these solutions are correct.
Key Concepts
Equilibrium SolutionsConstant SolutionsRate of Change
Equilibrium Solutions
When dealing with differential equations, the term "equilibrium solution" refers to a specific type of constant solution where the rate of change is zero. In simple terms, it's a scenario where the function remains stable, not increasing or decreasing, because the derivative or the slope of the function is zero.
To find equilibrium solutions, you'll typically set the derivative equal to zero and solve for the variable in question. For the differential equation given, \( y' = y(y - 3) \), setting \( y' = 0 \) gives us \( y(y - 3) = 0 \).
This results in two potential solutions: \( y = 0 \) and \( y = 3 \). These solutions suggest where the function levels off because the product of \( y \) and \( y - 3 \) equals zero, halting any change.
To find equilibrium solutions, you'll typically set the derivative equal to zero and solve for the variable in question. For the differential equation given, \( y' = y(y - 3) \), setting \( y' = 0 \) gives us \( y(y - 3) = 0 \).
This results in two potential solutions: \( y = 0 \) and \( y = 3 \). These solutions suggest where the function levels off because the product of \( y \) and \( y - 3 \) equals zero, halting any change.
- Equilibrium solutions are often called "stationary points," because the function "sits still" at these points.
- Such solutions play a crucial role in understanding the behavior of a dynamic system over time.
- In applications, they can signify stable environments or consistent processes.
Constant Solutions
Constant solutions are those where the function maintains the same value indefinitely for all values of the independent variable, in this instance, \( x \). In the context of differential equations, a constant solution occurs when the derivative is zero, indicating no rate of change.
For the equation \( y' = y(y - 3) \), substituting \( y = 0 \) or \( y = 3 \) into the equation results in \( y' = 0 \), confirming that these constants are solutions. These show where \( y \) does not depend on \( x \), meaning the function remains flat across the domain.
For the equation \( y' = y(y - 3) \), substituting \( y = 0 \) or \( y = 3 \) into the equation results in \( y' = 0 \), confirming that these constants are solutions. These show where \( y \) does not depend on \( x \), meaning the function remains flat across the domain.
- Constant solutions provide insight into fixed points in dynamic systems.
- They are invaluable in modeling scenarios with fixed output or steady states, such as in population models or mechanical systems.
- Finding these solutions by setting the derivative to zero is a reliable method for quickly identifying them.
Rate of Change
In differential equations, the concept of the rate of change is fundamental. Essentially, it describes how a variable, like \( y \), changes with respect to another, usually \( x \). In calculus terms, this derivative is expressed as \( y' \) or \( \frac{dy}{dx} \).
The original equation given is \( y' = y(y - 3) \). Here, the rate of change of \( y \) depends directly on the values of \( y \) and \( y - 3 \). This means that for certain values of \( y \), particularly the equilibrium solutions \( y = 0 \) or \( y = 3 \), the rate of change or the slope is zero.
The original equation given is \( y' = y(y - 3) \). Here, the rate of change of \( y \) depends directly on the values of \( y \) and \( y - 3 \). This means that for certain values of \( y \), particularly the equilibrium solutions \( y = 0 \) or \( y = 3 \), the rate of change or the slope is zero.
- A positive or negative rate of change would indicate an increasing or decreasing function, respectively.
- The rate at which \( y \) changes is a central focus when analyzing the dynamics described by a differential equation.
- Understanding the rate of change can highlight how quickly a system responds to changes, which is crucial in fields like physics and economics.
Other exercises in this chapter
Problem 14
Verify that the indicated function is an explicit solution of the given differential equation. Assume an appropriate interval \(I\) of definition for each solut
View solution Problem 14
y=c_{1} e^{x}+c_{2} e^{-x}\( is a two-parameter family of solutions of the second-order DE \)y^{\prime \prime}-y=0$. Find a solution of the second-order IVP con
View solution Problem 15
Verify that the indicated function \(y=\phi(x)\) is an explicit solution of the given first-order differential equation. Proceed as in Example 5, by considering
View solution Problem 15
Determine by inspection at least two solutions of the given first-order IVP. $$ y^{\prime}=3 y^{2 / 3}, \quad y(0)=0 $$
View solution