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Explain what is wrong with the statement. The function \(y=x^{2}\) is an equilibrium solution to the differential equation \(d y / d x=y-x^{2}.\)

Short Answer

Expert verified
The function \(y=x^2\) is not an equilibrium solution since \(2x \neq 0\) for all \(x\).

Step by step solution

01

Understand the Problem

We need to verify whether the function \(y = x^2\) satisfies the differential equation \( \frac{dy}{dx} = y - x^2 \) as an equilibrium solution.
02

Find the Derivative

Calculate the derivative of the function \(y = x^2\). We get \( \frac{dy}{dx} = 2x \).
03

Substitute into the Differential Equation

Substitute \(y = x^2\) and \(\frac{dy}{dx} = 2x\) into the differential equation. We replace \(\frac{dy}{dx} = y - x^2\) to check if it holds true: \(2x = x^2 - x^2\).
04

Evaluate the Substitution

Simplify the expression from the substitution: \(2x = 0\). Check if this is true for all \(x\).
05

Identify the Incorrect Assumption

The equation \(2x = 0\) is not true for all \(x\); it only holds when \(x = 0\). Therefore, \(y = x^2\) is not an equilibrium solution of the differential equation because it doesn't satisfy \(y = y-x^2\) for all \(x\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Differential Equations
Differential equations are mathematical expressions involving derivatives, which represent the rate of change of variables. In essence, these equations show relationships between functions and their derivatives. They are used across numerous fields, including physics, biology, and engineering, to model the behavior of complex systems.

An equilibrium solution is a special kind of solution for a differential equation where the rate of change is zero. This means the system is in a stable state, with no fluctuations as the dependent variable maintains a constant value over time. It’s like water in a still pond – no ripples or movement. The exercise above involved checking if the function \(y = x^2\) was an equilibrium solution to the differential equation \(\frac{dy}{dx} = y - x^2\).
  • First, identify the given function and differential equation.
  • Focus on the properties of equilibrium in mathematics, which include stability and constant values.
Differential equations can become complex quickly, but understanding the types of solutions they yield is crucial to navigating their complexity.
Exploring Function Derivatives
The derivative of a function provides insight into the function's rate of change at any point – think of it as the slope of the function's graph at a specific moment. For instance, the derivative of \(y = x^2\), given by \(\frac{dy}{dx} = 2x\), indicates that the function's rate of change increases linearly with \(x\).

To solve the differential equation in the exercise, it was necessary to know the function's derivative. Differentiation calculators use algorithmic rules, such as the power rule – which states that for \(y = x^n\), the derivative is \(nx^{n-1}\).
  • The derivative helps determine how a change in one variable affects another.
  • It can reveal important properties of the function, like maxima or minima.
Understanding derivatives is slightly like understanding the speed of a vehicle; it tells you how fast something is moving at a particular instant.
Applying the Substitution Method
The substitution method involves replacing variables with known values or expressions to simplify equations or verify solutions. This technique is particularly useful when dealing with differential equations, as seen in the exercise.

In the problem, the function \(y = x^2\) and its derivative \(\frac{dy}{dx} = 2x\) were substituted into the equation \(\frac{dy}{dx} = y - x^2\). The goal was to see if the equality \(2x = x^2 - x^2\) held true.
  • The substitution method provides a straightforward way to test potential solutions.
  • It's often used alongside other methods, such as integration or separation of variables.
This method is akin to checking if puzzle pieces fit together; you’re verifying to see if everything matches perfectly.
Performing Mathematical Verification
Mathematical verification is the process of checking whether a solution satisfies a given equation or condition. This involves careful comparison and algebraic manipulation to confirm solutions’ validity. In the exercise, mathematical verification revealed that \(y = x^2\) does not satisfy the equation \(\frac{dy}{dx} = y - x^2\) for all values of \(x\). This is because after substitution, the simplification led to \(2x = 0\), which is only true when \(x = 0\).

Verification is crucial in all mathematical and scientific endeavors as it ensures reliability and accuracy of results.
  • Always double-check your substitutions and simplifications during verification.
  • Be thorough; sometimes discrepancies might occur due to simple algebraic errors.
Think of verification like proofreading a document; it ensures nothing has been overlooked.

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Most popular questions from this chapter

Give the solution to the logistic differential equation with initial condition. $$\frac{d P}{d t}=0.04 P(1-0.0001 P) \text { with } P_{0}=200$$

Use the idea of nullclines dividing the plane into sectors to analyze the equations describing the interactions of robins and worms: $$ \begin{array}{l} \frac{d w}{d t}=w-w r \\ \frac{d r}{d t}=-r+r w \end{array} $$

Sketch solution curves with a variety of initial values for the differential equations. You do not need to find an equation for the solution. \(\frac{d y}{d t}=\alpha-y,\) where \(\alpha\) is a positive constant.

An item is initially sold at a price of \(p\text{dollars}\) per unit. Over time, market forces push the price toward the equilibrium price, \(p \text{dollars}^{*},\) at which supply balances demand. The Evans Price Adjustment model says that the rate of change in the market price, \(p\text{dollars},\) is proportional to the difference between the market price and the equilibrium price. (a) Write a differential equation for \(p\) as a function of \(t\) (b) Solve for \(p\) (c) Sketch solutions for various different initial prices, both above and below the equilibrium price. (d) What happens to \(p\) as \(t \rightarrow \infty ?\)

A bank account earns \(5 \%\) annual interest, compounded continuously. Money is deposited in a continuous cash flow at a rate of 1200 dollars per year into the account. (a) Write a differential equation that describes the rate at which the balance \(B=f(t)\) is changing. (b) Solve the differential equation given an initial balance \(B_{0}=0\) (c) Find the balance after 5 years.

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