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Chapter 11: Problem 37

Let \(y=f(x)\) be a solution to the differential equation. Determine whether \(y=2 f(x)\) is a solution. $$d y / d x=x$$

Short Answer

Expert verified
No, \( y = 2f(x) \) is not a solution to the differential equation.

Step by step solution

01

Understand the Original Differential Equation

The given differential equation is \( \frac{dy}{dx} = x \). This means that the derivative of the function \( y \) with respect to \( x \) should equal \( x \).
02

Test the Function \( y = f(x) \)

Assume that \( y = f(x) \) satisfies the equation \( \frac{dy}{dx} = x \). Hence, the derivative \( \frac{d}{dx}[f(x)] = x \).
03

Analyze the Function \( y = 2f(x) \)

We need to determine if \( y = 2f(x) \) is also a solution to the differential equation \( \frac{dy}{dx} = x \).
04

Differentiate \( y = 2f(x) \)

Differentiate \( y = 2f(x) \) with respect to \( x \): \( \frac{d}{dx}[2f(x)] = 2 \cdot \frac{d}{dx}[f(x)] = 2x \).
05

Compare with the Initial Differential Equation

The derivative we found is \( 2x \), but the given differential equation states that \( \frac{dy}{dx} = x \). Since \( 2x eq x \), \( y = 2f(x) \) does not satisfy the differential equation.

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Key Concepts

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

Solutions to Differential Equations
When tackling differential equations, the goal is to find a function or set of functions known as solutions, that satisfy the equation. In the problem provided, we are given a differential equation, specifically \( \frac{dy}{dx} = x \). This simply means that the rate of change of \(y\) with respect to \(x\) is equal to \(x\). To verify if a function is a solution to this differential equation, we need to check if its derivative matches the right side of the equation.
  • Finding Solutions: Solving a differential equation involves determining the function \(y\) that makes the equation true. The function \(f(x)\) was given as a solution because its derivative \( \frac{d}{dx}[f(x)] \) is equal to \(x\), satisfying the equation.
  • Testing Other Functions: The exercise involved checking if \(y = 2f(x)\) meets the criteria. By differentiating, we found \( \frac{d}{dx}[2f(x)] = 2x \), which does not match the original equation's form \( \frac{dy}{dx} = x \). Hence, \(y = 2f(x)\) is not a solution.
Derivatives
Derivatives are a fundamental concept in calculus used to measure how a function changes as its input changes. They are especially crucial in differential equations, where they directly relate to the solution of the equation.
  • Understanding Derivatives: The derivative \( \frac{dy}{dx} \) of a function \(y = f(x)\) describes the rate at which \(y\) changes with respect to \(x\). It's like a slope, telling us how steep the curve of the function is at a given point.
  • Application in Differential Equations: In the context of our exercise, finding the derivative of \(y = f(x)\) helps us verify whether the function satisfies the differential equation \( \frac{dy}{dx} = x \).
  • Checking Work: When we computed \( \frac{d}{dx}[2f(x)] = 2x \), we compared this with the original equation. Since \(2x eq x\), this confirmed that \(y = 2f(x)\) is not a solution.
Mathematical Functions
Mathematical functions are expressions that relate inputs to outputs, often used to describe real-world relationships and phenomena. Understanding these functions is key to solving differential equations.
  • Functions as Solutions: A function like \(f(x)\) can be a solution to a differential equation if its derivative satisfies the equation. The function \(f(x)\) in our example did just that.
  • Modifying Functions: When we considered \(y = 2f(x)\), we essentially modified the function to test its validity as a solution. This highlights how slight changes in functions can affect whether they satisfy certain equations.
  • Generalization and Use: Functions provide a general solution to many equations and understanding their behavior, through derivatives and integrals, is essential in math, physics, engineering, and more.
Understanding these core concepts of functions, derivatives, and the solutions they offer to differential equations paves the way for mastering more complex problems in calculus. They are the backbone of learning how to manipulate and predict outcomes within mathematical frameworks.

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Most popular questions from this chapter

True or false? Give an explanation for your answer. If \(y=f(t)\) is a particular solution to a first-order differential equation, then the general solution is \(y=\) \(f(t)+C,\) where \(C\) is an arbitrary constant.

A differential equation for a quantity that is increasing and grows fastest when the quantity is small and grows more slowly as the quantity gets larger.

We analyze world oil production. \(^{25}\) When annual world oil production peaks and starts to decline. major economic restructuring will be needed. We investigate when this slowdown is projected to occur. We define \(P\) to be the total oil production worldwide since 1859 in billions of barrels. In \(1993,\) annual world oil production was 22.0 billion barrels and the total production was \(P=724\) billion barrels. In 2013 , annual production was 27.8 billion barrels and the total production was \(P=1235\) billion barrels. Let \(t\) be time in years since 1993 (a) Estimate the rate of production, \(d P / d t,\) for 1993 and 2013 (b) Estimate the relative growth rate, \((1 / P)(d P / d \imath)\) for 1993 and 2013 (c) Find an equation for the relative growth rate. \((1 / P)(d P / d t),\) as a function of \(P,\) assuming that the function is linear. (d) Assuming that \(P\) increases logistically and that all oil in the ground will ultimately be extracted, estimate the world oil reserves in 1859 to the nearest billion barrels. (e) Write and solve the logistic differential equation modeling \(P\)

A model for the population. \(P\), of carp in a landlocked lake at time \(t\) is given by the differential equation $$\frac{d P}{d t}=0.25 P(1-0.0004 P)$$ (a) What is the long-term equilibrium population of carp in the lake? (b) A census taken ten years ago found there were 1000 carp in the lake. Estimate the current population. (c) Under a plan to join the lake to a nearby river, the fish will be able to leave the lake. A net loss of \(10 \%\) of the carp each year is predicted, but the patterns of birth and death are not expected to change. Revise the differential equation to take this into account. Use the revised differential equation to predict the future development of the carp population.

The system of differential equations models the interaction of two populations \(x\) and \(y\) (a) What kinds of interaction (symbiosis, \(^{34}\) competition, predator-prey) do the equations describe? (b) What happens in the long run? Your answer may depend on the initial population. Draw a slope field. $$\begin{aligned} &\frac{1}{x} \frac{d x}{d t}=y-1\\\ &\frac{1}{y} \frac{d y}{d t}=x-1 \end{aligned}$$
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