/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 74 The following are differential e... [FREE SOLUTION] | 91Ó°ÊÓ

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The following are differential equations stated in words. Find the general solution of each. The derivative of a function at each point is \(0 .\)

Short Answer

Expert verified
The general solution is \( f(x) = C \), where \( C \) is a constant.

Step by step solution

01

Understand the problem statement

The problem states that the derivative of an unknown function \( f(x) \) is 0 at each point. This means we are looking for a function where its slope is always zero, regardless of the value of \( x \).
02

Write the differential equation

The worded problem translates to the differential equation \( \frac{df}{dx} = 0 \). This implies that the function has no change in value as \( x \) increases or decreases.
03

Solve the differential equation

To find the function \( f(x) \) satisfying \( \frac{df}{dx} = 0 \), we integrate both sides with respect to \( x \). The integral of 0 with respect to \( x \) is a constant, \( C \). Thus, \( f(x) = C \).
04

State the general solution

The general solution to the differential equation \( \frac{df}{dx} = 0 \) is \( f(x) = C \), where \( C \) is a constant. This implies that the function is a constant function regardless of the value of \( x \).

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Key Concepts

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

General Solution
When we're dealing with differential equations, one of the first objectives is to find the general solution. This general solution represents a family of functions that satisfy the given differential equation. Differential equations often describe various phenomena such as motion, heat conduction, or fluid dynamics. Hence, understanding them is crucial in fields like physics and engineering.
To find the general solution, we need to solve the differential equation. In our context, the differential equation is given as \( \frac{df}{dx} = 0 \). The general solution refers to all the possible functions that have a derivative of zero at every point in their domain. In simpler terms, the function doesn't change as the input, \( x \), changes.
Because the derivative is zero, it implies that the function itself must be constant, leading us to \( f(x) = C \). Here, \( C \) can be any real number since it represents all the functions with a zero slope. A critical part of mastering differential equations is being familiar with how to derive these solutions by integration.
Constant Function
In the realm of differential equations, a constant function plays a vital role in understanding solutions like the one we've discussed. A constant function is one where the output value does not depend on the input. It remains the same no matter what value \( x \) takes.
When the derivative of a function is zero, it implies that the rate of change of the function is zero. The function is not increasing or decreasing, but rather stays steady. This is what we see in the example \( f(x) = C \). Every point in \( f(x) \) is the value \( C \), no matter what \( x \) you choose.
Constant functions are significant because they represent equilibrium states in various systems. They can model unchanging conditions in real-world situations, such as a car moving at a constant speed or water in a perfectly still pond. Understanding constant functions is foundational because they represent the simplest solutions in differential equations.
Integration
Integration is a fundamental technique used to solve differential equations and is the inverse process of differentiation. When you integrate a function, you essentially find another function whose derivative yields the original function.
In this particular exercise, we started with the differential equation \( \frac{df}{dx} = 0 \). The process of solving it involves integrating both sides. For the right side, integrating a zero results in a constant, which is where the constant \( C \) in the solution \( f(x) = C \) originates. This step effectively reverses the differentiation.
Integration is an essential skill because it allows us to reconstruct functions from their rates of change. This process has practical applications in various fields, such as computing the area under curves, solving motion problems, and analyzing growth rates. Mastery of integration techniques is key to advancing in calculus and understanding the deeper applications of mathematics.

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Most popular questions from this chapter

For each exercise: a. Solve without using a graphing calculator. b. Verify your answer to part (a) using a graphing calculator. The population of a town is increasing at the rate of \(400 t e^{0.02 t}\) people per year, where \(t\) is the number of years from now. Find the total gain in population during the next 5 years.

For each exercise: a. Solve without using a graphing calculator. b. Verify your answer to part (a) using a graphing calculator. A company begins advertising a new product and finds that after weeks the product is gaining customer recognition at the rate of \(t^{2} \ln t\) thousand customers per week (for \(t \geq 1\) ). Find the total gain in recognition from the end of week 1 to the end of week 6 .

BUSINESS: Value of a Building Let \(y(t)\) be the value of a commercial building (in millions of dollars) after \(t\) years. a. Write a differential equation that says that the rate of growth of the value of the building is equal to two times the one-half power of its present value. b. Write an initial condition that says that at time zero the value of the building is 9 million dollars. c. Solve the differential equation and initial condition. d. Use your solution to find the value of the building at time \(t=5\) years.

Your company has developed a new product, and your marketing department has predicted how it will sell. Let \(y(t)\) be the (monthly) sales of the product after \(t\) months. a. Write a differential equation that says that the rate of growth of the sales will be four times the one-half power of the sales. b. Write an initial condition that says that at time \(t=0\) sales were 10,000 . c. Solve this differential equation and initial value. d. Use your solution to predict the sales at time \(t=12\) months.

$$ \text { Use integration by parts to find each integral. } $$ $$ \int x e^{2 x} d x $$

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