/*! 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 1117 Find a first-order differential ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 43: Problem 1117

Find a first-order differential equation for the family of all circles passing through the origin and having their centers on the \(\mathrm{x}\) -axis.

Short Answer

Expert verified
The short answer for the family of all circles passing through the origin and having their centers on the x-axis is given by the first-order differential equation: \[\frac{dy}{dx} = -\frac{(x - h)}{y}\]

Step by step solution

01

Set up the general equation of a circle with a center on the x-axis.

We know that the equation of a circle with center (h, k) and radius r is given by: \[(x - h)^2 + (y - k)^2 = r^2\] Since the center of the circle lies on the x-axis, we can set k = 0. The above equation now becomes: \[(x - h)^2 + y^2 = r^2\]
02

Make the circle pass through the origin.

We need to find the equation of the circle when it passes through the origin (0, 0). We can achieve this by substituting x = 0 and y = 0 in the circle equation: \[(0 - h)^2 + (0)^2 = r^2\] Simplifying, we get: \[h^2 = r^2\] Now, we can substitute this expression for r^2 back into the circle equation: \[(x - h)^2 + y^2 = h^2\]
03

Differentiate the circle equation to obtain the first-order differential equation.

Now, we will differentiate the equation obtained in Step 2 with respect to x: We first implicitly differentiate both sides of the equation with respect to \(x\): \[\frac{d}{dx} [(x - h)^2 + y^2] = \frac{d}{dx} [h^2]\] Applying the chain rule on the left-hand side, we get: \[2(x - h) + 2y \frac{dy}{dx} = 0\] Now, we can solve for the first-order differential equation by isolating \(\frac{dy}{dx}\) on one side: \[\frac{dy}{dx} = -\frac{(x - h)}{y}\] This first-order differential equation represents the family of all circles passing through the origin with their centers on the x-axis.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A point moves on a curve in the \(\mathrm{x}-\mathrm{y}\) plane in such a way that the angle made by the tangent to the curve with the \(\mathrm{x}\) - axis is three times the angle between the radius vector and the \(\mathrm{x}\) -axis. Find the Cartesian equation of the family of curves satisfying this condition.

The attraction of a spherical mass on a particle within the mass is directed toward the center of the sphere and is proportional to the distance from the center. Suppose a straight tube were bored through the center of the earth and a particle of mass \(\mathrm{m} \mathrm{lb}\). were dropped into the tube. If the radius of the earth is \(3960 \mathrm{mi}\)., find how long it will take a) to pass through the tube; b) to drop halfway to the center. Neglect resistance.

A substance in solution, for example, cane sugar, is decomposed in a chemical reaction into other substances through the presence of acids, and the rate at which the reaction takes place is proportional to the mass of sugar still unchanged. We then have \((\mathrm{dx} / \mathrm{dt})=\mathrm{k}(\mathrm{a}-\mathrm{x})\), where \(\mathrm{x}\) is the amount of sugar converted in time \(\mathrm{t}\) and \(\mathrm{a}\) is the original amount of sugar. Find the dependence of the sugar converted on time \(t\).

A body falls from a height of \(300 \mathrm{ft}\). What distance has it traveled after 4 sec. if subject to \(g\), the earth's acceleration?

A body falls with an initial velocity of \(1000 \mathrm{ft} / \mathrm{sec}\). and is subject to the acceleration of gravity \(\left(\mathrm{g} \approx 32 \mathrm{ft} . / \mathrm{sec}^{2}\right.\).). What distance does it fall in \(3 \mathrm{sec}\) ?
See all solutions

Recommended explanations on Math Textbooks

Pure Maths

Read Explanation

Discrete Mathematics

Read Explanation

Geometry

Read Explanation

Calculus

Read Explanation

Probability and Statistics

Read Explanation

Applied Mathematics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.