/*! 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 9 For \(y^{\prime}-\frac{1}{x} y=x... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 9

For \(y^{\prime}-\frac{1}{x} y=x \sin x\) an integrating factor is \(e^{-\int(1 / x) d x}=\frac{1}{x}\) so that \(\frac{d}{d x}\left[\frac{1}{x} y\right]=\sin x\) and \(y=c x-x \cos x\) for \(0< x<\infty\).

Short Answer

Expert verified
The integrating factor solution to the differential equation is \( y = Cx - x \cos x \).

Step by step solution

01

Identify the Differential Equation

The given differential equation is \[ y^{\prime} - \frac{1}{x} y = x \sin x \] which is a first-order linear differential equation.
02

Determine the Integrating Factor

For a differential equation of the form \( y^{\prime} + P(x)y = Q(x) \), the integrating factor is given by \[ e^{\int P(x)\,dx} \]In this problem, \( P(x) = -\frac{1}{x} \), so the integrating factor is: \[ e^{-\int \frac{1}{x}\,dx} = e^{\ln x^{-1}} = \frac{1}{x} \] as provided in the problem.
03

Multiply Through by the Integrating Factor

Multiply every term in the differential equation by the integrating factor \( \frac{1}{x} \) to get \[ \frac{1}{x} y^{\prime} - \frac{1}{x^2} y = \sin x \] which simplifies to \[ \frac{d}{dx} \left( \frac{1}{x} y \right) = \sin x \].
04

Integrate Both Sides

Integrate both sides with respect to \(x\): \[ \int \frac{d}{dx} \left( \frac{1}{x} y \right) \, dx = \int \sin x \, dx \] This implies: \[ \frac{1}{x} y = -\cos x + C \] where \(C\) is the constant of integration.
05

Solve for \( y \)

Multiply both sides by \(x\) to solve for \(y\): \[ y = -x \cos x + Cx \]Thus, the general solution is: \[ y = Cx - x \cos x \].

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!

Key Concepts

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

Integrating Factor
An integrating factor is a mathematical tool used in solving first-order linear differential equations like the one given in the exercise: \[ y^{\prime} - \frac{1}{x} y = x \sin x \]When you see a differential equation in the standard linear form, which is \( y^{\prime} + P(x)y = Q(x) \), your first task is to determine the integrating factor. The integrating factor is a function, typically denoted as \( \,\mu(x) \), that you multiply through the entire differential equation to simplify it. Here鈥檚 a quick guide to finding and applying the integrating factor:
  • Identify \( P(x) \) from your equation. In our example, \( P(x) = -\frac{1}{x} \).
  • The integrating factor, \( \,\mu(x) \), is \( e^{\int P(x) \, dx} \).
  • For \( P(x) = -\frac{1}{x} \), the integral \( \int -\frac{1}{x} \, dx \) results in \( -\ln x \), so the integrating factor is \( e^{-\ln x} = \frac{1}{x} \).
  • Multiply the original differential equation by this integrating factor to transform it into a simpler form.
This transformation allows the equation to express the left-hand side as the derivative of a product, making it much easier to integrate and solve.
Constant of Integration
The constant of integration, often denoted as \( C \), plays a crucial role in solving indefinite integrals in calculus, particularly when finding the general solution to differential equations. Let's connect this concept with the exercise at hand:After integrating both sides of the simplified differential equation, we have:\[\frac{1}{x}y = -\cos x + C\]The constant of integration \( C \) appears as a result of performing the integration process. Since the integration of a function can yield infinitely many antiderivatives, the constant \( C \) accounts for all these possible solutions.
  • Every time you solve an indefinite integral, remember to add \( C \), because the derivative of a constant is zero and it does not appear in the differentiation process.
  • In the context of differential equations, this constant ensures that the solution represents the family of all possible solutions.
  • To find a particular solution, more information like initial conditions would be needed to determine the specific value of \( C \).
Including the constant \( C \) ensures that our solution captures all the possibilities that satisfy the original differential equation.
General Solution
A general solution to a differential equation encompasses all possible solutions. In the context of first-order linear differential equations, once an intercepting factor simplifies the process, we aim to obtain this general solution by integrating and applying the constant of integration.In the provided exercise:\[ y = Cx - x \cos x\]This example illustrates the overall solution taking the following components into consideration:
  • The solution incorporates the particular integral or specific function satisfying the non-homogeneous part of the differential equation, which is \(-x \cos x\) here.
  • It also includes the homogeneous part of the solution reflecting the general behavior without the specific non-homogeneous factor, expressed as \(Cx\).
  • The parameter \( C \) allows for the representation of all solutions to the differential equation without specific boundary or initial conditions.
The general solution presents a complete family of functions that satisfy the differential equation, with each specific \( C \) value providing a unique function that solves the equation under certain conditions. This is essential in understanding the flexibility and breadth of potential solutions.

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

Assume that \(d T / d t=k(T-5)\) so that \(T=5+c e^{k t} .\) If \(T(1)=55^{\circ}\) and \(T(5)=30^{\circ}\) then \(k=-\frac{1}{4} \ln 2\) and \(c=59.4611\) so that \(T(0)=64.4611^{\circ}\).

(a) The equilibrium solutions \(y(x)=2\) and \(y(x)=-2\) satisfy the initial conditions \(y(0)=2\) and \(y(0)=-2\) respectively. Setting \(x=\frac{1}{4}\) and \(y=1\) in \(y=2\left(1+c e^{4 x}\right) /\left(1-c e^{4 x}\right)\) we obtain $$1=2 \frac{1+c e}{1-c e}, \quad 1-c e=2+2 c e, \quad-1=3 c e, \quad \text { and } \quad c=-\frac{1}{3 e}$$. (b) Separating variables and integrating yields \\[ \begin{aligned} \frac{1}{4} \ln |y-2|-\frac{1}{4} \ln |y+2|+\ln c_{1} &=x \\ \ln |y-2|-\ln |y+2|+\ln c &=4 x \\ \ln \left|\frac{c(y-2)}{y+2}\right| &=4 x \\ c \frac{y-2}{y+2} &=e^{4 x}.\end{aligned}\\] Solving for \(y\) we get \(y=2\left(c+e^{4 x}\right) /\left(c-e^{4 x}\right) .\) The initial condition \(y(0)=-2\) implies \(2(c+1) /(c-1)=-2\) which yields \(c=0\) and \(y(x)=-2 .\) The initial condition \(y(0)=2\) does not correspond to a value of \(c,\) and it must simply be recognized that \(y(x)=2\) is a solution of the initial-value problem. Setting \(x=\frac{1}{4}\) and \(y=1\) in \(y=2\left(c+e^{4 x}\right) /\left(c-e^{4 x}\right)\) leads to \(c=-3 e .\) Thus, a solution of the initial-value problem is $$y=2 \frac{-3 e+e^{4 x}}{-3 e-e^{4 x}}=2 \frac{3-e^{4 x-1}}{3+e^{4 x-1}}.$$

From \(\frac{1}{1+(2 y)^{2}} d y=\frac{-x}{1+\left(x^{2}\right)^{2}} d x\) we obtain \(\frac{1}{2} \tan ^{-1} 2 y=-\frac{1}{2} \tan ^{-1} x^{2}+c \quad\) or \(\quad \tan ^{-1} 2 y+\tan ^{-1} x^{2}=c_{1}\). Using \(y(1)=0\) we find \(c_{1}=\pi / 4\). Thus, an implicit solution of the initial-value problem is \(\tan ^{-1} 2 y+\tan ^{-1} x^{2}=\pi / 4 .\) Solving for \(y\) and using a trigonometric identity we get $$\begin{aligned} 2 y &=\tan \left(\frac{\pi}{4}-\tan ^{-1} x^{2}\right) \\ y &=\frac{1}{2} \tan \left(\frac{\pi}{4}-\tan ^{-1} x^{2}\right) \\ &=\frac{1}{2} \frac{\tan \frac{\pi}{4}-\tan \left(\tan ^{-1} x^{2}\right)}{1+\tan \frac{\pi}{4} \tan \left(\tan ^{-1} x^{2}\right)} \\ &=\frac{1}{2} \frac{1-x^{2}}{1+x^{2}}. \end{aligned}$$

(a) Integrating \(d^{2} s / d t^{2}=-g\) we get \(v(t)=d s / d t=-g t+c .\) From \(v(0)=300\) we find \(c=300,\) and we are given \(g=32,\) so the velocity is \(v(t)=-32 t+300\). (b) Integrating again and using \(s(0)=0\) we get \(s(t)=-16 t^{2}+300 t .\) The maximum height is attained when \(v=0,\) that is, at \(t_{a}=9.375 .\) The maximum height will be \(s(9.375)=1406.25 \mathrm{ft}\).

For \(0 \leq t \leq 20\) the differential equation is \(20 d i / d t+2 i=120 .\) An integrating factor is \(e^{t / 10},\) so \((d / d t)\left[e^{t / 10} i\right]=\) \(6 e^{t / 10}\) and \(i=60+c_{1} e^{-t / 10} .\) If \(i(0)=0\) then \(c_{1}=-60\) and \(i=60-60 e^{-t / 10} .\) For \(t>20\) the differential equation is \(20 d i / d t+2 i=0\) and \(i=c_{2} e^{-t / 10} .\) At \(t=20\) we want \(c_{2} e^{-2}=60-60 e^{-2}\) so that \(c_{2}=60\left(e^{2}-1\right)\). Thus $$i(t)=\left\\{\begin{array}{ll} 60-60 e^{-t / 10}, & 0 \leq t \leq 20 \\ 60\left(e^{2}-1\right) e^{-t / 10}, & t>20 \end{array}\right.$$
See all solutions

Recommended explanations on Physics Textbooks

Medical Physics

Read Explanation

Torque and Rotational Motion

Read Explanation

Classical Mechanics

Read Explanation

Modern Physics

Read Explanation

Solid State Physics

Read Explanation

Electromagnetism

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.