/*! 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 11 $$x y^{\prime \prime}+(x-1) y^{\... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Problem 11

$$x y^{\prime \prime}+(x-1) y^{\prime}-2 y=0$$

Short Answer

Expert verified
The general solution to the differential equation \(xy'' + (x-1)y' - 2y = 0\) is \(y = c_1 x^2 + c_2 x^{-1}\) where \(c_1\) and \(c_2\) are arbitrary constants.

Step by step solution

01

Transform into a Homogeneous Differential Equation

The given equation is \(xy'' + (x-1)y' - 2y = 0\). Divide all terms by \(x\) to get the equation in the standard form: \(y'' + \frac{(x-1)}{x}y' - \frac{2}{x}y = 0\)
02

Solve the Homogeneous Differential Equation

Find the corresponding homogeneous equation by setting the equation equal to zero: \(y'' + \frac{(x-1)}{x}y' - \frac{2}{x}y = 0\). This is a Cauchy-Euler differential equation. By substituting \(y = x^m\) into the homogeneous equation, where \(m\) is an unknown constant to be determined, a quadratic in \(m\) is obtained: \(m(m - 1) + m - 2 = 0\). Solving the quadratic gives two roots, \(m = 2, -1\). Hence, the general solution of the homogeneous differential equation is \(y_h = c_1 x^2 + c_2 x^{-1}\), where \(c_1\) and \(c_2\) are arbitrary constants.
03

Use Variation of Parameters to Solve the Particular Solution

The particular solution is typically found using variation of parameters or undetermined coefficients. In this case, as there is no forcing function, the particular solution is trivially \(y_p = 0\).
04

Combine Solutions to Obtain General Solution

The general solution is the sum of the homogeneous and particular solutions: \(y = y_h + y_p = c_1 x^2 + c_2 x^{-1}\).

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

$$\left(t^{2}-t-2\right)^{2} x^{\prime \prime}+\left(t^{2}-4\right) x^{\prime}-t x=0$$

$$6 x^{3} y^{\prime \prime \prime}+13 x^{2} y^{\prime \prime}-\left(x^{2}+3 x\right) y^{\prime}-x y=0$$

$$x y^{\prime \prime}+y^{\prime}-4 y=0$$

Duffing's Equation. In the study of a nonlinear spring with periodic forcing, the following equation arises: $$ y^{\prime \prime}+k y+r y^{3}=A \cos \omega t $$ Let \(k=r=A=1\) and \(\omega=10\). Find the first three nonzero terms in the Taylor polynomial approximations to the solution with initial values \(y(0)=0, y^{\prime}(0)=1\).

Variable Spring Constant. As a spring is heated, its spring "constant" decreases. Suppose the spring Is heated so that the spring "constant" at time \(t\) is \(k(t)=6-t \mathrm{N} / \mathrm{m}\) (see Figure \(8.6 ) .\) If the unforced mass-spring system has mass \(m=2 \mathrm{kg}\) and a damping constant \(b=1 \mathrm{N}\) -sec/m with initial conditions \(x(0)=3 \mathrm{m}\)and \(x^{\prime}(0)=0 \mathrm{m} / \mathrm{sec},\) then the displacement \(x(t)\) is governed by the initial value problem \(2 x^{\prime \prime}(t)+x^{\prime}(t)+(6-t) x(t)=0\) \(x(0)=3, \quad x^{\prime}(0)=0\) Find at least the first four nonzero terms in a power series expansion about \(t=0\) for the displacement.
See all solutions

Recommended explanations on Math Textbooks

Probability and Statistics

Read Explanation

Applied Mathematics

Read Explanation

Geometry

Read Explanation

Calculus

Read Explanation

Discrete Mathematics

Read Explanation

Pure Maths

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.