/*! 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 24 The auxiliary equation \(m(m-1)+... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 24

The auxiliary equation \(m(m-1)+m-1=m^{2}-1=0\) has roots \(m_{1}=-1, m_{2}=1,\) so \(y_{c}=c_{1} x^{-1}+c_{2} x .\) With \(y_{1}=x^{-1}, y_{2}=x,\) and the identification \(f(x)=1 / x^{2}(x+1),\) we get $$W=2 x^{-1}, \quad W_{1}=-1 / x(x+1), \quad \text { and } \quad W_{2}=1 / x^{3}(x+1).$$ Then \(\left.u_{1}^{\prime}=W_{1} / W=-1 / 2(x+1), \quad u_{2}^{\prime}=W_{2} / W=1 / 2 x^{2}(x+1), \text { and integration (by partial fractions for } u_{2}^{\prime}\right)\) gives $$\begin{array}{l} u_{1}=-\frac{1}{2} \ln (x+1) \\ u_{2}=-\frac{1}{2} x^{-1}-\frac{1}{2} \ln x+\frac{1}{2} \ln (x+1), \end{array}$$ so $$\begin{aligned} y_{p} &=u_{1} y_{1}+u_{2} y_{2}=\left[-\frac{1}{2} \ln (x+1)\right] x^{-1}+\left[-\frac{1}{2} x^{-1}-\frac{1}{2} \ln x+\frac{1}{2} \ln (x+1)\right] x \\ &=-\frac{1}{2}-\frac{1}{2} x \ln x+\frac{1}{2} x \ln (x+1)-\frac{\ln (x+1)}{2 x}=-\frac{1}{2}+\frac{1}{2} x \ln \left(1+\frac{1}{x}\right)-\frac{\ln (x+1)}{2 x} \end{aligned}$$ and $$y=y_{c}+y_{p}=c_{1} x^{-1}+c_{2} x-\frac{1}{2}+\frac{1}{2} x \ln \left(1+\frac{1}{x}\right)-\frac{\ln (x+1)}{2 x}, \quad x > 0.$$

Short Answer

Expert verified
The solution is \(y = c_1 x^{-1} + c_2 x - \frac{1}{2} + \frac{1}{2} x \ln \left(1+\frac{1}{x}\right) - \frac{\ln (x+1)}{2 x}\) for \(x > 0\).

Step by step solution

01

Solve Auxiliary Equation

The auxiliary equation is given as \( m(m-1)+m-1 = m^{2}-1 = 0 \). Solve this equation to find the roots. Rearrange and factor the equation: \( (m+1)(m-1) = 0 \), which gives the roots \( m_1 = -1 \) and \( m_2 = 1 \).
02

Construct General Solution

Using the found roots, the general solution (complementary function \( y_c \)) can be expressed as a combination of the independent solutions: \( y_c = c_1 x^{-1} + c_2 x \) where \( c_1 \) and \( c_2 \) are arbitrary constants.
03

Identify Terms for Non-Homogeneous Solution

Given \( f(x) = \frac{1}{x^2 (x+1)} \), identify the terms: \( W = 2x^{-1} \), \( W_1 = -\frac{1}{x(x+1)} \), and \( W_2 = \frac{1}{x^3(x+1)} \). These help in constructing the particular solution.
04

Determine Derivatives of Functions

Evaluate derivatives \( u_1' = \frac{W_1}{W} = -\frac{1}{2(x+1)} \) and \( u_2' = \frac{W_2}{W} = \frac{1}{2x^2(x+1)} \). These derivatives will be integrated to determine the functions \( u_1 \) and \( u_2 \).
05

Integrate Derivatives

Integrate to find \( u_1 \) and \( u_2 \): \( u_1 = -\frac{1}{2} \ln (x+1) \) by direct integration, and \( u_2 = -\frac{1}{2} x^{-1} - \frac{1}{2} \ln x + \frac{1}{2} \ln (x+1) \) using partial fraction decomposition to simplify \( u_2' \).
06

Construct Particular Solution

Using found \( u_1 \) and \( u_2 \), derive the particular solution \( y_p = u_1 y_1 + u_2 y_2 = -\frac{1}{2} - \frac{1}{2} x \ln x + \frac{1}{2} x \ln (x+1) - \frac{\ln (x+1)}{2x} \).
07

Combine Solutions for Final Answer

The final solution combines the complementary and particular solutions: \( y = y_c + y_p = c_1 x^{-1} + c_2 x - \frac{1}{2} + \frac{1}{2} x \ln \left(1+\frac{1}{x}\right) - \frac{\ln (x+1)}{2 x} \), valid for \( x > 0 \).

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.

Auxiliary Equation
When tackling differential equations, the auxiliary equation is essential for finding the complementary solution. In this exercise, the auxiliary equation is given as \( m(m-1)+m-1 = m^2 - 1 = 0 \). This is a quadratic equation in terms of \( m \). The goal here is to find the roots that can help you construct the general solution of the differential equation.
The process is simple:- **Re-arrange the Equation:** Simplify the expression so it can be factored easily. Here, \( (m+1)(m-1) = 0 \) was identified.- **Factor the Equation:** This step reveals the roots: \( m_1 = -1 \) and \( m_2 = 1 \).- **Formulate the General Solution:** Using these roots, we write the complementary function, \( y_c = c_1 x^{-1} + c_2 x \).
The roots represent different possible behaviors of the solutions to the differential equation, and the complementary solution usually captures the homogeneous part of the solution. The particular constants \( c_1 \) and \( c_2 \) will be determined by boundary conditions or initial values.
Particular Solution
The particular solution, \( y_p \), is the segment of the overall solution that fits a non-homogeneous differential equation. In this exercise, it is achieved through the formula \( y_p = u_1 y_1 + u_2 y_2 \), where \( u_1 \) and \( u_2 \) are functions determined by integrating their respective derivatives.
Here's how it works:- Start with expressions \( y_1 = x^{-1} \) and \( y_2 = x \), which are solutions to the homogeneous equation.- Calculate the Wronskian \( W \) and the terms \( W_1 \) and \( W_2 \) to articulate deriving expressions: \( W = 2 x^{-1} \), \( W_1 = -1 / x(x+1) \), and \( W_2 = 1 / x^3(x+1) \).- Determine derivatives \( u_1' = W_1 / W \) and \( u_2' = W_2 / W \). Integrating these derivatives will provide \( u_1 \) and \( u_2 \): - \( u_1 = -\frac{1}{2} \ln (x+1) \) - \( u_2 = -\frac{1}{2} x^{-1} - \frac{1}{2} \ln x + \frac{1}{2} \ln (x+1) \)
Combining \( u_1 \) and \( u_2 \) with \( y_1 \) and \( y_2 \), we get the particular solution that accounts for the non-homogeneous part of the differential equation.
Integration Techniques
In working through differential equations, integration often plays a key role, especially when constructing a particular solution. Two main integration techniques used in this exercise are direct integration and partial fraction decomposition.
**Direct Integration**- A straightforward process, applied here to find \( u_1 \).- Use for cases where an antiderivative is easily identifiable, such as finding \( u_1 = -\frac{1}{2} \ln (x+1) \).
**Partial Fraction Decomposition**- Utilized when dealing with rational functions, like \( u_2' = \frac{1}{2 x^2(x+1)} \).- Decompose the complex fraction into simpler parts, integrate term by term: - Split \( \frac{1}{x^2(x+1)} \) into parts manageable by basic integration formulas.
It's crucial to understand these techniques as they allow for solving integrals that contribute to the complete solution of the differential equation, helping to express complex solutions in a simpler form.

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

From \(\left(D^{2}+5\right) x+D y=0\) and \((D+1) x+(D-4) y=0\) we obtain \((D-5)\left(D^{2}+4\right) x=0\) and \((D-5)\left(D^{2}+4\right) y=0\) The solution is $$\begin{aligned} &x=c_{1} e^{5 t}+c_{2} \cos 2 t+c_{3} \sin 2 t\\\ &y=c_{4} e^{5 t}+c_{5} \cos 2 t+c_{6} \sin 2 t\end{aligned}$$ Substituting into \((D+1) x+(D-4) y=0\) gives $$\left(6 c_{1}+c_{4}\right) e^{5 t}+\left(c_{2}+2 c_{3}-4 c_{5}+2 c_{6}\right) \cos 2 t+\left(-2 c_{2}+c_{3}-2 c_{5}-4 c_{6}\right) \sin 2 t=0$$ so that \(c_{4}=-6 c_{1}, c_{5}=\frac{1}{2} c_{3}, c_{6}=-\frac{1}{2} c_{2},\) and $$y=-6 c_{1} e^{5 t}+\frac{1}{2} c_{3} \cos 2 t-\frac{1}{2} c_{2} \sin 2 t.$$

In the case when \(\lambda=-\alpha^{2}<0,\) the solution of the differential equation is \(y=c_{1} \cosh \alpha x+c_{2} \sinh \alpha x .\) The condition \(y(0)=0\) gives \(c_{1}=0\) The condition \(y(1)-\frac{1}{2} y^{\prime}(1)=0 \quad\) applied to \(y=c_{2} \sinh \alpha x\) gives \(c_{2}\left(\sinh \alpha-\frac{1}{2} \alpha \cosh \alpha\right)=0 \quad\) or \(\tanh \alpha=\frac{1}{2} \alpha . \quad\) As can be seen from the figure, the graphs of \(y=\tanh x\) and \(y=\frac{1}{2} x\) intersect at a single point with approximate \(x\) -coordinate \(\alpha_{1}=1.915 .\) Thus, there is a single negative eigenvalue \(\lambda_{1}=-\alpha_{1}^{2} \approx-3.667\) and the corresponding eigenfuntion is \(y_{1}=\sinh 1.915 x\). For \(\lambda=0\) the only solution of the boundary-value problem is \(y=0\) For \(\lambda=\alpha^{2}>0\) the solution of the differential equation is \(y=c_{1} \cos \alpha x+c_{2} \sin \alpha x .\) The condition \(y(0)=0\) gives \(c_{1}=0,\) so \(y=c_{2} \sin \alpha x .\) The condition \(y(1)-\frac{1}{2} y^{\prime}(1)=0\) gives \(c_{2}\left(\sin \alpha-\frac{1}{2} \alpha \cos \alpha\right)=0,\) so the eigenvalues are \(\lambda_{n}=\alpha_{n}^{2}\) when \(\alpha_{n}, n=2,3,4, \ldots,\) are the positive roots of \(\tan \alpha=\frac{1}{2} \alpha .\) Using a CAS we find that the first three values of \(\alpha\) are \(\alpha_{2}=4.27487, \alpha_{3}=7.59655,\) and \(\alpha_{4}=10.8127 .\) The first three eigenvalues are then \(\lambda_{2}=\alpha_{2}^{2}=18.2738, \lambda_{3}=\alpha_{3}^{2}=57.7075,\) and \(\lambda_{4}=\alpha_{4}^{2}=116.9139\) with corresponding eigenfunctions \(y_{2}=\sin 4.27487 x, y_{3}=\sin 7.59655 x,\) and \(y_{4}=\sin 10.8127 x\).

(a) When \(T(x)=x^{2}\) the given differential equation is the Cauchy-Euler equation \\[ x^{2} y^{\prime \prime}+2 x y^{\prime}+\rho \omega^{2} y=0 \\] The solutions of the auxiliary equation \\[ m(m-1)+2 m+\rho \omega^{2}=m^{2}+m+\rho \omega^{2}=0 \\] are \\[ m_{1}=-\frac{1}{2}-\frac{1}{2} \sqrt{4 \rho \omega^{2}-1} i, \quad m_{2}=-\frac{1}{2}+\frac{1}{2} \sqrt{4 \rho \omega^{2}-1} \\] when \(\rho \omega^{2}>0.25 .\) Thus \\[ y=c_{1} x^{-1 / 2} \cos (\lambda \ln x)+c_{2} x^{-1 / 2} \sin (\lambda \ln x) \\] where \(\lambda=\frac{1}{2} \sqrt{4 \rho \omega^{2}-1} .\) Applying \(y(1)=0\) gives \(c_{1}=0\) and consequently \\[ y=c_{2} x^{-1 / 2} \sin (\lambda \ln x) \\]. The condition \(y(e)=0\) requires \(c_{2} e^{-1 / 2} \sin \lambda=0 . \quad\) We obtain a nontrivial solution when \(\lambda_{n}=n \pi\) \(n=1,2,3, \ldots .\) But $$\lambda_{n}=\frac{1}{2} \sqrt{4 \rho \omega_{n}^{2}-1}=n \pi$$ Solving for \(\omega_{n}\) gives \\[ \omega_{n}=\frac{1}{2} \sqrt{\left(4 n^{2} \pi^{2}+1\right) / \rho} \\]. The corresponding solutions are \\[ y_{n}(x)=c_{2} x^{-1 / 2} \sin (n \pi \ln x) \\]. b.

Solving \(\frac{1}{20} q^{\prime \prime}+2 q^{\prime}+100 q=0\) we obtain \(q(t)=e^{-20 t}\left(c_{1} \cos 40 t+c_{2} \sin 40 t\right) .\) The initial conditions \(q(0)=5\) and \(q^{\prime}(0)=0\) imply \(c_{1}=5\) and \(c_{2}=5 / 2 .\) Thus \\[q(t)=e^{-20 t}\left(5 \cos 40 t+\frac{5}{2} \sin 40 t\right)=\sqrt{25+25 / 4} e^{-20 t} \sin (40 t+1.1071)\\] and \(q(0.01) \approx 4.5676\) coulombs. The charge is zero for the first time when \(40 t+1.1071=\pi\) or \(t \approx\) 0.0509 second.

From \(D x+D^{2} y=e^{3 t}\) and \((D+1) x+(D-1) y=4 e^{3 t}\) we obtain \(D\left(D^{2}+1\right) x=34 e^{3 t}\) and \(D\left(D^{2}+1\right) y=-8 e^{3 t}\) The solution is $$\begin{array}{l} y=c_{1}+c_{2} \sin t+c_{3} \cos t-\frac{4}{15} e^{3 t} \\ x=c_{4}+c_{5} \sin t+c_{6} \cos t+\frac{17}{15} e^{3 t} \end{array}.$$ Substituting into \((D+1) x+(D-1) y=4 e^{3 t}\) gives $$\left(c_{4}-c_{1}\right)+\left(c_{5}-c_{6}-c_{3}-c_{2}\right) \sin t+\left(c_{6}+c_{5}+c_{2}-c_{3}\right) \cos t=0$$ so that \(c_{4}=c_{1}, c_{5}=c_{3}, c_{6}=-c_{2},\) and $$x=c_{1}-c_{2} \cos t+c_{3} \sin t+\frac{17}{15} e^{3 t}.$$
See all solutions

Recommended explanations on Physics Textbooks

Famous Physicists

Read Explanation

Turning Points in Physics

Read Explanation

Wave Optics

Read Explanation

Physics of Motion

Read Explanation

Kinematics Physics

Read Explanation

Magnetism

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.