/*! 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 29 Solve the given initial-value pr... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Solve the given initial-value problem. Give the largest interval \(I\) over which the solution is defined. $$ (x+1) \frac{d y}{d x}+y=\ln x, \quad y(1)=10 $$

Short Answer

Expert verified
The solution is valid on the interval \((0, \infty)\).

Step by step solution

01

Recognize the Type of Differential Equation

The given differential equation is \((x+1) \frac{dy}{dx} + y = \ln x\). This is a first-order linear differential equation in the form \(P(x)\frac{dy}{dx} + Q(x)y = R(x)\) with \(P(x) = x + 1\), \(Q(x) = 1\), and \(R(x) = \ln x\).
02

Rewrite in Standard Form

To rewrite the equation in the standard form \(\frac{dy}{dx} + p(x)y = g(x)\), divide every term by \(x+1\), giving \(\frac{dy}{dx} + \frac{1}{x+1}y = \frac{\ln x}{x+1}\).
03

Find the Integrating Factor

Calculate the integrating factor \(\mu(x)\) using \(\mu(x) = e^{\int \frac{1}{x+1} \, dx}\). The integral of \(\frac{1}{x+1}\) is \(\ln|x+1|\), so \(\mu(x) = e^{\ln|x+1|} = |x+1|\).
04

Solve Using the Integrating Factor

Multiply each term in the equation by the integrating factor \(|x+1|\): \(|x+1| \frac{dy}{dx} + \frac{|x+1|}{x+1} y = \frac{|x+1| \ln x}{x+1}\). Simplifying gives \((x+1) \frac{dy}{dx} = \ln x\). Since the derivative of \(y(x+1)\) with respect to \(x\) is \(\ln x\), integrate both sides with respect to \(x\) to solve for \(y\).
05

Integrate Both Sides

Integrating the left side \(\int d(y(x+1))\) gives \(y(x+1)\). The right side becomes \(\int \ln x \, dx\). Using integration by parts: let \(u = \ln x\) and \(dv = dx\), then \(du = \frac{1}{x}dx\) and \(v = x\). Therefore, \(\int u \, dv = uv - \int v \, du = x\ln x - \int x\frac{1}{x} \, dx = x\ln x - x+C\). After simplifying, \(y(x+1) = x\ln x - x + C\).
06

Solve for y

Divide through by \(x+1\) to get \(y = \frac{x\ln x - x + C}{x+1}\).
07

Apply Initial Condition

Use the initial condition \(y(1) = 10\) to find \(C\). Substitute \(x = 1\) and \(y = 10\) into the equation: \(10 = \frac{1 \cdot \ln 1 - 1 + C}{1+1} = \frac{-1 + C}{2}\). Solving for \(C\), we have: \(2 \cdot 10 = -1 + C\) gives \(C = 21\).
08

Write the General Solution

Substituting \(C = 21\) back, the solution is \(y = \frac{x \ln x - x + 21}{x+1}\).
09

Determine the Interval of Validity

The solution is defined where both the differential equation and the initial condition are valid: Since \(\ln x\) requires \(x > 0\) and the denominator \(x+1\) cannot be zero, \(x eq -1\), the largest interval \(I\) where the solution holds, given \(x = 1\) is included, is \((0, \infty)\).

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 91Ó°ÊÓ!

Key Concepts

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

Initial-Value Problems
An initial-value problem in the context of differential equations involves finding a function that satisfies a given differential equation, while also meeting initial conditions specified at a particular point. This problem provides extra information, such as a specific value of the function at a certain point, that sets it apart from other types of differential equations. For example, if you're given the differential equation with the condition \( y(1) = 10 \), it means that when \( x = 1 \), \( y \) must equal \( 10 \). This initial condition helps in determining the constant of integration, ensuring a unique solution to the differential equation.

Understanding initial-value problems is essential because they model many real-world systems. Wherever the state of a system at a specific time or location is known, an initial-value problem can describe how the system evolves over time or space. For students, mastering these problems includes learning to apply conditions accurately to find specific solutions.
Integrating Factor
The integrating factor is a pivotal tool in solving linear differential equations. Its purpose is to simplify differential equations by making them easily integrable. In our specific problem, the integrating factor \( \mu(x) \) is determined as \( e^{\int \frac{1}{x+1} \, dx} \).

  • This integrating factor essentially transforms the left-hand side of the differential equation into the derivative of a product of functions.
  • For example, multiplying the differential equation by the integrating factor, we obtain a simplification where \( \frac{d}{dx}(y(x+1)) \) appears.
This simplification makes it possible to integrate both sides easily, leading to a solution. Using an integrating factor is a powerful technique because it provides a methodical approach to solving a broad class of linear differential equations.
Interval of Validity
The interval of validity for a solution to a differential equation is crucial. It determines within which values the solution holds true or is defined. The interval takes into account any restrictions placed by the differential equation itself and any initial conditions. For our example, there are two main conditions to consider:

  • The natural logarithm \( \ln x \) is only defined for \( x > 0 \).
  • The term \( x+1 \) shouldn't equal zero, thus \( x eq -1 \).
Therefore, considering the initial condition of \( x = 1 \), the largest interval where the solution is valid is \( (0, \infty) \).

Students must be careful when determining the interval of validity, as ignoring constraints can lead to incorrect conclusions about the behavior of the solution. Understanding and applying these constraints ensures the solution accurately describes the system modeled by the differential equation.
First-Order Differential Equation
First-order differential equations are those where the highest derivative involved is the first derivative. These types of equations are the simplest form of differential equations and are often used to model rates of change.

In our problem, the equation \((x+1) \frac{dy}{dx} + y = \ln x\) is in this category. It can be rewritten in the standard form \( \frac{dy}{dx} + p(x)y = g(x) \). This form is essential for employing an integrating factor.

First-order differential equations are pervasive in both theoretical and applied contexts due to their simplicity and relevance to real-world systems, such as modeling exponential growth, decay processes, and cooling laws. Mastery of solving these equations forms the foundation for tackling more complex higher-order differential equations in mathematics and engineering.

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

Comsider the competition model defined by $$ \begin{aligned} &\frac{d x}{d t}=x(1-0.1 x-0.05 y) \\ &\frac{d y}{d t}=y(1.7-0.1 y-0.15 x) \end{aligned} $$ where the populations \(x(t)\) and \(y(t)\) are measured in the thousands and \(t\) in years. Use a numerical solver to analyze the populations over a long period of time for each of the cases: (a) \(x(0)=1\), \(y(0)=1\) (b) \(x(0)-4, \quad y(0)=10\) (c) \(x(0)=9\), \(y(0)=4\) (d) \(x(0)=5.5\), \(y(0)=3.5\)

Solve the given differential equation by finding, as in Example 4 , an appropriate integrating factor. $$ \cos x d x+\left(1+\frac{2}{y}\right) \sin x d y=0 $$

Solve the given differential equation by using an appropriate substitution. $$ 3\left(1+t^{2}\right) \frac{d y}{d t}=2 t y\left(y^{3}-1\right) $$

SIR Model A communicable disease is spread throughout a small community, with a fixed population of \(n\) people, by contact between infected individuals and people who are susceptible to the disease. Suppose initially that everyone is susceptible to the disease and that no one leaves the community while the epidemic is spreading. At time \(t\), let \(s(t), i(t)\), and \(r(t)\) denote, in tum, the number of people in the community (measured in hundreds) who are susceptible to the disease but not yet infected with it, the number of people who are infected with the disease, and the number of people who have recovered from the disease. Explain why the system of differential equations $$ \begin{aligned} &\frac{d s}{d t}=-k_{1} s i \\ &\frac{d i}{d t}=-k_{2} i+k_{1} s i \\ &\frac{d r}{d t}=k_{2} i \end{aligned} $$ where \(k_{1}\) (called the infection rate) and \(k_{2}\) (called the removal rate) are positive constants, is a reasonable mathematical model, commonly called a SIR model, for the spread of the epidemic throughout the community. Give plausible initial conditions associated with this system of equations. Show that the system implies that $$ \frac{d}{d t}(s+i+r)=0 $$

Two chemicals \(A\) and \(B\) are combined to form a chemical \(C\). The rate, or velocity, of the reaction is proportional to the product of the instantaneous amounts of \(A\) and \(B\) not converted to chemical \(C\). Initially there are 40 grams of \(A\) and 50 grams of \(B\), and for each gram of \(B\), 2 grams of \(A\) is used. It is observed that 10 grams of \(C\) is formed in 5 minutes. How much is formed in 20 minutes? What is the limiting amount of \(C\) after a long time? How much of chemicals \(A\) and \(B\) remains after a long time?

See all solutions

Recommended explanations on Physics Textbooks

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.