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Find the general solution of the given differential equation. Give the largest interval over which the general solution is defined. Determine whether there are any transient terms in the general solution. $$ x^{2} y^{\prime}+x y=1 $$

Short Answer

Expert verified
The general solution is \(y = \frac{\ln|x|}{x} + \frac{C}{x}\). It is defined for \((0, \infty)\) with a transient term \(\frac{C}{x}\).

Step by step solution

01

Identify the Type of Differential Equation

The given equation is \(x^{2} y^{\prime}+x y=1\). This is a first-order linear differential equation in standard form \(a(x)y^{\prime} + b(x)y = c(x)\), where \(a(x) = x^2\), \(b(x) = x\), and \(c(x) = 1\).
02

Solve for y'

Divide the entire equation by \(x^2\) to isolate \(y'\). This gives \(y^{\prime} + \frac{1}{x}y = \frac{1}{x^2}\).
03

Find the Integrating Factor

To solve the equation, find the integrating factor \(\mu(x)\) by calculating \(e^{\int \frac{1}{x} \, dx} = e^{\ln|x|} = |x|\). The integrating factor is \(|x|\).
04

Multiply Equation by Integrating Factor

Multiply the entire differential equation by the integrating factor, \(x\) since \(|x| = x\) when \(x > 0\), resulting in \(x y^{\prime} + y = \frac{1}{x}\).
05

Recognize the Left Side as a Derivative

The left side of the equation is the derivative of \(xy\), which gives: \(\frac{d}{dx}(xy) = \frac{1}{x}\).
06

Integrate Both Sides

Integrate both sides with respect to \(x\). The integral of \(\frac{d}{dx}(xy)\) is \(xy\), and the integral of \(\frac{1}{x}\) is \(\ln|x|\). Therefore, \(xy = \ln|x| + C\), where \(C\) is the integration constant.
07

Solve for y

Divide both sides by \(x\) to solve for \(y\): \(y = \frac{\ln|x|}{x} + \frac{C}{x}\). This is the general solution.
08

Determine the Interval of Definition

Since \(\ln|x|\) is only defined for \(x > 0\), the largest interval over which the solution is defined is \((0, \infty)\).
09

Identify Transient Terms

In the general solution \(y = \frac{\ln|x|}{x} + \frac{C}{x}\), the transient term is \(\frac{C}{x}\), because as \(x \to \infty\), this term tends to zero.

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Key Concepts

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

Integrating Factor Method
The integrating factor method is a powerful technique used to solve first-order linear differential equations. In a differential equation of the form \( y' + P(x)y = Q(x) \), the integrating factor, denoted usually by \( \mu(x) \), is calculated to simplify the equation and make it easier to find the solution.

To find the integrating factor, calculate \( e^{\int P(x) \, dx} \). For the given equation \( y' + \frac{1}{x}y = \frac{1}{x^2} \), the integrating factor becomes \( e^{\int \frac{1}{x} \, dx} = e^{\ln|x|} = |x| \).

This step transforms the differential equation into a form where the left side is the derivative of the product of the integrating factor \(|x|\) and \(y\), helping to streamline the solution process. Once the integrating factor is determined, multiplying every term in the equation by it allows the equation to be rewritten as an easily integrable expression.
General Solution
The general solution of a differential equation is the family of all possible solutions, including arbitrary constants that are not yet specified. For the equation given, after using the integrating factor and integrating both sides, the solution \( xy = \ln|x| + C \) was derived, where \( C \) represents a constant resulting from the integration.

Solving for \( y \) gives the general solution as \( y = \frac{\ln|x|}{x} + \frac{C}{x} \). This solution describes the behavior of the function \( y \) for all \( x \) in the interval of definition. The constant \( C \) allows for an infinite variety of solutions that can satisfy different initial conditions or particular values.
Interval of Definition
The interval of definition specifies the range of \( x \) values for which the solution \( y(x) \) is valid. It is essential to consider the domain constraints of the functions involved.

In the case of the equation \( y = \frac{\ln|x|}{x} + \frac{C}{x} \), logarithmic functions like \( \ln|x| \) are defined when \( x > 0 \). Thus, the largest interval over which this general solution is defined and the functions involved are valid is \( (0, \infty) \). This interval reflects the physical or mathematical context within which the solution can correctly describe a system or phenomenon.
Transient Terms in Differential Equations
Transient terms in differential equations refer to components of the solution that diminish as the independent variable, often \( x \), approaches infinity. These terms are generally associated with the response of a system to initial conditions and tend to zero over time or as \( x \rightarrow \infty \).

For the general solution \( y = \frac{\ln|x|}{x} + \frac{C}{x} \), the term \( \frac{C}{x} \) is identified as the transient term. As \( x \) becomes very large, \( \frac{C}{x} \) approaches zero, meaning it has a negligible impact on the solution for large values of \( x \). Understanding transient terms helps in analyzing how a system reaches its steady state or long-term behavior.

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Most popular questions from this chapter

The population of a town grows at a rate proportional to the population present at time \(t\). The initial population of 500 increases by \(15 \%\) in 10 years. What will the population be in 30 years? How fast is the population growing at \(t=30 ?\)

The number \(N(t)\) of people in a community who are exposed to a particularadvertisement is governed by the logisticequation. Initially \(N(0)=500\), and it is observed that \(N(1)=1000\). Solve for \(N(t)\) if it is predicted that the limiting number of people in the community who will see the advertisement is 50,000 .

When interest is compounded continuously, the amount of money increases at a rate proportional to the amount \(S\) present at time \(t\), that is, \(d S / d t=r S\), where \(r\) is the annual rate of interest. (a) Find the amount of money accrued at the end of 5 years when \(\$ 5000\) is deposited in a savings account drawing \(5 \frac{3}{4} \%\) annual interest compounded continuously. (b) In how many years will the initial sum deposited bave doubled? (c) Use a calculator to compare the amount obtained in part (a) with the amount \(S=5000\left(1+\frac{1}{4}(0.0575)\right)^{5(4)}\) that is arcrued when interest is compounded quarterly.

Suppose an \(R C\) -series circuit has a variable resistor. If the resistance at time \(t\) is given by \(R=k_{1}+k_{2} t\), where \(k_{1}\) and \(k_{2}\) are known positive constants, then (9) becomes $$ \left(k_{1}+k_{2} t\right) \frac{d q}{d t}+\frac{1}{C} q=E(t) $$ If \(E(t)=E_{0}\) and \(q(0)=q_{0}\), where \(E_{0}\) and \(q_{0}\) are constants, show that $$ q(t)=E_{0} C+\left(q_{0}-E_{0} C\right)\left(\frac{k_{1}}{k_{1}+k_{2} t}\right)^{1 / C k_{2}} $$

A skydiver is equipped with a stopwatch and an altimeter. She opens her parachute 25 seconds after exiting a plane flying at an altitude of \(20,000 \mathrm{ft}\) and observes that her altitude is \(14,800 \mathrm{ft}\). Assume that air resistance is proportional to the square of the instantaneous velocity, her initial velocity upon leaving the plane is zero, and \(g=32 \mathrm{ft} / \mathrm{s}^{2}\) (a) Find the distance \(s(t)\), measured from the plane, that the skydiver has traveled during free fall in time \(t\). (b) How far does the skydiver fall and what is her velocity at \(t=15 \mathrm{~s} ?\)

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