91Ó°ÊÓ

A linear differential equation is an equation involving an unknown function and its derivatives which can be expressed in the form:

• \( y' + P(x)y = Q(x) \)

where \( P(x) \) and \( Q(x) \) are given functions of \( x \), and \( y' \) is the first derivative of the unknown function \( y \).
These equations are linear in nature because the function and its derivatives appear to the first power.
They are fundamental in understanding many physical phenomena. For example:

• Heat conduction
• Population growth
• Circuit analysis

Such equations can often be solved using a method called the integrating factor, which helps in transforming the equation into an easily integrable form.

Initial-value problems specify the value of the solution at a particular point. In the given exercise, the initial condition is \( y(1) = 2 \).
This means that when \( x = 1 \), the value of \( y \) must be 2.
The initial value is vital because it allows us to find a particular solution from a family of solutions.
This is crucial not only for mathematical completeness but also for practical applications, such as predicting future states in a physical system based on known starting conditions.
In practice, initial conditions help solve real-world problems, like determining the future position of a moving object given its initial speed and position.

The integrating factor, typically denoted \( \mu(x) \), is a function used to transform a linear differential equation into an integrated form.
The standard approach to finding the integrating factor involves calculating:

• \( \mu(x) = e^{\int P(x) \, dx} \)

where \( P(x) \) is the coefficient of the term \( y \) in the equation \( y' + P(x)y = Q(x) \).
For the problem we're dealing with, after converting it to standard form, \( P(x) = \frac{1}{x} \).
The integrating factor is therefore \( e^{\int \frac{1}{x} \, dx} = x \), and by multiplying the entire equation by this factor, it can be rewritten as an exact derivative.

• Example: \( \frac{d}{dx}(xy) = e^x \)

This crucial step simplifies finding the solution because the equation now reduces to a more straightforward integration process.

The interval of definition refers to the range of \( x \) values over which the solution to a differential equation is valid and continuous.
For the provided exercise, the particular solution is \( y = \frac{e^x + (2 - e)}{x} \).
Here, the function \( y \) will not be defined when the denominator is zero.
Thus, \( x \) must not be zero.
Given the initial condition \( x=1 \), and the general validity of the function for positive \( x \), the largest possible interval is when \( x > 0 \).

• This interval ensures that the solution is both meaningful and matches the initial condition.
• Confirmed by substituting initial condition: \( y(1) = 2 \)

This concept is vital in ensuring solutions are applicable and meaningful within given constraints of real-world problems.