Problem 25 - Advanced Engineering Mathematics

Question

Solve the given initial-value problem. Give the largest interval \(I\) over which the solution is defined. $$ x y^{\prime}+y=e^{x}, \quad y(1)=2 $$

Step-by-Step Solution

Answer

The solution is defined for \( x > 0 \) with the particular solution \( y = \frac{e^x - e + 2}{x} \).

Step 1: Identify and Rewrite the Differential Equation

The given differential equation is \( x y' + y = e^x \). We can recognize this as a linear first-order differential equation in the standard form \( y' + P(x)y = Q(x) \) by dividing all terms by \( x \), leading to \( y' + \frac{1}{x}y = \frac{e^x}{x} \).

Step 2: Find the Integrating Factor

The integrating factor \( \mu(x) \) is calculated as \( e^{\int P(x) \, dx} = e^{\int \frac{1}{x} \, dx} = x \). Multiply the entire equation by this integrating factor to make the left side an exact derivative: \( x \left( y' + \frac{1}{x}y \right) = \frac{e^x}{x}x \).

Step 3: Rewrite the Equation Using Integrating Factor

Upon multiplication by the integrating factor \( x \), the equation becomes \( \frac{d}{dx}(xy) = e^x \). This simplifies solving as it transforms it into the derivative of the product of the integrating factor and the solution \( y \).

Step 4: Integrate to Find the General Solution

Integrate both sides with respect to \( x \). The left side integrates to \( xy \), and the right integrates to \( \int e^x \, dx = e^x + C \). Hence, \( xy = e^x + C \).

Step 5: Solve for the General Solution

Express \( y \) in terms of \( x \) and the constant \( C \): \( y = \frac{e^x + C}{x} \).

Step 6: Apply Initial Condition

Use the initial condition \( y(1) = 2 \) to find \( C \): Substitute \( x = 1 \) and \( y = 2 \) into the equation, resulting in \( 2 = \frac{e + C}{1} \). From this, solve \( C = 2 - e \).

Step 7: Write the Particular Solution

Substitute \( C = 2 - e \) back into the general solution: \( y = \frac{e^x + (2 - e)}{x} \). Simplify to \( y = \frac{e^x - e + 2}{x} \).

Step 8: Determine the Largest Interval of Definition

The solution is defined wherever the denominator \( x eq 0 \). Therefore, the largest interval is \( x > 0 \) since the initial condition \( x = 1 \) is positive.

Key Concepts

Linear Differential Equations

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

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.

Integrating Factor

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.

Interval of Definition

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.