Problem 18 - Calculus: Early Transcendentals

Question

Solve the initial-value problem. \( xy' + y = x \ln x, y(1) = 0 \)

Step-by-Step Solution

Answer

The solution is \( y = \frac{x \ln x}{2} - \frac{x}{4} + \frac{1}{4x} \).

Step 1: Identify the Type of Differential Equation

The given differential equation is in the form \( xy' + y = x \ln x \). Recognize it as a linear first-order differential equation of the form \( y' + P(x)y = Q(x) \) after dividing through by \( x \) to isolate \( y' \).

Step 2: Rewrite and Identify P(x) and Q(x)

Divide every term by \( x \) to get \( y' + \frac{1}{x} y = \ln x \). Now, \( P(x) = \frac{1}{x} \) and \( Q(x) = \ln x \).

Step 3: Find an Integrating Factor

Calculate the integrating factor \( \mu(x) \) using \( \mu(x) = e^{\int P(x) \, dx} = e^{\int \frac{1}{x} \, dx} = e^{\ln|x|} = |x| \). For \( x > 0 \), this simplifies to \( x \).

Step 4: Multiply Through by the Integrating Factor

Multiply the entire differential equation by the integrating factor \( x \), changing the equation to \( x y' + y = x \ln x \), which simplifies to \( \frac{d}{dx}(x y) = x \ln x \).

Step 5: Solve the Resulting Equation

Integrate both sides with respect to \( x \): \( \int \frac{d}{dx}(x y) \, dx = \int x \ln x \, dx \). The left-hand side integrates to \( x y \). For the right-hand side, use integration by parts: Let \( u = \ln x \) and \( dv = x \, dx \), then \( du = \frac{1}{x} \, dx \) and \( v = \frac{x^2}{2} \). The integral becomes \[ x^2 \ln x / 2 - \int x^2/2 \cdot 1/x \, dx = \frac{x^2 \ln x}{2} - \frac{x^2}{4}. \]

Step 6: Solve for y

After integration, we obtain \( xy = \frac{x^2 \ln x}{2} - \frac{x^2}{4} + C \). Solve for \( y \) to get \( y = \frac{x \ln x}{2} - \frac{x}{4} + \frac{C}{x} \).

Step 7: Apply Initial Condition

Use the initial condition \( y(1) = 0 \): Substitute \( x = 1 \) and \( y = 0 \) into \( y = \frac{x \ln x}{2} - \frac{x}{4} + \frac{C}{x} \). This gives \( 0 = \frac{1 \cdot \ln 1}{2} - \frac{1}{4} + C/1 \). Since \( \ln 1 = 0 \), solve \( 0 = 0 - \frac{1}{4} + C \) to find \( C = \frac{1}{4} \).

Step 8: Write the Final Solution

Substitute the value of \( C \) back into the expression for \( y \), giving \( y = \frac{x \ln x}{2} - \frac{x}{4} + \frac{1}{4x} \).

Key Concepts

Linear First-order Differential Equations

Linear first-order differential equations are a specific type of differential equation with unique properties that make them distinct and meaningful in mathematical modeling.
These equations have the general form:

• \( y' + P(x)y = Q(x) \)

In this equation, \( y' \) represents the derivative of \( y \) with respect to \( x \), \( P(x) \) is a function of \( x \) multiplied by the unknown function \( y \), and \( Q(x) \) is another function of \( x \).In the given problem, the differential equation \( xy' + y = x \ln x \) is nonlinear as it appears initially. However, by dividing every term by \( x \), it can be rewritten in the standard linear form:

• \( y' + \frac{1}{x}y = \ln x \)

This transformation is key to solving linear differential equations as it allows the use of integrating factors.
Understanding this framework can help in recognizing and organizing differential equations in scientific and engineering problems.

Integrating Factor

An integrating factor is a vital tool for solving linear first-order differential equations. It is a function, usually denoted by \( \mu(x) \), which is used to simplify the equation into an easily integrable form.
To find the integrating factor, derive it from:

• \( \mu(x) = e^{\int P(x) \, dx} \)

In this exercise, \( P(x) = \frac{1}{x} \). Calculating the integrating factor involves integrating this function:

• \( \mu(x) = e^{\int \frac{1}{x} \, dx} = e^{\ln|x|} = |x| \)

Because \( x > 0 \) in this context, the integrating factor simplifies to \( x \).
Applying the integrating factor to the differential equation enables us to write it as a derivative of a product:

• \( \frac{d}{dx}(xy) = x \ln x \)

This form makes it straightforward to integrate both sides, illustrating the power of the integrating factor in simplifying the solution process of differential equations.

Initial Conditions

Initial conditions are essential when solving differential equations, as they allow us to find specific solutions from a family of general solutions.
In the context of initial-value problems, these conditions specify the value of the unknown function at a particular point.In our problem, the initial condition is given by \( y(1) = 0 \). This means that when \( x = 1 \), the corresponding value of \( y \) is 0.
After finding the general solution:

• \( y = \frac{x \ln x}{2} - \frac{x}{4} + \frac{C}{x} \)

Substitute the initial values (\( x = 1 \) and \( y = 0 \)) into the solution to determine \( C \):

• \( 0 = \frac{1 \cdot \ln 1}{2} - \frac{1}{4} + C \)
• \( 0 = 0 - \frac{1}{4} + C \)

Thus, \( C = \frac{1}{4} \).Finally, substituting \( C \) back yields the specific solution:

• \( y = \frac{x \ln x}{2} - \frac{x}{4} + \frac{1}{4x} \)

This process of using initial conditions combines calculus and algebra, making it foundational in solving differential equations for real-world applications.