Question

Draw the directional field associated with the differential equation, then solve the differential equation. Draw a sample solution on the directional field. $$ y^{\prime}=\frac{1}{x}+\ln x-y, \text { for } x>0 $$

Short answer

Solve by using an integrating factor and draw a directional field with a sample solution.

Step-by-step solution

Understand the Differential Equation

The given differential equation is \( y' = \frac{1}{x} + \ln x - y \). This equation is a first-order linear differential equation. The terms involved are \( \frac{1}{x} \), \( \ln x \), and \( -y \). Our task is to analyze this equation to draw both the directional field and find its solution.

Analyze the Directional Field

A directional field shows how the slope \( y' \) behaves at different points \((x, y)\). We need to substitute various values of \( x \) and \( y \) to determine the slope \( y' \). For example, if \( x = 1 \) and \( y = 0 \) then the slope \( y' = \frac{1}{1} + \ln 1 - 0 = 1 \). Repeat for different values of \( x \) and \( y \) to fill out the field.

Solve the Differential Equation

The differential equation \( y' = \frac{1}{x} + \ln x - y \) can be written in standard form \( y' + y = \frac{1}{x} + \ln x \). This can be solved using an integrating factor. Let \( \, I(x) = e^{\int 1\,dx} = e^x \, \). Multiply through by the integrating factor to obtain \( e^x y' + e^x y = e^x \left(\frac{1}{x} + \ln x\right) \), and integrate both sides.

Integrate Using Integration by Parts

The integration of \( e^x \left( \frac{1}{x} + \ln x \right) \) requires breaking it into two parts. Use integration by parts with \( u = \ln x \) and \( dv = \frac{1}{x} e^x dx \), leading to \( v = \ln x \cdot e^x - \int \frac{e^x}{x} dx \). Integrate this expression step by step.

Solve for y

After integrating, solve for \( y \) by dividing both sides by \( e^x \). The general solution involves constant terms which arise from the integration process, denoted as \( C \). Substitute back to verify the solution if necessary.

Draw a Sample Solution on the Directional Field

Using the solution from Step 5, plot a sample solution curve on the directional field. This will illustrate how a specific solution to the differential equation behaves, tracing one pathway through the field.

Key concepts

Differential Equations

Differential equations are mathematical equations that relate a function with its derivatives. In this context, we are dealing with a first-order linear differential equation of the form \( y' + P(x)y = Q(x) \). These equations are used in various fields such as physics, engineering, and economics to model real-world systems. Understanding the behavior of the solution, even without solving the equation analytically, can be achieved by creating a directional field.

A directional field provides a visual representation of the slopes of the solution curves at various points in the plane. By plotting the direction vectors, students can grasp the general behavior of solutions. These fields are particularly helpful when it might be difficult or impossible to directly integrate and solve the equation analytically.

• The slope at each point is calculated using the differential equation.
• These slopes suggest how solutions will behave over the plane.
• By observing these directions, one can draw possible solution curves.

Integrating Factor

To solve a linear first-order differential equation such as \( y' + y = \frac{1}{x} + \ln x \), an integrating factor is used. The integrating factor is a function that, when multiplied with the differential equation, simplifies it such that the left-side turns into a product rule, making integration straightforward.

The standard formula for finding an integrating factor for an equation \( y' + P(x)y = Q(x) \) is given by \( I(x) = e^{\int P(x) \, dx} \). This factor, when applied, transforms the equation:

• Multiply every term by the integrating factor \( I(x) \).
• The left side becomes \( (I(x)y)' \).
• Integrate both sides with respect to \( x \) to solve for \( y \).

In our example, where \( P(x) = 1 \), the integrating factor would be \( e^x \). This application simplifies our example and allows the solution to be expressed more readily in terms of \( x \).

Integration by Parts

Integration by parts is a technique used when faced with the product of two functions that makes straightforward integration challenging. It relies on the formula:\[\int u \, dv = uv - \int v \, du\]This technique can be particularly useful when dealing with expressions inside a differential equation that do not easily fall into basic integration formulas.

In the context of solving the differential equation \( y' = \frac{1}{x} + \ln x - y \), integration by parts was employed to manage the troublesome integral \( \int e^x \left( \frac{1}{x} + \ln x \right) \). A common strategy is to choose:

• \( u \) as a function that simplifies when differentiated.
• \( dv \) as a function that simplifies when integrated.

In this example:

• Let \( u = \ln x \) and \( dv = \frac{1}{x}e^x\, dx \). This results in simplifying the integration of the product.
• The selection of \( u \) and \( dv \) is crucial for simplifying the integral.
• Compute \( du \) and \( v \), and substitute into the formula to solve.

Through integration by parts, complex components can be broken down, thus aiding the overall process of integrating and solving the differential equation.