Question

Solve the following differential equations by using integrating factors. $$ y^{\prime}+\ln x=\frac{y}{x} $$

Short answer

The solution to the differential equation is \( y(x) = -\frac{x}{2} (\ln x)^2 + Cx \).

Step-by-step solution

Rewrite the Equation in the Standard Form

We start with the given differential equation \( y' + \ln x = \frac{y}{x} \). The goal is to rewrite it in the standard linear form \( y' + P(x)y = Q(x) \). First, simplify the equation by rearranging it as \( y' - \frac{y}{x} = -\ln x \). Here, we can see that \( P(x) = -\frac{1}{x} \) and \( Q(x) = -\ln x \).

Find the Integrating Factor

The integrating factor \( \mu(x) \) is given by the formula \( e^{\int P(x) \, dx} \). In this case, \( \int P(x) \, dx = \int -\frac{1}{x} \, dx = -\ln x \). Therefore, the integrating factor is \( \mu(x) = e^{-\ln x} = \frac{1}{x} \).

Multiply the Equation by the Integrating Factor

Multiply the entire differential equation \( y' - \frac{y}{x} = -\ln x \) by the integrating factor \( \frac{1}{x} \), obtaining \( \frac{y'}{x} - \frac{y}{x^2} = -\frac{\ln x}{x} \). This step ensures that the left-hand side becomes the derivative of the product \( \frac{y}{x} \).

Write the Left Side as a Derivative

The left-hand side of the equation \( \frac{y'}{x} - \frac{y}{x^2} \) can now be written as \( \frac{d}{dx} \left( \frac{y}{x} \right) \). Thus, our equation becomes \( \frac{d}{dx} \left( \frac{y}{x} \right) = -\frac{\ln x}{x} \).

Integrate Both Sides

Integrate both sides with respect to \( x \): \( \int \frac{d}{dx} \left( \frac{y}{x} \right) \, dx = \int -\frac{\ln x}{x} \, dx \). The left-hand side simplifies to \( \frac{y}{x} \). The integral of \( -\frac{\ln x}{x} \) can be solved using integration by parts, yielding \( \frac{y}{x} = -\frac{1}{2} (\ln x)^2 + C \), where \( C \) is the constant of integration.

Solve for \( y \)

To find \( y \), multiply both sides of the equation \( \frac{y}{x} = -\frac{1}{2} (\ln x)^2 + C \) by \( x \): \( y(x) = -\frac{x}{2} (\ln x)^2 + Cx \). This expression for \( y \) is the solution to the original differential equation.

Key concepts

Integrating Factors

When given a linear first-order differential equation, one powerful tool we use is the integrating factor. This technique simplifies solving the equation by transforming it into a format that's easier to manage. The goal is to multiply the entire differential equation by a carefully chosen function, known as the integrating factor.

• The integrating factor is derived using the formula \( \mu(x) = e^{\int P(x) \ dx} \), where \( P(x) \) is a function from the rearranged form of the differential equation.
• In our exercise, once the equation was rearranged to \( y' - \frac{y}{x} = -\ln x \), we identified \( P(x) = -\frac{1}{x} \).
• Finding the integral \( \int -\frac{1}{x} \ dx \) resulted in \( -\ln x \), leading to an integrating factor \( \mu(x) = e^{- \ln x} = \frac{1}{x} \).

After finding \( \mu(x) \), it's used to multiply every term in the equation, which helps in consolidating parts of the equation into a derivative of a single term. This step is essential for simplifying and ultimately solving the differential equation.

Linear Differential Equations

Linear differential equations are a type of differential equation where the unknown function and its derivatives appear linearly (i.e., not multiplied together or raised to a power). These equations are crucial in mathematical modeling since they appear frequently in many applications, such as physics and engineering.

• The standard form of a linear first-order differential equation is \( y' + P(x)y = Q(x) \).
• In the given exercise, after rearranging, the equation became \( y' - \frac{y}{x} = -\ln x \), fitting the linear equation form with \( P(x) = -\frac{1}{x} \) and \( Q(x) = -\ln x \).
• By working through the standard form and applying the integrating factor, the solution is systematically derived, allowing effective use of calculus tools.

Understanding the structure of linear differential equations ensures we can identifiy the right strategy and techniques, such as using integrating factors, to reach a solution.

Integration by Parts

Integration by parts is a technique used in calculus to integrate the product of two functions. It is derived from the product rule of differentiation and is a valuable tool when dealing with more complex integrals.

• The formula for integration by parts is \( \int u \ dv = uv - \int v \ du \), where \( u \) and \( dv \) are parts of the original integral being solved.
• In our differential equation exercise, we used integration by parts to solve \( \int -\frac{\ln x}{x} \ dx \).
• This step resulted in the integral solution: \( -\frac{1}{2} (\ln x)^2 \). Here, \( u = \ln x \) and \( dv = -\frac{1}{x} \ dx \) were chosen, leading to an easier integration process.

Integration by parts can sometimes involve several steps, so it's important to appropriately choose \( u \) and \( dv \) to simplify the problem. This process is particularly useful when integrating expressions involving logarithms, exponentials, or polynomials.