Question

Solve the following initial-value problems by using integrating factors. $$ x y^{\prime}=y+2 x \ln x, y(1)=5 $$

Short answer

\(y = x(\ln x)^2 + 5x\).

Step-by-step solution

Rewrite the Equation

Rewrite the given differential equation in the standard linear form. Divide every term by \(x\) to obtain:\[y' - \frac{1}{x}y = 2\ln x\]This is in the form \(y' + P(x)y = Q(x)\) where \(P(x) = -\frac{1}{x}\) and \(Q(x) = 2\ln x\).

Compute the Integrating Factor

To find the integrating factor, \(\mu(x)\), calculate \(e^{\int P(x) \, dx}\), which is:\[\mu(x) = e^{-\int \frac{1}{x} \, dx} = e^{- ext{ln} |x|} = \frac{1}{x}.\]The integrating factor is \(\mu(x) = \frac{1}{x}\).

Multiply the Equation by Integrating Factor

Multiply every term in the differential equation by the integrating factor \(\frac{1}{x}\):\[\frac{1}{x}y' - \frac{1}{x^2}y = \frac{2\ln x}{x}.\]This simplifies to:\[\left(\frac{y}{x}\right)' = \frac{2\ln x}{x}.\]

Integrate Both Sides

Integrate both sides of the equation:\[\int \left(\frac{y}{x}\right)' \, dx = \int \frac{2\ln x}{x} \, dx.\]The left side integrates to \(\frac{y}{x}\), while the right side integrates to:\[2\int \ln x \, \frac{1}{x} \, dx = 2\cdot \left(\ln x\right)^2/2 = (\ln x)^2.\]Thus, we have:\[\frac{y}{x} = (\ln x)^2 + C.\]

Solve for the General Solution

Multiply through by \(x\) to solve for \(y\):\[y = x(\ln x)^2 + Cx.\]This is the general solution.

Apply Initial Condition

Use the initial condition \(y(1) = 5\) to find \(C\):\[5 = 1(\ln 1)^2 + C\cdot 1.\]Since \(\ln 1 = 0\), the equation simplifies to \(5 = C\).

Write the Specific Solution

Substitute \(C = 5\) back into the general solution to get the particular solution:\[y = x(\ln x)^2 + 5x.\]

Key concepts

Linear Differential Equations

Linear differential equations are a type of differential equation in which the unknown function and its derivatives appear linearly. This means that each term is either a constant, the derivative of the unknown function, or the unknown function itself multiplied by some function of the independent variable.
For example, in the given exercise, we have a linear differential equation in the form

• \( x y' = y + 2x \ln x \)

Here, the equation is not initially in standard linear form, so we first need to divide every term by the coefficient of \(y'\), which is \(x\), to make it look like:

• \( y' - \frac{1}{x}y = 2\ln x \)

This format is crucial because it prepares the equation for solving through certain methods, such as the integrating factor method.

Initial Value Problems

Initial value problems (IVPs) are differential equations that come with a specified value for the unknown function at a particular point. This added information allows us to not only write a general solution but also find a specific solution that matches the initial conditions.
In this exercise, we have:

• \( y(1) = 5 \)

The initial condition helps pin down the constant of integration after we've obtained the general solution. This step is essential because differential equations can often have multiple solutions, and the initial condition allows us to find the unique solution that fits the real-world scenario being modeled. Using the initial condition inputs like \( x = 1 \) and \( y = 5 \), we solved for constant \( C \) to finalize our solution.

Integrating Factor Method

The integrating factor method is a powerful technique used to solve linear first-order differential equations. It simplifies the equation into a form that is easily integrable.
This method involves:

• Identifying the function \( P(x) \) from the standard linear equation format \( y' + P(x) y = Q(x) \).
• Calculating the integrating factor \( \mu(x) \) as: \[ \mu(x) = e^{\int P(x) \, dx} \] . In our case, \( P(x) = -\frac{1}{x} \), so the integrating factor \( \mu(x) \) becomes \( \frac{1}{x} \).
• Multiplying through the entire original differential equation by this integrating factor to transform the left side into a derivative of a product.
• Integrating both sides, leading to solutions for \( y \).

The integrating factor method is a map from complexity toward simplicity, allowing one to solve equations thoroughly and efficiently.

Ordinary Differential Equations

Ordinary differential equations (ODEs) involve unknown functions of a single independent variable and their derivatives. Unlike partial differential equations, which involve multiple independent variables, ODEs deal with one dimension of change and are foundational to many areas of mathematics and science.
Our journey through the given problem showcases a first-order linear ODE:

• Where only the first derivative, \( y' \), and the function \( y \) itself appear in the linear ways alongside the terms involving \( x \), the independent variable.

ODEs are widely used to represent relations like population growth, heat transfer, and in this specific case, could represent various real applications where logarithmic growth under constraints occurs.
By learning the techniques like the integrating factor, we gain tools to crack any first-order linear ordinary differential equation, solving them while bearing in mind any initial conditions provided.