Question

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

Short answer

The solution is \( y = \frac{1}{2x} \ln x - \frac{1}{4x} + \frac{5}{4}x \).

Step-by-step solution

Identify the form of the equation

The given differential equation is \( x^2 y' = x y - \ln x \). This can be rewritten in the standard linear form as \( y' + P(x) y = Q(x) \). Start by dividing every term by \( x^2 \).

Rewrite the equation

Divide the entire equation by \( x^2 \) to make it a linear differential equation: \( y' - \frac{1}{x} y = - \frac{\ln x}{x^2} \). Now it is in the standard form \( y' + P(x) y = Q(x) \) with \( P(x) = -\frac{1}{x} \) and \( Q(x) = -\frac{\ln x}{x^2} \).

Find the integrating factor

The integrating factor \( \mu(x) \) is given by \( e^{\int P(x) \: dx} \). Calculate \( \mu(x) = e^{-\int \frac{1}{x} \: dx} = e^{-\ln|x|} = \frac{1}{x} \).

Multiply through by the integrating factor

Multiply the whole differential equation by the integrating factor \( \frac{1}{x} \): \(\frac{1}{x} y' - \frac{1}{x^2} y = -\frac{\ln x}{x^3} \). This allows the equation to be rewritten in a convenient form.

Simplify using the integrating factor property

The left side of the equation \( \frac{1}{x} y' - \frac{1}{x^2} y \) becomes the derivative of \( \frac{y}{x} \), i.e., \( \frac{d}{dx} \left( \frac{y}{x} \right) = -\frac{\ln x}{x^3} \).

Integrate both sides

Integrate both sides of the equation: \( \int \frac{d}{dx} \left( \frac{y}{x} \right) \: dx = \int -\frac{\ln x}{x^3} \: dx \). The left side integrates to \( \frac{y}{x} \).

Solve the integral

Solve the integral on the right, which involves integration by parts. Let \( u = \ln x \) and \( dv = -\frac{1}{x^3} dx \). Then, \( du = \frac{1}{x} dx \) and \( v = \frac{1}{2x^2} \). Hence, the integral is \( \frac{1}{2x^2} \ln x + \int \frac{1}{2x^3} dx = \frac{1}{2x^2} \ln x - \frac{1}{4x^2} \).

Write the general solution

Combine results to get \( \frac{y}{x} = \frac{1}{2x^2} \ln x - \frac{1}{4x^2} + C \), thus \( y = \frac{1}{2x} \ln x - \frac{1}{4x} + Cx \).

Apply the initial condition

Use the initial condition \( y(1)=1 \) to find the constant \( C \). Substituting \( x=1 \) and \( y=1 \), you obtain: \( 1 = 0 - \frac{1}{4} + C \), leading to \( C = \frac{5}{4} \).

Substitute the constant back

Substitute \( C \) back into the general solution to find the specific solution: \( y = \frac{1}{2x} \ln x - \frac{1}{4x} + \frac{5}{4}x \).

Key concepts

Initial-Value Problems

Initial-value problems in differential equations involve finding a specific solution to a differential equation that satisfies an initial condition. An initial condition provides a specific value for the function at a particular point. This condition helps us determine the exact form of the solution that corresponds to real-world situations or specific questions we are asked to solve.

For example, if you have a differential equation and a condition like \( y(1) = 1 \), it means that when \( x = 1 \), the value of \( y \) should be 1. In our problem, the initial condition allowed us to find the constant \( C \) in the solution \( y = \frac{1}{2x} \ln x - \frac{1}{4x} + Cx \).

This initial condition is essential because without it, the solution would have an arbitrary constant, leading to infinite possible solutions. By applying the initial condition, we narrow down these infinitely many solutions to just one.

Linear Differential Equations

Linear differential equations are essential in mathematics as they represent many physical phenomena. These equations have a standard form \( y' + P(x)y = Q(x) \), where both \( P(x) \) and \( Q(x) \) are given functions of \( x \). The key feature here is that the highest derivative (\( y' \)) is multiplied by the function itself \( y \).

In practice:

• The differential equation is considered linear because \( y \) and its derivatives appear linearly (raised only to the first power).
• We use techniques like multiplying through by an integrating factor to simplify or solve the equation.

Our example equation \( y' - \frac{1}{x} y = -\frac{\ln x}{x^2} \) is in linear form. By manipulating the equation with integrating factors, we can turn it into a form where the solution is more visible, and therefore, it is easier to solve.

Integration by Parts

Integration by parts is a mathematical technique used to integrate products of functions. It is based on the formula \( \int u \, dv = uv - \int v \, du \). This method is particularly useful when we deal with functions multiplied together that are not easily integrable by standard means.

In our exercise, we solve \( \int -\frac{\ln x}{x^3} \, dx \) using integration by parts. Setting \( u = \ln x \) and \( dv = -\frac{1}{x^3} \, dx \), we calculate \( du = \frac{1}{x} \, dx \) and find \( v = \frac{1}{2x^2} \).

The resulting integration becomes \( \frac{1}{2x^2} \ln x \), minus the integral of \( \int \frac{1}{2x^3} \, dx \), which gives \( \frac{1}{4x^2} \). Combining these, we get the simplified expression needed for our solution.