In Exercises 1–22, solve the differential equation.
xy' - y = 2x ln x

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Rewrite the given differential equation in the standard linear form. Starting with \(xy' - y = 2x \ln x\), divide both sides by \(x\) (assuming \(x \neq 0\)) to get \(y' - \frac{1}{x} y = 2 \ln x\).

Identify the integrating factor \(\mu(x)\), which is given by \(\mu(x) = e^{\int P(x) \, dx}\) where \(P(x)\) is the coefficient of \(y\) in the standard form. Here, \(P(x) = -\frac{1}{x}\), so calculate \(\mu(x) = e^{\int -\frac{1}{x} \, dx}\).

Simplify the integrating factor \(\mu(x)\) by evaluating the integral \(\int -\frac{1}{x} \, dx\) and then exponentiating the result.

Multiply the entire differential equation by the integrating factor \(\mu(x)\) to transform the left side into the derivative of the product \(\mu(x) y\). This gives \(\frac{d}{dx} [\mu(x) y] = \mu(x) \cdot 2 \ln x\).

Integrate both sides with respect to \(x\) to find \(\mu(x) y = \int \mu(x) \cdot 2 \ln x \, dx + C\), where \(C\) is the constant of integration. Finally, solve for \(y\) by dividing both sides by \(\mu(x)\).

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주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

First-Order Linear Differential Equations

A first-order linear differential equation has the form y' + P(x)y = Q(x). Solving it involves finding an integrating factor to simplify the equation into an exact derivative, allowing integration to find the general solution.

Integrating Factor Method

The integrating factor is a function, usually denoted μ(x), used to multiply both sides of a linear differential equation to make the left side an exact derivative. It is typically μ(x) = e^(∫P(x) dx), facilitating straightforward integration.

Properties of Logarithmic Functions

Understanding the natural logarithm function ln(x) is essential, especially its domain (x > 0) and differentiation properties. This knowledge helps in manipulating terms like 2x ln x and ensuring the solution respects the function's domain.

In Exercises 1–22, solve the differential equation.xy' - y = 2x ln x