91Ó°ÊÓ

First-order linear differential equations are equations involving a function and its first derivative. These equations are highly prevalent in mathematics and physics, making them essential for students to understand. A standard way to express these equations is:\[ \frac{\mathrm{d} y}{\mathrm{d} x} + P(x)y = Q(x) \]Here, \( P(x) \) and \( Q(x) \) are functions of \( x \). In the given exercise, the original equation was \( x \frac{\mathrm{d} y}{\mathrm{~d} x} - y = x^{2} \). To convert it into the standard form, divide by \( x \):\[ \frac{\mathrm{d} y}{\mathrm{d} x} - \frac{y}{x} = x \]In this form, it is evident that:

• \( P(x) = -\frac{1}{x} \)
• \( Q(x) = x \)

This equation can now be approached using methods for solving first-order linear differential equations, one of which involves integrating factors.

An integrating factor is a crucial tool to simplify first-order linear differential equations. It transforms the differential equation into a format where the left-hand side becomes the derivative of a single term. This simplifies integration.To find the integrating factor \( \mu(x) \), use the following formula:\[ \mu(x) = e^{\int P(x) \, dx} \]In the exercise, \( P(x) = -\frac{1}{x} \), which leads to:\[ \mu(x) = e^{\int -\frac{1}{x} \, dx} = e^{\ln |x|^{-1}} = \frac{1}{x} \]Now, multiply the entire differential equation by this integrating factor:\[ \frac{1}{x} \frac{\mathrm{d} y}{\mathrm{d} x} - \frac{y}{x^2} = 1 \]Or:\[ \frac{d}{dx} \left( \frac{y}{x} \right) = 1 \]By using the integrating factor, the original differential equation has been transformed so its integration becomes straightforward.

The general solution of a differential equation represents all possible solutions, incorporating the constant of integration. After transforming the original problem via the integrating factor, we found:\[ \frac{d}{dx} \left( \frac{y}{x} \right) = 1 \]Integrating both sides with respect to \( x \):\[ \frac{y}{x} = x + C \]This expression leads to the general solution by multiplying both sides by \( x \):\[ y = x^2 + Cx \]In this equation, \( C \) is an arbitrary constant, allowing for infinitely many solutions based on initial conditions. It offers a comprehensive understanding of the behavior and solutions of the differential equation:

• The \( x^2 \) term indicates the part of the solution varying quadratically with \( x \).
• The \( Cx \) term represents a family of solutions dependent on the constant \( C \).

Understanding the general solution provides insight into how differential equations model dynamic systems and predict future outcomes.