91Ó°ÊÓ

Linear differential equations are a fundamental part of calculus and are used to describe a wide range of phenomena in science and engineering. A first-order linear differential equation has the form \( y' + P(x)y = Q(x) \). Here, \( y' \) is the derivative of \( y \) with respect to \( x \), \( P(x) \) and \( Q(x) \) are functions of \( x \). This structure indicates that \( y \) and its derivative appear linearly, meaning they are not raised to any power other than one.

In the given exercise, the equation \( x y' + 3y = \frac{\sin x}{x^2} \) is transformed into standard linear form by dividing the entire equation by \( x \), resulting in \( y' + \frac{3}{x}y = \frac{\sin x}{x^3} \). This transformation helps to identify \( P(x) = \frac{3}{x} \) and \( Q(x) = \frac{\sin x}{x^3} \).

• Always ensure the equation is in standard form to apply solution techniques.
• Linear differential equations can be solved using integrating factors, as shown in subsequent steps.

The integrating factor is a powerful tool used to solve linear first-order differential equations. The goal is to make the left-hand side of the equation into the exact derivative of a product, usually involving the function \( y \) and some function of \( x \).

To compute the integrating factor \( \mu(x) \), use the formula \( e^{\int P(x) \ dx} \). For our exercise, this becomes \( e^{\int \frac{3}{x} \ dx} \). Evaluate the integral to get \( 3\ln|x| = \ln|x^3| \). Thus, the integrating factor is \( x^3 \).

• The integrating factor \( \mu(x) \) is always exponential.
• Multiplying through by \( \mu(x) \) transforms the equation's left side into a single derivative.

When you multiply the entire differential equation by the integrating factor \( x^3 \), the equation becomes \( x^3 y' + 3x^2 y = \sin x \). This equation shows that the left-hand side is the derivative of \( x^3 y \). This is a crucial step as it allows for simpler integration of the differential equation.

The general solution of a differential equation contains all possible solutions, including constants of integration that appear after integration. In this exercise, once the left side of the modified equation \( x^3 y' + 3x^2 y = \sin x \) is recognized as the derivative \( \frac{d}{dx}(x^3 y) \), we can integrate both sides easily.

With \( \frac{d}{dx}(x^3 y) = \sin x \), integrate with respect to \( x \) to solve for \( x^3 y \). The integral of \( \sin x \) is \( -\cos x \), yielding \( x^3 y = -\cos x + C \), where \( C \) is a constant. To express \( y \) explicitly, divide both sides by \( x^3 \), resulting in \( y = \frac{-\cos x + C}{x^3} \).

• Always include the constant of integration \( C \) in your general solution.
• The general solution provides a family of solutions that satisfy the differential equation for any value of \( C \).

Understanding that \( C \) allows for flexibility in fitting specific initial conditions or constraints makes the general solution a fundamental tool in mathematical modeling.