91Ó°ÊÓ

Understanding first-order linear differential equations is essential in calculus because they are common in various applications such as physics, engineering, and economics. A first-order linear differential equation has the general form:

• \( y' + P(x)y = Q(x) \)

In this structure:

• \( y' \) represents the derivative of \( y \) with respect to \( x \).
• \( P(x) \) and \( Q(x) \) are functions of \( x \), which can be constants or any expression involving \( x \).

Recognizing the equation in this form is crucial, as it dictates the approach to solve it. In the original exercise, we identify the equation:

• \( y' + \frac{2y}{x+1} = (x+1)^3 \)

as being first-order linear by comparing it to the general format, where \( P(x) = \frac{2}{x+1} \) and \( Q(x) = (x+1)^3 \). By solving such equations, we can predict behaviors, such as population growth in biology or heat distribution in a rod.

An integrating factor is a special function used to simplify the process of solving first-order linear differential equations. It is denoted as \( \mu(x) \) and calculated using the formula:\\[ \mu(x) = e^{\int P(x) \, dx} \]The purpose of the integrating factor is to transform the differential equation into an easier form that can be solved by straightforward integration. In the given problem, with \( P(x) = \frac{2}{x+1} \), we find the integrating factor as:

• \( \mu(x) = e^{\int \frac{2}{x+1} \, dx} = (x+1)^2 \)

This integrating factor is then used to multiply the entire differential equation, which turns it into a derivative of a product on the left-hand side. This maneuver simplifies solving the equation down the line because it allows us to express the left part as a single derivative of the product \( (x+1)^2 y \). This makes the equation much more manageable and prepares it for integration.

The constant of integration, often denoted as \( C \) in calculus, is a critical component when integrating differential equations. It arises when calculating indefinite integrals, acknowledging the fact that there can be infinitely many antiderivatives differing by a constant.
In the final stages of solving the given differential equation, after integrating both sides, we get:

• \( (x+1)^2 y = \frac{(x+1)^6}{6} + C \)

This constant \( C \) represents an arbitrary integration constant that can potentially be determined if initial conditions or additional constraints are provided. If no such information is available, \( C \) is left as an arbitrary constant, showing the general solution.
When we further solve for \( y \), the solution becomes:

• \( y = \frac{(x+1)^4}{6} + \frac{C}{(x+1)^2} \)

The presence of \( C \) in the solution reflects the flexibility and range of particular solutions the general solution represents. It's essential in ensuring that the solution can fit various initial values or boundary conditions defined by different real-world problems.