91Ó°ÊÓ

In the context of first-order linear differential equations, the integrating factor is a crucial tool that simplifies the process of solving these equations. The method works especially well for equations of the form \[ y' + P(x) y = Q(x). \]
This structure helps systematically find solutions by transforming the equation into a more manageable form.

• The purpose of the integrating factor, often denoted as \( \mu(x) \), is to enable the left-hand side of the differential equation to be written as the derivative of a product, simplifying the integration.
• The integrating factor is calculated as \( \mu(x) = e^{\int P(x) \, dx} \).

In our exercise:
The integrating factor is \( \mu(x) = e^{\frac{(r^2 + 1)x^2}{2}} \).
Multiplying the entire differential equation by this factor enables us to write the left side as the derivative \( \frac{d}{dx} \left( e^{\frac{(r^2 + 1)x^2}{2}} y \right) \), making it easier to solve.

Initial conditions are fundamental in determining the specific solution to a differential equation from a set of possible solutions. An initial condition provides a specific value for the solution at a particular point.

• For our problem, the initial condition is given as \( y(0) = 0 \).
• This condition allows us to solve for any constants that arise during the integration process.

During the solution of our differential equation, after finding the general solution involving an unknown constant \( C \), we used the initial condition to determine that \( C = 0 \). By applying the initial condition, we narrowed down the solution to a single, specific answer out of the infinite family of solutions. This step is vital, as it provides the exact solution relevant to the given conditions.

The standard form of a first-order linear differential equation is crucial for recognizing how to apply various solving techniques. The standard form is typically written as \[ y' + P(x) y = Q(x). \]
In this setup:

• \( P(x) \) is a function that multiplies the dependent variable.
• \( Q(x) \) is a function on the other side of the equation.

For our exercise, the original equation \[ \frac{1}{r^{2}+1} y^{\prime}+x y=3 \] needed to be rewritten into standard form.
We achieved this by multiplying through by \( r^2 + 1 \), which eliminated the fraction, yielding \[ y' + (r^2 + 1) x y = 3 (r^2 + 1). \] Having it in standard form allowed us to correctly identify \( P(x) = (r^2 + 1) x \) and apply the integrating factor method effectively. It's a pivotal step that often defines the success of solving the equation.