91Ó°ÊÓ

A first-order linear differential equation is one of the simplest types of differential equations that students encounter in calculus. Its general form is \( y' + P(x)y = Q(x) \), where:

• \( y' \) is the derivative of \( y \) with respect to \( x \)
• \( P(x) \) and \( Q(x) \) are functions of \( x \)

These equations are called "first-order" because they involve the first derivative of the function but not higher derivatives like the second or third.
To recognize a first-order linear differential equation, look for the structure: a derivative term \( y' \) added to a product of a function \( P(x) \) with the unknown function \( y \), set equal to another function \( Q(x) \). You don't have to memorize the form as long as you understand that it represents a linear relationship involving the derivative.

The integrating factor method is a technique used to solve first-order linear differential equations. It involves finding a special function, called the integrating factor, which simplifies the process of solving the equation.
In the formula \( y' + P(x)y = Q(x) \), the integrating factor \( \mu(x) \) is computed as \( e^{\int P(x)\, dx} \). For many problems, especially like the one in the exercise where \( P(x) = 0 \), the integrating factor becomes very simple: \( \mu(x) = e^{0} = 1 \).
With the integrating factor, you can transform the differential equation into a format that's easier to integrate and solve:

• Multiply every term in the equation by \( \mu(x) \) to simplify the integration.
• In this scenario, since \( \mu(x) = 1 \), the equation remains \( y' = 2 - 4x \).
• Direct integration becomes possible, leading you to the general solution.

After integrating and finding the general solution for a differential equation, determining a particular solution requires applying the given initial conditions.

• The general solution typically contains an arbitrary constant, \( C \), which represents a family of solutions.

To find the particular solution, use the initial condition in the problem, which often looks like \( y(x_0) = y_0 \). Substitute \( x_0 \) and \( y_0 \) into the general equation to solve for \( C \).
In the exercise, the initial condition is \( y(0) = 3 \). By substituting \( x = 0 \) and \( y = 3 \) into the equation \( y = 2x - 2x^2 + C \), we find:\[ 3 = C \]This gives \( C = 3 \), allowing us to write the particular solution: \( y = 2x - 2x^2 + 3 \).
The particular solution satisfies both the differential equation and the initial condition, making it the specific answer to the initial value problem.