91Ó°ÊÓ

A first-order linear differential equation is an equation involving a function and its first derivative. These types of equations have the general form: \[ a(x) y' + b(x) y = c(x) \] where \( a(x), b(x), \text{ and } c(x) \) are given functions of \( x \). In our exercise, the equation is: \[ 2x y' + y = 6x \]
This equation is first-order because it involves only the first derivative \( y' \), and it's linear because both \( y \) and \( y' \) appear to the power of one.
Solving these equations often requires recognizing them in a particular standard form to apply methods like integrating factor, which we'll explore next.

The integrating factor is a useful tool for solving first-order linear differential equations. It simplifies the process by converting the equation into a form that can be easily integrated.
To find an integrating factor, we usually look at the differential equation after it's been rearranged to standard form: \[ y' + p(x) y = q(x) \]
For our specific example, that becomes: \[ y' + \frac{1}{2x} y = 3 \]
The integrating factor is decided using: \[ \mu(x) = e^{\int p(x) \, dx} \] For our problem, this is: \[ \mu(x) = e^{\int \frac{1}{2x} \, dx} = x^{1/2} \]
Once the integrating factor \( x^{1/2} \) is applied, it transforms the equation so that we can integrate both sides easily.

The particular solution of a differential equation addresses the specific condition of the problem. This is achieved by applying an initial condition, which yields a specific \( C \) in the general solution.
For our problem, after finding the general solution: \[ y = 2x + \frac{C}{x^{1/2}} \] the initial condition \( y(4) = 20 \) is used.
Substituting \( x = 4 \) and \( y = 20 \): \[ 20 = 2\times4 + \frac{C}{4^{1/2}} \] simplifies to \( C = 24 \).
Thus, the particular solution is: \[ y = 2x + \frac{24}{x^{1/2}} \] which satisfies the initial condition provided. It is specific to the given problem with its exact context.

Initial conditions are crucial in determining the particular solution of a differential equation. They allow us to find the exact solution that fits specific criteria.
In the original exercise, the initial condition given is: \( y(4) = 20 \) This means when \( x = 4 \), \( y \) should equal 20.
To apply this: start by solving the general solution until the constant \( C \) appears. Then substitute the \( x \) and \( y \) values from the initial condition to solve for \( C \).
This approach integrates the boundary condition into the equation, producing a unique solution tailored to the problem's context.