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An integrating factor is a mathematical tool used to solve first-order linear ordinary differential equations (ODEs). The main purpose is to simplify the ODE, making it easier to integrate and find a solution.
In the equation provided, we first rewrite it in the standard form: \[ x' + P(t)x = Q(t) \]
Here, our equation is: \[ x' + 5x = 500(2 - \sin t) \]
We can identify that \( P(t) = 5 \) and \( Q(t) = 500(2 - \sin t) \).
The integrating factor \( I(t) \) is found by computing \( e^{\int P(t)dt} \). For this specific case, it is: \[ I(t) = e^{\int 5dt} = e^{5t} \]
Multiplying both sides of the original equation by the integrating factor \( e^{5t} \) transforms our ODE into a form that can be directly integrated.

Ordinary differential equations (ODEs) are equations that involve functions and their derivatives. They are classified based on linearity and order. The given equation is a first-order linear ODE because it involves the first derivative of \( x \) and \( x \) itself.
Here's the general form of a first-order linear ODE: \[ x' + P(t)x = Q(t) \]
In our example: \[ x' + 5x = 500(2 - \sin t) \]
To solve it, we use integrating factors, separation of variables, substitution, or other analytical methods. Understanding that each term in the equation represents a relationship between the function \( x \) and its rate of change (denoted \( x' \)) is crucial for solving the problem.

An initial value problem (IVP) specifies the value of the function at a particular point, providing a specific solution to a differential equation. In our equation, we are given: \[ x(0) = 5000 \]
This initial condition helps determine the unique solution for the ODE.
After finding the general solution using an integrating factor, we plug in \( t = 0 \) and \( x(0) = 5000 \) to solve for the constant of integration.
Using initial conditions allows us to precisely match the solution to real-world scenarios, wherever such constraints or starting points are necessary.