91Ó°ÊÓ

To solve a linear first-order differential equation, we often use the concept of an integrating factor. This is a function that simplifies the equation, making it easier to solve by integrating.
In our problem, the differential equation is:
\(x^{'} - x = f(t)\).
The standard linear form is \(x^{'} + p(t)x = q(t)\). Here, \(p(t) = -1\) and \(q(t) = f(t)\).

The integrating factor (IF) is found using the formula:
\(IF = e^{\int p(t)dt} = e^{-\int 1 dt} = e^{-t}\).

Once we have the integrating factor, we multiply both sides of the differential equation by it:
\(e^{-t}x^{'} - e^{-t}x = e^{-t}f(t)\).
This simplifies our differential equation, and we can recognize the left-hand side as the derivative of \(e^{-t}x\).

A periodic function repeats its values in regular intervals. In mathematical terms, a function \(f\) is periodic if there exists a positive number \(T\) such that \(f(t+T) = f(t)\) for all values of \(t\). In our problem, the emigration function \(f(t)\) is described as follows:

\( f(t) = \left\{ \begin{array}{ll} 5000(1 + \cos t), & 0 \leq t < 1 \ 0, & 1 \leq t < 2 \)
and this pattern repeats every 2 units, i.e., \(f(t)=f(t+2)\).

Understanding the nature of this periodic function is crucial because it helps us set up the integrals correctly over one period \(0 \le t < 2\). Solving the differential equation involves integrating this piecewise-defined periodic function.

An initial condition lets us find a specific solution to a differential equation. It provides a value at a particular point, which is essential to determine the constant of integration in the solution. In our case, the initial condition is:
\(x(0) = 5000\).

After solving the integral equation, we substitute \(t = 0\) to find the value of the constant \(C\). For example, if after integrating both sides we have:
\(e^{-t} x = \text{integral of } e^{-t} f(t) \text{ dt} + C\),
then applying \(x(0) = 5000\), we can solve:
\( e^{0} \cdot 5000 = \text{integral of} \ e^{0} f(t) \ \text{dt} + C\).
This ensures our solution fits the specific scenario described in the problem.

A piecewise function is one that is defined by different expressions over different intervals. In our problem, the piecewise function is:
\( f(t) = \left\{ \begin{array}{ll} 5000(1 + \cos t), & 0 \leq t < 1 \ 0, & 1 \leq t < 2 \right.\)
and this pattern repeats every 2 units.

When solving the differential equation, we need to break down the integral over each interval where the function has different expressions. For instance, we integrate:
\( \int_{0\leq t < 1} e^{-t} \cdot 5000(1 + \cos t) dt\)
and \( \int_{1\leq t < 2} e^{-t} \cdot 0 dt\)

Integration is the process of finding the integral of a function, which is essentially the reverse operation of differentiation. When solving differential equations, integration helps us find the general solution.

We need to integrate the right-hand side of our equation:
\( \int e^{-t} f(t) dt\).

Given the periodic piecewise function \(f(t)\), our integral is split over the intervals:
\int_{0\leq t < 1} e^{-t} \cdot 5000(1+\cos t) dt\
and
\int_{1\leq t < 2} e^{-t} \cdot 0 dt\.

For the first interval (0 \le\, t < 1):
\int e^{-t} \cdot 5000(1+\cos t) dt\ is broken into:
\int_{0}^{1} 5000e^{-t} dt + \int_{0}^{1} 5000e^{-t}\cos t dt\.
Each of these can be integrated separately and their sum gives the integral over that interval.