91Ó°ÊÓ

First-order linear differential equations are a fundamental topic in calculus and differential equations. These equations are linear, meaning they can be expressed as a first-degree polynomial. In general, a first-order linear differential equation can be written in the form:\[ y' + P(t) y = Q(t) \]Where:

• \( y' \) is the derivative of \( y \) with respect to \( t \).
• \( P(t) \) is a function of \( t \) that multiplies the dependent variable \( y \).
• \( Q(t) \) is a function of \( t \) representing the non-homogeneous part of the equation.

A key feature of these equations is that they can model many real-world scenarios where the rate of change of a quantity is proportional to the quantity itself plus some external factor. Understanding their structure helps determine the best solution method, such as the integrating factor method.

The integrating factor method is a technique used to solve first-order linear differential equations. It helps simplify the process of finding solutions by making the left-hand side of the equation easily integrable. Here's how it works:1. **Standard Form:** Ensure the equation is in the standard form:\[ y' + P(t) y = Q(t) \]2. **Finding the Integrating Factor:** The integrating factor, denoted by \( \mu(t) \), is given by:\[ \mu(t) = e^{\int P(t) \, dt} \]This factor, when multiplied with the entire differential equation, transforms it into a form where the left-hand side becomes the derivative of the product of the integrating factor and \( y \).3. **Solving the Equation:** After applying the integrating factor, the equation looks like:\[ \frac{d}{dt} [\mu(t) y] = \mu(t) Q(t) \]Integrate both sides to find \( y \). The integration on the right side can often be straightforward after applying the integrating factor.

An initial value problem (IVP) is a differential equation that comes with a specified value at an initial point, often to find a unique solution. The standard form of an initial value problem is:\[ y' + P(t) y = Q(t), \quad y(t_0) = y_0 \]Where:

• \( t_0 \) is the initial time.
• \( y_0 \) is the value of \( y \) at \( t_0 \).

Using the initial condition helps determine the constant of integration, \( C \), found after solving the differential equation. In the exercise above, the initial condition \( y(1) = \frac{1}{e} \) was used to solve for \( C \), confirming the specific solution among possible general solutions.

The general solution of a differential equation refers to the family of solutions that includes an arbitrary constant, often denoted by \( C \). After solving a first-order linear differential equation, you obtain:\[ y = f(t, C) \]The solution represents all possible curves that satisfy the differential equation. The constant \( C \) allows these solutions to represent different initial conditions.By applying the initial condition from an initial value problem, you can find a particular solution by solving for \( C \). This process completes the solution, specifying a curve that fits the initial data. In the example given, solving for \( C \) with the initial condition yielded the unique curve \( y = \frac{t}{e^t} \) that fits the scenario described.