91Ó°ÊÓ

First-order linear differential equations are a fundamental part of calculus used to model various phenomena in engineering, physics, biology, and economics. They are called "first-order" because they involve the first derivative of the unknown function, which usually represents some change over time or space.
These equations have a standard form, given by \( y' + P(x)y = Q(x) \).

• \( y' \) represents the first derivative of the function \( y \), with respect to \( x \).
• \( P(x) \) is a function of \( x \) that multiplies the dependent variable \( y \).
• \( Q(x) \) is a function of \( x \) on the right side of the equation.

The main goal when solving such equations is to find \( y \) as a function of \( x \), allowing us to predict or interpret real-world scenarios accurately.
Understanding this form is crucial because it allows the use of specific techniques, like integrating factors, to find solutions.

One powerful tool to solve first-order linear differential equations is the integrating factor method. The integrating factor, denoted \( \mu(x) \), is a function that simplifies the equation by making the left-hand side into an exact derivative.
To calculate the integrating factor, we use the formula:\[ \mu(x) = e^{\int P(x) \, dx} \]The integrating factor for our example, where \( P(x) = 3 \), becomes \( e^{3x} \).
The role of multiplying the entire equation by this integrating factor is to transform it into a form that can be integrated easily. After multiplying through by \( e^{3x} \), the original differential equation turns into:\[ e^{3x} y' + 3e^{3x} y = e^{5x} \]
This equation can now be written as a total derivative, \( \frac{d}{dx}(e^{3x} y) = e^{5x} \), which is much easier to solve.
Using this method simplifies the integration process and helps us efficiently find the general solution.

Initial conditions are values given at a certain point to find a specific solution to a differential equation. For our problem, the initial condition provided is \( y = 1 \) when \( x = 0 \).
Initial conditions are crucial because:

• They allow us to determine any unknown constants arising from integration.
• They help in finding unique solutions which fit real-world problems correctly.
• They ensure that the solution not only satisfies the differential equation but also aligns with given real data.

In the problem at hand, after integrating, we obtained a general solution containing a constant \( C \):\[ y = \frac{1}{5} e^{2x} + Ce^{-3x} \]Applying the initial condition \( (x = 0, y = 1) \), we calculated \( C \) by substituting these values:\[ 1 = \frac{1}{5} + C \]This gives \( C = \frac{4}{5} \).
Thus, the particular solution becomes \( y = \frac{1}{5} e^{2x} + \frac{4}{5} e^{-3x} \), fulfilling both the differential equation and the initial condition provided.
In the broader sense, initial conditions help to ensure that mathematical models are relevant to real-world applications, offering solutions that fit and predict individualized scenarios.