91Ó°ÊÓ

Differential equations are mathematical equations that involve functions and their derivatives. They describe how a particular quantity changes with respect to another. In essence, they provide a tool for modeling rates of change, which is why they are frequently used in a wide range of scientific fields such as physics, biology, and economics. The simplest form is the first-order differential equation, which involves the first derivative of the unknown function.
In this exercise, the equation given is a first-order linear differential equation, which can be written in the standard form as \( y' + p(x)y = g(x) \). The aim is to find the function \( y \) that satisfies this equation. Understanding the nature of the equation and recognizing its type is crucial in selecting the appropriate method for solving it.

The integrating factor is a technique used to solve first-order linear differential equations. The main goal is to transform the differential equation into a form that can be easily integrated.
To find the integrating factor \( \mu(x) \), set it as \( \mu(x) = e^{\int p(x) \, dx} \). Here, \( p(x) \) represents the coefficient of \( y \) in the standard form.
For the given problem, \( p(x) = 4 \), leading to the integrating factor \( \mu(x) = e^{4x} \). By multiplying every part of the original equation by this factor, the left-hand side of the equation becomes a perfect derivative.
The process simplifies solving the differential equation by allowing one to leverage the properties of derivatives, turning the equation into one involving an easily integrable expression.

Initial value problems are a form of differential equation where the solution must satisfy a specific initial condition. This condition is generally given for the value of the solution at a particular point. In our problem, the initial condition is \( y(0) = 4 \).
Specifying initial conditions helps us determine a unique solution among the infinite solutions that a differential equation might have.
Once the general solution to the differential equation is found, it often includes a constant of integration \( C \). By substituting the initial condition into the general solution, this constant can be explicitly solved, yielding the specific solution. In this exercise, substituting the initial condition allowed for solving \( C = 3 \).

Calculus is fundamentally important in solving differential equations. In this exercise, calculus helps in several ways:

• Integrating factors are derived using integration, an essential calculus operation.
• Once the equation is transformed, integration is used again to find the general solution.
• Derivatives, another major concept in calculus, form the very core of differential equations, expressing rates of change.

These operations not only give us solutions to differential equations but also help us understand the behavior and characteristics of different systems. Calculus provides the tools necessary for abstract reasoning and problem-solving, making it an indispensable field in mathematics and sciences.