91Ó°ÊÓ

First-order linear differential equations are a type of differential equation where the highest derivative is the first derivative.
These equations commonly appear in the format \( y' + P(x)y = Q(x) \), where \( P(x) \) and \( Q(x) \) are functions of \( x \). In the given problem, the equation is \( y' - 3y = e^x \).
This matches the form with \( P(x) = -3 \) and \( Q(x) = e^x \). When dealing with these types of equations, the goal is to find a solution \( y \) that satisfies the equation for all values of \( x \).

• They are called 'linear' because they form a straight line when plotted.
• Such equations model numerous physical processes, including cooling and electrical circuits.

Understanding this allows you to structure the process of finding solutions effectively.

The integrating factor simplifies the process of solving first-order linear differential equations.
It is a special multiplier that, when applied, helps express the equation's left side as a simple derivative of a product. This transformation is crucial because
it allows us to integrate easily. In our example, the integrating factor \( \mu(x) \) is determined by the formula \( e^{\int P(x) dx} \). For the equation \( y' - 3y = e^x \), we have \( P(x) = -3 \).
Hence, the integrating factor becomes \( e^{\int -3 \, dx} = e^{-3x} \).

• This factor converts the entire differential equation into a form that can be worked with more easily.
• Successfully applying the factor is pivotal in progressing towards the solution.

Using the integrating factor enhances the functional symmetry of the equation, making it simpler to derive further steps and eventually solve the equation.

Solving a differential equation involves finding the function \( y(x) \) that satisfies the given equation. In the process, once you've applied the integrating factor, the differential equation can be rearranged into a derivative form that is easier to integrate.
In the problem example, the equation becomes \( \frac{d}{dx}(e^{-3x} y) = e^{-2x} \). By integrating both sides, we obtain
\( e^{-3x} y = -\frac{1}{2} e^{-2x} + C \), where \( C \) is the constant of integration. To solve for \( y \), multiply both sides by \( e^{3x} \), which results in the general solution: \( y = -\frac{1}{2} e^x + Ce^{3x} \).

• This shows that the solution involves both the particular solution to the non-homogeneous part and the general solution to the homogeneous part of the equation.
• The constant \( C \) represents an infinite set of solutions, signifying that each value of \( C \) will conform to different initial conditions.

This step ensures the complete solution is versatile and accommodates various real-world scenarios where initial conditions may vary.