91Ó°ÊÓ

First-order linear differential equations like the one given, \( y' + 3y = 2xe^{-3x} \), greatly benefit from the use of an integrating factor. The integrating factor is a clever tool that simplifies the process of solving these equations. In this context, an integrating factor is a function, typically denoted as \( \mu(x) \), which we multiply with the entire differential equation to transform it into a form that can be easily integrated.

To find the integrating factor \( \mu(x) \), we use the formula:

• \( \mu(x) = e^{\int P(x) \, dx} \)

Here, \( P(x) \) is the coefficient of \( y \) in the differential equation. For our equation, \( P(x) = 3 \). Therefore, we calculate:

• \( \mu(x) = e^{\int 3 \, dx} = e^{3x} \)

By multiplying the entire differential equation by this integrating factor, we align the equation for further simplification and ensure that the left-hand side becomes the derivative of a product involving \( y \). This trick is instrumental in converting our initially challenging differential equation into a simpler one.

When solving differential equations, integrating plays a key role. While doing so, we encounter the constant of integration denoted by \( C \). This constant accounts for the indefinite nature of integration, where every indefinite integral represents not just a single function, but a whole family of functions.

In our differential equation:

• \( \frac{d}{dx}(e^{3x} y) = 2x \)

After integrating both sides, we have:

• \( e^{3x} y = \int 2x \, dx = x^2 + C \)

Here, \( C \) symbolizes our constant of integration. Its value could be any real number and depends on the specific solution to the differential equation based on initial or boundary conditions provided in a problem.

The constant of integration is essential because solutions to differential equations often describe continuous phenomena, like the position of a moving object or the growth of a population. Different values of \( C \) can significantly alter these descriptions, showing the versatility and breadth of differential equation solutions.

One ingenious step in solving the differential equation, \( y' + 3y = 2xe^{-3x} \), is rewriting the left-hand side using the product rule for differentiation. The product rule states that for functions \( u(x) \) and \( v(x) \), the derivative \( \frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x) \). This allows us to reframe the problem by recognizing a pattern.

For our equation, notice:

• The left side, \( e^{3x}y' + 3e^{3x}y \), can be rewritten as \( \frac{d}{dx}(e^{3x}y) \).

This simplification exploits the product rule directly:

• View \( u(x) = e^{3x} \) and \( v(x) = y \).
• Then differentiating gives us, \( \frac{d}{dx}(e^{3x}y) = e^{3x}y' + 3e^{3x}y \).

By recognizing that the left-hand side of the differential equation can be expressed as a single derivative, we streamline our pathway to finding the solution by simply integrating both sides. This use of the product rule not only advances us to the next critical step of integration, but it also highlights the elegance and efficiency achievable through understanding and applying fundamental calculus principles.