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To effectively solve a first-order linear differential equation like \(y^{\prime}+4y=e^{-x}\), we begin by finding the integrating factor. This is a crucial function that helps simplify the differential equation into an easily solvable form. The standard form of a first-order linear differential equation is \(y^{\prime}+P(x)y=Q(x)\). Here, since our function is \(y^{\prime}+4y=e^{-x}\), \(P(x)\) equals 4. The integrating factor, denoted as \(\mu(x)\), is computed with the formula:

• \(\mu(x)=e^{\int P(x)\,dx}=e^{\int 4\,dx}=e^{4x}\)

This exponential function \(e^{4x}\) is what we call the integrating factor. By multiplying the entire differential equation by this factor, we can transform it into a form where the left-hand side becomes the derivative of a product, which is much easier to handle.The integrating factor streamlines the process, allowing us to solve these equations step by step with precision.

Once we have the integrating factor, the process of solving the differential equation becomes straightforward. Let's revisit our equation: \(y^{\prime}+4y=e^{-x}\), which through the integrating factor becomes:

• \(e^{4x}y^{\prime} + 4e^{4x}y = e^{4x}e^{-x}\)

When simplified, the left side can be rewritten as the derivative of a product:

• \((e^{4x}y)^{\prime} = e^{3x}\)

This transformation is because the integrating factor turns our equation into a perfect derivative, making integration doable.To solve this, we simply integrate both sides with respect to \(x\):

• \(\int (e^{4x}y)^{\prime} \, dx = \int e^{3x} \, dx\)

The integration of the right side gives \(\frac{1}{3}e^{3x}\), and thus we arrive at:

• \(e^{4x}y = \frac{1}{3}e^{3x} + C\)

Where \(C\) is the constant of integration, added when evaluating indefinite integrals.From here, solving for \(y\) in terms of \(x\) is simplified by dividing through by the integrating factor \(e^{4x}\).

In differential equations, the constant of integration \(C\) plays a significant role. It arises when we compile our solution that involves an indefinite integral, essentially implying an entire family of possible solutions.When we integrated

• \((e^{4x}y)^{\prime} = e^{3x}\)

we obtained the solution

• \(e^{4x}y = \frac{1}{3}e^{3x} + C\)

Here, the \(C\) represents any constant value, which depends on initial conditions or boundary values affixed to the problem.To fully determine \(C\), you would need additional information about the function \(y\), such as its value at a specific point \(x\). Without this extra data, \(C\) remains arbitrary. Conclusively, the constant \(C\) reflects a fundamental principle of calculus: integration undoes differentiation to yield an infinite set of functions differing only by a constant. This embodies the rich complexity and flexibility inherent in solving differential equations.