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The integrating factor is key when solving first-order linear differential equations. It is a special function that, once determined, simplifies the differential equation into an easily solvable form. This factor relates to the equation's structure, specifically the term associated with the dependent variable, often expressed as \( P(x) \) in the equation format \( y' + P(x)y = Q(x) \).

To find the integrating factor, denoted as \( \mu(x) \), you compute the exponential of the integral of \( P(x) \). In this exercise, \( P(x) \) is \( \tan x \). Thus, the integrating factor is calculated by \( \mu(x) = e^{\int \tan x \, dx} \). This integral leads to an expression involving a logarithm: \( e^{\ln |\sec x|} \).

On the specified interval \( -\pi/2 < x < \pi/2 \), \( \sec x \) remains positive. Hence, \( \mu(x) = \sec x \). This integrating factor transforms the original equation into a format conducive to straightforward integration.

Understanding the interval over which a differential equation is defined is crucial for both the existence and uniqueness of solutions. In this exercise, the solution is sought on the interval \( -\pi/2 < x < \pi/2 \).

Why is this interval important? First, certain functions, like \( \tan x \) or \( \sec x \), may behave differently outside this interval, such as having discontinuities or undefined points at \( \pm \pi/2 \). On this open interval, however, \( \sec x \) and \( \tan x \) are well-behaved and continuous, making the equation solvable without complications from undefined points.

This interval also assures us that any expressions involving \( \sec x \) are straightforward—specifically here, \( \sec x \) being positive simplifies the integration and the subsequent steps.

The method of solving a first-order linear differential equation involves several clear steps, which effectively transform the given problem into a more manageable form. To achieve this, the integrating factor plays a central role.

After determining the integrating factor, the entire differential equation is multiplied by it. In this example, multiplying by \( \sec x \) simplifies the equation into \( \frac{d}{dx}(y \sec x) = \cos x \).

This simplified expression is the derivative of the product \( y \sec x \). By equating this to \( \cos x \) and integrating both sides with respect to \( x \), you obtain \( y \sec x = \sin x + C \).

The final step involves solving for \( y \) explicitly. Multiplying through by \( \cos x \) resolves the expression into \( y = \sin x \cos x + C \cos x \). This final expression represents the solution to the differential equation within the specified interval, with \( C \) being an arbitrary constant reflecting the general solution.