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To solve a first-order linear differential equation, identifying the integrating factor is a crucial step. This factor simplifies the equation and makes it easier to solve. A first-order linear differential equation is typically expressed in the form: \[ \frac{dy}{dx} + P(x)y = Q(x) \]The integrating factor \( \mu(x) \) is calculated using the formula:\[ \mu(x) = e^{\int P(x) \, dx} \]In the given equation, \( P(x) = \cot x \). To find the integrating factor, we integrate \( \cot x \):- Recall that \( \cot x = \frac{\cos x}{\sin x} \).- The integral becomes: \[ \int \frac{\cos x}{\sin x} \, dx = \ln |\sin x| \]Raising \( e \) to this result gives us:\[ \mu(x) = e^{\ln |\sin x|} = |\sin x| \]The integrating factor \( |\sin x| \) is then used to multiply through the equation, allowing us to integrate and find the solution.

The goal of finding a general solution is to express the solution to the differential equation in its most comprehensive form, often involving constants of integration. Using the integrating factor, the given differential equation:\[ \frac{dy}{dx} + y \cot x = 2 \cos x \]is multiplied by \( |\sin x| \), leading to:\[ \frac{d}{dx}(y \cdot |\sin x|) = 2 |\sin x| \cos x \]Integrate both sides:- The left side becomes: \[ y \cdot |\sin x| \]- For the right side, use the double-angle identity: \[ 2 \sin x \cos x = \sin(2x) \]Integrate:\[ \int 2 |\sin x| \cos x \, dx = \int |\sin(2x)| \, dx \]Assuming \( \sin x > 0 \), the integral resolves to:\[ -\frac{1}{2} \cos(2x) + C \]This gives the general solution as:\[ y = \frac{-\frac{1}{2} \cos(2x) + C}{\sin x} \] where \( C \) is a constant applied from the integration.

The interval of definition for the solution of a differential equation refers to the range of values \( x \) can take, given constraints like division by zero or undefined expressions.The solution expressed as:\[ y = \frac{-\frac{1}{2} \cos(2x) + C}{\sin x} \]implies that the denominator \( \sin x \) should never be zero. Thus, we avoid values where \( \sin x = 0 \):- These occur at multiples of \( \pi \), i.e., \( x = n\pi \), where \( n \) is an integer.Therefore, the solution is defined on intervals that exclude these points. A typical safe interval is:\[ x \in (0, \pi) \]On this interval, \( \sin x > 0 \), simplifying the analysis and ensuring no division by zero occurs, thereby giving a clear and valid solution.