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(A) Given equation: $x \frac{dy}{dx} + y - x + xy \cot x = 0$Divide by $x$: $\frac{dy}{dx} + \frac{y}{x} + y \cot x = 1$$\frac{dy}{dx} + y(\frac{1}{x}

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$y = -\cot x + \frac{1}{x} + \frac{C}{x \sin x}$

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(A) Given equation: $x \frac{dy}{dx} + y - x + xy \cot x = 0$Divide by $x$: $\frac{dy}{dx} + \frac{y}{x} + y \cot x = 1$$\frac{dy}{dx} + y(\frac{1}{x} + \cot x) = 1$This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \frac{1}{x} + \cot x$ and $Q = 1$.Integrating Factor $(I.F.)$ $= e^{\int P dx} = e^{\int (\frac{1}{x} + \cot x) dx} = e^{\ln|x| + \ln|\sin x|} = e^{\ln|x \sin x|} = x \sin x$.The general solution is $y(I.F.) = \int (Q \cdot I.F.) dx + C$.$y(x \sin x) = \int (1 \cdot x \sin x) dx + C$.Using integration by parts: $\int x \sin x dx = x(-\cos x) - \int (1)(-\cos x) dx = -x \cos x + \sin x$.So,$y(x \sin x) = -x \cos x + \sin x + C$.Dividing by $x \sin x$: $y = \frac{-x \cos x}{x \sin x} + \frac{\sin x}{x \sin x} + \frac{C}{x \sin x}$.$y = -\cot x + \frac{1}{x} + \frac{C}{x \sin x}$.

Find the general solution of the differential equation: $x \frac{dy}{dx} + y - x + xy \cot x = 0$ $(x \neq 0)$.

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