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(A) Given differential equation: $(1+x^{2}) dy + 2xy dx = \cot x dx$Divide by $(1+x^{2}) dx$:$\frac{dy}{dx} + \frac{2xy}{1+x^{2}} = \frac{\cot x}{1+x^

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$y(1+x^{2}) = \log |\sin x| + C$

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(A) Given differential equation: $(1+x^{2}) dy + 2xy dx = \cot x dx$Divide by $(1+x^{2}) dx$:$\frac{dy}{dx} + \frac{2xy}{1+x^{2}} = \frac{\cot x}{1+x^{2}}$This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \frac{2x}{1+x^{2}}$ and $Q = \frac{\cot x}{1+x^{2}}$.Integrating Factor $(I.F.)$ = $e^{\int P dx} = e^{\int \frac{2x}{1+x^{2}} dx} = e^{\log(1+x^{2})} = 1+x^{2}$.The general solution is $y(I.F.) = \int (Q 	imes I.F.) dx + C$.$y(1+x^{2}) = \int \left( \frac{\cot x}{1+x^{2}} 	imes (1+x^{2}) ight) dx + C$.$y(1+x^{2}) = \int \cot x dx + C$.$y(1+x^{2}) = \log |\sin x| + C$.

Find the general solution of the differential equation: $(1+x^{2}) dy + 2xy dx = \cot x dx$ where $x \neq 0$.

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