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(A) Given the differential equation: $\sin^{2} y \frac{dx}{dy} + x = \cot y$. Dividing by $\sin^{2} y$,we get: $\frac{dx}{dy} + (\operatorname{cosec}^

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$x = 1 + \cot y$

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(A) Given the differential equation: $\sin^{2} y \frac{dx}{dy} + x = \cot y$. Dividing by $\sin^{2} y$,we get: $\frac{dx}{dy} + (\operatorname{cosec}^{2} y)x = \cot y \operatorname{cosec}^{2} y$. This is a linear differential equation of the form $\frac{dx}{dy} + P(y)x = Q(y)$,where $P(y) = \operatorname{cosec}^{2} y$ and $Q(y) = \cot y \operatorname{cosec}^{2} y$. The Integrating Factor ($I$.$F$.) is $e^{\int P(y) dy} = e^{\int \operatorname{cosec}^{2} y dy} = e^{-\cot y}$. The solution is given by $x \cdot (I.F.) = \int Q(y) \cdot (I.F.) dy + C$. $x e^{-\cot y} = \int \cot y \operatorname{cosec}^{2} y e^{-\cot y} dy + C$. Let $t = -\cot y$,then $dt = \operatorname{cosec}^{2} y dy$. $x e^{-\cot y} = \int (-t) e^{t} dt + C = -(t e^{t} - e^{t}) + C = e^{t}(1 - t) + C$. Substituting back $t = -\cot y$: $x e^{-\cot y} = e^{-\cot y}(1 + \cot y) + C$. Given $x = 0$ at $y = \frac{3\pi}{4}$,we have $0 = e^{-\cot(3\pi/4)}(1 + \cot(3\pi/4)) + C$. Since $\cot(3\pi/4) = -1$,$0 = e^{1}(1 - 1) + C \implies C = 0$. Thus,$x e^{-\cot y} = e^{-\cot y}(1 + \cot y)$,which simplifies to $x = 1 + \cot y$.

The particular solution of the differential equation $\sin^{2} y \frac{dx}{dy} + x = \cot y$ when $x = 0$ and $y = \frac{3\pi}{4}$ is

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