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

We have,
\(\sin ^{2} \mathrm{y} \frac{\mathrm{d} x}{\mathrm{dy}}+\mathrm{x}=\cot \mathrm{y}\)
\(\therefore \frac{\mathrm{d} \mathrm{x}}{\mathrm{dy}}+\left(\operatorname{cosec}^{2} \mathrm{y}\right) \mathrm{x}=\cot \mathrm{y} \cdot \operatorname{cosec}^{2} \mathrm{y}\)
\(\therefore\) I.F. \(=\mathrm{e}^{\int \operatorname{cosec}^{2} \mathrm{ydy}}=\mathrm{e}^{-\cot y}\)
\(\therefore \mathrm{xe}^{-\cot y}=\int \mathrm{e}^{-\cot y} \cot \mathrm{y} \cos \mathrm{ec}^{2} \mathrm{y} \mathrm{dy}\)
In RHS, put \(-\cot \mathrm{y}=\mathrm{t} \Rightarrow \operatorname{cosec}^{2} \mathrm{y} \mathrm{dy}=\mathrm{dt}\)
\(\therefore \quad \mathrm{xe}^{-\cot y}=\int \mathrm{e}^{t}(-\mathrm{t}) \mathrm{dt}=-\int \mathrm{t} \mathrm{e}^{t} \mathrm{dt}\)
\(=-\left[\mathrm{t} \mathrm{e}^{t}-\int \mathrm{e}^{t} \mathrm{dt}\right]=-\mathrm{t} \mathrm{e}^{t}+\mathrm{e}^{t}=\mathrm{e}^{t}(1-\mathrm{t})\)
\(\quad x \mathrm{e}^{-\cot y}=\mathrm{e}^{-\cot y}(1+\cot y)+C\)...(2)
At \(x=0, y=\frac{3 \pi}{4}\), we get \(0=\mathrm{e}(1-1)+C\)
From equation \((2)\), required solution is \(x \mathrm{e}^{-\cot y}=\mathrm{e}^{-\cot y}(1+\cot y)\)
\(\quad x \quad=1+\cot y\) \(x=1+\cot y\)