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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
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$\sin ^{2} y \frac{d x}{d y}+x=\cot y$ when $x=0$ and $y=\frac{3 \pi}{4}$ is
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Verified Answer
The correct answer is:
$x=1+\cot y$
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 $x \mathrm{e}^{-\cot y}=\mathrm{e}^{-\cot y}(1+\cot y)$
$\quad x \quad=1+\cot y$ $x=1+\cot y$
$\sin ^{2} \mathrm{y} \frac{\mathrm{d} x}{\mathrm{dy}}+\mathrm{x}=\cot \mathrm{y}$
$\therefore \frac{\mathrm{d} \mathrm{x}}{\mathrm{dy}}+\left(\operatorname{cosec}^{2} \mathrm{y}\right) \mathrm{x}=\cot \mathrm{y} \cdot \operatorname{cosec}^{2} \mathrm{y}$
$\therefore$ 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 $x \mathrm{e}^{-\cot y}=\mathrm{e}^{-\cot y}(1+\cot y)$
$\quad x \quad=1+\cot y$ $x=1+\cot y$
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