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If the gradient of the tangent at any point $(x, y)$ of a curve which passes through the point $\left(1, \frac{\pi}{4}\right)$ is $\left\{\frac{y}{x}-\sin ^{2}\left(\frac{y}{x}\right)\right\}$, then equation of the curve is
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Verified Answer
The correct answer is:
$y=x \cot ^{-1}\left(\log _{e} e x\right)$
$\frac{d y}{d x}=\frac{y}{x}-\sin ^{2}\left(\frac{y}{x}\right)$
Put $y=v x \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}$
$\begin{array}{l}
v+x \frac{d v}{d x}=v-\sin ^{2} v \\
\Rightarrow-\operatorname{cosec}^{2} v d v=\frac{d x}{x}
\end{array}$
Integrating both sides, we get
$\begin{array}{l}
-\int \operatorname{cosec}^{2} v d v=\int \frac{d x}{x} \\
\Rightarrow \cot v=\log x+C \\
\cot \frac{y}{x}=\log x+C
\end{array}$
Curve passes through the point $\left(1, \frac{\pi}{4}\right)$
$\begin{array}{l}
\therefore \mathrm{C}=1 \\
\Rightarrow \cot \frac{\mathrm{y}}{\mathrm{x}}=\log \mathrm{x}+\log _{\mathrm{e}} \mathrm{e} \\
\Rightarrow \cot \frac{\mathrm{y}}{\mathrm{x}}=\log \mathrm{xe} \\
\Rightarrow \mathrm{y}=\mathrm{x} \cot ^{-1}(\log \mathrm{xe})
\end{array}$
Put $y=v x \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}$
$\begin{array}{l}
v+x \frac{d v}{d x}=v-\sin ^{2} v \\
\Rightarrow-\operatorname{cosec}^{2} v d v=\frac{d x}{x}
\end{array}$
Integrating both sides, we get
$\begin{array}{l}
-\int \operatorname{cosec}^{2} v d v=\int \frac{d x}{x} \\
\Rightarrow \cot v=\log x+C \\
\cot \frac{y}{x}=\log x+C
\end{array}$
Curve passes through the point $\left(1, \frac{\pi}{4}\right)$
$\begin{array}{l}
\therefore \mathrm{C}=1 \\
\Rightarrow \cot \frac{\mathrm{y}}{\mathrm{x}}=\log \mathrm{x}+\log _{\mathrm{e}} \mathrm{e} \\
\Rightarrow \cot \frac{\mathrm{y}}{\mathrm{x}}=\log \mathrm{xe} \\
\Rightarrow \mathrm{y}=\mathrm{x} \cot ^{-1}(\log \mathrm{xe})
\end{array}$
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