The slope of tangent at $(x, y)$ to a curve passing through $\left(1, \frac{\pi}{4}\right)$ is $\frac{y}{x}-\cos ^2 \frac{y}{x}$, then the equation of curve is

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Aicharya · Accepted Answer

$\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{y}{x}-\cos ^2 \frac{y}{x}...(i)$
Put $y=v x$...(ii) $\Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=\mathrm{v}+x \frac{\mathrm{dv}}{\mathrm{~d} x}...(iii)$
Substituting (ii) and (iii) in (i), we get. $\mathrm{v}+x \frac{\mathrm{~d} v}{\mathrm{~d} x}=\mathrm{v}-\cos ^2 \mathrm{v} \quad \Rightarrow x \frac{\mathrm{dv}}{\mathrm{~d} x}=-\cos ^2 \mathrm{v}$
Integrating on both sides, we get $\begin{aligned} & \int \sec ^2 v \cdot d v=-\int \frac{d x}{x}+c \\ & \Rightarrow \tan v=-\log x+c \\ & \Rightarrow \tan \frac{y}{x}=-\log x+c...(iv) \end{aligned}$ Since the required curve passes through $\left(1, \frac{\pi}{4}\right)$, $\begin{aligned} & \tan \frac{\pi}{4}=-\log 1+\mathrm{c} \Rightarrow \mathrm{c}=1 \\ \therefore \quad & \tan \frac{y}{x}=-\log x+1 \\ & \Rightarrow \tan \frac{y}{x}=-\log x+\log \mathrm{e} \\ & \Rightarrow y=x \tan ^{-1}\left[\log \left(\frac{\mathrm{e}}{x}\right)\right] \end{aligned}$ ...[From (iv)]

The slope of tangent at $(x, y)$ to a curve passing through $\left(1, \frac{\pi}{4}\right)$ is…

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