A curve passes through the point $\left(1, \frac{\pi}{6}\right)$. Let the slope of the curve at each point $(x, y)$ be given by $\frac{y}{x}+\sec \left(\frac{y}{x}\right), x0$, then the equation of the curve is

Question

A curve passes through the point $\left(1, \frac{\pi}{6}\right)$. Let the slope of the curve at each point $(x, y)$ be given by $\frac{y}{x}+\sec \left(\frac{y}{x}\right), x>0$, then the equation of the curve is

Aicharya · Accepted Answer

Here $\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{y}}{\mathrm{x}}+\sec \left(\frac{\mathrm{y}}{\mathrm{x}}ight)$ Let $\frac{y}{x}=v \Rightarrow y=v x \Rightarrow \frac{d y}{d x}=v+x \cdot \frac{d v}{d x}$ $\begin{aligned} & \Rightarrow v+x \cdot \frac{d v}{d x}=v+\sec v \ & \Rightarrow x \cdot \frac{d v}{d x}=\operatorname{secv} \ & \Rightarrow \int \cos v \cdot d v=\int \frac{d x}{x} \ & \Rightarrow \operatorname{sinv}=\log _e x+C\end{aligned}$ $\Rightarrow \sin \left(\frac{y}{x}ight)=\log _e x+C$ Putting $\mathrm{x}=1, \mathrm{y}=\frac{\pi}{6} \Rightarrow \mathrm{c}=\frac{1}{2}$ $\Rightarrow \sin \left(\frac{y}{x}ight)=\log _e^x+\frac{1}{2}$

A curve passes through the point $\left(1, \frac{\pi}{6}\right)$. Let the slope of the curve at each point…

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