Question

The solution of the differential equation $\frac{dy}{dx}=\sec \left(\frac{y}{x}\right)+\frac{y}{x}$ is:

• (A) $\cos \left(\frac{y}{x}\right)=\log (cx)$
• (B) $\sin \left(\frac{x}{y}\right)=\log (cx)$
• (C) $\sin \left(\frac{y}{x}\right)=\log (cx)$
• (D) None of these

Answer 1

The Correct Option is C

Explanation:
Given,$\frac{dy}{dx}=\sec \left(\frac{y}{x}\right)+\frac{y}{x}\dots \dots (1)$ Let,$\frac{y}{x}=t$ $\Rightarrow y=xt$Differentiating with respect to $x$, we get$\frac{dy}{dx}=x\times \frac{dt}{dx}+t\times \frac{d}{dx}x$We known that:$\frac{d}{dx}xy=y\times \frac{d}{dx}x+x\times \frac{d}{dx}y$So,$\frac{dy}{dx}=x\times \frac{dt}{dx}+t$Now, Putting these value in equation $(1)$, we get$x\frac{dt}{dx}+t=\sec t+t$$\Rightarrow x\frac{dt}{dx}=\sec t$$\Rightarrow \frac{dt}{\sec t}=\frac{dx}{x}$Integrating both sides, we get$\int \frac{dt}{\sec t}=\int \frac{dx}{x}$$\Rightarrow \int \cos tdt=\int \frac{dx}{x}$$\Rightarrow \sin t=\log x+\log c$$\Rightarrow \sin t=\log (cx)(\because \log m+\log n=\log (mn))$$\therefore \sin \left(\frac{y}{x}\right)=\log (cx)$Hence, the correct option is (C).