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Question: Answered & Verified by Expert
The general solution of the differential equation $\left(x \sin \frac{y}{x}\right) d y=\left(y \sin \frac{y}{x}-x\right) d x$ is
MathematicsDifferential EquationsTS EAMCETTS EAMCET 2023 (12 May Shift 2)
Options:
  • A $\sin ^{-1}\left(\frac{y}{x}\right)=\frac{x}{2}+c$
  • B $\sin \left(\frac{x}{y}\right)=\frac{x^2}{2}+c$
  • C $\sin \left(\frac{y}{x}\right)=\log |x|+c$
  • D $\cos \left(\frac{y}{x}\right)=\log |x|+c$
Solution:
2002 Upvotes Verified Answer
The correct answer is: $\cos \left(\frac{y}{x}\right)=\log |x|+c$
$\begin{aligned} & \text { (x) } \\ & \left(x \sin \frac{y}{x}\right) d y=\left(y \sin \frac{y}{x}-x\right) d x \\ & \Rightarrow \frac{d y}{d x}=\frac{y \sin \frac{y}{x}-x}{x \sin \frac{y}{x}} \Rightarrow \frac{d y}{d x}=\frac{y}{x}-\operatorname{cosec} \frac{y}{x} \\ & \text { Let } y=v x \\ & \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x} \Rightarrow v+x \frac{d v}{d x}=v-\operatorname{cosec} v \\ & \Rightarrow \quad x \frac{d v}{d x}=-\operatorname{cosec} v \Rightarrow \frac{d x}{x}=-\sin v d v \\ & \Rightarrow \quad \frac{d x}{x}=-\int \sin v d v \\ & \Rightarrow \log x+C=\cos v \\ & \therefore \cos \left(\frac{y}{x}\right)=\log |x|+C .\end{aligned}$

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