Question

Solution of$x\sin \left(\frac{y}{x}\right)dy=\left(y\sin \left(\frac{y}{x}\right)-x\right)dx$ is

• A. $\cos \left(\frac{y}{x}\right)=c$
• B. $\log \left(\frac{y}{x}\right)+\log x=c$
• C. $\log \left(\frac{y}{x}\right)-\log x=c$
• D. None of these

Answer 1

The correct option is D

None of these

Explanation for the correct answer:

Simplifying the given expression:

$$\begin{gathered} x\sin \left(\frac{y}{x}\right)dy=\left(y\sin \left(\frac{y}{x}\right)-x\right)dx \\ \Rightarrow \sin \left(\frac{y}{x}\right)\left(\frac{dy}{dx}\right)=\left(\frac{y}{x}\right)\sin \left(\frac{y}{x}\right)-1...\left(1\right) \end{gathered}$$

Substitute $\left(\frac{y}{x}\right)=v$

$$\begin{gathered} \Rightarrow \frac{c\left({\displaystyle \frac{dy}{dx}}\right)-y}{x^2}=\frac{dv}{dx} \\ \Rightarrow c\left(\frac{dy}{dx}\right)-y=x^2\left(\frac{dv}{dx}\right)\left[From\left(1\right)\right] \\ \Rightarrow \left(\frac{1}{x}\right)\left(x^2\left(\frac{dv}{dx}\right)+y\right)·\sin v=v\sin v-1 \\ \Rightarrow x\left(\frac{dv}{dx}\right)·\sin v=-1 \end{gathered}$$

Integrating on both the sides

$$\begin{gathered} \int dv\sin v=–\int \left(\frac{dx}{x}\right) \\ \Rightarrow –\cos \left(\frac{y}{x}\right)=–\log x+c \\ \Rightarrow \cos \left(\frac{y}{x}\right)+c=\log x \end{gathered}$$

Therefore, the correct answer is option (D).