The solution of the differential equation $x \cdot \sin \left(\frac{y}{x}\right) d y=\left[y \cdot \sin…
The solution of the differential equation $x \cdot \sin \left(\frac{y}{x}\right) d y=\left[y \cdot \sin \left(\frac{y}{x}\right)-x\right] d x$ is
- $\cos \left(\frac{x}{y}\right)=\log |x|+c$
- $\cos \left(\frac{y}{x}\right)=\log |y|+c$
- $\cos \left(\frac{y}{x}\right)=\log |x|+c$
- $\cos \left(\frac{x}{y}\right)=\log |y|+c$
Solution
We have $x \cdot \sin \left(\frac{y}{x}\right) d y=\left[y \cdot \sin \left(\frac{y}{x}\right)-x\right] d x$
$\therefore \frac{d y}{d x}=\frac{y \cdot \sin \left(\frac{y}{x}\right)-x}{x \sin \left(\frac{y}{x}\right)}=\left(\frac{y}{x}\right)-\frac{1}{\sin \left(\frac{y}{x}\right)}$
$\therefore \frac{d y}{d x}-\left(\frac{1}{x}\right) y=-\operatorname{cosec}\left(\frac{y}{x}\right)$
Put $\frac{y}{x}=v \Rightarrow y=v x \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}$
$\therefore\left(\mathrm{v}+\frac{\mathrm{xdv}}{\mathrm{dx}}\right)-\mathrm{v}=-\operatorname{cosec} \mathrm{v} \Rightarrow \mathrm{x} \frac{\mathrm{dv}}{\mathrm{dx}}=-\operatorname{cosec} \mathrm{v}$
$\therefore \int \frac{\mathrm{dv}}{\operatorname{cosec} \mathrm{v}}=-\int \frac{\mathrm{d} \mathrm{x}}{\mathrm{x}} \Rightarrow \int \sin \mathrm{v} \mathrm{dv}=-\log |\mathrm{x}|$
$-\cos v=-\log |x|+c_{1} \Rightarrow \log x+c=\cos v$
$\therefore \cos \left(\frac{y}{x}\right)=\log |x|+c$
Asked in: MHT CET 2020 (20 Oct Shift 2)
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