Solution of the differential equation $$xsinyxdy=ysinyx$$−xdx is

Correct option is 1logx=cosyx+c

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Solution of the differential equation $x \sin \frac{y}{x} d y = (y \sin \frac{y}{x} - x) d x$ is

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logx=cosyx+c

logy=cosyx+c

logx=cosxy+c

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detailed solution

Correct option is A

We have dydx=ysin⁡yx−xxsin⁡yx put y=Vx So that dydx=V+xdVdx Hence, V+xdVdx=Vsin⁡V−1sin⁡V=V−1sin⁡V⇒xdVdx=−1sin⁡V⇒∫dxx+∫sin⁡VdV=c⇒log⁡x−cos⁡V=c⇒log⁡x=cos⁡yx+c

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Solution of the differential equation xsinyxdy=ysinyx−xdx is

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Solution of the differential equation $x \sin \frac{y}{x} d y = (y \sin \frac{y}{x} - x) d x$ is