The solution of the differential equation x f | vedclass

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(B) Given differential equation is $x \frac{dy}{dx} = y - x 	an \left(\frac{y}{x}ight)$.Dividing by $x$,we get $\frac{dy}{dx} = \frac{y}{x} - 	an

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$y = x \sin^{-1}\left(\frac{k}{x}\right)$

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(B) Given differential equation is $x \frac{dy}{dx} = y - x 	an \left(\frac{y}{x}ight)$.Dividing by $x$,we get $\frac{dy}{dx} = \frac{y}{x} - 	an \left(\frac{y}{x}ight)$.Let $y = vx$,then $\frac{dy}{dx} = v + x \frac{dv}{dx}$.Substituting these into the equation: $v + x \frac{dv}{dx} = v - 	an v$.This simplifies to $x \frac{dv}{dx} = -	an v$,or $\frac{dv}{	an v} = -\frac{dx}{x}$.Integrating both sides: $\int \cot v \, dv = -\int \frac{1}{x} \, dx$.This gives $\ln |\sin v| = -\ln |x| + \ln |k|$.Using logarithm properties,$\ln |\sin v| = \ln \left|\frac{k}{x}ight|$,so $\sin v = \frac{k}{x}$.Substituting $v = \frac{y}{x}$,we get $\sin \left(\frac{y}{x}ight) = \frac{k}{x}$,which implies $y = x \sin^{-1} \left(\frac{k}{x}ight)$.Thus,option $B$ is correct.

The solution of the differential equation $x \frac{dy}{dx} = y - x \tan \left(\frac{y}{x}\right)$ is (Here,$k$ is an arbitrary constant)

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