The solution of the equation xfrac{dy}{dx} = | vedclass

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

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$x\sin \left( \frac{y}{x} \right) = c$

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(C) Given the differential equation: $x\frac{dy}{dx} = y - x	an \left( \frac{y}{x} ight)$.Dividing by $x$,we get: $\frac{dy}{dx} = \frac{y}{x} - 	an \left( \frac{y}{x} ight)$.This is a homogeneous differential equation. Let $y = vx$,then $\frac{dy}{dx} = v + x\frac{dv}{dx}$.Substituting these into the equation: $v + x\frac{dv}{dx} = v - 	an v$.Simplifying: $x\frac{dv}{dx} = -	an v$.Separating the variables: $\frac{dv}{	an v} = -\frac{dx}{x}$,which is $\cot v \, dv = -\frac{dx}{x}$.Integrating both sides: $\int \cot v \, dv = -\int \frac{dx}{x}$.$\ln |\sin v| = -\ln |x| + \ln |c|$.$\ln |\sin v| + \ln |x| = \ln |c|$.$\ln |x \sin v| = \ln |c|$.$x \sin v = c$.Substituting $v = \frac{y}{x}$ back: $x \sin \left( \frac{y}{x} ight) = c$.

The solution of the equation $x\frac{dy}{dx} = y - x\tan \left( \frac{y}{x} \right)$ is

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