The general solution of x frac{dy}{dx} = y - | vedclass

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(C) The 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} -

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

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(C) The given differential equation is $x \frac{dy}{dx} = y - x 	an \left( \frac{y}{x} ight)$.Dividing by $x$,we get $\frac{dy}{dx} = \frac{y}{x} - 	an \left( \frac{y}{x} ight)$.This is a homogeneous differential equation. Let $y = vx$,then $\frac{dy}{dx} = v + x \frac{dv}{dx}$.Substituting these into the equation: $v + x \frac{dv}{dx} = v - 	an v$.Subtracting $v$ from both sides: $x \frac{dv}{dx} = - 	an v$.Separating the variables: $\frac{dv}{	an v} = - \frac{dx}{x}$,which is $\cot v \, dv = - \frac{dx}{x}$.Integrating both sides: $\int \cot v \, dv = - \int \frac{1}{x} \, dx$.This gives $\ln |\sin v| = - \ln |x| + \ln |c|$.Using logarithmic properties: $\ln |\sin v| + \ln |x| = \ln |c| \Rightarrow \ln |x \sin v| = \ln |c|$.Thus,$x \sin v = c$. Substituting $v = \frac{y}{x}$ back,we get $x \sin \left( \frac{y}{x} ight) = c$.

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

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