The general solution of the differential equa | vedclass

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(B) Given the differential equation: $\left(x \sin \frac{y}{x}\right) dy = \left(y \sin \frac{y}{x} - x\right) dx$. Rearranging the terms,we get: $\fr

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$\log |x| - \cos \frac{y}{x} = c$

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(B) Given the differential equation: $\left(x \sin \frac{y}{x}\right) dy = \left(y \sin \frac{y}{x} - x\right) dx$. Rearranging the terms,we get: $\frac{dy}{dx} = \frac{y \sin(y/x) - x}{x \sin(y/x)}$. Since this is a homogeneous differential equation,let $y = vx$,then $\frac{dy}{dx} = v + x \frac{dv}{dx}$. Substituting these into the equation: $v + x \frac{dv}{dx} = \frac{vx \sin v - x}{x \sin v} = \frac{v \sin v - 1}{\sin v} = v - \frac{1}{\sin v}$. Subtracting $v$ from both sides: $x \frac{dv}{dx} = -\frac{1}{\sin v}$. Separating the variables: $\sin v \, dv = -\frac{1}{x} dx$. Integrating both sides: $\int \sin v \, dv = -\int \frac{1}{x} dx$. $-\cos v = -\log |x| + C$. Rearranging gives: $\log |x| - \cos v = C$. Substituting $v = \frac{y}{x}$ back,we get: $\log |x| - \cos \frac{y}{x} = C$.

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

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