The general solution of the differential equa | vedclass

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

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$\cos\left(\frac{y}{x}\right) = \log |x| + c$

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(B) Given differential equation: $\left(x \sin \frac{y}{x}ight) dy = \left(y \sin \frac{y}{x} - xight) dx$ Rearranging the terms: $\frac{dy}{dx} = \frac{y \sin \frac{y}{x} - x}{x \sin \frac{y}{x}} = \frac{y}{x} - \frac{1}{\sin(y/x)} = \frac{y}{x} - 	ext{cosec}\left(\frac{y}{x}ight)$ Let $y = vx$, then $\frac{dy}{dx} = v + x \frac{dv}{dx}$. Substituting these into the equation: $v + x \frac{dv}{dx} = v - 	ext{cosec}(v)$ $x \frac{dv}{dx} = -	ext{cosec}(v)$ Separating the variables: $\sin(v) dv = -\frac{1}{x} dx$ Integrating both sides: $\int \sin(v) dv = -\int \frac{1}{x} dx$ $-\cos(v) = -\log |x| + C$ $\cos(v) = \log |x| + C'$ Substituting $v = \frac{y}{x}$ back: $\cos\left(\frac{y}{x}ight) = \log |x| + C$.

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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