The general solution of the differential equa | vedclass

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Question

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

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$\cos \frac{y}{x} = \log_e x + c$

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

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

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