The solution of the differential equation x c | vedclass

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Question

(C) Given the differential equation: $x \sin \left(\frac{y}{x}\right) dy = \left[y \sin \left(\frac{y}{x}\right) - x\right] dx$ Dividing both sides by

Vedclass · Accepted Answer

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

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(C) Given the differential equation: $x \sin \left(\frac{y}{x}\right) dy = \left[y \sin \left(\frac{y}{x}\right) - x\right] dx$ Dividing both sides by $dx \cdot x \sin \left(\frac{y}{x}\right)$,we get: $\frac{dy}{dx} = \frac{y \sin \left(\frac{y}{x}\right) - x}{x \sin \left(\frac{y}{x}\right)} = \frac{y}{x} - \frac{1}{\sin \left(\frac{y}{x}\right)}$ Let $v = \frac{y}{x}$,so $y = vx$. Then $\frac{dy}{dx} = v + x \frac{dv}{dx}$. Substituting these into the equation: $v + x \frac{dv}{dx} = v - \frac{1}{\sin v}$ $x \frac{dv}{dx} = -\frac{1}{\sin v} = -\operatorname{cosec} v$ 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: $\cos \left(\frac{y}{x}\right) = \log |x| + c$

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

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