If cos frac{y}{x} = A log x + C is the genera | vedclass

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

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$1$

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(B) Given the differential equation: $(x \sin \frac{y}{x}) dy = (y \sin \frac{y}{x} - x) dx$ Rearranging the terms to find $\frac{dy}{dx}$: $\frac{dy}{dx} = \frac{y \sin(y/x) - x}{x \sin(y/x)} = \frac{y}{x} - \frac{1}{\sin(y/x)}$ Let $v = \frac{y}{x}$,then $y = vx$ and $\frac{dy}{dx} = v + x \frac{dv}{dx}$ Substituting these into the equation: $v + x \frac{dv}{dx} = v - \frac{1}{\sin v}$ $x \frac{dv}{dx} = -\frac{1}{\sin v}$ Separating the variables: $-\sin v \, dv = \frac{1}{x} \, dx$ Integrating both sides: $-\int \sin v \, dv = \int \frac{1}{x} \, dx$ $\cos v = \log x + C$ Substituting $v = \frac{y}{x}$ back: $\cos \frac{y}{x} = \log x + C$ Comparing this with the given solution $\cos \frac{y}{x} = A \log x + C$,we find $A = 1$.

If $\cos \frac{y}{x} = A \log x + C$ is the general solution of $(x \sin \frac{y}{x}) dy = (y \sin \frac{y}{x} - x) dx$,then $A =$

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