If x sin left(frac{y}{x}right) dy = left[y si | vedclass

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(B) Given the differential equation: $x \sin \left(\frac{y}{x}\right) \frac{dy}{dx} = y \sin \left(\frac{y}{x}\right) - x$. Let $y = vx$, then $\frac{

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$\log x$

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(B) Given the differential equation: $x \sin \left(\frac{y}{x}\right) \frac{dy}{dx} = y \sin \left(\frac{y}{x}\right) - x$. Let $y = vx$, then $\frac{dy}{dx} = v + x \frac{dv}{dx}$. Substituting these into the equation: $x \sin(v) (v + x \frac{dv}{dx}) = vx \sin(v) - x$. Dividing by $x$: $\sin(v) (v + x \frac{dv}{dx}) = v \sin(v) - 1$. $v \sin(v) + x \sin(v) \frac{dv}{dx} = v \sin(v) - 1$. $x \sin(v) \frac{dv}{dx} = -1$. Separating variables: $\sin(v) dv = -\frac{1}{x} dx$. Integrating both sides: $\int \sin(v) dv = -\int \frac{1}{x} dx$. $-\cos(v) = -\log|x| + C$. Since $y(1) = \frac{\pi}{2}$, at $x = 1$, $v = \frac{y}{x} = \frac{\pi/2}{1} = \frac{\pi}{2}$. $-\cos(\frac{\pi}{2}) = -\log(1) + C \Rightarrow 0 = 0 + C \Rightarrow C = 0$. Thus, $-\cos(v) = -\log x$, which implies $\cos(\frac{y}{x}) = \log x$.

If $x \sin \left(\frac{y}{x}\right) dy = \left[y \sin \left(\frac{y}{x}\right) - x\right] dx$, $x > 0$ and $y(1) = \frac{\pi}{2}$, then the value of $\cos \left(\frac{y}{x}\right)$ is

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