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(D) Given differential equation is $y \sin \left(\frac{x}{y}\right) dx = \left\{x \sin \left(\frac{x}{y}\right) - y\right\} dy$.Dividing by $dy \cdot

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(D) Given differential equation is $y \sin \left(\frac{x}{y}\right) dx = \left\{x \sin \left(\frac{x}{y}\right) - y\right\} dy$.Dividing by $dy \cdot y \sin \left(\frac{x}{y}\right)$,we get $\frac{dx}{dy} = \frac{x}{y} - \frac{1}{\sin(x/y)}$.Let $v = \frac{x}{y}$,then $x = vy$,so $\frac{dx}{dy} = v + y \frac{dv}{dy}$.Substituting this into the equation: $v + y \frac{dv}{dy} = v - \frac{1}{\sin v}$.This simplifies to $y \frac{dv}{dy} = -\frac{1}{\sin v}$.Separating variables: $\int \sin v \, dv = -\int \frac{dy}{y}$.Integrating both sides: $-\cos v = -\log_e |y| + C$,which simplifies to $\cos v = \log_e |y| + C$.Substituting $v = \frac{x}{y}$: $\cos \left(\frac{x}{y}\right) = \log_e |y| + C$.Given $y(\pi/4) = 1$,so at $x = \pi/4, y = 1$: $\cos(\pi/4) = \log_e(1) + C \Rightarrow \frac{1}{\sqrt{2}} = 0 + C \Rightarrow C = \frac{1}{\sqrt{2}}$.Thus,the solution is $\cos \left(\frac{x}{y}\right) = \log_e y + \frac{1}{\sqrt{2}}$.

The solution of the differential equation $y \sin \left(\frac{x}{y}\right) dx = \left\{x \sin \left(\frac{x}{y}\right) - y\right\} dy$ satisfying $y\left(\frac{\pi}{4}\right) = 1$ is

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