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(A) Given differential equation: $\left[x \sin ^{2}\left(\frac{y}{x}\right)-y\right] dx + x dy = 0$.Rearranging the terms: $x dy = -\left[x \sin ^{2}\

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$\cot \left(\frac{y}{x}\right)=\log |ex|$

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(A) Given differential equation: $\left[x \sin ^{2}\left(\frac{y}{x}\right)-y\right] dx + x dy = 0$.Rearranging the terms: $x dy = -\left[x \sin ^{2}\left(\frac{y}{x}\right)-y\right] dx$.$\frac{dy}{dx} = \frac{y - x \sin ^{2}\left(\frac{y}{x}\right)}{x} = \frac{y}{x} - \sin ^{2}\left(\frac{y}{x}\right)$ ... $(1)$.This is a homogeneous differential equation. Let $y = vx$,then $\frac{dy}{dx} = v + x \frac{dv}{dx}$.Substituting into $(1)$: $v + x \frac{dv}{dx} = v - \sin ^{2} v$.$x \frac{dv}{dx} = -\sin ^{2} v$.Separating variables: $\frac{dv}{\sin ^{2} v} = -\frac{dx}{x}$.$\int \csc ^{2} v dv = -\int \frac{dx}{x}$.$-\cot v = -\log |x| - C$.$\cot v = \log |x| + C$.Substituting $v = \frac{y}{x}$: $\cot \left(\frac{y}{x}\right) = \log |x| + C$.Given $y = \frac{\pi}{4}$ when $x = 1$: $\cot \left(\frac{\pi/4}{1}\right) = \log |1| + C$.$1 = 0 + C \Rightarrow C = 1$.Thus,$\cot \left(\frac{y}{x}\right) = \log |x| + 1 = \log |x| + \log e = \log |ex|$.The particular solution is $\cot \left(\frac{y}{x}\right) = \log |ex|$.

Find the particular solution satisfying the given condition: $\left[x \sin ^{2}\left(\frac{y}{x}\right)-y\right] d x+x d y=0$; $y=\frac{\pi}{4}$ when $x=1$.

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