A curve passes through the point left( 1, fra | vedclass

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(A) Given the differential equation: $\frac{dy}{dx} = \frac{y}{x} - \cos^2 \left( \frac{y}{x} \right)$.Let $v = \frac{y}{x}$,then $y = vx$. Differenti

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$y = x 	an^{-1} \left( \ln \frac{e}{x} ight)$

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(A) Given the differential equation: $\frac{dy}{dx} = \frac{y}{x} - \cos^2 \left( \frac{y}{x} ight)$.Let $v = \frac{y}{x}$,then $y = vx$. Differentiating with respect to $x$,we get $\frac{dy}{dx} = v + x \frac{dv}{dx}$.Substituting into the equation: $v + x \frac{dv}{dx} = v - \cos^2 v$.This simplifies to $x \frac{dv}{dx} = -\cos^2 v$.Separating variables: $\sec^2 v \, dv = -\frac{dx}{x}$.Integrating both sides: $\int \sec^2 v \, dv = -\int \frac{dx}{x} \implies 	an v = -\ln |x| + C$.Since the curve passes through $\left( 1, \frac{\pi}{4} ight)$,we have $v = \frac{\pi}{4}$ when $x = 1$.$	an \left( \frac{\pi}{4} ight) = -\ln(1) + C \implies 1 = 0 + C \implies C = 1$.Thus,$	an v = 1 - \ln x = \ln e - \ln x = \ln \left( \frac{e}{x} ight)$.Therefore,$v = 	an^{-1} \left( \ln \frac{e}{x} ight)$.Substituting $v = \frac{y}{x}$,we get $y = x 	an^{-1} \left( \ln \frac{e}{x} ight)$.

$A$ curve passes through the point $\left( 1, \frac{\pi}{4} \right)$ and its slope at any point is given by $\frac{dy}{dx} = \frac{y}{x} - \cos^2 \left( \frac{y}{x} \right)$. Find the equation of the curve.

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