The slope of the tangent at (x, y) to a curve | vedclass

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(C) Given the differential equation: $\frac{dy}{dx} = \frac{y}{x} - \cos^2\left( \frac{y}{x} \right)$.This is a homogeneous differential equation. Let

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

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(C) Given the differential equation: $\frac{dy}{dx} = \frac{y}{x} - \cos^2\left( \frac{y}{x} ight)$.This is a homogeneous differential equation. Let $y = vx$,then $\frac{dy}{dx} = v + x\frac{dv}{dx}$.Substituting these into the equation: $v + x\frac{dv}{dx} = v - \cos^2(v)$.This simplifies to $x\frac{dv}{dx} = -\cos^2(v)$,or $\sec^2(v) dv = -\frac{dx}{x}$.Integrating both sides: $\int \sec^2(v) dv = -\int \frac{1}{x} dx$.$	an(v) = -\log|x| + C$.Substituting $v = \frac{y}{x}$: $	an\left( \frac{y}{x} ight) = -\log|x| + C$.The curve passes through $\left( 1, \frac{\pi}{4} ight)$,so $	an\left( \frac{\pi/4}{1} ight) = -\log(1) + C \implies 1 = 0 + C \implies C = 1$.Thus,$	an\left( \frac{y}{x} ight) = -\log(x) + 1 = \log(e) - \log(x) = \log\left( \frac{e}{x} ight)$.Therefore,$y = x	an^{-1}\left[ \log\left( \frac{e}{x} ight) ight]$.

The slope of the tangent at $(x, y)$ to a curve passing through $\left( 1, \frac{\pi}{4} \right)$ is given by $\frac{y}{x} - \cos^2\left( \frac{y}{x} \right)$. Then the equation of the curve is:

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