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

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(C) Given the differential equation: $\frac{dy}{dx} = \frac{y}{x} - \cos^2 \frac{y}{x} \dots (i)$Substitute $y = vx$,then $\frac{dy}{dx} = v + x \frac

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

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(C) Given the differential equation: $\frac{dy}{dx} = \frac{y}{x} - \cos^2 \frac{y}{x} \dots (i)$Substitute $y = vx$,then $\frac{dy}{dx} = v + x \frac{dv}{dx} \dots (ii)$Substituting $(ii)$ into $(i)$: $v + x \frac{dv}{dx} = v - \cos^2 v \Rightarrow x \frac{dv}{dx} = -\cos^2 v$Separating variables: $\sec^2 v \, dv = -\frac{1}{x} \, dx$Integrating both sides: $\int \sec^2 v \, dv = -\int \frac{1}{x} \, dx + C \Rightarrow 	an v = -\log |x| + C$Since $v = \frac{y}{x}$,we have $	an \frac{y}{x} = -\log x + C \dots (iii)$The curve passes through $(1, \frac{\pi}{4})$,so $	an \frac{\pi}{4} = -\log 1 + C \Rightarrow 1 = 0 + C \Rightarrow C = 1$Thus,$	an \frac{y}{x} = 1 - \log x = \log e - \log x = \log \frac{e}{x}$Therefore,$y = x 	an^{-1} \left( \log \frac{e}{x} ight)$.

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

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