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

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

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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 for the slope of the tangent:$\frac{dy}{dx} = \frac{y}{x} - \cos^2\left(\frac{y}{x}ight)$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)$$x\frac{dv}{dx} = -\cos^2(v)$Separating the variables:$\sec^2(v) dv = -\frac{1}{x} dx$Integrating both sides:$\int \sec^2(v) dv = -\int \frac{1}{x} dx$$	an(v) = -\log|x| + C$Substituting $v = \frac{y}{x}$ back:$	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$$	an\left(\frac{\pi}{4}ight) = 0 + C \Rightarrow C = 1$Thus,$	an\left(\frac{y}{x}ight) = 1 - \log(x) = \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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