To find the equation of the curve, we need to integrate the given slope of the tangent. The slope of the tangent at \((x,y)\) is given by:
\(\frac{dy}{dx} = \frac{y}{x} - \cos^2\left(\frac{y}{x}\right)\)
We need to find a function \(y = f(x)\) such that it satisfies the above differential equation and also passes through the point \(\left(1, \frac{\pi}{4}\right)\).
Let's assume a solution of the form \(y = vx\), where \(v\) is a function of \(x\). Therefore, \(\frac{y}{x} = v\) and \(\frac{dy}{dx} = v + x\frac{dv}{dx}\).
Substituting these into the differential equation gives:
\(v + x\frac{dv}{dx} = v - \cos^2(v)\)
Simplifying, we get:
\(x\frac{dv}{dx} + \cos^2(v) = 0\)
Separating variables:
\(\frac{dv}{\cos^2(v)} = -\frac{dx}{x}\)
Integrating both sides:
\(\int \sec^2(v) \, dv = -\int \frac{1}{x} \, dx\)
This leads to:
\(\tan(v) = -\log|x| + C_1\)
Substituting back \(v = \frac{y}{x}\):
\(\tan\left(\frac{y}{x}\right) = -\log|x| + C_1\)
From the initial condition \(\left(1, \frac{\pi}{4}\right)\), we substitute \(x = 1\) and \(y = \frac{\pi}{4}\):
\(\tan\left(\frac{\pi}{4}\right) = 1 = -\log|1| + C_1\)
This implies \(C_1 = 1\).
Thus, the equation becomes:
\(\tan\left(\frac{y}{x}\right) = -\log|x| + 1\)
Rearranging the terms, we can write:
\(\frac{y}{x} = \tan^{-1}\left(1 - \log|x|\right)\)
The general solution can be written as:
\(\tan\left(\frac{y}{x}\right) = -\log\left(\frac{x}{c}\right)\) where \(c\) is the integration constant determined by the initial condition.
Thus, this can be adjusted to match the form given in the options:
The correct answer is: \(y = x\tan^{-1}\left(\log \frac{c}{x}\right)\)