The particular solution of the differential equation \(x\displaystyle\frac{dy}{dx} = y - x\cos^2\left(\frac{y}{x}\right)\), \(x \gt 0\), satisfying \(y(1) = \displaystyle\frac{\pi}{4}\), is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\tan\left(\displaystyle\frac{y}{x}\right) = 1 - \log x\)
Dividing by x:
\( \displaystyle \frac{dy}{dx} = \frac{y}{x} - \cos^2\left(\frac{y}{x}\right) \)
The equation is homogeneous. Let \(y = vx\):
\( \displaystyle v + x\frac{dv}{dx} = v - \cos^2 v \)
\( \displaystyle \sec^2 v\,dv = -\frac{dx}{x} \)
Integrating:
\( \tan v = -\log x + C \)
At \(x = 1\), \(v = \frac{y}{x} = \frac{\pi}{4}\):
\( 1 = 0 + C \Rightarrow C = 1 \)
\( \displaystyle \tan\left(\frac{y}{x}\right) = 1 - \log x \)
Why the other options are incorrect
Option (1) \(\tan\left(\frac{y}{x}\right) = 1 + \log x\) — Incorrect. It fits \(y(1) = \frac{\pi}{4}\), but it gives \(x\frac{dv}{dx} = +\cos^2 v\); the minus sign of the equation is lost.
Option (2) \(\cot\left(\frac{y}{x}\right) = 1 - \log x\) — Incorrect. It gives \(x\frac{dv}{dx} = \sin^2 v\), not \(-\cos^2 v\).
Option (3) \(\tan\left(\frac{y}{x}\right) = 1 - 2\log x\) — Incorrect. It gives \(x\frac{dv}{dx} = -2\cos^2 v\); the integral of \(\frac{dx}{x}\) is \(\log x\), with no factor 2.