The general solution of the differential equation \(x\displaystyle\frac{dy}{dx} = y + \sqrt{x^2 - y^2}\), where \(x \gt 0\) and \(|y| \lt x\), is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\sin^{-1}\left(\displaystyle\frac{y}{x}\right) = \log x + C\)
Since \(x \gt 0\), \(\sqrt{x^2 - y^2} = x\sqrt{1 - \frac{y^2}{x^2}}\). Dividing by x:
\( \displaystyle \frac{dy}{dx} = \frac{y}{x} + \sqrt{1 - \frac{y^2}{x^2}} \)
This is a homogeneous equation. With \(y = vx\), \(\frac{dy}{dx} = v + x\frac{dv}{dx}\):
\( \displaystyle v + x\frac{dv}{dx} = v + \sqrt{1 - v^2} \)
\( \displaystyle \frac{dv}{\sqrt{1 - v^2}} = \frac{dx}{x} \)
\( \sin^{-1} v = \log x + C \)
Putting \(v = \frac{y}{x}\) back gives \(\sin^{-1}\left(\frac{y}{x}\right) = \log x + C\).
Why the other options are incorrect
Option (2) \(\cos^{-1}\left(\displaystyle\frac{y}{x}\right) = \log x + C\) — Incorrect. The derivative of \(\cos^{-1} v\) is \(-\frac{1}{\sqrt{1 - v^2}}\); this would need \(x\frac{dv}{dx} = -\sqrt{1 - v^2}\).
Option (3) \(\tan^{-1}\left(\displaystyle\frac{y}{x}\right) = \log x + C\) — Incorrect. \(\tan^{-1} v\) comes from \(\int \frac{dv}{1 + v^2}\), not from \(\int \frac{dv}{\sqrt{1 - v^2}}\).
Option (4) \(\sin^{-1}\left(\displaystyle\frac{y}{x}\right) = x + C\) — Incorrect. The right side is \(\int \frac{dx}{x} = \log x\), not \(\int dx\).