The particular solution of the differential equation \(x\displaystyle\frac{dy}{dx} = y^2 - 1\), \(x \gt 0\), satisfying \(y(1) = 0\), is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(y = \displaystyle\frac{1 - x^2}{1 + x^2}\)
Separating the variables:
\( \displaystyle \frac{dy}{y^2 - 1} = \frac{dx}{x} \)
Using the standard integral \(\int \frac{dy}{y^2 - 1} = \frac{1}{2}\log\left|\frac{y - 1}{y + 1}\right|\):
\( \displaystyle \frac{1}{2}\log\left|\frac{y - 1}{y + 1}\right| = \log x + C \)
\( \displaystyle \log\left|\frac{y - 1}{y + 1}\right| = \log x^2 + 2C \)
\( \displaystyle \frac{y - 1}{y + 1} = Kx^2 \)
Putting x = 1, y = 0:
\( \displaystyle \frac{0 - 1}{0 + 1} = K \)
\(K = -1\)
So
\( \displaystyle \frac{y - 1}{y + 1} = -x^2 \)
\(y - 1 = -x^2y - x^2\)
\(y(1 + x^2) = 1 - x^2\)
\( \displaystyle y = \frac{1 - x^2}{1 + x^2} \)
Why the other options are incorrect
Option (1) \(y = \frac{x^2 - 1}{x^2 + 1}\) — Incorrect. It gives y = 0 at x = 1, but it is the negative of the solution: its \(y^2 - 1\) is the same while \(x\frac{dy}{dx}\) changes sign, so the two sides differ (at x = 1 they are 1 and −1).
Option (2) \(y = \frac{1 - x}{1 + x}\) — Incorrect. Here \(x\frac{dy}{dx} = -\frac{2x}{(1 + x)^2}\), while \(y^2 - 1 = -\frac{4x}{(1 + x)^2}\); at x = 1 these are \(-\frac{1}{2}\) and −1.
Option (4) \(y = \frac{1 - x^2}{2x}\) — Incorrect. At x = 1, \(x\frac{dy}{dx} = -\frac{1}{2} - \frac{1}{2} = -1\) matches, but at x = 2, \(x\frac{dy}{dx} = -\frac{5}{4}\) while \(y^2 - 1 = -\frac{7}{16}\).