The particular solution of the differential equation \(\displaystyle\frac{dy}{dx} = \frac{1 + y^2}{1 + x^2}\) satisfying \(y(0) = 1\) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(y = \displaystyle\frac{1 + x}{1 - x}\)
Separating the variables:
\( \displaystyle \frac{dy}{1 + y^2} = \frac{dx}{1 + x^2} \)
Integrating both sides:
\(\tan^{-1} y = \tan^{-1} x + C\)
Putting x = 0, y = 1:
\( \displaystyle \frac{\pi}{4} = 0 + C \)
So
\( \displaystyle \tan^{-1} y = \tan^{-1} x + \frac{\pi}{4} \)
Taking tan of both sides and using \(\tan(\alpha + \beta) = \frac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta}\) with \(\tan\frac{\pi}{4} = 1\):
\( \displaystyle y = \frac{x + 1}{1 - x \cdot 1} = \frac{1 + x}{1 - x} \)
Why the other options are incorrect
Option (1) \(y = \frac{1 - x}{1 + x}\) — Incorrect. This comes from \(\tan^{-1} y = \frac{\pi}{4} - \tan^{-1} x\); its slope at x = 0 is \(-2\), but the equation gives \(\frac{1 + 1}{1 + 0} = 2\) there.
Option (2) \(y = 1 + x\) — Incorrect. Its slope is always 1, but at x = 1 the equation needs \(\frac{1 + 4}{1 + 1} = \frac{5}{2}\).
Option (4) \(y = \frac{x - 1}{x + 1}\) — Incorrect. At x = 0 it gives y = −1, not 1.