The particular solution of the differential equation \(\displaystyle\frac{dy}{dx} = (2x + 1)y^2\) satisfying \(y(0) = 1\) is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(y = \displaystyle\frac{1}{1 - x - x^2}\)
Separating the variables:
\( \displaystyle \frac{dy}{y^2} = (2x + 1)\,dx \)
Integrating both sides, with \(\int y^{-2}\,dy = -\frac{1}{y}\):
\( \displaystyle -\frac{1}{y} = x^2 + x + C \)
Putting x = 0, y = 1:
\(-1 = C\)
So
\( \displaystyle -\frac{1}{y} = x^2 + x - 1 \)
\( \displaystyle y = \frac{1}{1 - x - x^2} \)
Why the other options are incorrect
Option (1) \(y = \frac{1}{1 + x + x^2}\) — Incorrect. This takes \(\int y^{-2}\,dy\) as \(+\frac{1}{y}\); the minus sign is lost. Its slope at x = 0 is −1, but the equation gives \((0 + 1) \times 1^2 = 1\) there.
Option (2) \(y = 1 + x + x^2\) — Incorrect. Its slope is \(2x + 1\), but the equation needs \((2x + 1)y^2\); at x = 1 these are 3 and 27.
Option (3) \(y = \frac{1}{1 - x^2}\) — Incorrect. Its slope at x = 0 is 0, but the equation gives 1 there.