The general solution of the differential equation \(\displaystyle\frac{dy}{dx} = \frac{1}{x + y^2}\) is (C is an arbitrary constant)
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(x + y^2 + 2y + 2 = Ce^{y}\)
The equation is not linear in y, but taking the reciprocal makes it linear in x:
\( \displaystyle \frac{dx}{dy} = x + y^2 \)
\( \displaystyle \frac{dx}{dy} - x = y^2 \)
Here \(P_1 = -1\), so the integrating factor is \(e^{-y}\):
\( \displaystyle x\,e^{-y} = \int y^2e^{-y}\,dy \)
Integrating by parts twice:
\( \displaystyle \int y^2e^{-y}\,dy = -y^2e^{-y} + \int 2y\,e^{-y}\,dy \)
\( \displaystyle \int 2y\,e^{-y}\,dy = -2ye^{-y} - 2e^{-y} \)
So
\( x\,e^{-y} = -(y^2 + 2y + 2)e^{-y} + C \)
\( x + y^2 + 2y + 2 = Ce^{y} \)
Why the other options are incorrect
Option (1) \(x + y^2 + 2y + 2 = Ce^{-y}\) — Incorrect. Multiplying \(x\,e^{-y} = \ldots + C\) by \(e^{y}\) gives \(Ce^{y}\), not \(Ce^{-y}\).
Option (2) \(x + y^2 - 2y + 2 = Ce^{y}\) — Incorrect. The sign of the 2y term is wrong; it gives \(\frac{dx}{dy} = x + y^2 - 4y + 4\).
Option (4) \(x + y^2 + 2y = Ce^{y}\) — Incorrect. The last step of the integration by parts, \(-2e^{-y}\), is dropped; it gives \(\frac{dx}{dy} = x + y^2 - 2\).