The general solution of the differential equation \(\displaystyle\frac{dy}{dx} = \frac{x + 2}{y - 1}\), \(y \ne 1\), is (C is an arbitrary constant)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(y^2 - x^2 - 2y - 4x = C\)
The equation is variables separable:
\((y - 1)\,dy = (x + 2)\,dx\)
Integrating both sides:
\( \displaystyle \frac{y^2}{2} - y = \frac{x^2}{2} + 2x + C_1 \)
Multiplying by 2 and writing \(C = 2C_1\):
\(y^2 - 2y = x^2 + 4x + C\)
\(y^2 - x^2 - 2y - 4x = C\)
Why the other options are incorrect
Option (2) \(x^2 + y^2 + 4x - 2y = C\) — Incorrect. Differentiating gives \(\frac{dy}{dx} = -\frac{x + 2}{y - 1}\), the wrong sign.
Option (3) \(y^2 - x^2 - y - 2x = C\) — Incorrect. Only the squared terms were doubled; differentiating gives \(\frac{dy}{dx} = \frac{2x + 2}{2y - 1}\).
Option (4) \(y^2 - x^2 + 2y + 4x = C\) — Incorrect. Differentiating gives \(\frac{dy}{dx} = \frac{x - 2}{y + 1}\), a different equation.