The general solution of the differential equation \(x\sqrt{1 + y^2}\,dx + y\sqrt{4 + x^2}\,dy = 0\) is (C is an arbitrary constant)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\sqrt{4 + x^2} + \sqrt{1 + y^2} = C\)
Dividing by \(\sqrt{4 + x^2}\,\sqrt{1 + y^2}\) separates the variables:
\( \displaystyle \frac{x}{\sqrt{4 + x^2}}\,dx + \frac{y}{\sqrt{1 + y^2}}\,dy = 0 \)
Since the derivative of \(\sqrt{4 + x^2}\) is \(\frac{x}{\sqrt{4 + x^2}}\), and the derivative of \(\sqrt{1 + y^2}\) is \(\frac{y}{\sqrt{1 + y^2}}\), integrating gives
\( \sqrt{4 + x^2} + \sqrt{1 + y^2} = C \)
Why the other options are incorrect
Option (1) \(\sqrt{4 + x^2} - \sqrt{1 + y^2} = C\) — Incorrect. Both terms are on the same side of the equation with a plus sign, so their integrals are added. This family solves the equation with \(-y\sqrt{4 + x^2}\,dy\).
Option (3) \(x^2 + y^2 = C\) — Incorrect. This solves \(x\,dx + y\,dy = 0\); the square-root factors are ignored.
Option (4) \(\sqrt{4 + x^2}\,\sqrt{1 + y^2} = C\) — Incorrect. Differentiating gives \(\frac{dy}{dx} = -\frac{x(1 + y^2)}{y(4 + x^2)}\), but the equation needs \(\frac{dy}{dx} = -\frac{x\sqrt{1 + y^2}}{y\sqrt{4 + x^2}}\).