Solution of the differential equation
\(\displaystyle\frac{dy}{dx} = \sqrt{1 + x^2 + y^2 + x^2y^2}\) is: (Here C is an arbitrary constant)
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\displaystyle\log\left|\frac{y + \sqrt{1 + y^2}}{\sqrt{x + \sqrt{1 + x^2}}}\right| = \frac{x}{2}\sqrt{1 + x^2} + C\)
The expression under the root factorises:
\(1 + x^2 + y^2 + x^2y^2 = (1 + x^2)(1 + y^2)\)
\(\displaystyle \frac{dy}{dx} = \sqrt{1 + x^2}\,\sqrt{1 + y^2}\)
So the variables separate:
\(\displaystyle \int \frac{dy}{\sqrt{1 + y^2}} = \int \sqrt{1 + x^2}\,dx\)
Using the standard integrals
\(\displaystyle \int \frac{dy}{\sqrt{1 + y^2}} = \log\left|y + \sqrt{1 + y^2}\right|\)
\(\displaystyle \int \sqrt{1 + x^2}\,dx = \frac{x}{2}\sqrt{1 + x^2} + \frac{1}{2}\log\left|x + \sqrt{1 + x^2}\right|\)
we get
\(\displaystyle \log\left|y + \sqrt{1 + y^2}\right| - \frac{1}{2}\log\left|x + \sqrt{1 + x^2}\right| = \frac{x}{2}\sqrt{1 + x^2} + C\)
Since \(\frac{1}{2}\log t = \log\sqrt{t}\):
\(\displaystyle \log\left|\frac{y + \sqrt{1 + y^2}}{\sqrt{x + \sqrt{1 + x^2}}}\right| = \frac{x}{2}\sqrt{1 + x^2} + C\)
Why the other options are incorrect
Option (1) \(\displaystyle\log\left|\frac{y + \sqrt{1 + y^2}}{x + \sqrt{1 + x^2}}\right| = C\) — Incorrect. It leaves out the term \(\frac{x}{2}\sqrt{1 + x^2}\), and it drops the factor \(\frac{1}{2}\) of the log term, which gives the square root in the denominator.
Option (2) \(\displaystyle\log\left|\frac{y + \sqrt{1 + y^2}}{\sqrt{x + \sqrt{1 + x^2}}}\right| = C\) — Incorrect. It leaves out the term \(\frac{x}{2}\sqrt{1 + x^2}\) of \(\int \sqrt{1 + x^2}\,dx\).
Option (4) \(\displaystyle\log\left|\frac{y + \sqrt{1 + y^2}}{x + \sqrt{1 + x^2}}\right| = \frac{x}{2}\sqrt{1 + x^2} + C\) — Incorrect. It drops the factor \(\frac{1}{2}\) of the log term in \(\int \sqrt{1 + x^2}\,dx\), so the denominator lacks the square root.