The general solution of the differential equation \((1 - x^2)\displaystyle\frac{dy}{dx} - xy = 1\), \(-1 \lt x \lt 1\), is (C is an arbitrary constant)
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(y\sqrt{1 - x^2} = \sin^{-1}x + C\)
Dividing by \(1 - x^2\) gives a linear equation:
\( \displaystyle \frac{dy}{dx} - \frac{x}{1 - x^2}\,y = \frac{1}{1 - x^2} \)
With \(t = 1 - x^2\), \(dt = -2x\,dx\):
\( \displaystyle \int -\frac{x}{1 - x^2}\,dx = \frac{1}{2}\log(1 - x^2) \)
So the integrating factor is
\( e^{\frac{1}{2}\log(1 - x^2)} = \sqrt{1 - x^2} \)
Then
\( \displaystyle y\sqrt{1 - x^2} = \int \frac{1}{1 - x^2}\cdot\sqrt{1 - x^2}\,dx = \int \frac{dx}{\sqrt{1 - x^2}} \)
\( y\sqrt{1 - x^2} = \sin^{-1}x + C \)
Why the other options are incorrect
Option (1) \(y(1 - x^2) = \sin^{-1}x + C\) — Incorrect. The factor \(\frac{1}{2}\) in front of the logarithm makes the integrating factor a square root, \(\sqrt{1 - x^2}\), not \(1 - x^2\).
Option (2) \(\frac{y}{\sqrt{1 - x^2}} = \sin^{-1}x + C\) — Incorrect. This uses \(\frac{1}{\sqrt{1 - x^2}}\), which comes from the wrong sign of P; \(\int -\frac{x}{1 - x^2}\,dx = +\frac{1}{2}\log(1 - x^2)\).
Option (3) \(y\sqrt{1 - x^2} = x + C\) — Incorrect. The integrating factor is right, but \(Q\cdot\sqrt{1 - x^2} = \frac{1}{\sqrt{1 - x^2}}\), whose integral is \(\sin^{-1}x\), not x.