The integrating factor of the differential equation \((1 + y^2)dx - (\tan^{-1}y - x)dy = 0\), is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(e^{\tan^{-1}y}\)
The equation is linear in x (not in y), so it is written in the form \(\frac{dx}{dy} + Px = Q\) with P and Q functions of y:
\((1 + y^2)\,dx = (\tan^{-1}y - x)\,dy\)
\( \displaystyle \frac{dx}{dy} = \frac{\tan^{-1}y - x}{1 + y^2} \)
\( \displaystyle \frac{dx}{dy} + \frac{1}{1 + y^2}\,x = \frac{\tan^{-1}y}{1 + y^2} \)
So
\( \displaystyle P = \frac{1}{1 + y^2} \)
The integrating factor is
\( \displaystyle \text{I.F.} = e^{\int P\,dy} = e^{\int \frac{dy}{1 + y^2}} \)
\( = e^{\tan^{-1}y} \)
Why the other options are incorrect
Option (1) \(\tan^{-1}y\) — Incorrect. This is only \(\int P\,dy\); the integrating factor is e raised to this power.
Option (3) \(\displaystyle\frac{1}{1 + y^2}\) — Incorrect. This is P itself, the coefficient of x, not the integrating factor.
Option (4) \(\displaystyle\frac{1}{x\left(1 + y^2\right)}\) — Incorrect. For an equation in dx/dy the integrating factor depends on y only; it cannot contain x.