Solution of a differential equation \((1 + y^2)\,dx = (\tan^{-1}y - x)\,dy\) is:
(where C is the constant of Integration)
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(xe^{\tan^{-1}y} = e^{\tan^{-1}y}(\tan^{-1}y - 1) + C\)
Dividing by \((1 + y^2)\,dy\), the equation is linear in x (x as a function of y):
\( \displaystyle \frac{dx}{dy} + \frac{1}{1 + y^2}\,x = \frac{\tan^{-1}y}{1 + y^2} \)
The integrating factor is
\( \displaystyle \text{I.F.} = e^{\int\frac{dy}{1 + y^2}} = e^{\tan^{-1}y} \)
So
\( \displaystyle x\,e^{\tan^{-1}y} = \int\frac{\tan^{-1}y}{1 + y^2}\,e^{\tan^{-1}y}\,dy \)
Let \(t = \tan^{-1}y\), so \(dt = \frac{dy}{1 + y^2}\):
\( \displaystyle x\,e^{t} = \int t\,e^{t}\,dt \)
Integrating by parts:
\( \displaystyle \int t\,e^{t}\,dt = t\,e^{t} - e^{t} + C = e^{t}(t - 1) + C \)
Putting back \(t = \tan^{-1}y\):
\(xe^{\tan^{-1}y} = e^{\tan^{-1}y}(\tan^{-1}y - 1) + C\)
Why the other options are incorrect
Option (1) \(ye^{\tan^{-1}y} = e^{\tan^{-1}y}(\tan^{-1}y + 1) + C\) — Incorrect. The dependent variable here is x, so the left side is \(x\,e^{\tan^{-1}y}\), and the bracket has −1, not +1.
Option (2) \(ye^{\tan^{-1}y} = e^{\tan^{-1}y}(\tan^{-1}y - 1) + C\) — Incorrect. The bracket is right, but the left side must be \(x\,e^{\tan^{-1}y}\): the equation is linear in x, not in y.
Option (4) \(xe^{\tan^{-1}y} = e^{\tan^{-1}y}(\tan^{-1}y + 1) + C\) — Incorrect. \(\int t\,e^{t}\,dt = e^{t}(t - 1) + C\); the sign in the bracket is minus.
NCERT (2026-27 edition):
Differential Equations
