Q72 / 150 · MHT CET 2026 (16 April Shift 1) · Mathematics
The integrating factor of the differential equation $(1 + t^2) + \left(x - e^{\tan^{-1}t}\right)\dfrac{dt}{dx} = 0$ is
- Correct answerA$e^{\tan^{-1}t}$
- B$-e^{\tan^{-1}t}$
- C$e^{-\tan^{-1}t}$
- D$-e^{-\tan^{-1}t}$
Solution
The given differential equation is $(1 + t^2) + \left(x - e^{\tan^{-1}t}\right)\dfrac{dt}{dx} = 0$
Rearranging the terms, we get:
$(1 + t^2)\dfrac{dx}{dt} + x - e^{\tan^{-1}t} = 0$
$\dfrac{dx}{dt} + \dfrac{1}{1 + t^2}x = \dfrac{e^{\tan^{-1}t}}{1 + t^2}$
This is a linear differential equation of the form $\dfrac{dx}{dt} + Px = Q$, where $P = \dfrac{1}{1 + t^2}$ and $Q = \dfrac{e^{\tan^{-1}t}}{1 + t^2}$
The integrating factor is given by $e^{\int P dt}$
$IF = e^{\int \dfrac{1}{1 + t^2} dt} = e^{\tan^{-1}t}$
Answer: $e^{\tan^{-1}t}$
Rearranging the terms, we get:
$(1 + t^2)\dfrac{dx}{dt} + x - e^{\tan^{-1}t} = 0$
$\dfrac{dx}{dt} + \dfrac{1}{1 + t^2}x = \dfrac{e^{\tan^{-1}t}}{1 + t^2}$
This is a linear differential equation of the form $\dfrac{dx}{dt} + Px = Q$, where $P = \dfrac{1}{1 + t^2}$ and $Q = \dfrac{e^{\tan^{-1}t}}{1 + t^2}$
The integrating factor is given by $e^{\int P dt}$
$IF = e^{\int \dfrac{1}{1 + t^2} dt} = e^{\tan^{-1}t}$
Answer: $e^{\tan^{-1}t}$
Hence, the correct answer is option: 'A'.