Question

The integrating factor of the differential equation \((1+t^2)+(x-e^{tan^{-1}t})\frac{dt}{dx} = 0\) is

• A. \(e^{tan^{-1}t}\)
• B. \(-e^{tan^{-1}t}\)
• C. \(e^{-tan^{-1}t}\)
• D. \(-e^{-tan^{-1}t}\)

Hint:

Rewrite so x is the dependent variable: dx/dt + P x = Q.

Answer 1

The Correct Option is A

Step 1: Standard form:
A linear equation in $x$ with independent variable $t$ looks like $\dfrac{dx}{dt} + Px = Q$.

Step 2: Identify P:
From $\dfrac{dx}{dt} = \dfrac{e^{\tan^{-1}t} - x}{1+t^2}$ we get $P = \dfrac{1}{1+t^2}$.

Step 3: Integrate:
$\int P\,dt = \tan^{-1}t$, so the integrating factor is $e^{\tan^{-1}t}$, option (A).

Final Answer:
The integrating factor is e to the arctan t. \[ \boxed{\text{(A) }e^{\tan^{-1}t}} \]