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: Understanding the Concept:
The equation is \((1 + t^2) + (x - e^{\tan^{-1}t})\dfrac{dt}{dx} = 0\). Rewrite to get \(\dfrac{dx}{dt}\).

Step 2: Rearrange:
\[ (x - e^{\tan^{-1}t})\frac{dt}{dx} = -(1+t^2) \Rightarrow \frac{dx}{dt} = -\frac{x - e^{\tan^{-1}t}}{1 + t^2} \]
\[ \frac{dx}{dt} + \frac{x}{1+t^2} = \frac{e^{\tan^{-1}t}}{1+t^2} \]
This is linear in \(x\) with \(P = \dfrac{1}{1+t^2}\).

Step 3: Integrating factor:
\[ \text{I.F.} = e^{\int P\,dt} = e^{\int\frac{dt}{1+t^2}} = e^{\tan^{-1}t} \]

Step 4: Result:
Option (A). Options (B) and (D) carry a minus sign that cannot come from an exponential, and (C) has the wrong sign inside the exponent.

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