Question

The integrating factor of the differential equation \[ (1 + x^2) \, dt = \left( \tan^{-1} x - t \right) \, dx \text{ is} \]

• A. \( -e^{\frac{(\tan^{-1} x)^2}{2}} \)
• B. \( -e^{\tan^{-1} x} \)
• C. \( e^{\frac{(\tan^{-1} x)^2}{2}} \)
• D. \( e^{\tan^{-1} x} \)

Hint:

To find the integrating factor for a linear first-order differential equation, use the formula \( \mu(x) = e^{\int P(x) dx} \), where \( P(x) \) is the coefficient of \( y \).

Answer 1

The Correct Option is D

Step 1: Identifying the form of the equation.
The given differential equation is of the form: \[ (1 + x^2) \, dt = \left( \tan^{-1} x - t \right) \, dx \] We can rewrite this as: \[ \frac{dt}{dx} + t = \tan^{-1} x \] This is a first-order linear differential equation of the form: \[ \frac{dy}{dx} + P(x)y = Q(x) \] where \( P(x) = -1 \) and \( Q(x) = \tan^{-1} x \).
Step 2: Finding the integrating factor.
The integrating factor \( \mu(x) \) is given by: \[ \mu(x) = e^{\int P(x) dx} \] In this case, \( P(x) = -1 \), so the integrating factor is: \[ \mu(x) = e^{\int -1 \, dx} = e^{-\tan^{-1} x} \] Thus, the correct integrating factor is \( e^{\tan^{-1} x} \).

Step 3: Conclusion.
Therefore, the integrating factor is \( e^{\tan^{-1} x} \), which makes option (D) the correct answer.