Question

The general solution of the differential equation given by: \[ \tan^{-1} x + \tan^{-1} y = c \]

• A. \( \frac{dy}{dx} = \frac{1 + y^2}{1 + x^2} \)
• B. \( \frac{dy}{dx} = \frac{1 + x^2}{1 + y^2} \)
• C. \( (1 + x^2) dy + (1 + y^2) dx = 0 \)
• D. \( (1 + x^2) dx + (1 + y^2) dy = 0 \)

Hint:

For equations involving inverse trigonometric functions, differentiate both sides using chain rule carefully.

Answer 1

The Correct Option is C

Step 1: Differentiating both sides.
Given: \[ \tan^{-1} x + \tan^{-1} y = c \] Differentiating both sides with respect to \( x \): \[ \frac{d}{dx} (\tan^{-1} x) + \frac{d}{dx} (\tan^{-1} y) = 0 \] Step 2: Using derivative formulas.
We use the derivative of inverse tangent: \[ \frac{d}{dx} (\tan^{-1} x) = \frac{1}{1 + x^2} \] Since \( y \) is also a function of \( x \), we apply the chain rule: \[ \frac{1}{1 + x^2} + \frac{1}{1 + y^2} \cdot \frac{dy}{dx} = 0 \] Step 3: Rewriting the equation.
Rearranging, \[ (1 + x^2) dy + (1 + y^2) dx = 0 \] Thus, the correct answer is (C).