Question

If \[ \tan^{-1} x + \tan^{-1} y = c \text{ is the general solution of the differential equation} \]

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

Hint:

When differentiating inverse trigonometric functions, use the chain rule and the derivatives of \( \tan^{-1} x \) and \( \tan^{-1} y \).

Answer 1

The Correct Option is A

Step 1: Differentiate the given equation.
We are given \( \tan^{-1} x + \tan^{-1} y = c \). Differentiating both sides with respect to \( x \) gives: \[ \frac{1}{1 + x^2} \frac{dx}{dx} + \frac{1}{1 + y^2} \frac{dy}{dx} = 0 \]
Step 2: Solve for \( \frac{dy}{dx} \).
Solving for \( \frac{dy}{dx} \), we get: \[ \frac{dy}{dx} = \frac{1 + y^2}{1 + x^2} \]
Step 3: Conclusion.
Thus, the general solution of the differential equation is \( \frac{dy}{dx} = \frac{1 + y^2}{1 + x^2} \), corresponding to option (A).