Question

Find the general solution of the differential equation \(\frac{dy}{dx} = \frac{1 + y^2}{1 + x^2}\)

• A. \(\tan^{-1} x + \tan^{-1} y + \text{constant} = 0\)
• B. \(-\tan^{-1} x - \tan^{-1} y + \text{constant} = 0\)
• C. \(\tan x - \tan y + \text{constant} = 0\)
• D. \(\tan^{-1} y - \tan^{-1} x + \text{constant} = 0\)

Hint:

This can also be expressed compactly as \(\frac{y - x}{1 + xy} = C\) using the identity \(\tan^{-1} y - \tan^{-1} x = \tan^{-1}\left(\frac{y - x}{1 + xy}\right)\).

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
A first-order ordinary differential equation is separable if all terms involving \(y\) can be gathered with \(dy\) on one side and all terms involving \(x\) with \(dx\) on the other side.
Key Formula or Approach:
\[ \frac{dy}{dx} = g(x)h(y) \implies \frac{dy}{h(y)} = g(x)\,dx \implies \int \frac{dy}{h(y)} = \int g(x)\,dx + C \]

Step 2: Detailed Explanation:
Given the differential equation:
\[ \frac{dy}{dx} = \frac{1 + y^2}{1 + x^2} \]
Separating the variables:
\[ \frac{dy}{1 + y^2} = \frac{dx}{1 + x^2} \]
Integrating both sides:
\[ \int \frac{dy}{1 + y^2} = \int \frac{dx}{1 + x^2} + C' \]
\[ \tan^{-1} y = \tan^{-1} x + C' \]
Rearranging the terms to standard form:
\[ \tan^{-1} y - \tan^{-1} x + \text{constant} = 0 \]

Step 3: Final Answer:
Thus, the general solution is \(\tan^{-1} y - \tan^{-1} x + \text{constant} = 0\), matching option (D).