Question

The general solution of the differential equation \[ (1 + y^2) \, dx + (1 + x^2) \, dy = 0 \] is:

• A. \( x - y = C(1 - xy) \)
• B. \( x - y = C(1 + xy) \)
• C. \( x + y = C(1 - xy) \)
• D. \( x + y = C(1 + xy) \)

Hint:

To solve separable differential equations, separate the variables, integrate both sides, and then simplify to find the general solution.

Answer 1

The Correct Option is C

We are given the differential equation: \[ (1 + y^2) \, dx + (1 + x^2) \, dy = 0. \] Step 1: Separation of variables. We can rewrite the given differential equation as: \[ \frac{dx}{dy} = -\frac{1 + x^2}{1 + y^2}. \] This is a separable differential equation, and we can separate the variables: \[ \frac{dx}{1 + x^2} = -\frac{dy}{1 + y^2}. \] Step 2: Integrating both sides. Now, integrate both sides: \[ \int \frac{dx}{1 + x^2} = -\int \frac{dy}{1 + y^2}. \] The integral of \( \frac{1}{1 + x^2} \) is \( \tan^{-1}(x) \), and the integral of \( \frac{1}{1 + y^2} \) is \( \tan^{-1}(y) \). So, we get: \[ \tan^{-1}(x) = -\tan^{-1}(y) + C, \] where \( C \) is the constant of integration. Step 3: Solving for the general solution. Now, solve for the general solution: \[ \tan^{-1}(x) + \tan^{-1}(y) = C. \] Using the identity for the sum of inverse tangents, we have: \[ \tan^{-1} \left( \frac{x + y}{1 - xy} \right) = C. \] Taking the tangent of both sides: \[ \frac{x + y}{1 - xy} = \tan(C). \] Let \( C_1 = \tan(C) \), where \( C_1 \) is a new constant, so the equation becomes: \[ x + y = C_1(1 - xy). \] Thus, the general solution is: \[ x + y = C(1 - xy), \] where \( C \) is a constant. Step 4: Conclusion. Therefore, the correct answer is (c) \( x + y = C(1 - xy) \).