Question

Find the general solution of the differential equation \( (x - y -1) dy = (x + y + 1) dx \).

• A. \( \tan^{-1} \left( \frac{y+1}{x} \right) - \frac{1}{2} \log(x^2 + y^2 + 2y + 1) = c \)
• B. \( (x - y) + \log(x + y) = c \)
• C. \( y^2 - x^2 + xy - 3y - x = c \)
• D. \( (x - y -1)^2 (x + y + 1)^3 = c \)

Hint:

For solving first-order differential equations, substitution methods simplify non-linear forms into solvable integrable expressions.

Answer 1

The Correct Option is A

Step 1: Given differential equation. We start with the equation: \[ (x - y -1) dy = (x + y + 1) dx \] Step 2: Expressing in separable form. Rewriting the equation in the standard form: \[ \frac{dy}{dx} = \frac{x + y + 1}{x - y -1} \] Using the substitution: \[ v = y + 1, \quad \text{so that} \quad dv = dy. \] Rewriting: \[ \frac{dv}{dx} = \frac{x + v}{x - v}. \] Step 3: Solving using separation of variables. Separating terms: \[ \frac{x - v}{x + v} dv = dx. \] Integrating both sides, we get: \[ \int \frac{x - v}{x + v} dv = \int dx. \] Step 4: Integrating both sides. Solving the integration: \[ \tan^{-1} \left( \frac{v}{x} \right) - \frac{1}{2} \log(x^2 + v^2) = c. \] Step 5: Substituting back \( v = y+1 \). \[ \tan^{-1} \left( \frac{y+1}{x} \right) - \frac{1}{2} \log(x^2 + y^2 + 2y + 1) = c. \]