Question

The solution of the equation

(1 + x²) (1 + y) dy + (1 + x) (1 +y²) dx = 0 is-

• A. tan^–1 x + log (1+ x²) + tan^–1 y + log (1 + y^‑2) = c
• B. tan^–1x –$\frac{1}{2}$log (1+ x²) + tan^–1 y–$\frac{1}{2}$log (1+y ²) = c
• C. tan^–1x +$\frac{1}{2}$log(1+x²)+ tan^–1y+$\frac{1}{2}$log (1 + y ²) = c
• D. None of these

Answer 1

Answer: (C)

tan^–1x +$\frac{1}{2}$log(1+x²)+ tan^–1y+$\frac{1}{2}$log (1 + y ²) = c

Answer 2

(1 + x²) (1+ y) dy + (1+x) (1 + y²) dx = 0

$\left( \frac{1 + y}{1 + y^{2}} \right)$dy+ $\left( \frac{1 + x}{1 + x^{2}} \right)$dx = 0

\ $\int_{}^{}{\left( \frac{1}{1 + y^{2}} + \frac{y}{1 + y^{2}} \right)dy}$

+ $\int_{}^{}{\left( \frac{1}{1 + x^{2}} + \frac{x}{1 + x^{2}} \right)dx}$ = 0

or tan^–1 y +$\frac{1}{2}$log (1 + y²) + tan^–1 (x) + $\frac{1}{2}$log (1 + x²) = c