Question

Solution of the equation x dx + y dy + $\frac{xdy - ydx}{x^{2} + y^{2}}$ = 0 is –

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

Answer 1

Answer: (C)

y = x tan $\left( \frac{c–x^{2}–y^{2}}{2} \right)$

Answer 2

We have, x dx + y dy + $\frac{xdy - ydx}{x^{2} + y^{2}} = 0$

Ž $\frac{1}{2}$ d (x² + y²) + d tan^–1 $\left( \frac{y}{x} \right)$ = 0

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

Ž x² + y² + 2 tan^–1 $\frac{y}{x}$ = c.

\ y = x tan $\left( \frac{c - x^{2} - y^{2}}{2} \right)$ is the required solution.

Hence (3) is the correct answer