Q33 / 90 · JEE Main 2019 (10 Jan Shift 2) · Mathematics

A curve amongst the family of curves represented by the differential equation, $(x^{2} - y^{2}) d x + 2 x y d y = 0$ which passes through $(1, 1)$ , is

Solution

$x^{2} d x - y^{2} d x + 2 x y d y = 0$

$x^{2} d x = y^{2} d x - 2 x y d y$

$\Rightarrow - d x = d (\frac{y^{2}}{x})$ (exact differential equation)

$\Rightarrow - x = \frac{y^{2}}{x} + C$

$\Rightarrow x^{2} + y^{2} + C x = 0$

$∵$ It passes through $(1, 1)$ .

Hence, $C = - 2$

$\Rightarrow x^{2} + y^{2} - 2 x = 0$

Hence, the correct answer is option: 'A'.