Q. The equation of curve through point (1,0) which satisfies the differential equation $\left( 1 + y^{2} \right)dx - xy\ dy = 0$ is
A) $x^{2} + y^{2} = 4$
B) $x^{2} - y^{2} = 1$
C) $2x^{2} + y^{2} = 2$
D) None of these
Explanation
(b)
Given, $\frac{dx}{x} = \frac{ydy}{1 + y^{2}}$
$\Rightarrow$ $\log x = \frac{1}{2}\log{(1 + y^{2})} + \log c$
$\Rightarrow$ $x = c\sqrt{1 + y^{2}}$
But it passes through (1,0), so we get $c = 1$
$\therefore$ Solution is $x^{2} - y^{2} = 1$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.