Q. If $x^{2} + y^{2} = 1$, then
A) $yy^{''} - \left( 2y^{'} \right)^{2} + 1 = 0$
B) $yy^{''} + \left( y^{'} \right)^{2} + 1 = 0$
C) $yy^{''} - \left( y^{'} \right)^{2} - 1 = 0$
D) $yy^{''} + 2\left( y^{'} \right)^{2} + 1 = 0$
Explanation
(b)
Given, $x^{2} + y^{2} = 1$
On differentiating $w.r.t.\ x$, we get
$2x + 2yy^{'} = 0$
$\Rightarrow$ $x + yy^{'} = 0$
Again , differentiating, we get
$1 + yy^{''} + \left( y^{'} \right)^{2} = 0$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.