If x2 + y2 = 1, then

Question

If x2 + y2 = 1, then (A)  yy'' - 2 (y' )2 + 1 = 0  (B)  yy'' + (y' )2 + 1 = 0  (C)  yy'' + (y' )2 - 1 = 0  (D)  yy'' + 2 (y' )2 + 1 = 0

Tardigrade Admin · Accepted Answer

Given, x2 + y2 = 1 On differentiating w.r.t. x, we get 2 x + 2 yy ' = 0 ⇒ x + yy' = 0 Again, differentiating w.r.t. x, we get 1 + y ' y' + yy" = 0 ⇒ 1 + (y')2 + yy" = 0

Q. If $x^{2} + y^{2} = 1$ , then

$y y^{''} - 2 (y^{'})^{2} + 1 = 0$

$y y^{''} + (y^{'})^{2} + 1 = 0$

$y y^{''} + (y^{'})^{2} - 1 = 0$

$y y^{''} + 2 (y^{'})^{2} + 1 = 0$

Given, $x^{2} + y^{2} = 1$
On differentiating w.r.t. $x$ , we get
$2 x + 2 y y^{'} = 0$
$\Rightarrow x + y y^{'} = 0$
Again, differentiating w.r.t. $x$ , we get
$1 + y^{'} y^{'} + yy " = 0$
$\Rightarrow 1 + (y^{'})^{2} + yy " = 0$

Q. If x2+y2=1, then

yy′′−2(y′)2+1=0

yy′′+(y′)2+1=0

yy′′+(y′)2−1=0

yy′′+2(y′)2+1=0