Practicing Success
Practicing Success
CUET Preparation Today
If $x^2+y^2=1$, then |
$y y''-2\left(y'\right)^2+1=0$ $y y''+\left(y'\right)^2+1=0$ $y y''+\left(y'\right)^2-1=0$ $y y''+2\left(y'\right)^2+1=0$ |
$y y''+\left(y'\right)^2+1=0$ |
We have, $x^2+y^2=1$ 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 $1+\left(y'\right)^2+y y''=0$ |
