Practicing Success
Practicing Success
CUET Preparation Today
If $y=\sqrt{x+1}+\sqrt{x-1}$ then $(x^2-1)y_2+xy_1=$ [where $y_2=\frac{d^2y}{dx^2}$ & $y_1=\frac{dy}{dx}$] |
$\frac{1}{2}y$ $\frac{3}{2}y$ $(2x+1)y$ $\frac{1}{4}y$ |
$\frac{1}{4}y$ |
