Question

If $y = \left[x+\sqrt{x^{2}-1}\right]^{15} + \left[x-\sqrt{x^{2}-1}\right]^{15}$, then $ \left(x^{2} - 1\right) \frac{d^{2}y}{dx^{2}} + x \frac{dy}{dx} $ is equal to :

• A. $125\, y$
• B. $224\,y^2$
• C. $225 \,y^2$
• D. $225\, y$

Answer 1

Answer: (D)

$y = \left\{x+\sqrt{x^{2}-1}^{15}\right\}+\left\{x-\sqrt{x^{2}-1}\right\}^{15}$
$\frac{dy}{dx} = 15\left(x+\sqrt{x^{2}-1}\right)^{14} 15\left(x-\sqrt{x^{2}-1}\right)^{14} \left(1-\frac{x}{\sqrt{x^{2}-1}}\right)$
$\frac{dy}{dx} = \frac{15}{\sqrt{x^{2}-1}}.y\quad\quad...\left(i\right)$
$\sqrt{x^{2}-1}\cdot\frac{dy}{dx} = 15y$
$\frac{x}{\sqrt{x^{2}-1}}. \frac{dy}{dx} + \sqrt{x^{2}-1} \frac{d^{2}y}{dx^{2}} = 15 \frac{dy}{dx}$
$x \frac{dy}{dx}+\left(x^{2}-1\right) \frac{d^{2}y}{dx^{2}} = 15\sqrt{x^{2}-1}. \frac{15}{\sqrt{x^{2}-1}}.y = 225y$