Question

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

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

Answer 1

Answer: (D)

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