Question

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

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

Answer 1

Answer: (C)

$225y$

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