The value is=?

Mar 15, 2018

Explanation:

$y = {\left(x + \sqrt{{x}^{2} - 1}\right)}^{15} + {\left(x - \sqrt{{x}^{2} - 1}\right)}^{15}$

Hence $\frac{\mathrm{dy}}{\mathrm{dx}} = 15 {\left(x + \sqrt{{x}^{2} - 1}\right)}^{14} \cdot \left(1 + \frac{2 x}{2 \sqrt{{x}^{2} - 1}}\right) + 15 {\left(x - \sqrt{{x}^{2} - 1}\right)}^{14} \cdot \left(1 - \frac{2 x}{2 \sqrt{{x}^{2} - 1}}\right)$

= $15 {\left(x + \sqrt{{x}^{2} - 1}\right)}^{14} \cdot \left(1 + \frac{x}{\sqrt{{x}^{2} - 1}}\right) + 15 {\left(x - \sqrt{{x}^{2} - 1}\right)}^{14} \cdot \left(1 - \frac{x}{\sqrt{{x}^{2} - 1}}\right)$

= $15 {\left(x + \sqrt{{x}^{2} - 1}\right)}^{14} \cdot \left(\frac{\sqrt{{x}^{2} - 1} + x}{\sqrt{{x}^{2} - 1}}\right) + 15 {\left(x - \sqrt{{x}^{2} - 1}\right)}^{14} \cdot \left(\frac{\sqrt{{x}^{2} - 1} - x}{\sqrt{{x}^{2} - 1}}\right)$

= $15 {\left(x + \sqrt{{x}^{2} - 1}\right)}^{15} / \sqrt{{x}^{2} - 1} - 15 {\left(x - \sqrt{{x}^{2} - 1}\right)}^{15} / \sqrt{{x}^{2} - 1}$

or $\frac{\sqrt{{x}^{2} - 1}}{15} \frac{\mathrm{dy}}{\mathrm{dx}} = {\left(x + \sqrt{{x}^{2} - 1}\right)}^{15} - {\left(x - \sqrt{{x}^{2} - 1}\right)}^{15}$

and similarly we have

$\frac{\sqrt{{x}^{2} - 1}}{15} \frac{{d}^{2} y}{{\mathrm{dx}}^{2}} + \frac{\mathrm{dy}}{\mathrm{dx}} \cdot \frac{2 x}{30 \sqrt{{x}^{2} - 1}} = 15 {\left(x + \sqrt{{x}^{2} - 1}\right)}^{15} / \sqrt{{x}^{2} - 1} + 15 {\left(x - \sqrt{{x}^{2} - 1}\right)}^{15} / \sqrt{{x}^{2} - 1}$

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

or $\left({x}^{2} - 1\right) \frac{{d}^{2} y}{{\mathrm{dx}}^{2}} + x \frac{\mathrm{dy}}{\mathrm{dx}} = 225 y$

Additional Information $-$ Observe that for $y = {\left(x + \sqrt{{x}^{2} - 1}\right)}^{n} + {\left(x - \sqrt{{x}^{2} - 1}\right)}^{n}$, $\left({x}^{2} - 1\right) \frac{{d}^{2} y}{{\mathrm{dx}}^{2}} + x \frac{\mathrm{dy}}{\mathrm{dx}}$ would be ${n}^{2} y$