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Put $n = 1$ then, $y = (x + \sqrt{x^2 - 1})^1 + (x - \sqrt{x^2 - 1})^1 = 2x$

Then, $\dfrac {dy}{dx} = 2(1) = 2$ and $\dfrac {d^2y}{dx^2} = 2(0) = 0$

Therefore, we get $(x^2 - 1)\dfrac {d^2y}{dx^2} + x\dfrac {dy}{dx} + n^2y = (x^2 - 1)(0) + x (2) + (1)^2 (2x) = 2\times 2x = 2(y)$

We can write $2(y)$ as $2(1)^2(y) = 2(n^2)(y)$ .Since, $n= 1$

Therefore the above expression can be simplified as $\boxed{2n^2y}$

$\begin{aligned} y & = \left(x + \sqrt{x^2-1}\right)^n + \left(x - \sqrt{x^2-1}\right)^n & \small \color{#3D99F6} \text{Differentiate both sides w.r.t }x \\ \frac {dy}{dx} & = \frac n{\sqrt{x^2-1}}\left[ \left(x + \sqrt{x^2-1}\right)^n - \left(x - \sqrt{x^2-1}\right)^n\right] & \small \color{#3D99F6} \text{Multiply both sides with }\sqrt{x^2-1} \\ \sqrt{x^2-1}\frac {dy}{dx} & = n \left[ \left(x + \sqrt{x^2-1}\right)^n - \left(x - \sqrt{x^2-1}\right)^n\right] & \small \color{#3D99F6} \text{Differentiate both sides w.r.t }x \\ \sqrt{x^2-1}\frac {d^2y}{dx^2} + \frac x{\sqrt{x^2-1}}\frac {dy}{dx} & = \frac {n^2}{\sqrt{x^2-1}} \left[ \left(x + \sqrt{x^2-1}\right)^n + \left(x - \sqrt{x^2-1}\right)^n\right] & \small \color{#3D99F6} \text{Multiply both sides with }\sqrt{x^2-1} \\ (x^2-1)\frac {d^2y}{dx^2} + x\frac {dy}{dx} & = n^2y & \small \color{#3D99F6} \text{Add }n^2y \text{ on both sides} \\ \implies (x^2-1)\frac {d^2y}{dx^2} + x\frac {dy}{dx} + n^2y & = \boxed{2n^2y} \end{aligned}$