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

First, we rewrite the given expression so that it becomes easier to differentiate. We notice that the form $$x \pm \sqrt{x^2-1}$$ suggests the substitution $$x=\cosh t$$ because

$$\cosh^2 t-\sinh^2 t = 1 \;\; \Rightarrow \;\; \sqrt{x^2-1}= \sqrt{\cosh^2 t-1}= \sinh t.$$

Hence

$$x+\sqrt{x^2-1}= \cosh t+\sinh t= e^{\,t}, \qquad x-\sqrt{x^2-1}= \cosh t-\sinh t= e^{-\,t}.$$

Substituting these in the definition of $$y$$, we obtain

$$y=\left[e^{\,t}\right]^{15}+\left[e^{-\,t}\right]^{15}=e^{15t}+e^{-15t}=2\cosh 15t.$$

We now need the first and second derivatives of $$y$$ with respect to $$x$$. Because $$x=\cosh t$$, we first find

$$\frac{dx}{dt}=\sinh t.$$

Using $$y=2\cosh 15t$$, we have

$$\frac{dy}{dt}=2\cdot15\sinh 15t=30\sinh 15t.$$

By the chain rule,

$$\frac{dy}{dx}= \frac{dy/dt}{dx/dt}= \frac{30\sinh 15t}{\sinh t}.$$

Next we differentiate $$\dfrac{dy}{dx}$$ with respect to $$x$$ to obtain $$\dfrac{d^2y}{dx^2}$$. First differentiate with respect to $$t$$:

We write $$\dfrac{dy}{dx}=30\bigl(\sinh 15t\bigr)\bigl(\sinh t\bigr)^{-1}$$ and use the quotient rule

$$\frac{d}{dt}\!\left[\frac{\sinh 15t}{\sinh t}\right]= \frac{(15\cosh 15t)(\sinh t)-(\sinh 15t)(\cosh t)}{(\sinh t)^2}.$$

Multiplying by the constant 30,

$$\frac{d}{dt}\!\left[\frac{dy}{dx}\right]= 30\;\frac{15\cosh 15t\,\sinh t-\sinh 15t\,\cosh t}{(\sinh t)^2}.$$

Now divide again by $$\dfrac{dx}{dt}=\sinh t$$ (chain rule) to get

$$\frac{d^2y}{dx^2}=30\;\frac{15\cosh 15t\,\sinh t-\sinh 15t\,\cosh t}{(\sinh t)^3}.$$

We are asked to evaluate $$(x^2-1)\dfrac{d^2y}{dx^2}+x\dfrac{dy}{dx}.$$ Because $$x=\cosh t$$ and $$x^2-1=\cosh^2 t-1=\sinh^2 t$$, we substitute:

$$\bigl(x^2-1\bigr)\frac{d^2y}{dx^2}= \sinh^2 t\; \frac{30\bigl(15\cosh 15t\,\sinh t-\sinh 15t\,\cosh t\bigr)}{(\sinh t)^3} =30\;\frac{15\cosh 15t\,\sinh t-\sinh 15t\,\cosh t}{\sinh t}.$$

Similarly,

$$x\frac{dy}{dx}= \cosh t\;\frac{30\sinh 15t}{\sinh t}=30\;\frac{\cosh t\sinh 15t}{\sinh t}.$$

Adding the two results:

$$\begin{aligned} (x^2-1)\frac{d^2y}{dx^2}+x\frac{dy}{dx} &=30\;\frac{15\cosh 15t\,\sinh t-\sinh 15t\,\cosh t+\cosh t\sinh 15t}{\sinh t}\\[4pt] &=30\;\frac{15\cosh 15t\,\sinh t}{\sinh t}\\[4pt] &=30\cdot15\cosh 15t\\[4pt] &=450\cosh 15t. \end{aligned}$$

Recall that $$y=2\cosh 15t$$, so $$\cosh 15t=\dfrac{y}{2}$$. Substituting this back, we get

$$450\cosh 15t=450\left(\frac{y}{2}\right)=225y.$$

Hence, the correct answer is Option C.