If $$\cos^{-1}\left(\frac{y}{2}\right) = \log_e\left(\frac{x}{5}\right)^5, |y| < 2$$, then

We are given: $$\cos^{-1}\left(\frac{y}{2}\right) = \log_e\left(\frac{x}{5}\right)^5$$, where $$|y| < 2$$.

Simplifying the equation yields $$\cos^{-1}\left(\frac{y}{2}\right) = 5\ln\left(\frac{x}{5}\right) = 5(\ln x - \ln 5)$$. Letting $$\theta = 5\ln\left(\frac{x}{5}\right)$$ gives $$\frac{y}{2} = \cos\theta$$, so $$y = 2\cos\left(5\ln\frac{x}{5}\right)$$.

To find the first derivative, differentiate with respect to $$x$$: $$y' = -2\sin\left(5\ln\frac{x}{5}\right) \cdot \frac{5}{x}$$, which simplifies to $$y' = \frac{-10\sin\left(5\ln\frac{x}{5}\right)}{x}$$. Multiplying both sides by $$x$$ gives $$xy' = -10\sin\left(5\ln\frac{x}{5}\right) \quad \cdots (1)$$.

Next, differentiating equation (1) with respect to $$x$$ leads to $$xy'' + y' = -10\cos\left(5\ln\frac{x}{5}\right) \cdot \frac{5}{x}$$, or equivalently $$xy'' + y' = \frac{-50\cos\left(5\ln\frac{x}{5}\right)}{x}$$. Multiplying through by $$x$$ then gives $$x^2y'' + xy' = -50\cos\left(5\ln\frac{x}{5}\right)$$.

Since $$y = 2\cos\left(5\ln\frac{x}{5}\right)$$, it follows that $$\cos\left(5\ln\frac{x}{5}\right) = \frac{y}{2}$$. Substituting this into the previous result yields $$x^2y'' + xy' = -50 \cdot \frac{y}{2}$$, which simplifies to $$x^2y'' + xy' = -25y$$ and hence $$x^2y'' + xy' + 25y = 0$$.

The correct answer is Option D: $$x^2y'' + xy' + 25y = 0$$.