1. Given relation

   $\cos^{-1}\left(\frac y2\right)=\log_e\left(\left(\frac x5\right)^5\right)$

   Since $\log_e\left(\left(\frac x5\right)^5\right)=5\ln\left(\frac x5\right),$ let $u=5\ln\left(\frac x5\right).$ Then $\cos^{-1}\left(\frac y2\right)=u.$

2. Express $y$ explicitly in terms of $u$

   Taking cosine on both sides, $\frac y2=\cos u$ so $y=2\cos u=2\cos\left(5\ln\frac x5\right).$

3. Differentiate once

   $y=2\cos u,\qquad u=5\ln\frac x5$

   First, $\frac{du}{dx}=5\cdot \frac1x=\frac5x.$

   Therefore, $y' = 2(-\sin u)\cdot \frac{du}{dx}$ $y'=-2\sin u\cdot \frac5x=-\frac{10}{x}\sin u.$

4. Differentiate again

   $y'=-10x^{-1}\sin u$

   Using product rule, $y''=-10\left[\frac{d}{dx}(x^{-1})\sin u+x^{-1}\frac{d}{dx}(\sin u)\right].$

   Now, $\frac{d}{dx}(x^{-1})=-x^{-2},$ and $\frac{d}{dx}(\sin u)=\cos u\cdot \frac{du}{dx}=\cos u\cdot \frac5x.$

   Hence, $y''=-10\left[-\frac{1}{x^2}\sin u+\frac1x\cdot \cos u\cdot \frac5x\right]$ $y''=\frac{10}{x^2}\sin u-\frac{50}{x^2}\cos u.$

5. Form the required differential expression

   Multiply by $x^2$: $x^2y''=10\sin u-50\cos u.$

   Also, $xy'=x\left(-\frac{10}{x}\sin u\right)=-10\sin u.$

   Adding, $x^2y''+xy'=(10\sin u-50\cos u)+(-10\sin u)=-50\cos u.$

   Since $y=2\cos u,$ we get $-50\cos u=-25(2\cos u)=-25y.$

   Therefore, $x^2y''+xy'=-25y$ or $x^2y''+xy'+25y=0.$

6. Match with the options

   This is exactly Option D.

Final Answer: D