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

$\begin{array}{l}{\cos }^{-1}\left(\frac{y}{2}\right)={\log }_e{\left(\frac{x}{5}\right)}^5\\ \\ {\cos }^{-1}\left(\frac{y}{2}\right)=5{\log }_e\left(\frac{x}{5}\right)\\ \\ \frac{-1}{\sqrt{1-\frac{y^2}{4}}}·\frac{y^{\text{'}}}{2}=5·\frac{1}{x}\\ \\ \Rightarrow \frac{-y^{\text{'}}}{\sqrt{4-y^2}}=\frac{5}{x}\\ \\ -x{y}^{\text{'}}=5\sqrt{4-y^2}\\ \\ -x{y}^{\text{'}\text{'}}-y^{\text{'}}=5·\frac{1}{2\sqrt{4-y^2}}\left(-2y{y}^{\text{'}}\right)\\ \\ \Rightarrow x{y}^{\text{'}\text{'}}+y^{\text{'}}=\frac{5y^{\text{'}}·y}{\sqrt{4-y^2}}\\ \\ x{y}^{\text{'}\text{'}}+y^{\text{'}}=5·\left(\frac{-5}{x}\right)y\\ \\ x^2{y}^{\text{'}\text{'}}+x{y}^{\text{'}}=-25y\end{array}$