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

Given,

${\cos }^{-1}\left(\frac{y}{2}\right)={\log }_e{\left(\frac{x}{5}\right)}^5$

Using property $\log x^m=m\log x$ we get,

${\cos }^{-1}\left(\frac{y}{2}\right)=5{\log }_e\left(\frac{x}{5}\right)$

Now differentiating both side w.r.t $x$ we get,

$\frac{-1}{\sqrt{1-\frac{y^2}{4}}}·\frac{y^{\text{'}}}{2}=5·\frac{1}{\frac{x}{5}}·\left(\frac{1}{5}\right)$

$\Rightarrow \frac{-y^{\text{'}}}{\sqrt{4-y^2}}=\frac{5}{x}$

$\Rightarrow -x{y}^{\text{'}}=5\sqrt{4-y^2}$

Again differentiating w.r.t $x$ we get,

$\Rightarrow -x{y}^{\text{'}\text{'}}-y^{\text{'}}=5·\frac{1}{2\sqrt{4-y^2}}\left(-2y{y}^{\text{'}}\right)$

$\Rightarrow x{y}^{\text{'}\text{'}}+y^{\text{'}}=\frac{5yy\text{'}}{\sqrt{4-y^2}}$

$\Rightarrow x{y}^{\text{'}\text{'}}+y^{\text{'}}=\left(5y\right)·\left(-\frac{5}{x}\right)$

$\Rightarrow x^2{y}^{\text{'}\text{'}}+x{y}^{\text{'}}=-25y$

$\Rightarrow x^2{y}^{\text{'}\text{'}}+x{y}^{\text{'}}+25y=0$