If ^ - 1 ((y)/(2)) = _ e ((x)/(5)) ^ 5 , |y| < 2 , then

Given,

$\cos^{- 1}\left(\dfrac{y}{2}\right) = \log_{e}{\left(\dfrac{x}{5}\right)}^{5}$

Using property $log{x}^{m} = mlogx$ we get,

$\cos^{- 1}\left(\dfrac{y}{2}\right) = 5\log_{e}\left(\dfrac{x}{5}\right)$

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

$\dfrac{- 1}{\sqrt{1 - \dfrac{{y}^{2}}{4}}} \cdot \dfrac{{y}^{'}}{2} = 5 \cdot \dfrac{1}{\dfrac{x}{5}} \cdot \left(\dfrac{1}{5}\right)$

$⇒ \dfrac{- {y}^{'}}{\sqrt{4 - {y}^{2}}} = \dfrac{5}{x}$

$⇒ - x{y}^{'} = 5\sqrt{4 - {y}^{2}}$

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

$⇒ - x{y}^{''} - {y}^{'} = 5 \cdot \dfrac{1}{2\sqrt{4 - {y}^{2}}}\left(- 2y{y}^{'}\right)$

$⇒ x{y}^{''} + {y}^{'} = \dfrac{5yy '}{\sqrt{4 - {y}^{2}}}$

$⇒ x{y}^{''} + {y}^{'} = \left(5y\right) \cdot \left(- \dfrac{5}{x}\right)$

$⇒ {x}^{2}{y}^{''} + x{y}^{'} = - 25y$

$⇒ {x}^{2}{y}^{''} + x{y}^{'} + 25y = 0$