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

Question

Accepted Answer

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

Differentiating on both side

$$ - {1 \over {\sqrt {1 - {{\left( {{y \over 2}} \right)}^2}} }} \times {{y'} \over 2} = {5 \over {{x \over 5}}} \times {1 \over 5}$$

${{ - xy'} \over 2} = 5\sqrt {1 - {{\left( {{y \over 2}} \right)}^2}}$

Square on both side

${{{x^2}y{'^2}} \over 4} = 25\left( {{{4 - {y^2}} \over 4}} \right)$

Diff on both side

$2xy{'^2} + 2y'y''{x^2} = - 25 \times 2yy'$

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

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