If cos ^{-1}left(frac{y}{2}right)=log _{e}lef | vedclass

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(A) Given: $\cos ^{-1}\left(\frac{y}{2}\right)=5 \log _{e}\left(\frac{x}{5}\right)$.Differentiating both sides with respect to $x$:$\frac{-1}{\sqrt{1-

Vedclass · Accepted Answer

$x^{2} y^{\prime \prime}+x y^{\prime}+25 y=0$

Vedclass · Answer

(A) Given: $\cos ^{-1}\left(\frac{y}{2}ight)=5 \log _{e}\left(\frac{x}{5}ight)$.Differentiating both sides with respect to $x$:$\frac{-1}{\sqrt{1-\frac{y^{2}}{4}}} \cdot \frac{y^{\prime}}{2} = 5 \cdot \frac{1}{x/5} \cdot \frac{1}{5} = \frac{5}{x}$.$\frac{-y^{\prime}}{\sqrt{4-y^{2}}} = \frac{5}{x} \implies -x y^{\prime} = 5 \sqrt{4-y^{2}}$.Squaring both sides: $x^{2} (y^{\prime})^{2} = 25(4-y^{2}) = 100 - 25y^{2}$.Differentiating again with respect to $x$:$x^{2} \cdot 2 y^{\prime} y^{\prime \prime} + 2x (y^{\prime})^{2} = -50 y y^{\prime}$.Dividing by $2 y^{\prime}$ (assuming $y^{\prime} 
eq 0$):$x^{2} y^{\prime \prime} + x y^{\prime} = -25 y$.Therefore,$x^{2} y^{\prime \prime} + x y^{\prime} + 25 y = 0$.

If $\cos ^{-1}\left(\frac{y}{2}\right)=\log _{e}\left(\frac{x}{5}\right)^{5}$ and $|y| < 2$,then which of the following is true?

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