Question

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

• A. $x^2{y}^{\prime \prime }+x{y}^{\prime }-25y=0$
• B. $x^2{y}^{\prime \prime }-x{y}^{\prime }-25y=0$
• C. $x^2{y}^{\prime \prime }-x{y}^{\prime }+25y=0$
• D. $x^2{y}^{\prime \prime }+x{y}^{\prime }+25y=0$

Answer 1

Answer: (D)

${\cos }^{-1}\left(\frac{y}{2}\right)={\log }_e{\left(\frac{x}{5}\right)}^5$
${\cos }^{-1}\left(\frac{y}{2}\right)=5{\log }_e\left(\frac{x}{5}\right)$
$\frac{-1}{\sqrt{1-\frac{y^2}{4}}}\cdot \frac{y^{\prime }}{2}=5\cdot \frac{1}{\frac{x}{5}}\times \frac{1}{5}$
$\Rightarrow \frac{-y^{\prime }}{\sqrt{4-y^2}}=\frac{5}{x}$
$-x{y}^{\prime }=5\sqrt{4-y^2}$
$-x{y}^{\prime \prime }-y^{\prime }=5\cdot \frac{1}{2\sqrt{4-y^2}}\left(-2y{y}^{\prime }\right)$
$\Rightarrow x{y}^{\prime \prime }+y^{\prime }=\frac{5y^{\prime }\cdot y}{\sqrt{4-y^2}}$
$x{y}^{\prime \prime }+y^{\prime }=5\cdot \left(\frac{-5}{x}\right)y$
$x^2{y}^{\prime \prime }+x{y}^{\prime }=-25y$