Question

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

• A. $x^2y′′ + xy′ – 25y = 0$
• B. $x^2y′′ – xy′ – 25y = 0$
• C. $x^2y′′ – xy′+ 25y = 0$
• D. $x^2y′′ + xy′+ 25y = 0$

Answer 1

Answer: (D)

$x^2y′′ + xy′+ 25y = 0$

Answer 2

$cos^{-1}(\frac{y}{2})=log_e(\frac{x}{5})^5, |y| < 2$

Differentiate on both side

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

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

Square on both side

$\frac{x^2y'^2}{4} = 25\bigg(\frac{4-y^2}{4}\bigg)$

Diff on both side

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

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

Hence, the correct option is (D): $x^2y′′ + xy′+ 25y = 0$