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

The Correct Option is D

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

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): \(x^2y′′ + xy′+ 25y = 0\)