JEE Mains · Maths · STD 12 - 5. continuity and differentiation
If \(\cos ^{-1}\left(\frac{y}{2}\right)=\log _{e}\left(\frac{x}{5}\right)^{5},|y|<2\), then
- A \(x^{2} y^{\prime \prime}+x y^{\prime}-25 y=0\)
- B \(x^{2} y^{\prime \prime}+x y^{\prime}-25 y=0\)
- C \(x^{2} y^{\prime \prime}-x y^{\prime}+25 y=0\)
- D \(x^{2} y^{\prime \prime}+x y^{\prime \prime}+25 y=0\)
Answer & Solution
Correct Answer
(D) \(x^{2} y^{\prime \prime}+x y^{\prime \prime}+25 y=0\)
Step-by-step Solution
Detailed explanation
\(\cos ^{-1}\left(\frac{y}{2}\right)=\log _{e}\left(\frac{x}{5}\right)^{5}\) \(\cos ^{-1}\left(\frac{y}{2}\right)=5 \log _{e}\left(\frac{x}{5}\right)\) \(\frac{-1}{\sqrt{1-\frac{y^{2}}{4}}} \cdot \frac{y^{\prime}}{2}=5 \cdot \frac{1}{\frac{x}{5}} \times \frac{1}{5}\)…
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