JEE Main On 8 April 2017 Question 18
Question: If $ y={{[ x+\sqrt{x^{2}-1} ]}^{15}}+y{{[ x-\sqrt{x^{2}-1} ]}^{15}}, $ then $ (x^{2}-1)\frac{d^{2}y}{dx^{2}}+x\frac{dy}{dx} $ is equal to: [JEE Online 08-04-2017]
Options:
A) $ 224y^{2} $
B) $ 125y $
C) $ 225y $
D) $ 225y^{2} $
Show Answer
Answer:
Correct Answer: C
Solution:
- $ y={ x+{{\sqrt{x^{2}-1}}^{15}} }+{{{ x-\sqrt{x^{2}-1} }}^{15}} $ $ \frac{dy}{dx}=15{{( x+\sqrt{x^{2}-1} )}^{14}}15{{( x-\sqrt{x^{2}-1} )}^{14}}( 1-\frac{x}{\sqrt{x^{2}-1}} ) $ $ \frac{dy}{dx}=\frac{15}{\sqrt{x^{2}-1}}.y $ ?(i) $ \sqrt{x^{2}-1}.\frac{dy}{dx}=15y $ $ \frac{x}{\sqrt{x^{2}-1}}.\frac{dy}{dx}+\sqrt{x^{2}-1}\frac{d^{2}y}{dx^{2}}=15\frac{dy}{dx} $ $ x\frac{dy}{dx}+( x^{2}-1 )\frac{d^{2}y}{dx^{2}}=15\sqrt{x^{2}-1}.\frac{15}{\sqrt{x^{2}-1}}.y=225y $
