If y= \left ( \sec ^{-1}x \right )^{2}, x> 0,  show that  x^{2}\left ( x^{2} -1\right )\frac{d^{2}y}{dx^{2}}+\left ( 2x^{3}-x \right )\frac{dy}{dx}-2= 0

 

 

 

 
 
 
 
 

Answers (1)

y= \left ( \sec ^{-1} x\right )^{2},x> 0\: \left [ given \right ]
   \Rightarrow \frac{dy}{dx}= 2\sec ^{-1}x\: \: \: \frac{d\: \: \sec ^{-1}x}{dx}
  \Rightarrow \frac{dy}{dx}= 2\sec ^{-1}x\: \frac{1}{x\sqrt{x^{2}-1}}---\left ( i \right ) 
 \Rightarrow \frac{d^{2}y}{dx^{2}}= 2\left [ \frac{1}{x^{2}\left ( x^{2}-1 \right )} \right ]+2\sec ^{-1}x\left [ \frac{-\sqrt{x^{2}-1}\: -x\left ( \frac{2x}{2\sqrt{x^{2}-1}} \right )}{x^{2}\left ( x^{2} -1\right )} \right ]
\Rightarrow \frac{d^{2}y}{dx^{2}}= 2\left [ \frac{1}{x^{2}\left ( x^{2}-1 \right )} \right ]+2\sec ^{-1}x\frac{1}{x\sqrt{x^{2}-1}}\left [ \frac{x\left ( 1-2x^{2} \right )}{x^{2}\left ( x^{2}-1 \right )} \right ]---\left ( ii \right )
\Rightarrow \frac{d^{2}y}{dx^{2}}= 2\left [ \frac{1}{x^{2}\left ( x^{2}-1 \right )} \right ]+\frac{dy}{dx}\left [ \frac{x\left ( 1-2x^{2} \right )}{x^{2}\left ( x^{2}-1 \right )} \right ]
\Rightarrow x^{2}\left ( x^{2}-1 \right )\frac{d^{2}y}{dx^{2}}+\left ( 2x^{3}-x \right )\frac{dy}{dx}-2= 0

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