Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa

if\small y=\left [ x+\sqrt{^{x2}-1} \right ]^{15} +\left [ x-\sqrt{^{x2}-1} \right ]^{15},

then\left ( x^{2}-1 \right )\tfrac{d^{2}y}{dx^{2}}+x\tfrac{dy}{dx}   is equal to :

  • Option 1)

    125 y

  • Option 2)

    224 y2

  • Option 3)

    225 y2

  • Option 4)

    225 y

 

Answers (1)

best_answer

As we learnt in 

Differential Equations -

An equation involving independent variable (x), dependent variable (y) and derivative of dependent variable with respect to independent variable 
\left (\frac{\mathrm{d} y}{\mathrm{d} x} \right )

- wherein

eg:

  \frac{d^{2}y}{dx^{2}}- 3\frac{dy}{dx}+5x=0

 

 y=\left [x+\sqrt{x^{2}-1} \right ]^{15}+\left [ x-\sqrt{x^{2}-1} \right ]^{15}

Since\:\left [x+\sqrt{x^{2}-1} \right ]\left [ x-\sqrt{x^{2}-1} \right ]=1

So that

y=\left [ x+\sqrt{x^{2}-1} \right ]^{15}+\left [ x+\sqrt{x^{2}-1} \right ]^{-15}

\frac{dy}{dx}= 15\left [ x+\sqrt{x^{2}-1} \right ]^{14}\times \left ( 1+\frac{2x}{2\sqrt{x^{2}-1}} \right )-15\left [ x+\sqrt{^{2}-1} \right ]^{-16} \left [ 1+\frac{2x}{2\sqrt{x^{2}-1}} \right ]

= 15\left [ x+\sqrt{x^{2}-1} \right ]^{15}\frac{1}{\sqrt{x^{2}-1}}-15\left [ x+\sqrt{x^{2}-1} \right ]^{-15}\frac{1}{\sqrt{x^{2}-1}}

\therefore \frac{15}{\sqrt{x^{2}-1}}\left [ \left [ x+\sqrt{x^{2}-1} \right ]^{15} -\left [ x+\sqrt{x^{2}-1} \right ]^{-15}\right ]

\frac{dy}{dx}\sqrt{x^{2}-1}=15\left [ \left ( x+\sqrt{x^{2}-1} \right )^{15}-\left ( x+\sqrt{x^{2}-1} \right )^{-15} \right ]

\frac{d^{2}y}{dx^{2}}\sqrt{x^{2}-1}+\frac{dy}{dx}.\frac{2x}{2\sqrt{x^{2}-1}}=\frac{225}{\sqrt{x^{2}-1}}\left [ \left ( x+\sqrt{x^{2}-1} \right )^{15}+\left ( x+\sqrt{x^{2}-1} \right )^{-15} \right ]

\therefore \left ( x^{2}-1 \right )\frac{d^{2}y}{dx^{2}}+x\frac{dy}{dx}=225y

 


Option 1)

125 y

Incorrect option

Option 2)

224 y2

Incorrect option

Option 3)

225 y2

Incorrect option

Option 4)

225 y

Correct option

Posted by

prateek

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Advanced 2024: Chemical engineering cut-offs at IITs; know opening, closing ranks
anam-khan

IIT JEE Mains Result 2024 Session 2: How is overall merit list prepared?
anam-khan

JEE Mains Session 2 result 2024 soon; List of top NITs in India

Latest Question