QJEEM17Ma6989: JEE Main 2017 Mathematics - Class 12 JEEM17Ma6989 Solution: If y = then (x 2 1) is equal to :

The given equation is

$y = {({x^2} + \sqrt {{x^2} - 1} )^{15}} + {(x - \sqrt {{x^2} - 1} )^{15}}$

Differentiating w.r.t. x, we get

${{dy} \over {dx}} = 15{(x + \sqrt {{x^2} - 1} )^{14}}\left( {1 + {{1(2x)} \over {2\sqrt {{x^2} - 1} }}} \right) + 15{(x - \sqrt {{x^2} - 1} )^{14}}\left( {1 - {{1(2x)} \over {2\sqrt {{x^2} - 1} }}} \right)$

Here, we have used the standard differentiatials

${d \over {dx}}{x^n} = n\,{x^{n - 1}}$

That is, ${d \over {dx}}(\sqrt {f(x)} ) = {1 \over {2\sqrt {f(x)} }} \times {d \over {dx}}(f(x))$

Therefore

${{dy} \over {dx}} = {{15{{(x + \sqrt {{x^2} - 1} )}^{14}}(\sqrt {{x^2} - 1} + x)} \over {\sqrt {{x^2} - 1} }} + {{15{{(x - \sqrt {{x^2} - 1} )}^{14}}(\sqrt {{x^2} - 1} - x)} \over {\sqrt {{x^2} - 1} }}$

$\Rightarrow \sqrt {{x^2} - 1} {{dy} \over {dx}} = 15{(x + \sqrt {{x^2} - 1} )^{15}} - 15{(x - \sqrt {{x^2} - 1} )^{15}}$

Differentiating w.r.t. x, we get

${{1(2x)} \over {2\sqrt {{x^2} + 1} }}{{dy} \over {dx}} + \sqrt {{x^2} - 1} {{{d^2}y} \over {d{x^2}}} = 15 \times 15{(x + \sqrt {{x^2} - 1} )^{14}}\left( {1 + {{1(2x)} \over {2\sqrt {{x^2} - 1} }}} \right) - 15 \times 15{(x - \sqrt {{x^2} - 1} )^{14}}\left( {{{1 - 1(2x)} \over {2\sqrt {{x^2} - 1} }}} \right)$

$\Rightarrow {x \over {\sqrt {{x^2} - 1} }}{{dy} \over {dx}} + \sqrt {{x^2} - 1} {{{d^2}y} \over {d{x^2}}} = 225{(x + \sqrt {{x^2} - 1} )^{14}}{{(\sqrt {{x^2} - 1} + x)} \over {\sqrt {{x^2} - 1} }} - {{225{{(x - \sqrt {{x^2} - 1} )}^{14}}(\sqrt {{x^2} - 1} - x)} \over {\sqrt {{x^2} - 1} }}$

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

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

Substituting ${(x + \sqrt {{x^2} - 1} )^{15}} + {(x - \sqrt {{x^2} - 1} )^{15}} = y$ , we get

$({x^2} - 1){{{d^2}y} \over {d{x^2}}} + {{x\,dy} \over {dx}} = 225y$