Question 31 of 67MediumMCQ
If y = \[{\left[ {x + \sqrt {{x^2} - 1} } \right]^{15}} + {\left[ {x - \sqrt {{x^2} - 1} } \right]^{15}},\]
then (x2 - 1) \[{{{d^2}y} \over {d{x^2}}} + x{{dy} \over {dx}}\] is equal to :
Solutions · 1
Official solution
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
\[({x^2} - 1){{{d^2}y} \over {d{x^2}}} + {{x\,dy} \over {dx}} = 225y\]
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