If 2x = y{^{{1 \over 5}}} + y{^{ - {1 \over 5}}} and (x2 - 1) {{{d^2}y} \over {d{x^2}}} + \lambda x {{dy} \over {dx}} + ky = 0, then \lambda + k is equal to :

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Answer: B. - 24 It is given that 2x = {y^{1/5}} + {y^{ - 1/5}} \Rightarrow 2x = {y^{1/5}} + 1/{y^{1/5}} Therefore, 2x = a + {1 \over a} \Rightarrow {a^2} - 2ax + 1 = 0 a = {{2x \pm \sqrt {4{x^2} - 4} } \over 2} \Rightarrow a = {{2x \pm 2\sqrt {{x^2} - 1} } \over 2} \Rightarrow a = x \pm \sqrt {{x^2} - 1} \Rightarrow {y^{1/5}} = x \pm \sqrt {{x^2} - 1} \Rightarrow y = {(x \pm \sqrt {{x^2} - 1} )^5} Therefore, {{dy} \over {dx}} = 5{(x \pm \sqrt {{x^2} - 1} )^4}\left( {1 \pm {{2x} \over {2\sqrt {{x^2} - 1} }}} \right) = 5{(x + \sqrt {{x^2} - 1} )^4}\left( {{{\sqrt {{x^2} - 1} \pm x} \over {\sqrt {{x^2} - 1} }}} \right) \Rightarrow {{dy} \over {dx}} = {{ - 5y} \over {\sqrt {{x^2} - 1} }} ...... (1) \Rightarrow {{{d^2}y} \over {d{x^2}}} = {{\left[ {\sqrt {{x^2} - 1} \left( { - 5{{dy} \over {dx}}} \right) - 5( - 5y){1 \over 2}{{2x} \over {\sqrt {{x^2} - 1} }}} \right]} \over {({x^2} - 1)}} Therefore, ({x^2} - 1){{{d^2}y} \over {d{x^2}}} = - 5\sqrt {{x^2} - 1} {{dy} \over {dx}} + 5y{x \over {\sqrt {{x^2} - 1} }} \Rightarrow ({x^2} - 1){{{d^2}y} \over {d{x^2}}} = 25y - x{{dy} \over {dx}} \Rightarrow ({x^2} - 1){{{d^2}y} \over {d{x^2}}} + 1x{{dy} \over {dx}} - 25y = 0 Therefore, \lambda = 1, k = -25; hence, \lambda + k = - 24

If 2x = y$${^{{1 \over 5}}}$$ + y$${^{ - {1 \over 5}}}$$ and (x2 $$-$$ 1) $${{{d^2}y} \over {d{x^2}}}$$ + $$\lambda $$x $${{dy} \over {dx}}$$ + ky = 0, then $$\lambda $$ + k is equal to :

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