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,

$$\lambda + k = - 24$$