If 2x = y… + y… and (x² … 1) … + …x … + ky = 0, then … + k is equal to :

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Differential Equations

JEE Main 2017 · 9 Apr, Morning shift · Q3512

MathematicsMedium38×since 2002Q3512

If 2x = y

{^{{1 \over 5}}}

+ y

{^{ - {1 \over 5}}}

and (x²

-

{{{d^2}y} \over {d{x^2}}}

\lambda

{{dy} \over {dx}}

+ ky = 0, then

\lambda

+ k is equal to :

JEE Main 2017 · 9 Apr, Morning shift · Q3512

MathematicsMedium38×since 2002Q3512

If 2x = y

{^{{1 \over 5}}}

+ y

{^{ - {1 \over 5}}}

and (x²

-

{{{d^2}y} \over {d{x^2}}}

\lambda

{{dy} \over {dx}}

+ ky = 0, then

\lambda

+ k is equal to :