MCQ
If $2x = {y^{\frac{1}{5}}} + {y^{ - \frac{1}{5}}}$ and $(x^2 -1) \frac{{{d^2}y}}{{d{x^2}}} + \lambda x\frac{{dy}}{{dx}} + ky = 0$ , then $ \lambda + k$ is equal to
  • A
    $-23$
  • $-24$
  • C
    $26$
  • D
    $-26$

Answer

Correct option: B.
$-24$
b
${y^{1/5}} + {y^{ - 1/5}} = 2x$

$\left( {\frac{1}{5}{y^{ - 4/5}} - \frac{1}{5}{y^{ - 6/5}}} \right).\frac{{dy}}{{dx}} = 2$

$y'\left( {{y^{1/5}} - {y^{ - 1/5}}} \right) = 10y$

${y^{1/5}} + {y^{ - 1/5}} = 2x$

${y^{1/5}} - {y^{ - 1/5}} = \sqrt {4{x^2} - 4} $

$y'\left( {2\sqrt {{x^2} - 1} } \right) = 10y$

$y''\left( {2\sqrt {{x^2} - 1} } \right) + y'2\frac{{2x}}{{2\sqrt {{x^2} - 1} }} = 10y'$

                   $y''\left( {{x^2} - 1} \right) + xy' = 5\sqrt {{x^2} - 1} \left( {y'} \right)$

$\boxed{y''\left( {{x^2} - 1} \right) + xy' - 25y = 0}$

$\lambda  = 1,k =  - 25$

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