If 2 x = y 1 5 + y - 1 5 and x 2 - 1 d 2 y d x 2 + λ x d y d x + k y = 0 , then λ + k is equal to

Given $y^{\frac{1}{5}}+y^{-\frac{1}{5}}=2x$

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

$\Rightarrow y^{\prime }\left(y^{\frac{1}{5}}-y^{-\frac{1}{5}}\right)=10y$

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

$\Rightarrow y^{\prime } \sqrt{x^2-1}=5y$

Differentiating again

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

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

$\Rightarrow y^{\prime \prime } \left(x^2-1\right)+x{y}^{\prime }-25y=0$

$\Rightarrow \lambda =1, k=-25$