If $2x=y^{\frac{1}{5}}+y^{-\frac{1}{5}}$ and $\left(x^2-1\right)\frac{d^{2}y}{d{x}^2}+\lambda x\frac{dy}{dx}+ky=0$, then $\lambda +k$ 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

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

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

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

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