If 2x=y^ 1 5+y^- 1 5 and (x^2-1) d^2y dx^2+λ x dy dx+ky=0, then λ +k is equal to

Given ${y}^{\frac{1}{5}}+{y}^{-\frac{1}{5}}=2x$y51​+y−51​=2x

$$\Rightarrow (\frac{1}{5}{y}^{-\frac{4}{5}}-\frac{1}{5}{y}^{-\frac{6}{5}})\frac{dy}{dx}=2$$⇒(51​y−54​−51​y−56​)dxdy​=2

$$\Rightarrow {y}^{'}({y}^{\frac{1}{5}}-{y}^{-\frac{1}{5}})=10y$$⇒y′(y51​−y−51​)=10y

$$\Rightarrow {y}^{'} (2\sqrt{{x}^{2}-1})=10y$$⇒y′(2x2−1​)=10y

$$\Rightarrow {y}^{'} \sqrt{{x}^{2}-1}=5y$$⇒y′x2−1​=5y

Differentiating again

$${y}^{''}(2\sqrt{{x}^{2}-1})+{y}^{'}2\frac{2x}{2\sqrt{{x}^{2}-1}}=10y'$$y′′(2x2−1​)+y′22x2−1​2x​=10y′

$$\Rightarrow {y}^{''}({x}^{2}-1)+x{y}^{'}=5\sqrt{{x}^{2}-1} ({y}^{'})$$⇒y′′(x2−1)+xy′=5x2−1​(y′)

$$\Rightarrow {y}^{''} ({x}^{2}-1)+x{y}^{'}-25y=0$$⇒y′′(x2−1)+xy′−25y=0

$$\Rightarrow \lambda =1, k=-25$$⇒λ=1,k=−25