If y:[1,∞] →[1,∞] is a function of x such that 2x=y1/5+y−1/5 and(x2-1) d2ydx2+xdydx=ky (k∈R) Then k =

2x=y1/5+y−1/5→y=x+x2−15dydx=5yx2−1x2−1d2ydx+xdydx=25y

If y:[1,∞] →[1,∞] is a function of x such that 2x=y1/5+y−1/5 and(x2-1) d2ydx2+xdydx=ky (k∈R) Then k =

2x=y1/5+y−1/5→y=x+x2−15dydx=5yx2−1x2−1d2ydx+xdydx=25y

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If y:[1,∞] →[1,∞] is a function of x such that $2 x = y^{1 / 5} + y^{- 1 / 5}$ and
(x²-1) $\frac{d^{2} y}{d x^{2}} + x \frac{d y}{d x} = k y (k \in R)$ Then k =

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$\begin{array}{l} 2 x = y 1 / 5 + y - 1 / 5 \to y = (x + \sqrt{x^{2} - 1}) 5 \\ \frac{dy}{dx} = \frac{5 y}{\sqrt{x^{2} - 1}} \\ (x^{2} - 1) \frac{d^{2} y}{dx} + x \frac{dy}{dx} = 25 y \end{array}$

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If y:[1,∞] →[1,∞] is a function of x such that 2x=y1/5+y−1/5 and(x2-1) d2ydx2+xdydx=ky (k∈R) Then k =

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If y:[1,∞] →[1,∞] is a function of x such that $2 x = y^{1 / 5} + y^{- 1 / 5}$ and
(x²-1) $\frac{d^{2} y}{d x^{2}} + x \frac{d y}{d x} = k y (k \in R)$ Then k =