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Find the constants in a relation like (1 - x^2)y'' - xy' + ky = 0 for a given y

This type of question, asked 2 times: 2021 Paper 20, 9 Apr 2024. All 2 questions below, oldest first.

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  1. Question 1

    2021 Paper 20

    90
    If y1/4+y−1/4=2xy^{1/4}+y^{- 1/4}= 2x, and (x2−1)d2ydx2+αxdydx+βy=0\left( x^{2}- 1 \right)\frac{d^{2}y}{dx^{2}}+\alpha x\frac{dy}{dx}+\beta y= 0, then ∣α−β∣|\alpha-\beta| is equal to

  2. Question 2

    9 Apr 2024

    164
    If log⁡ey=3sin⁡−1x\log_{e}y= 3\sin^{- 1}x, then (1−x2)y′′−xy′(1 -x^{2})y^{''}-xy^{'} at x=12x=\frac{1}{2} is equal to :

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