<s>[INST] If $2x = y^{\frac{1}{m}} + y^{\frac{-1}{m}}(x≥1) $ then prove that $(x^2-1)y^{"}+xy^{'} = m^{2}y$ How do I prove following?
If $2x = y^{\frac{1}{m}} + y^{\frac{-1}{m}},(x≥1)$, then prove that $(x^2-1)y^{"}+xy^{'} = m^{2}y$
 [/INST] Complete the square:
$2x = y^\frac{1}{m} + y^\frac{-1}{m}$
$2xy^\frac{1}{m} = y^\frac{2}{m} + 1$
$0 = y^\frac{2}{m} - 2xy^\frac{1}{m} + 1$
$0 = (y^\frac{2}{m} - 2xy^\frac{1}{m} + x^2) - x^2 + 1$
$0 = (y^\frac{1}{m} - x)^2 - x^2 + 1$
$(y^\frac{1}{m} - x)^2 = x^2 - 1$
$y^\frac{1}{m} - x = \sqrt{x^2 - 1}$
$y^\frac{1}{m} = x + \sqrt{x^2 - 1}$
$y = (x+\sqrt{x^2 - 1})^m$
Differentiate from there and plug in.
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