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 $(x^2-1)y^{"}+xy^{'} = m^{2}y$

Let $y=e^{mu}$. Then $x=e^u+e^{-u}=\cosh u$. Note that 
$$y'=mu'e^{mu}=mu'y.$$ 
But $x=\cosh u$, so $1=u'\sinh u$, and therefore 
$$u'=\frac{1}{\sinh u}=\frac{1}{\sqrt{\cosh^2 u-1}}=\frac{1}{\sqrt{x^2-1}}.$$
It follows that 
$$y'=mu'y=\frac{my}{\sqrt{x^2-1}},\quad\text{and therefore}\quad \sqrt{x^2-1}\,y'=my.$$
 Differentiate again. We get 
$$y''\sqrt{x^2-1}+\frac{xy'}{\sqrt{x^2-1}}=my'.$$ 
Multiply by $\sqrt{x^2-1}$. We get
$$(x^2-1)y'' +xy' =my'\sqrt{x^2-1}.$$
But the right-hand side is just $m^2y$. 
