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 $(x^2-1)y^{"}+xy^{'} = m^{2}y$
 A: 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.
A: 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 $m^2y$. 
A: $2x = y^{\frac{1}{m}} + y^{\frac{-1}{m}}(x≥1) $ then prove that $(x^2-1)y^{"}+xy^{'} = m^{2}y.........(1)$
Differntiating w.r.t  x
$2 = \frac{1}{m}y^{\frac{1}{m}-1}.y^{'} -\frac{-1}{m}y^{\frac{-1}{m}-1}.y^{'}$
$2 = \frac{1}{m}\left(\frac{y^\frac{1}{m}}{y}-\frac{y^{\frac{-1}{m}}}{y}\right)y^{'}$
$2my = \left(y^\frac{1}{m}-y^\frac{-1}{m}\right)y^{'}$
$4m^2y^2 = \left(y^\frac{1}{m}-y^\frac{-1}{m}\right)^2(y^{'})^2$
$4m^2y^2 = (y^{'})^2\left((y^\frac{1}{m}+y^\frac{-1}{m})^2-4\right)$
$4m^2y^2 = (y^{'})^2\left(4x^2-4\right)$......(By -(1))
$m^2y^2 = (y^{'})^2\left(x^2-1\right)$
Differentiating w.r.t x
$m^2(2yy^{'}) = (y^{'})^2(2x) + \left(x^2-1\right)(2y^{'}y^{''})$
canceling $2y^{'}$ on both side
$(x^2-1)y^{"}+xy^{'} = m^{2}y$
