If $y = {e^{a{{\cos }^{ - 1}}x}}, - 1 \le x \le 1$, show that $(1 - {x^2})\cfrac{{{d^2}y}}{{d{x^2}}} - x\cfrac{{dy}}{{dx}} - {a^2}y = 0.$

We have, $y = {e^{a{{\cos }^{ - 1}}x}}$ …(i)

Differentiating (i) on both sides w.r.t. x, we get
$\cfrac{{dy}}{{dx}} = {e^{a{{\cos }^{ - 1}}x}}\cfrac{d}{{dx}}(a{\cos ^{ - 1}}x)$

$= {e^{a{{\cos }^{ - 1}}x}}\left( {\cfrac{{ - a}}{{\sqrt {1 - {x^2}} }}} \right) = \cfrac{{ - ay}}{{\sqrt {1 - {x^2}} }}$ …(ii)

Differentiating (ii) on both sides w.r.t. x, we get

$\cfrac{{{d^2}y}}{{d{x^2}}} = - a\left[ {\cfrac{{\sqrt {1 - {x^2}} \cfrac{{dy}}{{dx}} - y\cfrac{d}{{dx}}\sqrt {1 - {x^2}} }}{{(1 - {x^2})}}} \right]$

$\Rightarrow$ $\cfrac{{{d^2}y}}{{d{x^2}}} = - a\left[ {\cfrac{{\sqrt {1 - {x^2}} \cfrac{{dy}}{{dx}} - \cfrac{y}{{2\sqrt {1 - {x^2}} }} \cdot ( - 2x)}}{{{{(1 - x)}^2}}}} \right]$

$\Rightarrow$ $(1 - {x^2})\cfrac{{{d^2}y}}{{d{x^2}}} = - a\left[ { - ay + \cfrac{{xy}}{{\sqrt {1 - {x^2}} }}} \right]$ [From (ii)]

$\Rightarrow$ $(1 - {x^2})\cfrac{{{d^2}y}}{{d{x^2}}} = - a\left[ { - ay + x \cdot \left( {\cfrac{{ - 1}}{a} \cdot \cfrac{{dy}}{{dx}}} \right)} \right]$

$\Rightarrow$ $(1 - {x^2})\cfrac{{{d^2}y}}{{d{x^2}}} = {a^2}y + x\cfrac{{dy}}{{dx}}$

$\Rightarrow$ $(1 - {x^2})\cfrac{{{d^2}y}}{{d{x^2}}} - x\cfrac{{dy}}{{dx}} - {a^2}y = 0$