If $x = \sin t$ and $y = \sin pt,$ then prove that
$\left( {1 - {x^2}} \right)\frac{{{d^2}y}}{{d{x^2}}} - x\frac{{dy}}{{dx}} + {p^2}y = 0$

We have, $x = \sin t$ and $y = \sin pt$
therefore,$\frac{{dx}}{{dt}} = \cos t$ and $\frac{{dy}}{{dt}} = \cos pt \cdot p$ …….(i)

$\Rightarrow$ $\frac{{dy}}{{dx}} = \frac{{dy/dt}}{{dx/dt}} = \frac{{p \cdot \cos pt}}{{\cos t}}$

Again, differentiating both sides w.r.t. $x$, we get $\frac{{{d^2}y}}{{d{x^2}}} = \frac{{\cos t \cdot \frac{d}{{dt}}(p \cdot \cos pt)\frac{{dt}}{{dx}} - p\cos pt \cdot \frac{d}{{dt}}\cos t \cdot \frac{{dt}}{{dx}}}}{{{{\cos }^2}t}}$

$= \frac{{[\cos t \cdot p \cdot ( - \sin pt) \cdot p - p\cos pt \cdot ( - \sin t)]\frac{{dt}}{{dx}}}}{{{{\cos }^2}t}}$

$= \frac{{\left[ { - {p^2}\sin pt \cdot \cos t + p\sin t \cdot \cos pt} \right] \cdot \frac{1}{{\cos t}}}}{{{{\cos }^2}t}}$

$\Rightarrow$ $\frac{{{d^2}y}}{{d{x^2}}} = \frac{{ - {p^2}\sin pt \cdot \cos t + p\cos pt \cdot \sin t}}{{{{\cos }^3}t}}$ …..(ii)

Since, we have to prove

$\left( {1 - {x^2}} \right)\frac{{{d^2}y}}{{d{x^2}}} - x\frac{{dy}}{{dx}} + {p^2}y = 0$

therefore,${\rm{LHS}} = \left( {1 - {{\sin }^2}t} \right)\frac{{\left[ { - {p^2}\sin pt \cdot \cos t + p\cos pt \cdot \sin t} \right]}}{{{{\cos }^3}t}}$

$- \sin t \cdot \frac{{p\cos pt}}{{\cos t}} + {p^2}\sin pt$

$= \frac{1}{{{{\cos }^3}t}}\left[ {\begin{array}{llllllllllllllllllll}{\left( {1 - {{\sin }^2}t} \right)\left( { - {p^2}\sin pt \cdot \cos t + p\cos pt \cdot \sin t} \right)}\\{ - p\cos pt \cdot \sin t \cdot {{\cos }^2}t + {p^2}\sin pt \cdot {{\cos }^3}t}\end{array}} \right]$

$= \frac{1}{{{{\cos }^3}t}} \cdot 0$

$= 0$

Hence proved.