Given, m $sin^{-1} x = log_{e} \,y$
$\Rightarrow y = e^{m\,sin^{-1}\,x}$
On differentiating w.r.t. x, we get
$y' = e^{m\,sin^{-1}\,x} X \frac{m}{\sqrt{1-x^{2}}}$
$\Rightarrow \sqrt{1-x^{2}}\cdot y' = my$
On squaring both sides, we get
$\left(1-x^{2}\right)y'^{2} = m^{2}y^{2}$
Again differentiating, we get
$\left(1-x^{2}\right) 2y' X y'' + y'^{2}\left(-2x\right) = m^{2} \cdot 2yy'$
$\therefore \left(1-x^{2}\right)y''-xy' = m^{2}y$