Question

If $y=e^{a{\sin }^{-1}x}$ then prove that $(1-x^2)y_2-x{y}_1-a^{2}y=0$, where $y_1$ and $y_2$ are first and second order derivatives of $y$ respectively.

Answer 1

$y=e^{a{\sin }^{-1}x}$

$\therefore y_1=\frac{d\left(e^{a{\sin }^{-1}x}\right)}{dx}=e^{a{\sin }^{-1}x}\frac{d\left(a{\sin }^{-1}x\right)}{dx}=a{e}^{a{\sin }^{-1}x}\frac{1}{\sqrt{1-x^2}}=a\frac{y}{(1-x^2{)}^{0.5}}$

$\therefore y_2=\frac{y_1}{dx}=\frac{\left(a{e}^{a{\sin }^{-1}x}\frac{1}{\sqrt{1-x^2}}\right)}{dx}$

$\therefore y_2=a[e^{a{\sin }^{-1}x}\frac{d\frac{1}{\sqrt{1-x^2}}}{dx}+\frac{1}{\sqrt{1-x^2}}\frac{d{e}^{a{\sin }^{-1}x}}{dx}]$

$\therefore y_2=a\left[e^{a{\sin }^{-1}x}\frac{-1}{2(\sqrt{1-x^2}{)}^3}\right(-2x)+\frac{y_1}{\sqrt{1-x^2}}]$

$\therefore y_2=a[\frac{xy}{(1-x^2{)}^{1.5}}+\frac{y_1}{\sqrt{1-x^2}}]$

Now, multiply by $(1-x^2)$ on both sides

$\therefore (1-x^2)y_2=x{y}_1+a{y}_1\sqrt{1-x^2}$ (substituting $y_1$)

$\therefore (1-x^2)y_2=x{y}_1+a^2\frac{y}{(1-x^2{)}^{0.5}}\sqrt{1-x^2}$

$\therefore (1-x^2)y_2=x{y}_1+a^{2}y$

$\therefore (1-x^2)y_2-x{y}_1-a^{2}y=0$

Hence, proved.