Question

If $y={\sin }^{-1}x$, prove that $(1-x^2)y_2-x{y}_1=0$

Answer 1

$y=si{n}^{-1}x$

$y_1=\frac{1}{\sqrt{1-x^2}}-\left(1\right)(y_1=\frac{dy}{dx})$

$y_2=\frac{\sqrt{1-x^2}\frac{d}{dx}\left(1\right)-1\frac{d}{dx}\sqrt{1-x^2}}{(\sqrt{1-x^2}{)}^2}$

$y_2=\frac{\sqrt{1-x^2}\times 0-1\times (1-x^2{)}^{-\frac{1}{2}}.-2x.\frac{1}{2}}{1-x^2}$

$y_2=\frac{x}{\sqrt{1-x^2}.(1-x^2)}$

$y_2.(1-x^2)=\frac{x}{\sqrt{1-x^2}}$

$y_2(1-x^2)=x.y_1$ (From equation (1))

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

Hence proved