Question
If $=\text{y}=(\sin^{-1}\text{x})^2,$ prove that $(1-\text{x}^2)\frac{\text{d}^2\text{y}}{\text{dx}^2}-\text{x}\frac{\text{dy}}{\text{dx}}-2=0.$

Answer

Here $=\text{y}=(\sin^{-1}\text{x})^2$
$\Rightarrow\text{y}'=2(\sin^{-1}\text{x})\times\frac{1}{\sqrt{1-\text{x}^2}}$
$\Rightarrow\text{y}'\sqrt{1-\text{x}^2}=2(\sin^{-1}\text{x})$
$\Rightarrow\text{y}''\sqrt{1-\text{x}^2}+\text{y}'\times\frac{-2\text{x}}{2\sqrt{1-\text{x}^2}}=\frac{2}{\sqrt{1-\text{x}^2}}$
$\Rightarrow(1-\text{x}^2)\text{y}''-\text{xy}'=2$
$\therefore(1-\text{x}^2)\frac{\text{d}^2\text{y}}{\text{dx}^2}-\text{x}\frac{\text{dy}}{\text{dx}}-2=0.$

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