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,
$\text{y}=(\sin^{-1}\text{x})^2$
Now,
$\text{y}_1=2\sin^{-1}\text{x}\frac{1}{\sqrt{1-\text{x}^2}}$
$\Rightarrow\text{y}_2=\frac{2}{1-\text{x}^2}+\frac{2\text{x}\sin^{-1}\text{x}}{(1-\text{x}^2)^\frac{3}{2}}$
$\Rightarrow\text{y}_2=\frac{2}{1-\text{x}^2}+\frac{2\text{x}\sin^{-1}\text{x}}{(1-\text{x}^2)\sqrt{1-\text{x}^2}}$
$\Rightarrow\text{y}_2=\frac{2}{1-\text{x}^2}+\frac{\text{xy}_1}{(1-\text{x}^2)}$
$\Rightarrow\text{y}_2(1-\text{x}^2)=2+\text{xy}_1$
$\text{y}_2(1-\text{x}^2)-\text{xy}_1-2=0$
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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