If y = x + 1 - x - 1 , Prove that (x^ 2 - 1) d^ 2 y dx^ 2 + x dy … - Vidyadip

Question

$\text{If y} = \sqrt{\text{x + 1}} - \sqrt{\text{x - 1}}, \ \text{Prove that}\ (\text{x}^{2} - 1) \frac{\text{d}^{2}\text{y}}{\text{dx}^{2}} + \text{x} \frac{\text{dy}}{\text{dx}} - \frac{1}{4} \text{y}= 0.$

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Answer

$\text{y = }\sqrt{\text{x + 1}} - \sqrt{\text{x - 1}}$
$\frac{\text{dy}}{\text{dx}} = \frac{1}{2\sqrt{\text{x + 1}}} - \frac{1}{2\sqrt{\text{x - 1}}}$
$= \frac{\sqrt{\text{x - 1}} - \sqrt{\text{x + 1}}}{2\sqrt{\text{x}^{2} - 1}}$
$4(\text{x}^{2} - 1)\bigg(\frac{\text{dy}}{\text{dx}}\bigg)^{2} = \text{y}^{2}$
$4(\text{x}^{2} - 1) 2\frac{\text{dy}}{\text{dx}}.\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}} + 8\text{x}\bigg(\frac{\text{dy}}{\text{dx}}\bigg)^{2} = \text{2y}\frac{\text{dy}}{\text{dx}}$
$\text{(x}^{2} - 1) \frac{\text{d}^{2}\text{y}}{\text{dx}^{2}} + \text{x}\frac{\text{dy}}{\text{dx}} = \frac{\text{y}}{\text{4}}$
$\text{(x}^{2} - 1)\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}} + \text{x}\frac{\text{dy}}{\text{dx}} - \frac{\text{y}}{\text{4}} = 0$

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