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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY4 Marks
Question
If y $ = \log\bigg[\text{x+}\sqrt{\text{x}^{2} + \text{a}^{2}}\bigg],\text {show that } (\text{x}^{2} + \text{a}^{2})\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}} + \text{x}\frac{\text{dy}}{\text{dx}} = 0.$
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Answer

Given y $ = \log\bigg[\text{x} +\sqrt{\text{x}^{2} + \text{a}^{2}}\bigg]$
$\Rightarrow\frac{\text{dy}}{\text{dx}} =\frac{1}{\text{x} + \sqrt{\text{x}^{2} + \text{a}^{2}}}.\bigg[1+ \frac{2\text{x}}{2\sqrt{\text{x}^{2} + \text{a}^{2}}}\bigg]\Rightarrow\frac{\text{dy}}{\text{dx}} = \frac{\text{x} + \sqrt{\text{x}^{2} + \text{a}^{2}}}{\bigg(\text{x} + \sqrt{\text{x}^{2} + \text{a}^{2}}\bigg)\bigg(\sqrt{\text{x}^{2} + \text{a}^{2}}\bigg)}$
$\Rightarrow\frac{\text{dy}}{\text{dx}} = \frac{1}{\sqrt{\text{x}^{2} + \text{a}^{2}}}$
Differentiating again w.r.t. x we get
$\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}} = -\frac{1}{2}(\text{x}^{2} + \text{a}^{2})^{-\frac{3}{2}}.2\text{x} = \frac{-\text{x}}{(\text{x}^{2} + \text{a}^{2})^{\frac{3}{2}}}$
$\Rightarrow\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}} =\frac{-\text{x}}{(\text{x}^{2} + \text{a}^{2}).\sqrt{\text{x}^{2} +\text{a}^{2}}}\Rightarrow(\text{x}^{2} + \text{a}^{2})\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}} = -\frac{\text{x}}{\sqrt{\text{x}^{2} + \text{a}^{2}}}$
$\Rightarrow(\text{x}^{2} + \text{a}^{2}) \frac{\text{d}^{2}\text{y}}{\text{dx}^{2}} +\text{x}.\frac{\text{dy}}{\text{dx}} = 0 .$ [from (i)].

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