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Differential Equations — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDifferential Equations4 Marks
Question
verify that $\text{y}=\log(\text{x}+\sqrt{\text{x}^2+\text{a}^2})^2$ is a solution of the differential equation $(\text{a}^2+\text{x}^2)\frac{\text{d}^2\text{y}}{\text{dx}^2}+\text{x}\frac{\text{dy}}{\text{dx}}=0$
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Answer

$\text{y}=\log(\text{x}+\sqrt{\text{x}^2+\text{a}^2})^2$
Differentiating both sides of (1) with respect to x, we get
$\frac{\text{dy}}{\text{dx}}\frac{1}{(\text{x}+\sqrt{\text{a}^2+\text{x}^2})^2}\times2(\text{x}+\sqrt{\text{x}^2+\text{a}^2})\frac{\text{d}}{\text{dx}}(\text{x}+\sqrt{\text{x}^2+\text{a}^2})$
$=\frac{2}{(\text{x}+\sqrt{\text{a}^2+\text{x}^2})}\times\Big(1+\frac{1}{2\sqrt{\text{x}^2+\text{a}^2}}(2\text{x})\Big)$
$=\frac{2}{(\text{x}+\sqrt{\text{a}^2+\text{x}^2})}\Big(\frac{\sqrt{\text{x}2+\text{a}^2}+\text{x}}{2\sqrt{\text{x}^2+\text{a}^2}}\Big)$
$\frac{\text{dy}}{\text{dx}}=\frac{1}{\sqrt{\text{a}^2+\text{x}^2}}$
$\sqrt{\text{a}^2+\text{x}^2}\frac{\text{dy}}{\text{dx}}=1$
Again differentiating it with respect to x,
$\sqrt{1-\text{x}^2}\frac{\text{d}^2\text{y}^2}{\text{dx}^2}+\frac{1}{2\sqrt{1-\text{x}^2}}(-2\text{x})\frac{\text{dy}}{\text{dx}}=-\text{m}\frac{\text{dy}}{\text{dx}}$
$\sqrt{1-\text{x}^2}\frac{\text{d}^2\text{y}^2}{\text{dx}^2}-\frac{\text{x}}{\sqrt{1-\text{x}^2}}\frac{\text{dy}}{\text{dx}}-\Big(\frac{-\text{e}^{\text{m}^{\cos^{-1}}}\text{m}}{\sqrt{1-\text{x}^2}}\Big)=0$
Using equation (1)
$\sqrt{\text{a}^2+\text{x}^2}\frac{\text{d}^2\text{y}}{\text{dx}^2}+\frac{2\text{x}}{2\sqrt{\text{a}^2+\text{x}^2}}\frac{\text{dy}}{\text{dx}}=0$
$(\text{a}^2+\text{x}^2)\frac{\text{d}^2\text{y}}{\text{dx}^2}+\text{x}\frac{\text{dy}}{\text{dx}}=0$

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