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DIFFERENTIAL EQUATIONS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDIFFERENTIAL EQUATIONS5 Marks
Question
verify that $\text{y}^2=4\text{a}(\text{x}+\text{a})$ is a solution of the differential equation $\Big\{1-\Big(\frac{\text{dy}}{\text{dx}}\Big)^2\Big\}=2\text{x}\frac{\text{dy}}{\text{dx}}.$
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Answer

$\text{y}^2=4\text{a}(\text{x}+\text{a})\ ...(1)$
Differentiating both sides of (1) with respect to x, we get
$2\text{y}\frac{\text{dy}}{\text{dx}}=4\text{a}$
$\frac{\text{dy}}{\text{dx}}=\frac{2\text{a}}{\text{y}}\ ...(2)$
Now,
$\text{y}\Big\{1-\Big(\frac{\text{dy}}{\text{dx}}\Big)^2\Big\}$
$=\Big[\text{y}^2\Big\{1-\Big(\frac{\text{dy}}{\text{dx}}\Big)^2\Big\}\Big]\frac{1}{\text{y}}$
$=\Big[4\text{a}(\text{x}+\text{a})-4\text{a}(\text{x}+\text{a})\Big(\frac{2\text{a}}{\text{y}}\Big)^2\Big]\frac{1}{\text{y}}$
Using equation (1) and (2)
$=\Big[4\text{ax}+4\text{a}^2-\frac{16\text{a}3\text{x}}{\text{y}^2}-\frac{16\text{a}^4}{\text{y}^2}\Big]\frac{1}{\text{y}}$
$=\frac{4\text{a}}{\text{y}^3}[\text{xy}^2+\text{ay}^2-4\text{a}^2\text{x}-4\text{a}^3\Big]$
$=\frac{4\text{a}}{\text{y}^3}[\text{y}^2(\text{a}+\text{x})-4\text{a}^2(\text{x}+\text{a})]$
$\frac{4\text{a}}{\text{y}^3}(\text{a}+\text{x})(\text{y}^2-4\text{a}^2)$

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