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Differential Equations — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDifferential Equations2 Marks
Question
Write the order of the differential equation $\text{y}=\text{x}\frac{\text{dy}}{\text{dx}}+\text{a}\sqrt{1+\Big(\frac{\text{dy}}{\text{dx}}\Big)^{2}}.$
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Answer

We have,
$\text{y}=\text{x}\frac{\text{dy}}{\text{dx}}+\text{a}\sqrt{1+\Big(\frac{\text{dy}}{\text{dx}}\Big)^{2}}$
$\text{y}-\text{x}\frac{\text{dy}}{\text{dx}}=\text{a}\sqrt{1+\Big(\frac{\text{dy}}{\text{dx}}\Big)^{2}}$
Squaring both sides, we get
$\Big(\text{y}-\text{x}\frac{\text{dy}}{\text{dx}}\Big)^{2}=\left\{\text{a}\sqrt{1+\Big(\frac{\text{dy}}{\text{dx}}\Big)^{2}}\right\}^{2}$
$\text{y}^{2}-\text{x}^{2}\Big(\frac{\text{dy}}{\text{dx}}\Big)^{2}-2\text{xy}\frac{\text{dy}}{\text{dx}}=\text{a}^{2}\left\{{1+\Big(\frac{\text{dy}}{\text{dx}}\Big)^{2}}\right\}^{2}$
$\text{y}^{2}+\Big(\frac{\text{dy}}{\text{dx}}\Big)^{2}(\text{x}^{2}-\text{a}^{2})-2\text{xy}\frac{\text{dy}}{\text{dx}}=\text{a}^{2}$
From the above equation, we see that the highest order is 1.
So, its order is 1 and the power of the order is 2.
Thus, it is differential equation of order 1 and degree 2.

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