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What are the degree and order respectively of the
differential equation $\mathrm{y}=\mathrm{x}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^{2}+\left(\frac{\mathrm{dx}}{\mathrm{dy}}\right)^{2} ? \quad$
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differential equation $\mathrm{y}=\mathrm{x}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^{2}+\left(\frac{\mathrm{dx}}{\mathrm{dy}}\right)^{2} ? \quad$
Solution:
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Verified Answer
The correct answer is:
4,1
$y=x\left(\frac{d y}{d x}\right)^{2}+\left(\frac{d x}{d y}\right)^{2}$
$=\mathrm{x}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^{2}+\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^{-2}$
$\Rightarrow \mathrm{y}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^{2}=\mathrm{x} \cdot\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^{4}+1$
So, degree $=4$, order $=1$
$=\mathrm{x}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^{2}+\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^{-2}$
$\Rightarrow \mathrm{y}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^{2}=\mathrm{x} \cdot\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^{4}+1$
So, degree $=4$, order $=1$
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