The degree of the differential equation
$y_{2}^{3 / 2} - y_{1}^{1 / 2} - 4 = 0$ is:

6

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3

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2

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4

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Solution

The correct option is A $6$
$y_{2}^{\frac{3}{2}} = y_{1}^{\frac{1}{2}} + 4$
Squaring both sides, we have
$y_{2}^{3} = y_{1} + 16 + 8 y_{1}^{\frac{1}{2}}$
$\Rightarrow {(y_{2}^{3} - y_{1} - 16)}^{2} = 64 y_{1} \Rightarrow y_{2}^{6} - 32 y_{2}^{3} - 2 y_{2}^{3} y_{1} + y_{1}^{2} - 32 y_{1} + 256 = 0$
Hence the degree of the given equation is $6$

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