Explanation:
(B): Given, \(\left(\mathrm{y}_{2}\right)^{2}-\sqrt{\mathrm{y}_{1}}=\mathrm{y}^{3}\)
\(\mathrm{y}_{2}=\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}, \mathrm{y}_{1}=\frac{\mathrm{dy}}{\mathrm{dx}}\)
Degree \(=\) Power of highest order derivative
\(\left(\mathrm{y}_{2}\right)^{2}=\mathrm{y}^{3}+\sqrt{\mathrm{y}_{1}}\)
Squaring both the side, we have-
\(\left(\mathrm{y}_{2}\right)^{4}=\mathrm{y}^{6}+2 \mathrm{y}_{1}^{1 / 2} \mathrm{y}^{3}+\mathrm{y}_{1}\)
So, \(\quad\) degree \(=4\)