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Q.
The degree of the differential equation $ y_{2}^{3/2}-y_{1}^{1/2}-4=0 $ is
JamiaJamia 2013
Solution:
Given, $ y_{2}^{3/2}=y_{1}^{1/2}+4 $ Squaring on both sides, we get $ y_{2}^{3}={{y}_{1}}+16+8y_{1}^{1/2} $ $ \Rightarrow $ $ {{(y_{2}^{3}-{{y}_{1}}-16)}^{2}}=64{{y}_{1}} $ $ \Rightarrow $ $ y_{2}^{6}-32y_{2}^{3}-2y_{2}^{3}.{{y}_{1}}+y_{1}^{2}-32{{y}_{1}}+256=0 $ Hence, the degree of the given equation is 6.