You have not asked a question here!(y''')^ (4/3) +(y')^ (1/5) + y = 0
...To get rid of the fractional powers, you would have to take the 15th power of both sides. What is the power of the highest order derivative then? You do not need to actually do the entire multiplication to answer that.)
Hmmm, why not cube then, if the remaining powers don't matter? \(\displaystyle ((y''')^{4/3})^{3}= (y''')^{4}\).Fractional powers will remain but that is not relevant. The highest power of the highest derivative, y''', will be \(\displaystyle ((y''')^{4/3})^{15}= (y''')^{4(5)}= (y''')^{20}\).
(y'''^4 + y^3)/(3*y) = -y'^(1/5) * [ (y')^(1/5)+y] ................................??? ....................... uncle
I give up ...... but I think HoI is correct