# Find the degree of a ODE $(y''')^{\frac{4}{3}}+(y')^{\frac{1}{5}}+ y = 0$

Find the degree of the differential equation.

$$(y''')^{\frac{4}{3}}+(y')^{\frac{1}{5}}+ y = 0$$

The answer is available (order $$= 3$$; degree $$= 60$$). I need help with the steps. I'm stuck in eliminating the radicals of the differential coefficients.

• Use LaTeX please. Jun 17, 2019 at 18:20
• Thank you! This was my first time.
– NVS
Jun 17, 2019 at 18:34
• The degree of a differential equation is the degree of the highest derivative which occurs in it, after the differential equation has been made free from radicals and fractions as far as the derivatives are concerned. Jun 18, 2019 at 3:01
• The degree is meaningless if the equation is not a polynomial. And even then, the only useful degree is that of the leading derivative, as that tells you how many solution curves can originate (maximally) in each initial point. Jun 18, 2019 at 9:31
• I've tried different pairings and taken powers of 3 and 5 on each side. Nothing has worked. Perhaps I'm missing something subtle here. Any help is welcome. Sep 4, 2020 at 17:43

Went through @Lutz's answer Find the degree of the differential equation $\left( \frac{d^3y}{dx^3} \right)^{\frac{4}{3}} + \left( \frac{dy}{dx} \right)^{\frac{1}{5}} + y = 0.$. But despite his elaborate explanations, I found it really hard to understand.
Then, I tried it the hard way. It's only a 2-step solution. Nothing fancy. $$(y''')^{4/3}+y=-(y')^{1/5}$$ First, take power $$5$$ both sides, $$5y^4(y''')^{4/3}+10y^{3}(y''')^{8/3}+5y(y''')^{16/3}+(y''')^{20/3}=-y'-y^5-10y^2(y''')^{4}$$
$$(y''')^{4/3}[5y^4+5y(y''')^{4}]+(y''')^{8/3}[10y^{3}+(y''')^{4}]=-y'-y^5-10y^2(y''')^{4}$$ This is equivalent to $$A+B=C$$. $$A$$ is $$O(\frac{16}{3})$$. $$B$$ is $$O(\frac{20}{3})$$. $$C$$ is $$O(4)$$.
Cubing both sides $$A^3+B^3 +3AB(A+B)=C^3$$, where $$A+B=C$$ can be substituted from the above equation. $$A^3+B^3 +3ABC=C^3$$ $$(y''')^{4}[5y^4+5y(y''')^{4}]^{3}+(y''')^{8}[10y^{3}+(y''')^{4}]^{3}+3(y''')^{4}[5y^4+5y(y''')^{4}][10y^{3}+(y''')^{4}][-y'-y^5-10y^2(y''')^{4}]=[-y'-y^5-10y^2(y''')^{4}]^{3}$$ $$O(16)+O(20)+O(16)=O(12)$$
So, the degree comes out to be $$20$$. Evidently, the bookish answer of $$60$$ must be wrong.
If you raise both sides to the $$15$$th power, the equation is polynomial in $$y,\, y'$$ and $$y''',$$ with the latter clearly being of degree $$20.$$