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.
 A: 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.
A: 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 $20.$
