Degree of differential equation What is the degree of this differential equtaion:
$$(y'')^{2/3}+(y'')^{3/2}=0$$
Solution
$$(y'')^{2/3}=-(y'')^{3/2}$$
Raising both sides with power 6 we get
$$(y'')^4=(y'')^9$$
Now, the degree of highest derivative is both $4$ and $9$.
Then what is the degree? Is it $4$ or $9$?
 A: Your equation is not even a differential equation, as only the second derivative but no other derivatives (or the function itself) appears. You can simply integrate twice.
However, in my opinion:
$(y'')^{2/3}=-(y'')^{3/2}$
$(y'')^{3/2}\ (y'')^{-2/3} = (y'')^{5/6}=-1$
so $(y'')^{5/6}+1=0$
A: I think you mixed up the degree of the differential equation and the degree of the polynomials.
Example,
$$\frac{d^3y}{dx^3}+\frac{dy}{dx}+y=0$$
is called a third order differential equation. The highest derivative inside the differential equation is $3$. So, it is a third order.
Consider this
$$(\frac{d^3y}{dx^3})^2+y=0$$
This is still considered a third order differential equation.
$$(\frac{dy}{dx})^{10}+y=0$$
This is called the first order differential equation despite that it is raised to the power of 10.
It is very different from the polynomial 
Like,
$$y=x^2+x+3$$
This is called a second degree polynomial.
Conclusion:
The highest derivative that exists inside the differential equation let's say 2 is known as the '2' order differential equation.
The highest power that exists inside the polynomial say 10 $(x^{10}) $is called the  10th degree polynomial.
A: 
Polynomials: These are functions which can be written in the form $f(x) = a_0 +a_1x+···+a_dx^d$ for some numbers $a_j, 0 ≤ j ≤ d, d ∈ N$. If $a_d \neq 0$, we say that f is a polynomial of degree d with leading coefficient $a_d$. The maximal domain of any polynomial is R. We say that r is a root or a zero of the polynomial f if $f(r) = 0$.

Where as

A differential equation is an equation of the form $F(x,y,y',...,y^{n})= 0, x ∈ I$, with the following properties:
(1) n ∈ N is the order of the equation.
(2) F is a given function of n+2 variables.
(3) y is the unknown function of the independent variable x.
(4) I is the interval in which the equation is to hold.

Therefore $\sum^n_{j=1} (y'')^{j}=0$ is still of degree/order 2
(Both extracts are lifted from lecture notes)
Further if were talking about degrees of a differential equation we talk about the highest derivative. If we talk about degrees of a polynomial we talk about the exponent.
Btw for the equation $x^{4}=x^{9}$ where $x=y''$it is not in its simplist form
Thus $1 = x^{5} \Rightarrow x=1$
A: I dont think any existing answers answer your question .
The degree of the differential equation you asked is 9 because when you try to convert it to a polynomial form , you get the form as $(y")^9-(y")^4=0$ which is similar to finding degree of polynomial $(x)^9-(x)^4=0$ which is clearly 9 .
Hope this answers your doubt .
