what is degree of given PDE... 
When the given differential eqn is completely free from radicals the the final exponent on the highest order derivative amounts degree of given differential eqn. 
In present case it is 3 or 6. i.e. we should square 1 time or 2 times?
 A: 
Degree of a differential equation: 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.

Note: The above definition of degree does not require variables $~x,~t,~u~$etc. to be free from radicals and frictions.
Given differential equation is $$(y'')^{3/2}-(y')^{1/2}-4=0$$
$$\implies (y'')^{3/2}=(y')^{1/2}+4$$
squaring both side we have,$$(y'')^3=\{(y')^{1/2}+4\}^2$$
$$\implies (y'')^3=y'+8(y')^{1/2}+16$$
$$\implies (y'')^3-y'-16=8(y')^{1/2}$$
To get rid of radicals, again squaring both side we  have
$$\{(y'')^3-y'-16\}^2=8y'$$
which is of $~2^{\text{th}}~$ order and $~6^{\text{th}}~$ degree because the order of the highest differential coefficient $~y''~$ is $~2~$ and highest degree of $~y''~$ is $~6~$.
A: The differential equation $$(y'')^\frac{2}{3} = 2 + 3y'$$ can be rationalized by cubing both sides, to obtain $$(y'')^2 = (2 + 3y' )^3.$$ 
Thus it is of degree two. This example will help you to understand your question.
