Problem: Determine the order and degree of $(\frac{dr}{d\theta})^{4/3} + r = 2\theta$.
Attempt: From what I know, order means the highest derivative of the equation while degree is the exponent of that order. So in the equation, $\frac{dr}{d\theta}$ is the highest and is in first derivative, in which its exponent is $\frac{4}{3}$. So answer should be "Order : $1$ ; Degree : $\frac{4}{3}$".
Inquiry: However, the answer key on the book shows that the degree is actualy $4$. So I want to ask whether if I'm wrong or the book is wrong. Thank you.
The reason is cause if you want to identify the order of the ODE like in polynomials you need to have "something" elevated to an integer $n\ge 0$ so
$$\left(\frac{dr}{d\theta}\right)^{4/3}+r=2\theta\rightarrow \left(\frac{dr}{d\theta}\right)^{4/3}=2\theta-r$$
$$\left(\frac{dr}{d\theta}\right)^4=(2\theta-r)^3$$ Then $n=4$; therefore is a First Order ODE of $4^{th}$ degree.