Example 1: Order and degree of $y' - \cos x = 0$⁠, $xyy'' + x(y')^2 - yy' = 0$ and $y''' + y^2 + e^{y'} = 0$

Basics you need

Part (i): $\frac{dy}{dx} - \cos x = 0$

1. Find the highest derivative: the order

   The only derivative in the equation is $\frac{dy}{dx}$⁠, a first derivative. So the highest order derivative present is $\frac{dy}{dx}$⁠, and

   $$\text{order} = 1.$$
   The order counts how many times $y$ has been differentiated in the most differentiated term. Here that is once. The term $\cos x$ is a function of $x$ only; it is not a derivative and plays no part in the order.
   Section 9.2.1: the order is the order of the highest order derivative present. Lesson: order and degree

2. Check the polynomial condition, then read the degree

   Write $p = \frac{dy}{dx}$⁠. The equation becomes $p - \cos x = 0$⁠, which is a polynomial in $p$ (of the form $1\cdot p^1 + (\text{a function of } x)$⁠). So the degree is defined. The highest order derivative is $p = \frac{dy}{dx}$⁠, and its highest power is 1:

   $$\text{degree} = 1.$$
   "Polynomial in the derivatives" means: replace each derivative by a letter, and the left side must be a sum of terms (function of $x$ and $y$⁠) $\times$ (whole-number powers of those letters). Here $\cos x$ is allowed because it contains no derivative; only the derivatives have to appear polynomially.
   Section 9.2.2: the degree is the highest power of the highest order derivative. Lesson: order and degree

Part (ii): $xy\frac{d^2y}{dx^2} + x\left(\frac{dy}{dx}\right)^2 - y\frac{dy}{dx} = 0$

1. Find the highest derivative: the order

   The derivatives present are $\frac{dy}{dx}$ (order 1) and $\frac{d^2y}{dx^2}$ (order 2). The highest is $\frac{d^2y}{dx^2}$⁠, so

   $$\text{order} = 2.$$
   The square on $\frac{dy}{dx}$ is a power, not an order: $\left(\frac{dy}{dx}\right)^2$ is the first derivative multiplied by itself, while $\frac{d^2y}{dx^2}$ is the derivative of the derivative. Only the second kind raises the order.
   $\left(\frac{dy}{dx}\right)^2 \ne \frac{d^2y}{dx^2}$⁠: a power and an order are different things. Basics: y′ and y″

2. Check the polynomial condition, then read the degree

   Write $p = \frac{dy}{dx}$ and $q = \frac{d^2y}{dx^2}$⁠. The equation becomes

   $$xy\,q + x\,p^2 - y\,p = 0,$$

   a polynomial in $q$ and $p$ whose coefficients $xy$⁠, $x$ and $-y$ are functions of $x$ and $y$⁠. So the degree is defined. The highest order derivative is $q = \frac{d^2y}{dx^2}$⁠, and it appears only to the power 1:

   $$\text{degree} = 1.$$
   The degree is read from the highest order derivative only. The power 2 on $\frac{dy}{dx}$ does not count, because $\frac{dy}{dx}$ is not the highest derivative here. The factors $x$ and $y$ in the coefficients are allowed: the condition is on the derivatives, not on $y$⁠.
   Section 9.2.2: degree = highest power of the highest order derivative, once the equation is a polynomial in the derivatives. Lesson: order and degree

Part (iii): $y''' + y^2 + e^{y'} = 0$

1. Find the highest derivative: the order

   The derivatives present are $y'$ (inside $e^{y'}$⁠) and $y'''$⁠. The highest is $y'''$⁠, so

   $$\text{order} = 3.$$
   A derivative counts for the order wherever it sits, even inside an exponential. $y^2$ is a power of $y$ itself, not a derivative, so it does not matter for the order either.
   Section 9.2.1; $y''' = \frac{d^3y}{dx^3}$ (Section 9.2, Note 1). Basics: y′ and y″

2. Check the polynomial condition: it fails

   Write $p = y'$ and $r = y'''$⁠. The equation becomes $r + y^2 + e^{p} = 0$⁠. The term $e^{p}$ is not a polynomial in $p$⁠. Here is why: if $e^{p}$ were equal to a polynomial in $p$ of some degree $n$⁠, then differentiating both sides $n + 1$ times with respect to $p$ would give $e^{p} = 0$⁠, because a polynomial of degree $n$ differentiated $n + 1$ times is 0. But $e^{p} > 0$ for every $p$⁠. So no polynomial equals $e^{p}$⁠, and the equation is not a polynomial equation in its derivatives. Therefore

   $$\text{degree: not defined.}$$
   The degree is only defined for equations that are polynomials in the derivatives. Once a derivative sits inside $e$ (or sin, cos, log, $\tan^{-1}$⁠, …), there is no "highest power" to read. The order is still defined: it is 3. The term $y^2$ is harmless; it is the $e^{y'}$ that spoils the degree.
   Section 9.2.2, equation (11): a derivative inside a sine has no degree; the same holds inside an exponential. Basics: e^{kt} and its rate

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