Miscellaneous · Q1: Order and degree of $y'' + 5x(y')^2 - 6y = \log x$⁠, $(y')^3 - 4(y')^2 + 7y = \sin x$ and $y'''' - \sin(y''') = 0$

Basics you need

Part (i): $\frac{d^2y}{dx^2} + 5x\left(\frac{dy}{dx}\right)^2 - 6y = \log x$

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 in $\left(\frac{dy}{dx}\right)^2$ is a power of the first derivative, not a second derivative, so it does not raise the order. The terms $6y$ and $\log x$ contain no derivative at all.
   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 $p = \frac{dy}{dx}$ and $q = \frac{d^2y}{dx^2}$⁠. The equation becomes

   $$q + 5x\,p^2 - 6y - \log x = 0,$$

   a polynomial in $q$ and $p$ whose coefficients $1$⁠, $5x$ and the remaining part $-6y - \log x$ are functions of $x$ and $y$ only. So the degree is defined. The highest order derivative $q = \frac{d^2y}{dx^2}$ appears only to the power 1:

   $$\text{degree} = 1.$$
   The degree is read from the highest order derivative only, so the power 2 on $\frac{dy}{dx}$ does not count. $\log x$ is allowed: the polynomial condition is about the derivatives, and $\log x$ contains none.
   Section 9.2.2: degree = highest power of the highest order derivative. Lesson: Example 1, the same reading

Part (ii): $\left(\frac{dy}{dx}\right)^3 - 4\left(\frac{dy}{dx}\right)^2 + 7y = \sin x$

1. Find the highest derivative: the order

   The only derivative present is $\frac{dy}{dx}$⁠, which appears twice, with powers 3 and 2. So the highest order derivative is $\frac{dy}{dx}$⁠, and

   $$\text{order} = 1.$$
   The cube and the square are powers of the same first derivative. However high the power, $y$ has been differentiated only once.
   A power and an order are different things: $\left(\frac{dy}{dx}\right)^3 \ne \frac{d^3y}{dx^3}$⁠. Basics: y′ and y″

2. Check the polynomial condition, then read the degree

   Write $p = \frac{dy}{dx}$⁠. The equation becomes

   $$p^3 - 4p^2 + (7y - \sin x) = 0,$$

   a polynomial in $p$ (a cubic) whose last coefficient $7y - \sin x$ is a function of $x$ and $y$⁠. So the degree is defined, and the highest power of the highest order derivative $p$ is 3:

   $$\text{degree} = 3.$$
   Here the highest order derivative and the only derivative are the same, $\frac{dy}{dx}$⁠, so its highest power, 3, is the degree. $\sin x$ does not spoil the degree, because it is a function of $x$⁠, not of a derivative.
   Section 9.2.2; compare the book's equation (10), a polynomial in $y'$ alone, of degree 2. Lesson: order and degree

Part (iii): $\frac{d^4y}{dx^4} - \sin\left(\frac{d^3y}{dx^3}\right) = 0$

1. Find the highest derivative: the order

   The derivatives present are $\frac{d^4y}{dx^4}$ (order 4) and $\frac{d^3y}{dx^3}$ (order 3, inside the sine). The highest is $\frac{d^4y}{dx^4}$⁠, so

   $$\text{order} = 4.$$
   A derivative counts for the order wherever it sits, even inside a sine. Here the one inside the sine is lower, so the order is set by the fourth derivative.
   Section 9.2.1; $\frac{d^3y}{dx^3} = y'''$ (Section 9.2, Note 1). Basics: y′ and y″

2. Check the polynomial condition: it fails

   Write $r = \frac{d^3y}{dx^3}$ and $s = \frac{d^4y}{dx^4}$⁠. The equation becomes $s - \sin r = 0$⁠. The term $\sin r$ is not a polynomial in $r$⁠:

   $\sin r$ always lies between $-1$ and $1$⁠; a polynomial in $r$ that is not constant takes values as large as we like when $r$ is large, because its highest power wins; and $\sin r$ is not constant ($\sin 0 = 0$⁠, $\sin\frac{\pi}{2} = 1$⁠). So no polynomial equals $\sin r$⁠, and the equation is not a polynomial equation in its derivatives. Therefore

   $$\text{degree: not defined.}$$
   The definition of degree has two parts: first the polynomial condition, then the highest power of the highest derivative. The first part fails, so the second is never reached. Answering "degree 1" because $\frac{d^4y}{dx^4}$ appears to the first power skips the condition.
   Section 9.2.2, equation (11): a derivative inside a sine has no degree. The same test in Exercise 9.1 Q1

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