Chapter 9 · Differential Equations

Exercise 9.1 · Q1: Order and degree of $\frac{d^4y}{dx^4} + \sin(y''') = 0$

Exercise 9.1Order and degree · Sections 9.2.1 and 9.2.2

The question

Determine order and degree (if defined) of differential equations given in Exercises 1 to 10.

1. $\dfrac{d^4y}{dx^4} + \sin(y''') = 0$

Exam answer

The highest order derivative present in the equation is $\dfrac{d^4y}{dx^4}$ ⁠, so its order is 4.

The equation contains the term $\sin(y''')$ ⁠. So it is not a polynomial equation in its derivatives $y'''$ and $\dfrac{d^4y}{dx^4}$ ⁠.

Hence its degree is not defined.

$\therefore$ Order 4; degree not defined.

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

Order and degree are two separate questions. The order is the order of the highest derivative, here $\frac{d^4y}{dx^4}$ ⁠, so the order is 4. The degree exists only when the equation is a polynomial equation in its derivatives. Here $y'''$ sits inside a sine, and $\sin(y''')$ is not a polynomial in $y'''$ ⁠. So the degree is not defined, even though $\frac{d^4y}{dx^4}$ itself appears only to the first power.

Given, to find, and the plan

Given: The differential equation $\frac{d^4y}{dx^4} + \sin(y''') = 0$ ⁠, with $y$ the dependent variable and $x$ the independent variable.
To find: Its order, and its degree if the degree is defined.
Plan: (1) List every derivative in the equation. (2) The order is the highest of them. (3) Test whether the equation is a polynomial equation in its derivatives. (4) Decide the degree.

Basics you need

1Derivatives and their orders

$y' = \frac{dy}{dx}$ is the first derivative, $y'' = \frac{d^2y}{dx^2}$ the second and $y''' = \frac{d^3y}{dx^3}$ the third. $\frac{d^4y}{dx^4}$ is the fourth: $y$ differentiated four times (Note 1, Section 9.2). Basics: y′ and y″

2Order

The order of a differential equation is the order of the highest order derivative in it (Section 9.2.1). Lesson: order and degree

3Degree, and its condition

The degree is defined only when the equation is a polynomial equation in its derivatives. Then it is the highest power (a positive integer) of the highest order derivative (Section 9.2.2). Lesson: order and degree

4What a polynomial in $p$ is

A finite sum of terms $c\,p^k$ with whole-number powers $k = 0, 1, 2, \ldots$ ⁠. $\sin p$ ⁠, $\cos p$ ⁠, $e^p$ and $\log p$ are not of this form. Lesson: Example 1, order and degree

Solution

List the derivatives in the equation

The equation contains two derivatives of $y$ with respect to $x$ ⁠:

$\dfrac{d^4y}{dx^4}$ ⁠, of order 4, and $y''' = \dfrac{d^3y}{dx^3}$ ⁠, of order 3.

There is no other derivative. The $y'''$ is easy to miss because it sits inside the sine.

Both the order and the degree are about the derivatives of the dependent variable $y$ ⁠, so we first find all of them, wherever they sit in the equation.

Notation $y'$ ⁠, $y''$ ⁠, $y'''$ ⁠: Note 1, Section 9.2. Basics: y′ and y″
Read off the order

The highest order derivative present is $\dfrac{d^4y}{dx^4}$ (order 4 is more than 3). So
$\text{order} = 4.$
Section 9.2.1 defines the order as the order of the highest order derivative in the equation. Its power, its coefficient and the functions around the other terms do not matter for the order.

Definition of order, Section 9.2.1. Lesson: order and degree
Is it a polynomial equation in its derivatives?

Write $p = y'''$ and $q = \frac{d^4y}{dx^4}$ ⁠. The equation reads $q + \sin p = 0$ ⁠.

The term $q$ is a polynomial in $q$ ⁠. But $\sin p$ is not a polynomial in $p$ ⁠. Suppose it were. A polynomial in $p$ that is not constant takes values as large in size as we like when $p$ is large, because its highest power wins. But $\sin p$ always lies between $-1$ and $1$ ⁠, so that polynomial would have to be a constant. And $\sin p$ is not constant: $\sin 0 = 0$ but $\sin\frac{\pi}{2} = 1$ ⁠. This contradiction shows that $\sin p$ is not a polynomial in $p$ ⁠.

So the equation is not a polynomial equation in its derivatives.

The degree is only defined for a polynomial equation in the derivatives, so this test must come before any talk of powers. The situation is the same as the book's equation (11) in Section 9.2.2, $\frac{dy}{dx} + \sin\left(\frac{dy}{dx}\right) = 0$ ⁠, whose degree is not defined.

Condition for the degree, Section 9.2.2. Lesson: Example 1, a derivative inside a function
Decide the degree

The equation is not a polynomial equation in its derivatives, so
$\text{degree: not defined}.$
It is tempting to answer "degree 1", because the highest derivative $\frac{d^4y}{dx^4}$ appears to the first power. That reading skips the condition: the highest power is asked for only after the whole equation is a polynomial in all its derivatives, and $\sin(y''')$ breaks this.

The definition in Section 9.2.2 has two parts: first the polynomial condition, then the highest power of the highest order derivative. When the first part fails, the second is never reached.

Definition of degree and the Note "order and degree (if defined) are always positive integers", Section 9.2.2. Lesson: Exercise 9.1 player

Result

The equation $\frac{d^4y}{dx^4} + \sin(y''') = 0$ has order 4, and its degree is not defined, because it is not a polynomial equation in its derivatives.

(Textbook answer: Order 4; Degree not defined.)

Check it with numbers

Give $y'''$ larger and larger values: $\sin 10 \approx -0.5440$ ⁠, $\sin 100 \approx -0.5064$ ⁠, $\sin 1000 \approx 0.8269$ ⁠. The values stay between $-1$ and $1$ ⁠, while any polynomial in $y'''$ that is not constant would run off to very large sizes. This is the boundedness argument of Step 3 in numbers.

Now hold $y''' = 1$ fixed. The left side is $\frac{d^4y}{dx^4} + \sin 1$ ⁠, with $\sin 1 \approx 0.841471$ ⁠. At $\frac{d^4y}{dx^4} = 0, 1, 2$ it is $0.841471$ ⁠, $1.841471$ ⁠, $2.841471$ ⁠: equal steps of 1, so $\frac{d^4y}{dx^4}$ by itself enters to the first power. That is where the wrong answer "degree 1" comes from.

Common mistakes

Answering "degree 1". The highest derivative does appear to the first power, but the polynomial condition fails first, so no degree is defined.
Overlooking the derivative inside the sine. $y'''$ is a derivative even inside $\sin(\;)$ ⁠. It is lower than the fourth derivative here, so it does not change the order, but it decides the degree.
Thinking any sine kills the degree. A sine spoils the degree only when a derivative sits inside it. $\sin y$ or $\sin 3x$ is fine; see Exercise 9.1 Q10.

What to take away

Order: find the highest derivative anywhere in the equation.
Degree: first ask "is this a polynomial equation in the derivatives?". A derivative inside sin, cos, tan, log or e means the degree is not defined.

Every fact used, and where it comes from

Fact	Used in	Source
$y''' = \frac{d^3y}{dx^3}$ ⁠; $\frac{d^4y}{dx^4}$ is the fourth derivative	Step 1	Note 1, Section 9.2
Order = order of the highest order derivative	Step 2	Section 9.2.1 (Lesson: order and degree)
Degree is defined only for a polynomial equation in the derivatives	Steps 3, 4	Section 9.2.2
$\frac{dy}{dx} + \sin\left(\frac{dy}{dx}\right) = 0$ has no degree	Step 3	Equation (11), Section 9.2.2
$-1 \le \sin p \le 1$ for every real $p$	Step 3	Trigonometric functions (Class XI)
Degree = highest power of the highest order derivative	Step 4	Section 9.2.2

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Exercise 9.1 · Q1: Order and degree of d4ydx4+sin⁡(y′′′)=0\frac{d^4y}{dx^4} + \sin(y''') = 0dx4d4y​+sin(y′′′)=0

Basics you need

Solution

List the derivatives in the equation

Read off the order

Is it a polynomial equation in its derivatives?

Decide the degree

Every fact used, and where it comes from

Exercise 9.1 · Q1: Order and degree of $\frac{d^4y}{dx^4} + \sin(y''') = 0$