Chapter 9 · Differential Equations

Exercise 9.1 · Q9: Order and degree of $y'' + (y')^2 + 2y = 0$

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.

9. $y'' + (y')^2 + 2y = 0$

Exam answer

The highest order derivative present in the equation is $y''$ ⁠, so its order is 2.

The equation is a polynomial equation in its derivatives $y''$ and $y'$ ⁠.

The highest power of $y''$ is 1, so its degree is 1.

$\therefore$ Order 2; degree 1.

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

The one idea this problem is built on

The highest derivative is $y''$ ⁠, so the order is 2. The equation is a polynomial in $y''$ and $y'$ ⁠, so the degree is defined. The square in $(y')^2$ is on the lower derivative $y'$ ⁠, so it does not count: the power of $y''$ is 1, and the degree is 1.

Given, to find, and the plan

Given: The differential equation $y'' + (y')^2 + 2y = 0$ ⁠, 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 with its power. (2) The order is the highest derivative. (3) Test whether the equation is a polynomial equation in its derivatives. (4) Read off the power of the highest derivative only.

Basics you need

1Dashes and powers

$y'$ and $y''$ are the first and second derivatives (Note 1, Section 9.2). $(y')^2$ is the first derivative squared: one dash, power 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

Solution

List the derivatives and their powers

$y''$ ⁠: the second derivative, to the power 1.

$(y')^2$ ⁠: the first derivative, to the power 2.

$2y$ ⁠: $y$ itself, not a derivative.

Keeping "which derivative" apart from "what power" for each term is what the definitions need.

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

The highest order derivative present is $y''$ ⁠. So
$\text{order} = 2.$
Section 9.2.1: the order is the order of the highest order derivative. The square on $y'$ makes it a squared first derivative, not a second derivative.

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

Write $p = y'$ and $q = y''$ ⁠. The equation reads
$q + p^2 + 2y = 0,$
a polynomial in $p$ and $q$ ⁠. So the equation is a polynomial equation in its derivatives, and its degree is defined.

The degree is only defined for a polynomial equation in the derivatives (Section 9.2.2). Whole-number powers of derivatives, like $p^2$ ⁠, are allowed; only derivatives inside functions such as sin, cos, tan, log or e are not.

Condition for the degree, Section 9.2.2. Lesson: Example 1, order and degree
Read off the degree

The highest order derivative is $q = y''$ ⁠, and its highest power in the equation is 1. So
$\text{degree} = 1.$
By Section 9.2.2, the degree is the highest power of the highest order derivative. The power 2 belongs to $y'$ ⁠, the lower derivative, so it is not the degree. The book's Example 1 (ii) is the same: $\left(\frac{dy}{dx}\right)^2$ is present, yet the degree is 1.

Definition of degree, Section 9.2.2. Lesson: Exercise 9.1 player

Result

The equation $y'' + (y')^2 + 2y = 0$ has order 2 and degree 1. It is a polynomial equation in its derivatives, so the degree is defined.

(Textbook answer: Order 2; Degree 1.)

Check it with numbers

Hold $y' = 1.5$ and $y = 0.25$ fixed, so $(y')^2 + 2y = 2.25 + 0.5 = 2.75$ ⁠. The left side is $y'' + 2.75$ ⁠. At $y'' = 0, 1, 2$ it is $2.75$ ⁠, $3.75$ ⁠, $4.75$ ⁠: equal steps of 1, so $y''$ enters to the first power, and the degree is 1.

Common mistakes

Answering "degree 2". The square is on $y'$ ⁠, not on the highest derivative $y''$ ⁠.
Reading $(y')^2$ as $y''$ ⁠. A power outside the bracket is not an extra dash: $(y')^2 = y' \cdot y'$ ⁠.
Letting $2y$ count. $y$ is not a derivative; its coefficient and power never matter here.

What to take away

Find the highest derivative first, then look only at its power.
A squared lower derivative makes the equation harder to solve, but it does not change the degree.

Every fact used, and where it comes from

Fact	Used in	Source
$y'$ ⁠, $y''$ are the first and second derivatives	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	Step 3	Section 9.2.2
Degree = highest power of the highest order derivative	Step 4	Section 9.2.2
An equation with $\frac{d^2y}{dx^2}$ and $\left(\frac{dy}{dx}\right)^2$ can have degree 1	Step 4	Example 1 (ii)

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Exercise 9.1 · Q9: Order and degree of y′′+(y′)2+2y=0y'' + (y')^2 + 2y = 0y′′+(y′)2+2y=0

Basics you need

Solution

List the derivatives and their powers

Read off the order

Is it a polynomial equation in its derivatives?

Read off the degree

Every fact used, and where it comes from

Exercise 9.1 · Q9: Order and degree of $y'' + (y')^2 + 2y = 0$