Chapter 9 · Differential Equations

Exercise 9.1 · Q2: Order and degree of $y' + 5y = 0$

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.

2. $y' + 5y = 0$

Exam answer

The highest order derivative present in the equation is $y'$ ⁠, so its order is 1.

The equation is a polynomial equation in its derivative $y'$ ⁠.

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

$\therefore$ Order 1; degree 1.

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

The one idea this problem is built on

The only derivative in $y' + 5y = 0$ is $y' = \frac{dy}{dx}$ ⁠, so the order is 1. The equation is a polynomial in $y'$ (it is $y'$ plus a term with no derivative at all), so the degree is defined, and it is the power of $y'$ ⁠, which is 1. This is a first order, first degree equation, the kind this chapter later learns to solve.

Given, to find, and the plan

Given: The differential equation $y' + 5y = 0$ ⁠, 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) Read off the power of the highest derivative.

Basics you need

1The first derivative

$y' = \frac{dy}{dx}$ is the slope of $y$ ⁠: the first derivative of $y$ with respect to $x$ (Note 1, Section 9.2). $y$ itself is not a derivative. 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 in the equation

The only derivative in $y' + 5y = 0$ is $y' = \dfrac{dy}{dx}$ ⁠, of order 1.

The term $5y$ contains $y$ itself, which is not a derivative.

Order and degree are about the derivatives of the dependent variable $y$ ⁠. The variables $x$ and $y$ themselves do not count as derivatives.

Notation $y' = \frac{dy}{dx}$ ⁠: Note 1, Section 9.2. Basics: y′ and y″
Read off the order

The highest order derivative present is $y'$ ⁠. So
$\text{order} = 1.$
Section 9.2.1 defines the order as the order of the highest order derivative in the equation. With only one derivative, that is the one.

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

Write $p = y'$ ⁠. The equation reads $p + 5y = 0$ ⁠.

This is a polynomial in $p$ ⁠: the term $p$ has power 1, and $5y$ is a term with no $p$ in it (power 0). 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. The test is about the derivatives only: how $y$ or $x$ appear does not matter for it.

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

The highest order derivative is $y'$ ⁠, and its highest power in the equation is 1. So
$\text{degree} = 1.$
By Section 9.2.2, once the equation is a polynomial in its derivatives, the degree is the highest power of the highest order derivative. Here $y'$ appears only as $y'$ ⁠, that is $(y')^1$ ⁠.

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

Result

The equation $y' + 5y = 0$ has order 1 and degree 1. It is a polynomial equation in its derivative $y'$ ⁠, so the degree is defined.

(Textbook answer: Order 1; Degree 1.)

Check it with numbers

Hold $y = 0.4$ fixed, so $5y = 2$ ⁠. The left side is $y' + 2$ ⁠. At $y' = 0, 1, 2, 3$ it is $2$ ⁠, $3$ ⁠, $4$ ⁠, $5$ ⁠: equal steps of 1 each time $y'$ goes up by 1. Equal steps mean $y'$ enters to the first power, so the degree is 1.

A solution also shows the equation is a sensible first order one. Take $y = e^{-5x}$ at $x = 0.1$ ⁠: $y = e^{-0.5} \approx 0.606531$ and $y' = -5e^{-0.5} \approx -3.032653$ ⁠, so $y' + 5y \approx -3.032653 + 3.032653 = 0$ ⁠.

Common mistakes

Counting $y$ as a derivative. $5y$ has no derivative in it; it does not affect the order or the degree.
Answering 0 for the order or the degree. Order and degree (if defined) are always positive integers (Note, Section 9.2.2).
Mixing up degree and order. The order is which derivative (here the first); the degree is its power (here 1).

What to take away

An equation like $y' + ky = 0$ is first order and first degree.
Terms with no derivative (in $x$ or $y$ only) never change the order or the degree.

Every fact used, and where it comes from

Fact	Used in	Source
$y' = \frac{dy}{dx}$	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
Order and degree (if defined) are positive integers	Common mistakes	Note, Section 9.2.2

Found an error, or want more?

Tell us if something on this page is wrong or unclear, or if you would like a fuller explanation, another method or a new feature. Send feedback on this problem →

Exercise 9.1 · Q2: Order and degree of y′+5y=0y' + 5y = 0y′+5y=0

Basics you need

Solution

List the derivatives in the equation

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 · Q2: Order and degree of $y' + 5y = 0$