Chapter 9 · Differential Equations

Exercise 9.1 · Q8: Order and degree of $y' + y = e^x$

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.

8. $y' + y = e^x$

Exam answer

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

The equation can be written as $y' + y - e^x = 0$ ⁠. The term $e^x$ contains no derivative, so this is a polynomial equation in $y'$ ⁠.

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 is $y'$ ⁠, so the order is 1. The exponential $e^x$ is a function of $x$ ⁠, not of a derivative, so it does not stop the equation from being a polynomial in $y'$ ⁠. The degree is the power of $y'$ ⁠, which is 1.

Given, to find, and the plan

Given: The differential equation $y' + y = e^x$ ⁠, 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) Bring everything to one side and test whether it 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 first derivative of $y$ with respect to $x$ (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

Solution

List the derivatives in the equation

The only derivative is $y'$ ⁠, of order 1. The terms $y$ and $e^x$ are not derivatives.

Order and degree are about the derivatives of $y$ ⁠. $e^x$ is a function of the independent variable alone.

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: the order is the order of the highest order derivative in the equation.

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

Bring every term to the left and write $p = y'$ ⁠:
$p + y - e^x = 0.$
As a polynomial in $p$ ⁠, this has the term $p$ (power 1) and the term $y - e^x$ ⁠, which has no $p$ in it (power 0). So the equation is a polynomial equation in its derivatives, and its degree is defined.

An exponential spoils the polynomial condition only when a derivative sits in it, as $e^{y'}$ does in the book's Example 1 (iii). Here the exponent is $x$ ⁠, which is not a derivative.

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, the degree is the highest power of the highest order derivative.

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

Result

The equation $y' + y = e^x$ has order 1 and degree 1. The exponential is of $x$ ⁠, not of a derivative, so it is a polynomial equation in its derivatives.

(Textbook answer: Order 1; Degree 1.)

Check it with numbers

Take $x = 0.5$ ⁠, so $e^{0.5} \approx 1.648721$ ⁠, and hold $y = 0.2$ ⁠. The left side minus the right side is $y' + 0.2 - 1.648721 = y' - 1.448721$ ⁠. At $y' = 0, 1, 2$ it is $-1.448721$ ⁠, $-0.448721$ ⁠, $0.551279$ ⁠: equal steps of 1, so $y'$ enters to the first power.

A solution: $y = \frac12 e^x$ has $y' = \frac12 e^x$ ⁠, so $y' + y = e^x$ ⁠, as required.

Common mistakes

Saying "degree not defined" because of $e^x$ ⁠. An exponential spoils the polynomial condition only when a derivative sits in it, like $e^{y'}$ ⁠.
Counting $y$ as a derivative. $y$ has no dash; the only derivative is $y'$ ⁠.
Answering 0 for the degree. Order and degree (if defined) are positive integers (Note, Section 9.2.2).

What to take away

$e^x$ ⁠, $\sin x$ ⁠, $\log x$ ⁠: functions of $x$ never affect the order or the degree.
$y' + y = e^x$ is first order, first degree; it is the linear type solved later in this chapter.

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
$y''' + y^2 + e^{y'} = 0$ has no degree	Step 3	Example 1 (iii)
Degree = highest power of the highest order derivative	Step 4	Section 9.2.2

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Exercise 9.1 · Q8: Order and degree of y′+y=exy' + y = e^xy′+y=ex

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 · Q8: Order and degree of $y' + y = e^x$