Chapter 9 · Differential Equations

Exercise 9.1 · Q10: Order and degree of $y'' + 2y' + \sin y = 0$

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.

10. $y'' + 2y' + \sin y = 0$

Exam answer

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

The term $\sin y$ contains no derivative. So 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 sine here is a sine of $y$ ⁠, and $y$ is not a derivative. The polynomial condition is about the derivatives only, so the equation is a polynomial in $y''$ and $y'$ (though not in $y$ ⁠), the degree is defined, and it is the power of $y''$ ⁠, which is 1.

Given, to find, and the plan

Given: The differential equation $y'' + 2y' + \sin y = 0$ ⁠, 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, looking at what sits inside the sine. (4) Read off the power of the highest derivative.

Basics you need

1Derivatives versus the variable

$y'$ and $y''$ are the first and second derivatives of $y$ (Note 1, Section 9.2). $y$ itself, with no dash, is the dependent variable, 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 derivatives present are $y''$ (order 2) and $y'$ (order 1, with coefficient 2), each to the first power. The term $\sin y$ is a sine of $y$ itself; there is no derivative inside it.

Whether a function such as sine is harmless depends on what is inside it. So we look inside: here it is $y$ ⁠, with no dash.

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 in the equation.

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
$q + 2p + \sin y = 0.$
As a polynomial in $p$ and $q$ ⁠, this has the terms $q$ and $2p$ ⁠, and the term $\sin y$ ⁠, which contains neither $p$ nor $q$ (power 0). So the equation is a polynomial equation in its derivatives, and its degree is defined.

The condition in Section 9.2.2 asks for a polynomial in the derivatives $y'$ ⁠, $y''$ ⁠, $\ldots$ ⁠, not in $y$ ⁠. The book says this directly about its equation (10), $\left(\frac{dy}{dx}\right)^2 + \frac{dy}{dx} - \sin^2 y = 0$ ⁠: it is a polynomial equation in $y'$ ⁠, "not a polynomial in $y$ though", and its degree is defined.

Condition for the degree, and equation (10), 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 $y'' + 2y' + \sin y = 0$ has order 2 and degree 1. The sine is of $y$ ⁠, not of a derivative, so it is a polynomial equation in its derivatives.

(Textbook answer: Order 2; Degree 1.)

Check it with numbers

Hold $y = 0.5$ and $y' = 0.3$ fixed: $\sin 0.5 \approx 0.479426$ and $2y' = 0.6$ ⁠, so the left side is $y'' + 1.079426$ ⁠. At $y'' = 0, 1, 2$ it is $1.079426$ ⁠, $2.079426$ ⁠, $3.079426$ ⁠: equal steps of 1, so $y''$ enters to the first power, and the degree is 1.

Common mistakes

Saying "degree not defined" because of the sine. That rule is for a derivative inside sin, cos, tan, log or e. Here only $y$ is inside.
Mixing this up with Exercise 9.1 Q4. There the cosine holds $\frac{dy}{dx}$ ⁠, so the degree is not defined; here the sine holds $y$ ⁠, so it is defined.
Thinking "not a polynomial in $y$ " matters. It does not: the condition is only about the derivatives.

What to take away

Always look inside the function: $\sin y$ is fine, $\sin y'$ is not.
The degree needs a polynomial in the derivatives, not in $y$ or $x$ ⁠.

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
A polynomial in $y'$ need not be a polynomial in $y$	Step 3	Equation (10), Section 9.2.2
Degree = highest power of the highest order derivative	Step 4	Section 9.2.2

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Exercise 9.1 · Q10: Order and degree of y′′+2y′+sin⁡y=0y'' + 2y' + \sin y = 0y′′+2y′+siny=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 · Q10: Order and degree of $y'' + 2y' + \sin y = 0$