Chapter 9 · Differential Equations

Exercise 9.1 · Q4: Order and degree of $\left(\frac{d^2y}{dx^2}\right)^2 + \cos\frac{dy}{dx} = 0$

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.

4. $\left(\dfrac{d^2y}{dx^2}\right)^2 + \cos\left(\dfrac{dy}{dx}\right) = 0$

Exam answer

The highest order derivative present in the equation is $\dfrac{d^2y}{dx^2}$ ⁠, so its order is 2.

The equation contains the term $\cos\left(\dfrac{dy}{dx}\right)$ ⁠. So it is not a polynomial equation in its derivatives $\dfrac{d^2y}{dx^2}$ and $\dfrac{dy}{dx}$ ⁠.

Hence its degree is not defined.

$\therefore$ Order 2; degree not defined.

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

The one idea this problem is built on

The highest derivative is $\frac{d^2y}{dx^2}$ ⁠, so the order is 2. The degree needs a polynomial equation in the derivatives, and $\cos\left(\frac{dy}{dx}\right)$ is not a polynomial in $\frac{dy}{dx}$ ⁠. So the degree is not defined. The square on $\frac{d^2y}{dx^2}$ looks like an answer, but it is never reached, because the polynomial condition fails first.

Given, to find, and the plan

Given: The differential equation $\left(\frac{d^2y}{dx^2}\right)^2 + \cos\left(\frac{dy}{dx}\right) = 0$ ⁠, 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

$\frac{dy}{dx} = y'$ is the first derivative and $\frac{d^2y}{dx^2} = y''$ the second (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 $\dfrac{d^2y}{dx^2}$ (order 2), squared, and $\dfrac{dy}{dx}$ (order 1), inside a cosine. There is no other derivative.

Order and degree are both about the derivatives of $y$ with respect to $x$ ⁠, so we first find all of them, including the one inside the cosine.

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

The highest order derivative present is $\dfrac{d^2y}{dx^2}$ ⁠. So
$\text{order} = 2.$
Section 9.2.1: the order is the order of the highest order derivative. Its power (here 2) does not affect the order; the order would be 2 even without the square.

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

Write $p = \frac{dy}{dx}$ and $q = \frac{d^2y}{dx^2}$ ⁠. The equation reads $q^2 + \cos p = 0$ ⁠.

The term $q^2$ is a polynomial in $q$ ⁠. But $\cos p$ 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 $\cos p$ always lies between $-1$ and $1$ ⁠, so that polynomial would have to be a constant. And $\cos p$ is not constant: $\cos 0 = 1$ but $\cos \pi = -1$ ⁠. This contradiction shows that $\cos p$ 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 (Section 9.2.2), so this test comes first. It is the same situation as the book's equation (11), $\frac{dy}{dx} + \sin\left(\frac{dy}{dx}\right) = 0$ ⁠, with cosine in place of sine.

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 does not help that the bad term involves only the lower derivative $\frac{dy}{dx}$ ⁠. The whole equation must be a polynomial in all its derivatives before the power of the highest one is read off.

The definition of degree in Section 9.2.2 has two parts: first the polynomial condition for the whole equation, then the highest power of the highest order derivative. The first part fails here, so the square on $\frac{d^2y}{dx^2}$ is never used.

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

Result

The equation $\left(\frac{d^2y}{dx^2}\right)^2 + \cos\left(\frac{dy}{dx}\right) = 0$ has order 2, and its degree is not defined, because it is not a polynomial equation in its derivatives.

(Textbook answer: Order 2; Degree not defined.)

Check it with numbers

Give $\frac{dy}{dx}$ larger and larger values: $\cos 10 \approx -0.8391$ ⁠, $\cos 100 \approx 0.8623$ ⁠, $\cos 1000 \approx 0.5624$ ⁠. The values stay between $-1$ and $1$ ⁠, while a polynomial that is not constant would run off to very large sizes. This is Step 3's argument in numbers.

Now hold $\frac{dy}{dx} = 0$ ⁠, so $\cos 0 = 1$ ⁠. The left side is $\left(\frac{d^2y}{dx^2}\right)^2 + 1$ ⁠. At $\frac{d^2y}{dx^2} = 0, 1, 2, 3$ it is $1, 2, 5, 10$ ⁠. The steps are $1, 3, 5$ ⁠, and the steps of those are $2, 2$ ⁠: constant second steps, the mark of a square. That square is where the wrong answer "degree 2" comes from.

Common mistakes

Answering "degree 2". The square on $\frac{d^2y}{dx^2}$ only counts if the whole equation is a polynomial in its derivatives, and $\cos\left(\frac{dy}{dx}\right)$ breaks that.
Ignoring a bad term because it holds a lower derivative. Any derivative inside cos spoils the polynomial condition, whichever derivative it is.
Changing the order because of the cosine. The order is still 2; only the degree is affected.

What to take away

Check for a derivative inside sin, cos, tan, log or e before reading any power.
The order is always defined; the degree may not be. Compare Exercise 9.1 Q1, with a derivative inside sine.

Every fact used, and where it comes from

Fact	Used in	Source
$\frac{dy}{dx} = y'$ ⁠, $\frac{d^2y}{dx^2} = y''$	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
$-1 \le \cos p \le 1$ for every real $p$	Step 3	Trigonometric functions (Class XI)

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Exercise 9.1 · Q4: Order and degree of (d2ydx2)2+cos⁡dydx=0\left(\frac{d^2y}{dx^2}\right)^2 + \cos\frac{dy}{dx} = 0(dx2d2y​)2+cosdxdy​=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 · Q4: Order and degree of $\left(\frac{d^2y}{dx^2}\right)^2 + \cos\frac{dy}{dx} = 0$