Chapter 9 · Differential Equations

Exercise 9.1 · Q12 (MCQ): Order of $2x^2y'' - 3y' + y = 0$

Exercise 9.1Order of a differential equation · Section 9.2.1

The question

Choose the correct answer.

The order of the differential equation

2x^2\frac{d^2y}{dx^2} - 3\frac{dy}{dx} + y = 0

(A) 2 (B) 1 (C) 0 (D) not defined

Exam answer

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

Answer: (A) 2

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

The one idea this problem is built on

The order is the order of the highest derivative in the equation, and that is $\frac{d^2y}{dx^2}$ ⁠. So the order is 2, option (A). The coefficient $2x^2$ in front of it, and the power it is raised to, play no part in the order. The order is always defined, so (D) can never be right for an order question.

Given, to find, and the plan

Given: The differential equation $2x^2\frac{d^2y}{dx^2} - 3\frac{dy}{dx} + y = 0$ ⁠, with $y$ the dependent variable and $x$ the independent variable.
To find: Its order, and which option gives it.
Plan: (1) List every derivative in the equation. (2) The order is the highest of them. (3) For completeness, test the polynomial condition and find the degree. (4) Examine every option.

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). It is always a positive integer (Note, Section 9.2.2). 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 of the highest order derivative (Section 9.2.2). Lesson: order and degree

Solution

List the derivatives in the equation

The derivatives present are $\dfrac{d^2y}{dx^2}$ (order 2, with coefficient $2x^2$ ⁠) and $\dfrac{dy}{dx}$ (order 1, with coefficient $-3$ ⁠). The term $y$ is not a derivative.

The order depends only on which derivatives are present, so we list them. The coefficients $2x^2$ and $-3$ are multipliers, not derivatives.

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
$\text{order} = 2.$
The correct option is (A).

Section 9.2.1 defines the order as the order of the highest order derivative. Its coefficient $2x^2$ is a function of $x$ and does not change which derivative it is.

Definition of order, Section 9.2.1. Lesson: order and degree
For completeness: the degree

Write $p = \frac{dy}{dx}$ and $q = \frac{d^2y}{dx^2}$ ⁠. The equation reads $2x^2 q - 3p + y = 0$ ⁠, a polynomial in $p$ and $q$ (the coefficients $2x^2$ ⁠, $-3$ and the term $y$ contain no derivative). So the degree is defined, and the power of $q$ is 1: degree 1.

The question asks only for the order, but the degree explains option (B): 1 is the degree of this equation, not its order.

Definition of degree, Section 9.2.2. Lesson: Example 1, order and degree

Examine every option

Option	Where the number comes from	Verdict
(A) 2	The highest derivative is the second	Correct
(B) 1	The degree, or the power of $\frac{d^2y}{dx^2}$	Wrong: that is the degree
(C) 0	No derivative counted	Wrong: the order is a positive integer
(D) not defined	Confusing order with degree	Wrong: the order is always defined

(B) answers a different question, the degree. (C) is impossible: the equation contains derivatives, and the Note in Section 9.2.2 says order and degree (if defined) are positive integers. (D) mixes up the two ideas: only the degree can fail to be defined, and here even the degree is defined. Only (A) is the order.

Note "order and degree (if defined) are always positive integers", Section 9.2.2. Lesson: the multiple-choice questions

Result

The highest derivative in $2x^2\frac{d^2y}{dx^2} - 3\frac{dy}{dx} + y = 0$ is $\frac{d^2y}{dx^2}$ ⁠, so its order is 2: option (A). (Its degree is 1.)

(Textbook answer: A.)

Check it with numbers

The equation is a rule for the second derivative. Take $x = 0.5$ ⁠, $y = 0.2$ ⁠, $\frac{dy}{dx} = 0.4$ ⁠: then $2(0.25)\frac{d^2y}{dx^2} - 1.2 + 0.2 = 0$ ⁠, so $0.5\,\frac{d^2y}{dx^2} = 1$ and $\frac{d^2y}{dx^2} = 2$ ⁠. Knowing $x$ ⁠, $y$ and the first derivative, the equation fixes the second: it is second order.

With the same values, the left side is $0.5\,\frac{d^2y}{dx^2} - 1$ ⁠. At $\frac{d^2y}{dx^2} = 0, 1, 2$ it is $-1$ ⁠, $-0.5$ ⁠, $0$ ⁠: equal steps of 0.5, so the second derivative enters to the first power (the degree 1 of Step 3).

Common mistakes

Answering with the degree. 1 is the power of the highest derivative (the degree), not its order.
Letting $2x^2$ count. The 2 in $x^2$ is a power of $x$ ⁠, which is not a derivative; it has nothing to do with the order.
Choosing "not defined". The order of a differential equation is always defined.

What to take away

Order: the highest derivative present. Coefficients, even ones with $x$ in them, never change it.
"Not defined" can only be the answer to a degree question, as in Exercise 9.1 Q11.

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	Step 3	Section 9.2.2
Order and degree (if defined) are positive integers	Step 4	Note, Section 9.2.2

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Exercise 9.1 · Q12 (MCQ): Order of 2x2y′′−3y′+y=02x^2y'' - 3y' + y = 02x2y′′−3y′+y=0

Basics you need

Solution

List the derivatives in the equation

Read off the order

For completeness: the degree

Examine every option

Every fact used, and where it comes from

Exercise 9.1 · Q12 (MCQ): Order of $2x^2y'' - 3y' + y = 0$