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Chapter 9 · Differential Equations

Exercise 9.1 · Q7: Order and degree of y′′′+2y′′+y′=0y''' + 2y'' + 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.

7. y′′′+2y′′+y′=0y''' + 2y'' + y' = 0

Exam answer

The highest order derivative present in the equation is y′′′y'''⁠, so its order is 3.

The equation is a polynomial equation in its derivatives y′′′y'''⁠, y′′y'' and y′y'⁠.

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

∴\therefore Order 3; 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′′′y'''⁠, so the order is 3. Every derivative appears to the first power and nothing sits inside a function, so the equation is a polynomial in its derivatives. The degree is the power of y′′′y'''⁠, which is 1. The coefficient 2 in 2y′′2y'' is a number in front, not a power.

Given, to find, and the plan
Given
The differential equation y′′′+2y′′+y′=0y''' + 2y'' + y' = 0⁠, with yy the dependent variable and xx 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

1Derivatives and their orders

y′y'⁠, y′′y''⁠, y′′′y''' are the first, second and third derivatives of yy with respect to xx (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

  1. List the derivatives in the equation

    The derivatives present are y′′′y''' (order 3), y′′y'' (order 2, with coefficient 2) and y′y' (order 1). Each appears to the first power.

    Order and degree are about the derivatives of yy⁠, so we first list them with their orders and powers.
    Notation y′y'⁠, y′′y''⁠, y′′′y'''⁠: Note 1, Section 9.2. Basics: y′ and y″

  2. Read off the order

    The highest order derivative present is y′′′y'''⁠. So

    order=3.\text{order} = 3.
    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

  3. Is it a polynomial equation in its derivatives?

    Write p=y′p = y'⁠, q=y′′q = y''⁠, r=y′′′r = y'''⁠. The equation reads

    r+2q+p=0,r + 2q + p = 0,

    a polynomial (in fact a first degree one) in pp⁠, qq⁠, rr⁠. 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 (Section 9.2.2). Number coefficients such as 2 are always allowed.
    Condition for the degree, Section 9.2.2. Lesson: Example 1, order and degree

  4. Read off the degree

    The highest order derivative is y′′′y'''⁠, and its highest power in the equation is 1. So

    degree=1.\text{degree} = 1.
    By Section 9.2.2, the degree is the highest power of the highest order derivative. y′′′y''' appears only as y′′′y'''⁠, that is (y′′′)1(y''')^1⁠.
    Definition of degree, Section 9.2.2. Lesson: Exercise 9.1 player
Result

The equation y′′′+2y′′+y′=0y''' + 2y'' + y' = 0 has order 3 and degree 1. It is a polynomial equation in its derivatives, so the degree is defined.

(Textbook answer: Order 3; Degree 1.)

Check it with numbers

Hold y′′=0.35y'' = 0.35 and y′=0.4y' = 0.4 fixed, so 2y′′+y′=0.7+0.4=1.12y'' + y' = 0.7 + 0.4 = 1.1⁠. The left side is y′′′+1.1y''' + 1.1⁠. At y′′′=0,1,2y''' = 0, 1, 2 it is 1.11.1⁠, 2.12.1⁠, 3.13.1⁠: equal steps of 1, so y′′′y''' enters to the first power.

A solution shows the order at work. y=e−xy = e^{-x} has y′=−e−xy' = -e^{-x}⁠, y′′=e−xy'' = e^{-x}⁠, y′′′=−e−xy''' = -e^{-x}⁠, so y′′′+2y′′+y′=(−1+2−1)e−x=0y''' + 2y'' + y' = (-1 + 2 - 1)e^{-x} = 0⁠. In fact y=A+(B+Cx)e−xy = A + (B + Cx)e^{-x} solves it for all constants AA⁠, BB⁠, CC⁠: three arbitrary constants for an equation of order 3.

Common mistakes
  1. Reading the coefficient 2 as a power. 2y′′2y'' is twice y′′y''⁠, not y′′y'' squared; and it is not the highest derivative anyway.
  2. Mixing up order and degree. Order 3 (three dashes), degree 1 (power on y′′′y'''⁠).
  3. Expecting yy to appear. An equation need not contain yy itself; the order and degree come from the derivatives present.
What to take away
  1. A sum of derivatives, each to the first power, with number coefficients, always has degree 1.
  2. The order tells how many arbitrary constants a general solution has: here 3. Basics: one constant per order

Every fact used, and where it comes from

FactUsed inSource
y′y'⁠, y′′y''⁠, y′′′y''' are the first, second, third derivativesStep 1Note 1, Section 9.2
Order = order of the highest order derivativeStep 2Section 9.2.1 (Lesson: order and degree)
Degree is defined only for a polynomial equation in the derivativesStep 3Section 9.2.2
Degree = highest power of the highest order derivativeStep 4Section 9.2.2
A general solution has as many arbitrary constants as the orderWhat to take awaySummary of Chapter 9; Section 9.3 (Lesson: general and particular solutions)
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