Chapter 9 · Differential Equations

Exercise 9.1 · Q11 (MCQ): Degree of $(y'')^3 + (y')^2 + \sin y' + 1 = 0$

Exercise 9.1Degree of a differential equation · Section 9.2.2

The question

Choose the correct answer.

The degree of the differential equation

\begin{aligned} &\left(\frac{d^2y}{dx^2}\right)^3 + \left(\frac{dy}{dx}\right)^2 \\ &\quad + \sin\left(\frac{dy}{dx}\right) + 1 = 0 \end{aligned}

(A) 3 (B) 2 (C) 1 (D) not defined

Exam answer

The equation contains the term $\sin\left(\dfrac{dy}{dx}\right)$ ⁠, so it is not a polynomial equation in its derivatives.

Hence its degree is not defined.

Answer: (D) not defined

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

The one idea this problem is built on

Before reading any power, ask whether the equation is a polynomial in its derivatives. Here $\frac{dy}{dx}$ sits inside a sine, and $\sin\left(\frac{dy}{dx}\right)$ is not a polynomial in $\frac{dy}{dx}$ ⁠. So the degree is not defined, option (D). The cube on $\frac{d^2y}{dx^2}$ is the trap: it would be the degree only if the sine term were not there.

Given, to find, and the plan

Given: The differential equation $\left(\frac{d^2y}{dx^2}\right)^3 + \left(\frac{dy}{dx}\right)^2 + \sin\left(\frac{dy}{dx}\right) + 1 = 0$ ⁠, with $y$ the dependent variable and $x$ the independent variable.
To find: Its degree, and which option gives it.
Plan: (1) List every derivative and how it appears. (2) Test whether the equation is a polynomial equation in its derivatives. (3) Decide 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″

2Degree, 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

3What 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 and how they appear

$\left(\dfrac{d^2y}{dx^2}\right)^3$ ⁠: the second derivative, cubed.

$\left(\dfrac{dy}{dx}\right)^2$ ⁠: the first derivative, squared.

$\sin\left(\dfrac{dy}{dx}\right)$ ⁠: the first derivative inside a sine.

The highest order derivative is $\frac{d^2y}{dx^2}$ ⁠, so the order is 2 (not asked, but it names the derivative whose power would be the degree).

The degree depends on how every derivative appears, so each term is read separately, including the one inside the sine.

Notation for derivatives: Note 1, Section 9.2; order: Section 9.2.1. Basics: y′ and y″
Is it a polynomial equation in its derivatives?

Write $p = \frac{dy}{dx}$ and $q = \frac{d^2y}{dx^2}$ ⁠. The equation reads
$q^3 + p^2 + \sin p + 1 = 0.$
The terms $q^3$ ⁠, $p^2$ and $1$ are polynomial terms, but $\sin p$ 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 $\sin p$ always lies between $-1$ and $1$ ⁠, so that polynomial would have to be a constant. And $\sin p$ is not constant: $\sin 0 = 0$ but $\sin\frac{\pi}{2} = 1$ ⁠. This contradiction shows that $\sin p$ 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, so this test comes first. It matches the book's equation (11) in Section 9.2.2, $\frac{dy}{dx} + \sin\left(\frac{dy}{dx}\right) = 0$ ⁠, whose degree is not defined.

Condition for the degree, Section 9.2.2. Lesson: Example 1, a derivative inside a function
Decide the degree

Since the equation is not a polynomial equation in its derivatives,
$\text{degree: not defined}.$
The correct option is (D).

The definition in Section 9.2.2 reads the highest power of the highest order derivative only after the polynomial condition holds. Here it fails, so no power is read.

Definition of degree, Section 9.2.2. Lesson: the multiple-choice questions

Examine every option

Option	Where the number comes from	Verdict
(A) 3	The power of $\frac{d^2y}{dx^2}$	Wrong: would be the degree only without $\sin\left(\frac{dy}{dx}\right)$
(B) 2	The power of $\frac{dy}{dx}$ ⁠, or the order	Wrong: neither is the degree
(C) 1	Reading $\sin\left(\frac{dy}{dx}\right)$ as a first-power term	Wrong: a sine is not a power
(D) not defined	The polynomial condition fails	Correct

(A) is the most tempting. If the term

\sin\left(\frac{dy}{dx}\right)

were removed, the equation

\left(\frac{d^2y}{dx^2}\right)^3 + \left(\frac{dy}{dx}\right)^2 + 1 = 0

would be a polynomial equation with degree 3; the sine term is exactly what changes the answer. (B) confuses the degree with the power of a lower derivative, or with the order 2. (C) has no source in the definition at all. Only (D) follows the definition.

Order and degree (if defined) are positive integers: Note, Section 9.2.2. Lesson: Exercise 9.1 player

Result

The equation is not a polynomial equation in its derivatives, because of $\sin\left(\frac{dy}{dx}\right)$ ⁠. So its degree is not defined: option (D). (Its order is 2.)

(Textbook answer: D.)

Check it with numbers

Give $\frac{dy}{dx}$ larger and larger values: $\sin 5 \approx -0.9589$ ⁠, $\sin 50 \approx -0.2624$ ⁠, $\sin 500 \approx -0.4678$ ⁠. The values stay between $-1$ and $1$ ⁠, as Step 2 says; a polynomial that is not constant would not stay bounded.

Now see where (A) comes from. Hold $\frac{dy}{dx} = 0$ ⁠, so $(0)^2 + \sin 0 = 0$ ⁠, and the left side is $\left(\frac{d^2y}{dx^2}\right)^3 + 1$ ⁠. At $\frac{d^2y}{dx^2} = 0, 1, 2, 3, 4$ it is $1, 2, 9, 28, 65$ ⁠. The steps are $1, 7, 19, 37$ ⁠, their steps $6, 12, 18$ ⁠, and the next steps $6, 6$ ⁠: constant third steps, the mark of a cube. That cube is real, but the sine term means it is not the degree.

Common mistakes

Choosing (A) by reading the power first. Always test the polynomial condition before reading any power.
Thinking the bad term must hold the highest derivative. Any derivative inside sin, cos, tan, log or e, even a lower one, makes the degree undefined.
Confusing degree and order. The order (2) is always defined; only the degree fails here.

What to take away

In an order-and-degree MCQ, look first for a derivative inside sin, cos, tan, log or e. If there is one, the degree is "not defined".
Same idea as Exercise 9.1 Q1 (sine of $y'''$ ⁠) and Exercise 9.1 Q4 (cosine of $\frac{dy}{dx}$ ⁠).

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 1	Section 9.2.1
Degree is defined only for a polynomial equation in the derivatives	Steps 2, 3	Section 9.2.2 (Lesson: order and degree)
$\frac{dy}{dx} + \sin\left(\frac{dy}{dx}\right) = 0$ has no degree	Step 2	Equation (11), Section 9.2.2
$-1 \le \sin p \le 1$ for every real $p$	Step 2	Trigonometric functions (Class XI)
Degree = highest power of the highest order derivative	Steps 3, 4	Section 9.2.2

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Exercise 9.1 · Q11 (MCQ): Degree of (y′′)3+(y′)2+sin⁡y′+1=0(y'')^3 + (y')^2 + \sin y' + 1 = 0(y′′)3+(y′)2+siny′+1=0

Basics you need

Solution

List the derivatives and how they appear

Is it a polynomial equation in its derivatives?

Decide the degree

Examine every option

Every fact used, and where it comes from

Exercise 9.1 · Q11 (MCQ): Degree of $(y'')^3 + (y')^2 + \sin y' + 1 = 0$