Classification — order, degree, linear vs nonlinear, autonomous vs non-autonomous: Question bank

Before we start, three vocabulary anchors so nothing here uses an unexplained word:

Definition The words we test

Order = which derivative is the deepest (a $y^{''}$ beats a $y^{'}$ ). Think "how many nested rates."
Degree = the power on that deepest derivative, but only after you've cleaned the equation into a polynomial in all the derivatives. "Cleaned" is precise: (a) clear every denominator that contains a derivative (multiply through), and (b) remove every radical sign over a derivative (raise both sides to the matching power), and only then read the exponent on the deepest derivative. If no such polynomial form exists (e.g. a $sin$ or $ln$ of a derivative survives), the degree is undefined.
Linear = the unknown $y$ and its derivatives enter gently: first power, never multiplied together, never wrapped in a nonlinear function. The coefficients in front (functions of the independent variable) may be as ugly as they like.
Autonomous = the independent variable never appears by itself on the right-hand side; the rule depends only on where you are, not on what the clock reads.

Spot the error

Each line contains a plausible-but-wrong claim. Explain the flaw.

Claim: "

x^{2} y^{''} + (sin x) y = 0

is nonlinear because of the

sin x

Wrong.

sin x

is a coefficient — a function of the independent variable — which linearity always permits. The forbidden case is

sin y

(a nonlinear function of the unknown). This one is linear.

Claim: "

(y^{'''})^{2} + y = 0

has order 2 because of the square."

Wrong. The square is the degree (power

2

), not the order. The deepest derivative is

y^{'''}

, so the order is

3

; degree is

2

Claim: "

y^{'} = sin y

is non-autonomous because of the sine."

Wrong. The sine wraps

y

, not the independent variable; no

t

(or

x

) appears explicitly, so it is autonomous. (It is nonlinear, because

sin y

breaks the no-nonlinear-function rule.)

Claim: "

y y^{'} = 1

is linear — both

y

and

y^{'}

appear only to the first power."

Wrong. Rule 2 also forbids the unknown and its derivatives being multiplied together. The product

y \cdot y^{'}

makes it nonlinear even though each factor is first power.

Claim: "

y^{''} + sin (y^{'}) = 0

has degree 1 since

y^{'}

appears once."

Wrong.

sin (y^{'})

is not a polynomial in

y^{'}

and no clearing of denominators or radicals removes it, so the degree is undefined. (Order is still

2

Claim: "

y^{'} = x y

is autonomous because it looks simple."

Wrong. The independent variable

x

appears explicitly as a factor on the right, so it is non-autonomous. Simplicity is irrelevant to the autonomy test.

Claim: "Since

y^{''} + ω^{2} y = 0

is autonomous, its arrows in the slope field must be flat lines."

Wrong. Autonomy means the rule is time-independent, so the slope pattern repeats along every horizontal (

t

) line — not that solutions are constant. This one oscillates (sine and cosine).

Claim: "

e^{y^{'}} + y = x

can be made polynomial by taking logs, so its degree is 1."

Wrong. Taking a log doesn't clear the exponential into a polynomial in

y^{'}

; the dependence on

y^{'}

is transcendental, so degree is undefined.

Claim: "

\frac{1}{y ^{''}} + y = x

has undefined degree because of the fraction."

Wrong. The rule says clear denominators containing a derivative: multiply through by

y^{''}

to get

1 + y y^{''} = x y^{''}

, a polynomial in derivatives. Degree is

1

(power of

y^{''}

); it is, however, nonlinear because of the product

y y^{''}

Why questions

These probe the reason behind each check, matching the parent's "WHY" emphasis. Two carry a step figure so the intuition is seen, not just told.

Why must we clear radicals and denominators before reading off the degree?

A radical or fraction hides the true power:

1 + (y^{'})^{2}

shows no honest exponent. Squaring (and clearing any denominator) exposes the polynomial, so degree is read only from the cleaned form.

Why does linearity, not autonomy, decide whether superposition applies?

Superposition needs the machine

L

to distribute over sums and scalars —

L [c_{1} y_{1} + c_{2} y_{2}] = c_{1} L [y_{1}] + c_{2} L [y_{2}]

— which is exactly linearity. Autonomy is about explicit-

t

dependence and never touches this distributivity. Figure below.

Why does the order tell you how many initial conditions to supply? ::: Undoing each derivative by integration introduces one free constant; an order- $n$ equation hides $n$ constants, so you need $n$ conditions to pin down a unique solution. Why can autonomous equations be studied with a phase line while non-autonomous ones generally can't? ::: In an autonomous $y^{'} = f (y)$ the slope depends only on $y$ , so the slope field's arrows are identical along every vertical column — you can collapse the whole picture onto one $y$ -axis (the phase line) and read the flow. With explicit $t$ the arrows change column by column, so a static 1-D picture no longer captures the motion — see Phase Line and Equilibria. Figure below.
Why is linearity called "the most important split"? ::: Linear ODEs obey superposition and have a complete solution theory (integrating factors, characteristic equations); nonlinear ones generally don't, so this single label most changes which tools are legal, e.g. Separable Equations vs Second-Order Linear ODEs with Constant Coefficients. Why don't ugly coefficients like $x^{2}$ or $e^{x}$ break linearity? ::: Linearity is a statement about the unknown $y$ : it must enter to first power, unmultiplied, unwrapped. Coefficients are the "landscape" the equation lives in and may be any function of the independent variable. Why are autonomous solutions "time-translates" of one another? ::: If the rule $f (y)$ never mentions the independent variable, then shifting a solution $y (t)$ to $y (t + c)$ still satisfies the same rule, because the equation can't tell the clock was moved. So every horizontal shift of a solution is again a solution — visible as identical arrow-columns in the figure above.

Edge cases

Boundary and degenerate inputs the naive checker forgets.

Classify the "equation"

y^{'} = 0

Order

1

, degree

1

, linear (a linear equation with zero forcing, i.e. homogeneous

g = 0

), and autonomous (no explicit independent variable). Its solutions are all constants — every value of

y

is an equilibrium.

Is a purely algebraic equation like

y^{2} = x

(no derivatives) an ODE at all?

No. With no derivative present there is no "highest derivative," so order is undefined and it isn't a differential equation — it's an algebraic relation.

What are the order and degree of

(\frac{d ^{2} y}{d x ^{2}})^{1/3} = y

Cube both sides first to clear the fractional power:

(\frac{d ^{2} y}{d x ^{2}}) = y^{3}

. Now order

2

, degree

1

. The exponent

1/3

was a disguise, not a genuine cube-root of a derivative.

y^{''} + ω^{2} y = 0

linear even though

ω^{2}

looks like a power?

Yes.

ω

is a constant, not the unknown;

ω^{2}

is just a number multiplying

y

. Only powers of

y

or its derivatives threaten linearity.

Classify

y^{'} + y = e^{t}

: linear and autonomous, or something else?

Linear and non-homogeneous (

g (t) = e^{t}

) but non-autonomous, because

e^{t}

makes the independent variable appear explicitly. Linear and autonomous are separate axes.

A "degree undefined" equation — can it still have a well-defined order?

Yes.

sin (y^{''}) + y = 0

has undefined degree yet its deepest derivative is clearly

y^{''}

, so order is

2

. Order survives even when degree collapses.

Can an equation be simultaneously nonlinear, non-autonomous, and of degree greater than one?

Yes — all four labels are independent.

x (y^{''})^{3} + (y^{'})^{2} = sin x

is order

2

, degree

3

, nonlinear (higher powers of derivatives), and non-autonomous (explicit

x

Recall One-line self-test before you leave

Cover the answers and re-derive: for $y y^{''} = sin x$ — order? degree? linear? autonomous? homogeneous? ::: Order $2$ , degree $1$ (the $y^{''}$ appears to first power), nonlinear (product $y \cdot y^{''}$ ), non-autonomous (explicit $x$ in $sin x$ ), and non-homogeneous ( $g (x) = sin x \neq = 0$ ).

Question bank — Classification — order, degree, linear vs nonlinear, autonomous vs non-autonomous

True or false — justify

Spot the error

Why questions

Edge cases