Classification — order, degree, linear vs nonlinear, autonomous vs non-autonomous: Visual walkthrough

Before we can classify, we must agree on what the pieces of an ODE even mean as pictures. So the first three steps build the vocabulary; the last steps do the classification.

Step 1 — What is a function, and what is a derivative? (a picture)

WHAT: We draw a curve and mark one point on it. WHY: Every symbol in the equation is a statement about this curve. If we cannot see the curve, the symbols are noise. PICTURE: In the figure, the black curve is $y (x)$ . The red dot sits at one chosen $x$ . Its height is $y$ .

Figure — Classification — order, degree, linear vs nonlinear, autonomous vs non-autonomous

The notation $\frac{d y}{d x}$ is literally "a tiny change in $y$ divided by the tiny change in $x$ that caused it" — the slope of that ramp squeezed down to a single point.

Step 2 — The second derivative $y^{''}$ as concavity

WHAT: We overlay two nearby ramps and compare their slopes. WHY: Our target equation contains $\frac{d ^{2} y}{d x ^{2}}$ . To classify it we must know which geometric feature that symbol names — it names concavity, and that it is the highest derivative present decides the order. PICTURE: The red arc shows the curve cupping upward; two black ramps at nearby points have different slopes, and $y^{''}$ measures how quickly the slope tips from one ramp to the next.

Why write $\frac{d ^{2} y}{d x ^{2}}$ and not just " $y$ twice"? Because we differentiated twice: once to get the slope $y^{'} = \frac{d y}{d x}$ , again to get the slope of the slope. The little "2"s are a tally of how many differentiations we performed — and that tally is exactly what order counts.

Step 3 — Reading the ORDER off the equation

WHAT: We scan our equation left to right and circle every derivative. $x 3 (\frac{d ^{2} y}{d x ^{2}}) + 2 (y^{'}) = sin x .$ WHY: Order tells us how many arbitrary constants the general solution carries, and therefore how many initial conditions we must be handed. It answers "how tall is this problem?" before we spend effort solving. PICTURE: We rank the derivatives on a vertical "depth ladder." $y^{'}$ sits on rung 1; $y^{''}$ sits on rung 2; the highest occupied rung is the order.

The highest occupied rung is 2 (because $\frac{d ^{2} y}{d x ^{2}}$ appears). The cube outside it changes nothing here — a power is not a differentiation. So:

Step 4 — Reading the DEGREE (the power on the top rung)

WHAT: We isolate the highest-order derivative $y^{''}$ and read its exponent. $x (\frac{d ^{2} y}{d x ^{2}})^{3} + (y^{'})^{2} = sin x .$ The little red $3$ is the exponent sitting on the order-2 derivative $y^{''}$ . Nothing needs cleaning — there is no radical hiding a power, and no derivative is trapped inside a transcendental function. So we read it straight off. WHY the "polynomial first" caveat: a power can hide. $1 + (y^{'})^{2}$ secretly contains a square; you must square-to-clear before an honest exponent is visible. And $sin (y^{'})$ can never be turned into a polynomial in $y^{'}$ (it is transcendental), so its degree is undefined — no amount of algebra exposes a single whole-number power. PICTURE: We stack copies of $y^{''}$ to show its exponent as a literal tower of height 3 — degree is the height of the tower on the top rung.

Step 5 — LINEAR vs NONLINEAR: does $y$ enter "gently"?

WHAT: We audit each term of our equation against the three rules, one term per line:

WHY it matters so much: linear equations obey the superposition principle — you can add solutions of the homogeneous equation to get new solutions, which is the engine behind integrating factors and characteristic equations. Powers like $(y^{''})^{3}$ break that additivity, so those tools are illegal and we must fall back on qualitative methods like the phase line. PICTURE: A "gentleness meter." A single straight line through the origin (output proportional to input, slope one power) is the linear picture; a curve that bends because of a squared/cubed term is nonlinear. Our terms land on the bending side.

Since two terms carry powers above one:

Step 6 — AUTONOMOUS vs NON-AUTONOMOUS: does the clock appear?

WHAT: We hunt for any explicit $x$ in our equation, ignoring the derivatives themselves. $explicit x x (\frac{d ^{2} y}{d x ^{2}})^{3} + (y^{'})^{2} = explicit x sin x .$ Two explicit $x$ 's stare back: the coefficient $x$ out front, and the $sin x$ on the right. WHY the distinction is physical: autonomous means "the landscape doesn't move with time." The slope-field arrows are identical along every horizontal shift, so any solution slid sideways is still a solution. Non-autonomous means the landscape itself changes as $x$ advances — like a rule that behaves differently in summer than winter. PICTURE: Two slope fields. In the autonomous one every column of arrows is a copy of its neighbour (no dependence on horizontal position). In ours, the red column of arrows differs from the black column because $x$ appears — the field genuinely changes as we move right.

Step 7 — Edge cases the picture must still handle

Classification must survive the weird inputs. Here are the degenerate ones and how each one reads.

Worked example Degenerate readings

Hidden power (radical). $\frac{d ^{2} y}{d x ^{2}} = 1 + (y^{'})^{2}$ . Before cleaning, degree is unreadable. Square both sides → $(\frac{d ^{2} y}{d x ^{2}})^{2} = 1 + (y^{'})^{2}$ → now honestly order 2, degree 2. The picture: the root was a locked door; squaring is the key.
Degree undefined. $y^{''} + sin (y^{'}) = 0$ . No algebra frees $y^{'}$ from the transcendental $sin$ . Order 2, degree undefined. Yet order is still perfectly readable — the depth ladder does not care about wrappers.
Linear yet non-autonomous together. $y^{'} = y + e^{x}$ is linear ( $y$ enters to first power, gently) and non-autonomous (explicit $e^{x}$ ). Linearity and autonomy are independent axes; do not let one decide the other. Contrast the autonomous linear constant-coefficient case $y^{''} + ω^{2} y = 0$ .
Nonlinear yet autonomous. $y^{'} = y (1 - y)$ (logistic): no explicit $x$ → autonomous, but $y^{2}$ → nonlinear. Its whole behaviour is read from a phase line; separation solves it, and Picard–Lindelöf guarantees the solution is unique.

Worked example Zero-coefficient collapse — a full worked example

Suppose you are handed $0 \cdot \frac{d ^{2} y}{d x ^{2}} + 3 \frac{d y}{d x} + 2 y = 0.$ WHAT: The coefficient in front of the highest-order derivative $\frac{d ^{2} y}{d x ^{2}}$ is literally $0$ , so that whole term vanishes: $0 \cdot \frac{d ^{2} y}{d x ^{2}} = 0 ⟹ 3 \frac{d y}{d x} + 2 y = 0.$ WHY simplify first: classification always reads the equation as it actually is after cancellation. Reading the original naively would call it order 2 — but there is genuinely no second derivative acting. RESULT: the surviving equation $3 y^{'} + 2 y = 0$ has highest derivative $y^{'}$ , so:

Order drops from an apparent $2$ down to $1$ .
It is a polynomial of power $1$ in $y^{'}$ → degree 1.
$y^{'}$ and $y$ enter to first power, unmultiplied, unwrapped → linear.
No explicit $x$ appears → autonomous.

Moral: a zero multiplier can quietly lower the order, so cancel before you classify.

Each of these confirms the four checks are truly independent — every combination occurs.

The one-picture summary

We collapse the entire walkthrough into a single "triage card": one glance, four labels.

Recall Feynman retelling of the whole walkthrough

Picture a mystery curve. Its steepness is $y^{'}$ ; how fast the steepness changes (which way it cups, its concavity) is $y^{''}$ . We differentiated twice to reach $y^{''}$ , so the deepest rung on our ladder is 2 — that's the order. On that top rung sits an exponent of 3 (the derivative is cubed), and since nothing was hidden inside a root or a transcendental function like sine, we read the degree straight off: 3. Then we asked whether $y$ shows up gently — first power, unmultiplied, un-wrapped. It doesn't: both $(y^{''})^{3}$ and $(y^{'})^{2}$ carry big powers, so the equation is nonlinear (superposition is off the table). Finally we asked whether the equation peeks at the clock — whether a bare $x$ appears. It does, twice (the leading $x$ and the $sin x$ ), so it's non-autonomous (its slope-field landscape shifts as you move right). Four cheap looks — depth, exponent, gentleness, clock — and the animal is fully tagged before we ever try to solve it.

Recall Quick self-check

In $x (y^{''})^{3} + (y^{'})^{2} = sin x$ , why is the order 2 and not 3? ::: The little "3" is a power, not a differentiation; order counts differentiations, and the deepest is $y^{''}$ (2). Why does $sin x$ not make the equation nonlinear? ::: Linearity inspects how $y$ enters; $sin x$ contains no $y$ at all, so it's an allowed forcing term. Give one equation that is linear but non-autonomous. ::: $y^{'} = y + e^{x}$ . What is the difference between $y^{''}$ and true curvature? ::: $y^{''}$ is the raw concavity; true curvature is the normalised $κ = y^{''} / (1 + (y^{'})^{2})^{3/2}$ .