Worked examples — Classification — order, degree, linear vs nonlinear, autonomous vs non-autonomous

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The scenario matrix

Every ODE-classification problem is really a point in a four-axis space: two of those axes — linear vs nonlinear and autonomous vs non-autonomous — are independent, so they form a genuine 2-D grid of four corners. The picture below draws that grid and drops each worked example into its corner (order and degree are then read off separately per example).

The table lists the distinct case-classes — the corners and edges where beginners slip. The last column names the example that nails that cell.

#	Case class (the "scenario")	What makes it tricky	Covered by
C1	Clean polynomial ODE, all four labels obvious	baseline — nothing hidden	Ex 1
C2	Radical hiding a power	must square before reading degree	Ex 2
C3	Transcendental function of a derivative ($\sin y^{\prime }$, $\ln y^{\prime \prime }$)	degree is undefined	Ex 3
C4	Ugly $x$-coefficients that mimic nonlinearity	still linear — look only at $y$	Ex 4
C5	Products/powers of $y$ or its derivatives	genuinely nonlinear	Ex 5
C6	Linear yet non-autonomous (explicit $t$ in forcing)	the two axes are independent	Ex 6
C7	Autonomous but nonlinear (no clock, still $y^2$)	autonomy ≠ linearity	Ex 7
C8	Degenerate / limiting — degree-zero-power, missing $y$, algebraic (order 0)	boundary of "is it an ODE?"	Ex 8
C9	Word problem → build the ODE, then classify	translation, not just labelling	Ex 9
C10	Exam twist — fractions of derivatives, must clear denominators	cleanup changes the visible degree	Ex 10

Read the matrix top to bottom: C1 is the easy warm-up, C8 is the "trap" row, C9 is application, C10 is the final boss.

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Example 1 — the clean baseline (cell C1)

Steps.

1. Order. Highest derivative is $y^{\prime \prime }$ → order $2$. Why this step? Order is always the first thing to read; it decides how many arbitrary constants the general solution carries — here, two.
2. Degree. The equation is already a polynomial in $y,y^{\prime },y^{\prime \prime }$; the power on $y^{\prime \prime }$ is $1$ → degree $1$. Why this step? No radicals, no fractions of derivatives → nothing to clean, so we can read the exponent directly.
3. Linear? Each of $y,y^{\prime },y^{\prime \prime }$ appears to the first power, none multiplied, none inside a nonlinear function → linear. Why this step? Linearity unlocks superposition and the characteristic-equation method.
4. Autonomous? No explicit independent variable $x$ anywhere → autonomous. Why this step? Autonomy means the slope field repeats along every horizontal line, so we could study equilibria on a phase line.

Answer: order 2, degree 1, linear, autonomous.

Verify: the general solution is $y=c_1{e}^{-2x}+c_2{e}^{-3x}$ (roots of $r^2+5r+6=0$ are $-2,-3$). Two arbitrary constants ✓ matches order 2. Plugging $y=e^{-2x}$: $4e^{-2x}-10e^{-2x}+6e^{-2x}=0$ ✓.

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Example 2 — a radical hides the true degree (cell C2)

Steps.

1. Spot the hidden power. A cube root is a fractional power; degree is defined only for a polynomial in the derivatives. So we must remove the root first. Why this step? You cannot honestly read an exponent while one derivative sits under a radical — the equation isn't yet a polynomial.
2. Cube both sides to clear $\sqrt[3]{}$: ${\left(\frac{d^{2}y}{d{x}^2}\right)}^3=1+{\left(\frac{dy}{dx}\right)}^2.$ Why this step? Cubing is the inverse of the cube root — it turns the fractional power into an integer power, making the equation a genuine polynomial in $y^{\prime }$ and $y^{\prime \prime }$.
3. Order. Highest is $y^{\prime \prime }$ → order $2$. Why this step? Order never changes under cleanup — squaring or cubing raises powers, not the identity of the highest derivative.
4. Degree. After cleanup, the power on $y^{\prime \prime }$ is $3$ → degree $3$. Why this step? Only now is the exponent of the highest derivative unambiguous.
5. Linear? $(y^{\prime \prime }{)}^3$ is a third power → nonlinear. Why this step? A power above 1 on any derivative breaks linearity rule 1, forbidding superposition.
6. Autonomous? No explicit $x$ → autonomous. Why this step? The independent variable $x$ appears nowhere on its own, so the "rules" don't depend on the input value.

Answer: order 2, degree 3, nonlinear, autonomous.

Verify: the figure below plots the two sides as functions of the curvature $y^{\prime \prime }$ while holding the slope $y^{\prime }$ fixed. Wherever the two graphs cross is a curvature that actually satisfies the equation — and there is exactly one crossing.

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Example 3 — transcendental function of a derivative (cell C3)

Steps.

1. Try to polynomialise. Can $\sin (y^{\prime })$ ever be written as a finite polynomial in $y^{\prime }$? No — its Taylor series $y^{\prime }-\frac{(y^{\prime }{)}^3}{6}+\dots$ never terminates, and no algebra removes it. Why this step? Degree is only defined once the equation is a polynomial in all derivatives. If that's impossible, we stop.
2. Conclude degree. Since it cannot be made polynomial in derivatives → degree undefined. Why this step? This is the honest answer, not "1". The presence of $\sin (y^{\prime })$ is the whole point.
3. Order. Highest derivative $y^{\prime \prime }$ → order $2$ (transcendental wrapping doesn't hide the order). Why this step? Order counts which derivative is highest, and $y^{\prime \prime }$ is still visibly present regardless of the $\sin$ around $y^{\prime }$.
4. Linear? $\sin (y^{\prime })$ puts a derivative inside a nonlinear function → nonlinear (breaks rule 3). Why this step? Rule 3 forbids any derivative living inside $\sin ,e^{(\cdot )},\ln$, etc.
5. Autonomous? $e^x$ on the right is explicit $x$ → non-autonomous. Why this step? The independent variable $x$ appears explicitly in the forcing, so the rule changes with the input.

Answer: order 2, degree undefined, nonlinear, non-autonomous.

Verify: contrast — $y^{\prime \prime }+(y^{\prime }{)}^2=e^x$ would be degree 1 (polynomial in derivatives; highest $y^{\prime \prime }$ to power 1). Swapping $(y^{\prime }{)}^2\to \sin (y^{\prime })$ is exactly what destroys "degree". The difference is polynomial vs transcendental, confirming step 1. ✓

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Example 4 — ugly coefficients that look nonlinear (cell C4)

Steps.

1. Isolate how $y$ enters. Mentally cover the coefficients and keep only the underlined unknown parts: $x^2\underline{y^{\prime \prime }}+(\sin x)\underline{y^{\prime }}+e^x\underline{y}=\tan x.$ Each underlined piece — $\underline{y^{\prime \prime }},\underline{y^{\prime }},\underline{y}$ — appears once, to the first power, not multiplied by another $y$-thing, and not inside a nonlinear function of $y$. Why this step? The linearity test inspects only the unknown $y$ and its derivatives (the underlined parts) — never the coefficients wrapped around them.
2. Check the coefficients. $x^2,\sin x,e^x,\tan x$ are functions of $x$ only — allowed to be as wild as they like. Why this step? Definition of linear ODE: coefficients $a_k(x)$ and forcing $g(x)$ may be any functions of $x$.
3. Verdict: all three linearity rules hold → linear. Why this step? Passing every rule from steps 1–2 is exactly the definition of linear, so we can now confidently label it.
4. Order / degree. Highest $y^{\prime \prime }$, first power → order $2$, degree $1$. Why this step? With no radicals or derivative-fractions to clean, we read order (which derivative) and degree (its power) straight off.
5. Autonomous? $x$ appears explicitly everywhere → non-autonomous. Why this step? The autonomy test only asks whether the independent variable shows up on its own — here $x$ is all over the equation.

Answer: order 2, degree 1, linear, non-autonomous.

Verify: put it in standard linear form $a_2(x)y^{\prime \prime }+a_1(x)y^{\prime }+a_0(x)y=g(x)$ with $a_2=x^2,a_1=\sin x,a_0=e^x,g=\tan x$. Every slot is a pure function of $x$ ✓ — the defining shape of a linear ODE.

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Example 5 — genuinely nonlinear via a product (cell C5)

Steps.

1. Scan for the three violations. Look for (i) powers $>1$, (ii) products of $y$-things, (iii) nonlinear functions of $y$. Why this step? One violation is enough to make it nonlinear; naming which one sharpens understanding.
2. Find them. $y\cdot y^{\prime \prime }$ is a product of two $y$-quantities (rule 2), and $(y^{\prime }{)}^2$ is a second power (rule 1). Two violations. Why this step? Products and powers of the unknown are the hallmark of nonlinearity.
3. Verdict: nonlinear. Why this step? Any single violation from step 2 is decisive; two make it unambiguous.
4. Order / degree. Highest $y^{\prime \prime }$ to power $1$ → order $2$, degree $1$. Why this step? Note the surprise: it's degree 1 (power on $y^{\prime \prime }$ is one) yet still nonlinear — degree and linearity are different questions.
5. Autonomous? No explicit $x$ → autonomous. Why this step? The independent variable never appears alone, so the rule only depends on where $y$ currently is.

Answer: order 2, degree 1, nonlinear, autonomous.

Verify: notice $y{y}^{\prime \prime }+(y^{\prime }{)}^2=\frac{d}{dx}(y{y}^{\prime })$. So the equation is $(y{y}^{\prime }{)}^{\prime }=0\Rightarrow y{y}^{\prime }=c_1\Rightarrow \frac{1}{2}(y^2{)}^{\prime }=c_1\Rightarrow y^2=2c_{1}x+c_2$. Two constants ✓ matches order 2. Why the product proves nonlinearity (superposition test): superposition means "if $y_1$ and $y_2$ are solutions, so is $y_1+y_2$." Take two solutions of our equation and add them: the left side of $y{y}^{\prime \prime }$ becomes $(y_1+y_2)(y_1+y_2{)}^{\prime \prime }=y_1{y}_1^{\prime \prime }+y_1{y}_2^{\prime \prime }+y_2{y}_1^{\prime \prime }+y_2{y}_2^{\prime \prime }$. The two cross terms $y_1{y}_2^{\prime \prime }+y_2{y}_1^{\prime \prime }$ do not cancel and were never present in either original equation — the sum fails to solve the ODE. Those uncancelled cross terms are born precisely from the product $y{y}^{\prime \prime }$; a linear equation (no product) would have no cross terms and the sum would survive. That is why a product forbids superposition.

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Example 6 — linear and non-autonomous (cell C6)

Steps.

1. Linearity. $y^{\prime \prime }$ and $y$ appear to first power, not multiplied, not wrapped; the constant ${\omega }^2$ is just a (constant) coefficient → linear. Why this step? A constant like ${\omega }^2$ is a special case of an allowed coefficient — its being a fixed number can never break linearity.
2. Autonomy. The forcing $\cos (\gamma t)$ depends explicitly on $t$ → non-autonomous. Why this step? Autonomy asks only whether the independent variable $t$ shows up explicitly; a linear equation can still peek at the clock. Note $\gamma$ inside is a constant — it is the presence of $t$ that matters, not $\gamma$.
3. Order / degree. Highest $y^{\prime \prime }$, first power → order $2$, degree $1$. Why this step? No cleanup is needed, so order (highest derivative) and degree (its power) read off directly.

Answer: order 2, degree 1, linear, non-autonomous — proving the two axes are independent.

Verify: the homogeneous partner $y^{\prime \prime }+{\omega }^{2}y=0$ is autonomous; adding the $t$-dependent forcing is what breaks autonomy. The slope-field picture below makes the difference visible.

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Example 7 — autonomous but nonlinear (cell C7)

Steps.

1. Expand the right side: $y^{\prime }=y-y^2$. Why this step? The hidden $y^2$ is easier to see expanded, so linearity can be judged honestly.
2. Linearity. The $y^2$ term is a second power of the unknown → nonlinear (rule 1 broken). Why this step? Even a single squared unknown forbids superposition.
3. Autonomy. No explicit $t$ on the right → autonomous. Why this step? The rate depends only on the current population $y$, not on the calendar — a classic phase-line candidate.
4. Order / degree. Highest $y^{\prime }$, first power → order $1$, degree $1$. Why this step? Only $y^{\prime }$ appears (order 1) and it is never raised to a power (degree 1) once expanded.

Answer: order 1, degree 1, nonlinear, autonomous.

Verify: its equilibria solve $y(1-y)=0\Rightarrow y=0$ and $y=1$. Two equilibria for a first-order autonomous ODE ✓ — exactly what a phase line predicts. This being separable (a nonlinear-but-solvable case) is why the logistic curve has a closed form despite nonlinearity.

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Example 8 — degenerate & limiting cases (cell C8)

Steps.

1. (a) $y^{\prime }=3$. Order 1 (one derivative), degree 1, linear (RHS is a constant $g(x)=3$, $y$ absent but that's fine), and autonomous (no explicit $x$). Why this step? A constant right side is the simplest linear forcing; missing $y$ doesn't break linearity.
2. (b) $(y^{\prime }{)}^0+y=x$. Any nonzero quantity to the power $0$ is $1$, so this collapses to $1+y=x$, i.e. $y=x-1$ — no derivative survives. It is an algebraic equation, order 0: not an ODE at all. Why this step? The limiting exponent $0$ is a trap; degenerate powers can erase the derivative entirely.
3. (c) $x^2+y^2=4$. No derivatives → order $0$ → an algebraic (implicit) relation, not an ODE. (Differentiate it and you'd get one: $2x+2y{y}^{\prime }=0$.) Why this step? "Order 0" is the degenerate floor of the order axis — worth naming so you don't mislabel it.
4. (d) $y^{\prime \prime }=0$. Order 2, degree 1, linear, autonomous. Limiting-simple case: general solution $y=c_{1}x+c_2$ (a straight line). Why this step? The all-zeros right side is the homogeneous limit; two constants confirm order 2.

Answers: (a) order 1, deg 1, linear, autonomous. (b) collapses to order 0 (algebraic, not an ODE). (c) order 0 (algebraic). (d) order 2, deg 1, linear, autonomous.

Verify: (a) solution $y=3x+c$, one constant ✓ order 1. (d) $y=c_{1}x+c_2$, $y^{\prime \prime }=0$ ✓, two constants ✓ order 2.

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Example 9 — word problem: build then classify (cell C9)

Steps.

1. Translate the words. "Rate of change of $T$" $=\frac{dT}{dt}$. "Proportional to how much hotter than room" $=-k(T-R)$ with $k>0$ a fixed constant (minus: it cools). Why this step? Each English phrase maps to one algebraic piece — this is the modelling half, and getting it right is prerequisite to any labelling.
2. Fixed-room ODE: $\frac{dT}{dt}=-k(T-20).$ Why this step? Substituting the constant $R=20$ gives the concrete equation we will classify.
3. Classify (fixed room). Order 1; degree 1; linear ($T$ to first power, constant coefficient); autonomous (no explicit $t$ — $20$ is a constant). Why this step? With $R$ constant the "landscape" doesn't move → autonomous, and we can find the equilibrium $T=20$ on a phase line.
4. Moving-room ODE: $\frac{dT}{dt}=-k(T-(20+5\cos t))=-kT+k(20+5\cos t).$ Why this step? Substituting the time-varying $R(t)$ and expanding exposes the explicit $\cos t$, which is what the autonomy test looks for.
5. Re-classify (moving room). Still order 1, degree 1, still linear in $T$ — but now $\cos t$ appears explicitly → non-autonomous. Why this step? Only the autonomy label flips; the $t$-dependent room temperature is external forcing, exactly like Example 6.

Answers: fixed room — order 1, degree 1, linear, autonomous; moving room — order 1, degree 1, linear, non-autonomous.

Verify (fixed room): solve $T^{\prime }=-k(T-20)$. Substituting $T=20$ gives $T^{\prime }=0$ ✓ equilibrium. General solution $T=20+C{e}^{-kt}\to 20$ as $t\to \infty$ ✓ coffee reaches room temperature. One constant ✓ order 1.

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Example 10 — exam twist: fractions of derivatives (cell C10)

Steps.

1. Spot the derivative in a denominator. $y^{\prime }$ sits in a denominator; the equation is not yet a polynomial in the derivatives. Why this step? Degree needs a polynomial form, so a derivative in a denominator must be cleared first (just like a radical).
2. Multiply through by $(1+y^{\prime })$: $\left(1+\frac{dy}{dx}\right)\frac{d^{2}y}{d{x}^2}=1,$ i.e. $y^{\prime \prime }+y^{\prime }{y}^{\prime \prime }=1.$ Why this step? Clearing the fraction turns the equation into an honest polynomial in $y^{\prime }$ and $y^{\prime \prime }$.
3. Order. Highest is $y^{\prime \prime }$ → order $2$. Why this step? Order counts which derivative is highest, unaffected by clearing the denominator.
4. Degree. Now read powers of the highest derivative $y^{\prime \prime }$: it appears as $y^{\prime \prime }$ and inside $y^{\prime }{y}^{\prime \prime }$ — both are first power in $y^{\prime \prime }$ → degree $1$. Why this step? Degree looks only at the power of the highest-order derivative; $y^{\prime \prime }$ is never squared, so degree stays $1$ despite the messy cleanup.
5. Linear? The term $y^{\prime }{y}^{\prime \prime }$ is a product of two derivatives → nonlinear (rule 2). Why this step? Cleanup revealed a product that the fraction was hiding.
6. Autonomous? No explicit $x$ → autonomous. Why this step? The independent variable $x$ never appears on its own — the rule depends only on the current $y^{\prime }$.

Answer: order 2, degree 1, nonlinear, autonomous.

Verify: the trap was expecting degree $>1$ from the fraction; instead the highest derivative stays first-power, but a hidden product $y^{\prime }{y}^{\prime \prime }$ makes it nonlinear. So cleanup can change the linearity verdict even when it leaves the degree at 1 ✓ — the exact skill this cell tests.

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Recall Quick self-test (reveal after guessing)

Give order, degree, linear?, autonomous? for each. $\sqrt{y^{\prime \prime }}=y^{\prime }+x$ ::: order 2; degree 2 (square both sides → $y^{\prime \prime }=(y^{\prime }+x{)}^2$); nonlinear ($(y^{\prime }+x{)}^2$ has $y^{\prime }$ squared); non-autonomous (explicit $x$). $y^{\prime \prime \prime }+y=0$ ::: order 3; degree 1; linear; autonomous. $y^{\prime }=\frac{y}{t}$ ::: order 1; degree 1; linear; non-autonomous (explicit $t$). $(y^{\prime \prime }{)}^2=y$ ::: order 2; degree 2; nonlinear (power 2 on $y^{\prime \prime }$); autonomous.