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

Reminders of the exact tests (so you never guess):

Level 1 — Recognition

L1.1

State the order of each: (a) $\frac{d y}{d x} = 3 y$ (b) $y^{''} + 4 y = 0$ (c) $(\frac{d ^{4} y}{d x ^{4}})^{7} + y = 0$ .

Recall Solution L1.1

WHAT we look for: the highest derivative, ignoring any power on it.

(a) highest is $y^{'}$ → order 1.
(b) highest is $y^{''}$ → order 2.
(c) highest is $y^{(4)}$ ; the exponent $7$ is a degree clue, not an order clue → order 4.

L1.2

State the degree of each: (a) $y^{''} + 5 y^{'} + 6 y = 0$ (b) $(y^{'''})^{2} + y^{'} = x$ (c) $(\frac{d y}{d x})^{3} = y$ .

Recall Solution L1.2

WHAT we look for: every equation is already a polynomial in its derivatives, so read the power on the highest derivative.

(a) highest $y^{''}$ , power $1$ → degree 1.
(b) highest $y^{'''}$ , power $2$ → degree 2 (the $y^{'}$ term's power is irrelevant).
(c) highest derivative is $y^{'}$ , raised to power $3$ → degree 3 (order is only 1!).

Level 2 — Application

L2.1

Classify linear vs nonlinear and give a one-word reason: (a) $y^{'} + x^{2} y = cos x$ (b) $y^{'} + y^{2} = x$ (c) $y y^{''} = 1$ (d) $y^{''} + sin (x) y = 0$ (e) $y^{'} = sin y$ .

Recall Solution L2.1

Apply the three linearity rules to how $y$ enters (coefficients in $x$ are always allowed).

(a) $y, y^{'}$ first power, coefficient $x^{2}$ is fine → linear.
(b) $y^{2}$ → power $2$ on $y$ breaks rule 1 → nonlinear.
(c) $y \cdot y^{''}$ → two unknowns multiplied breaks rule 2 → nonlinear.
(d) $sin x$ is a coefficient in the independent variable, $y$ is first power → linear.
(e) $sin y$ puts $y$ inside a nonlinear function, breaks rule 3 → nonlinear.

L2.2

Classify autonomous vs non-autonomous: (a) $y^{'} = y (1 - y)$ (b) $y^{'} = y + e^{t}$ (c) $y^{''} + 9 y = 0$ (d) $y^{''} + 9 y = cos (2 t)$ .

Recall Solution L2.2

Test: does the independent variable ( $t$ ) show up explicitly?

(a) RHS is a function of $y$ only → autonomous (logistic — see Phase Line and Equilibria).
(b) $e^{t}$ contains $t$ explicitly → non-autonomous.
(c) no $t$ anywhere → autonomous.
(d) forcing $cos (2 t)$ contains $t$ → non-autonomous.

Level 3 — Analysis

L3.1

For $\frac{d ^{2} y}{d x ^{2}} = 1 + (\frac{d y}{d x})^{2}$ , give order and degree, showing the cleanup.

Recall Solution L3.1

WHY square first: the radical hides a power on the derivatives; you cannot read an honest exponent while a is present. Square both sides: $(\frac{d ^{2} y}{d x ^{2}})^{2} = 1 + (\frac{d y}{d x})^{2} .$ Now it is a polynomial in $y^{'}, y^{''}$ . Highest derivative $y^{''}$ has power $2$ . Order 2, degree 2. (Note: order was 2 before squaring too — squaring never changes the order.)

L3.2

For $y^{''} + sin (y^{'}) = 0$ , give order and degree with justification.

Recall Solution L3.2

Highest derivative is $y^{''}$ → order 2. Degree: $sin (y^{'})$ is transcendental in the derivative $y^{'}$ ; no algebra turns the equation into a polynomial in $y^{'}, y^{''}$ . Therefore degree is undefined. (Contrast L3.1, where a radical could be removed by squaring — a genuine transcendental function of a derivative cannot.)

L3.3

The graph below is the slope field of an autonomous equation and of a non-autonomous one. Which is which, and why?

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

Recall Solution L3.3

What to look at: slide your eye horizontally (fixed $y$ , changing $t$ ).

Left panel: along every horizontal line the arrows have identical slope — the field is invariant under time-shift. This is the signature of an autonomous equation $y^{'} = f (y)$ : the rule ignores the clock, so a solution slid sideways is still a solution.
Right panel: arrows change as $t$ increases at fixed $y$ (they tilt with the wave). The rule peeks at the clock → non-autonomous $y^{'} = f (t, y)$ .

Level 4 — Synthesis

L4.1

Give all four labels for $x (\frac{d ^{3} y}{d x ^{3}})^{2} + (\frac{d y}{d x})^{4} = e^{x} .$

Recall Solution L4.1

Order: highest derivative $y^{'''}$ → 3.
Degree: already polynomial in derivatives; power on $y^{'''}$ is $2$ → 2 (the $(y^{'})^{4}$ is not the highest derivative).
Linear? $(y^{'''})^{2}$ and $(y^{'})^{4}$ are powers $> 1$ → nonlinear.
Autonomous? coefficient $x$ and forcing $e^{x}$ are explicit in the independent variable → non-autonomous.

L4.2

Build (or state impossible) an ODE that is simultaneously linear, non-autonomous, order 2, degree 1. Then one that is nonlinear yet autonomous.

Recall Solution L4.2

Linear + non-autonomous + order 2 + degree 1: $y^{''} + 3 y^{'} + 2 y = cos t .$ $y, y^{'}, y^{''}$ enter to first power (linear, degree 1), highest derivative $y^{''}$ (order 2), and $cos t$ makes it non-autonomous. Valid — linearity and autonomy are independent axes (compare Second-Order Linear ODEs with Constant Coefficients).

Nonlinear + autonomous: $y^{'} = y (1 - y) .$ No $t$ (autonomous) but $y^{2}$ term makes it nonlinear. The logistic equation is the classic witness.

Level 5 — Mastery

L5.1

Classify fully and explain why superposition may or may not apply: (a) $y^{''} + ω^{2} y = 0$ (b) $y^{''} + ω^{2} sin y = 0$ (pendulum).

Recall Solution L5.1

(a) Order 2, degree 1, linear, autonomous. Because it is linear and homogeneous, if $y_{1}, y_{2}$ solve it so does $c_{1} y_{1} + c_{2} y_{2}$ — the Superposition Principle applies. This is exactly why the solution space is spanned by two basis functions.

(b) Order 2, degree — $sin y$ is transcendental in $y$ (not in a derivative, so order is fine, but the equation is not polynomial in $y$ ; degree is a property of derivatives, so we say degree 1 in $y^{''}$ if we treat $sin y$ as a coefficient-free term — conventionally the equation is nonlinear, degree 1 in the highest derivative $y^{''}$ ). Nonlinear, autonomous. Superposition fails: $sin (y_{1} + y_{2}) \neq = sin y_{1} + sin y_{2}$ , so a sum of solutions is not a solution. This is why the large-swing pendulum has no simple closed form.

L5.2

For each, decide whether the Existence–Uniqueness machinery you would reach for treats it as an initial-value problem, and state the order = number of required initial conditions: (a) $y^{'} = y^{2}, y (0) = 1$ (b) $y^{'''} + y = 0, y (0) = y^{'} (0) = y^{''} (0) = 0$ .

Recall Solution L5.2

The number of arbitrary constants in the general solution equals the order, which equals the number of initial conditions needed for a unique solution (see Picard–Lindelöf).

(a) order 1 → one initial condition $y (0) = 1$ . It's autonomous, nonlinear ( $y^{2}$ ). Solving by Separable Equations gives $y = \frac{1}{1 - x}$ , valid only for $x < 1$ (finite-time blow-up).
(b) order 3 → three initial conditions, as supplied. Linear, autonomous; here all-zero data forces $y \equiv 0$ .

L5.3

Verify the solution claim in L5.2(a): show $y = \frac{1}{1 - x}$ solves $y^{'} = y^{2}$ with $y (0) = 1$ , and give its maximal interval.

Recall Solution L5.3

Check the ODE: $y = \frac{1}{1 - x} = (1 - x)^{- 1}$ , so $y^{'} = - (1 - x)^{- 2} \cdot (- 1) = (1 - x)^{- 2} = (\frac{1}{1 - x})^{2} = y^{2} . ✓$ Check the datum: $y (0) = \frac{1}{1 - 0} = 1. ✓$ Maximal interval: the formula explodes at $x = 1$ ; starting from $x = 0$ the solution lives on $(- \infty, 1)$ . This blow-up in finite time is a hallmark of nonlinear growth — a linear equation can never do it.

Recall One-line self-test

Which two of the four classifications are completely independent axes? ::: Linearity and autonomy — an equation can be any of the four combinations (linear/nonlinear × autonomous/non-autonomous).