Homogeneous with constant coefficients — characteristic equation: Worked examples

This page is a drill through every case the characteristic equation of a constant-coefficient linear ODE can throw at you. We start by listing all the shapes the roots can take (the "scenario matrix"), then work one example per cell. If you have not yet met the three cases, read the parent first: Homogeneous with constant coefficients — characteristic equation.

The scenario matrix

The discriminant is $== Δ = b^{2} - 4 a c ==$ — the thing under the square root in $r = \frac{- b \pm Δ}{2 a}$ . Its sign, and a few special coefficient values, split every possible problem into these cells. Throughout, $n$ means the order of the ODE (the highest derivative), which equals the degree of the characteristic polynomial:

Cell	Trigger	Root type	Answer shape
A Distinct real, both negative	$Δ > 0$ , $b > 0$ , $c > 0$	$r_{1}, r_{2} < 0$	decaying $c_{1} e^{r_{1} x} + c_{2} e^{r_{2} x}$
A2 Distinct real, both positive	$Δ > 0$ , $b < 0$ , $c > 0$	$r_{1}, r_{2} > 0$	pure growth $c_{1} e^{r_{1} x} + c_{2} e^{r_{2} x}$
B Distinct real, opposite signs	$Δ > 0$ , $c < 0$	one $+$ , one $-$	one term blows up
C One root exactly zero	$c = 0$	$r = 0$ and $r = - b / a$	constant $+$ exponential
D Repeated real root	$Δ = 0$	$r$ double	$(c_{1} + c_{2} x) e^{r x}$
E Pure imaginary ( $α = 0$ )	$b = 0$ , $c > 0$	$\pm i β$	undamped $cos, sin$
F Complex with decay	$Δ < 0$ , $b > 0$	$α \pm i β$ , $α < 0$	damped oscillation
G Complex with growth	$Δ < 0$ , $b < 0$	$α \pm i β$ , $α > 0$	growing oscillation
H Higher order ( $n \geq 3$ )	degree $\geq 3$	mix of above	product of factors
I Word problem (physics)	—	any	real-world reading
J Exam twist (given a solution)	reverse-engineer	—	build the ODE

Each example below is tagged with its cell(s). (Here $α$ is the real part and $β$ the imaginary part of a complex root $r = α \pm i β$ .)

Cell A — Distinct real roots, both negative (pure decay)

Worked example Cell A:

$y^{''} + 5 y^{'} + 6 y = 0$ Forecast first: all coefficients positive — do you expect growth or decay? Guess before reading.

Char. eq.: $r^{2} + 5 r + 6 = 0$ . Why this step? Replace $y^{''} \to r^{2}$ , $y^{'} \to r$ , $y \to 1$ ; the ODE collapses to algebra.
Discriminant: $Δ = 5^{2} - 4 (1) (6) = 25 - 24 = 1 > 0$ . Why this step? The sign of $Δ$ picks the case. $Δ > 0$ → two distinct real roots (Cell A/A2/B/C).
Roots: $(r + 2) (r + 3) = 0 \Rightarrow r = - 2, - 3$ . Why this step? Factoring finds where the polynomial is zero — those exponents are our two solutions.
Both roots negative → both $e^{- 2 x}, e^{- 3 x}$ shrink. So: $y = c_{1} e^{- 2 x} + c_{2} e^{- 3 x} .$ Why this step? Distinct real roots give independent exponentials (Case 1); combine linearly.

Verify: put $y = e^{- 2 x}$ back: $y^{''} = 4 e^{- 2 x}$ , $y^{'} = - 2 e^{- 2 x}$ , so $4 - 10 + 6 = 0$ . ✓ Your forecast should have been decay — matches negative exponents.

Cell A2 — Distinct real roots, both positive (pure growth)

Cell B — Distinct real, opposite signs (one term explodes)

Cell C — One root exactly zero ( $c = 0$ )

Cell D — Repeated real root ( $Δ = 0$ )

Worked example Cell D:

$y^{''} + 4 y^{'} + 4 y = 0$ Forecast: $Δ = 16 - 16 = 0$ . How many exponentials do you get, and what must you do about the "missing" second solution?

Char. eq.: $r^{2} + 4 r + 4 = (r + 2)^{2} = 0 \Rightarrow r = - 2$ (double). Why this step? $Δ = 0$ means the two roots have merged into one.
$e^{- 2 x}$ is only one solution — an $n = 2$ equation needs two independent ones. Why this step? A second-order ODE always needs two independent basic solutions (see Wronskian and Linear Independence of Solutions).
The second solution is $x e^{- 2 x}$ . Why this step? (summary of the extra $x$ ): When the polynomial factors as $a (D - r_{0})^{2}$ with $D = \frac{d}{d x}$ , solving $(D - r_{0}) y = e^{r_{0} x}$ forces you to integrate a constant (the exponential cancels against the integrating factor), and integrating a constant produces a term linear in $x$ — hence $x e^{r_{0} x}$ . Because $x$ is not a constant multiple of $1$ , this second solution is genuinely independent of $e^{r_{0} x}$ . (Full derivation: Second Order Linear ODE — Reduction of Order.) $y = (c_{1} + c_{2} x) e^{- 2 x} .$

Verify: put $y = x e^{- 2 x}$ . Then $y^{'} = e^{- 2 x} - 2 x e^{- 2 x}$ , $y^{''} = - 4 e^{- 2 x} + 4 x e^{- 2 x}$ . Sum: $(- 4 + 4 x) + 4 (1 - 2 x) + 4 x = - 4 + 4 x + 4 - 8 x + 4 x = 0$ ✓.

Cell E — Pure imaginary roots ( $b = 0$ , undamped)

Worked example Cell E:

$y^{''} + 9 y = 0$ Forecast: no $y^{'}$ term means no friction. Growth, decay, or forever-oscillation?

Char. eq.: $r^{2} + 9 = 0 \Rightarrow r^{2} = - 9 \Rightarrow r = \pm 3 i$ . Why this step? No real number squares to $- 9$ ; the roots are purely imaginary ( $α = 0, β = 3$ ).
Using Euler $e^{i β x} = cos β x + i sin β x$ (see Euler's Formula and Complex Exponentials), the real solutions are $cos 3 x, sin 3 x$ . Why this step? The ODE is real, so we take real and imaginary parts of $e^{3 i x}$ to get real solutions.
With $α = 0$ the envelope $e^{0 \cdot x} = 1$ never shrinks: $y = c_{1} cos 3 x + c_{2} sin 3 x .$

The figure below plots the $c_{1} = 1, c_{2} = 0$ solution $y = cos 3 x$ . Look at the two coral dashed lines at $y = \pm 1$ : they are the envelope, and because $α = 0$ they stay perfectly flat — the wave neither grows nor shrinks. Count the lavender wiggles: because $β = 3$ , exactly three full oscillations fit into each interval of length $2 π$ . This is the visual signature of Cell E — steady, undying oscillation.

Verify: $y = cos 3 x$ : $y^{''} = - 9 cos 3 x$ , so $- 9 cos 3 x + 9 cos 3 x = 0$ ✓. Pure oscillation — your forecast should be forever-wiggling.

Cell F — Complex with decay (damped oscillation)

Worked example Cell F:

$y^{''} + 2 y^{'} + 10 y = 0$ , $y (0) = 2, y^{'} (0) = 0$ Forecast: small friction ( $b = 2$ ), stiff spring ( $c = 10$ ). Wiggle that fades, or fade with no wiggle?

Char. eq.: $r^{2} + 2 r + 10 = 0$ . $Δ = 4 - 40 = - 36 < 0$ → complex roots. Why this step? Substitute $y = e^{r x}$ to turn the ODE into this polynomial; then read $Δ < 0$ to know the roots are complex.
Roots: $r = \frac{- 2 \pm - 36}{2} = - 1 \pm 3 i$ . So $α = - 1, β = 3$ . Why this step? $α < 0$ → decaying envelope; $β = 3$ → oscillation frequency.
General: $y = e^{- x} (c_{1} cos 3 x + c_{2} sin 3 x)$ .
Apply $y (0) = 2$ : at $x = 0$ , $e^{0} = 1, cos 0 = 1, sin 0 = 0$ , so $c_{1} = 2$ . Why this step? Evaluating at $x = 0$ kills the $sin$ term and pins $c_{1}$ .
Apply $y^{'} (0) = 0$ : differentiate (product rule!): $y^{'} = e^{- x} [(- c_{1} + 3 c_{2}) cos 3 x + (- c_{2} - 3 c_{1}) sin 3 x] .$ At $x = 0$ : $- c_{1} + 3 c_{2} = 0 \Rightarrow c_{2} = \frac{c _{1}}{3} = \frac{2}{3}$ . Why this step? Must differentiate the whole product before substituting (see the mistakes list in the parent).
$y = e^{- x} (2 cos 3 x + \frac{2}{3} sin 3 x) .$

Verify: at $x = 0$ , $y = 2$ ✓ and $y^{'} = (- 2 + 3 \cdot \frac{2}{3}) = 0$ ✓. This is a Damped Harmonic Oscillator — a fading wiggle.

Cell G — Complex with growth (negative damping)

Cell H — Higher order (a cubic that mixes cases)

Worked example Cell H:

$y^{'''} - y^{''} + y^{'} - y = 0$ Forecast: degree-3 polynomial → three roots. Can one equation combine a decay and a wiggle?

Char. eq.: $r^{3} - r^{2} + r - 1 = 0$ . Why this step? Same substitution $y = e^{r x}$ ; here $n = 3$ so the third derivative gives $r^{3}$ , producing a cubic.
Factor by grouping: $r^{2} (r - 1) + (r - 1) = (r - 1) (r^{2} + 1) = 0$ . Why this step? Grouping exposes the real root $r = 1$ and the complex pair.
Roots: $r = 1$ (real), and $r^{2} + 1 = 0 \Rightarrow r = \pm i$ ( $α = 0, β = 1$ ). Why this step? One real root → one exponential; one imaginary pair → $cos, sin$ .
$y = c_{1} e^{x} + c_{2} cos x + c_{3} sin x .$ Why this step? Superposition: an $n$ -th order equation needs $n = 3$ independent pieces (see Linear Differential Operators and Superposition).

Verify: $y = cos x$ : $y^{'''} = sin x, y^{''} = - cos x, y^{'} = - sin x$ . Then $sin x - (- cos x) + (- sin x) - cos x = sin x + cos x - sin x - cos x = 0$ ✓. And $y = e^{x}$ : $1 - 1 + 1 - 1 = 0$ ✓.

Cell I — Word problem (real-world reading)

Worked example Cell I: a shock absorber

A car suspension satisfies $m y^{''} + c y^{'} + k y = 0$ with mass $m = 1 kg$ , damping $c = 6 N\cdots/m$ , stiffness $k = 9 N/m$ . Here the independent variable is time $t$ (in seconds) instead of $x$ — only the name of the variable changes; the derivatives $y^{''} = \frac{d ^{2} y}{d t ^{2}}$ , $y^{'} = \frac{d y}{d t}$ and the whole method are identical. $y$ is displacement (metres). Describe the motion. Forecast: is this bounce-and-fade, or slide-back-with-no-bounce?

Char. eq.: $r^{2} + 6 r + 9 = 0$ . Why this step? Divide by $m = 1$ , then substitute $y = e^{r t}$ : each time-derivative $y^{(k)}$ becomes $r^{k}$ , giving this polynomial in $r$ .
Discriminant: $Δ = 36 - 36 = 0$ → repeated root (Cell D territory). Why this step? $Δ = 0$ is the critically damped boundary — the fastest return with no oscillation.
Root: $r = - 3$ (double). $y = (c_{1} + c_{2} t) e^{- 3 t} .$
Reading: no $cos / sin$ → no bouncing. The $e^{- 3 t}$ pulls displacement to zero; the $t$ lets it overshoot at most once. This is exactly critical damping — engineers tune $c$ to hit $Δ = 0$ .

Verify (units + math): $Δ = c^{2} - 4 mk = 36 - 4 (1) (9) = 0$ ✓ (units $N^{2} s^{2} / m^{2}$ consistent on both terms). Root check $y = t e^{- 3 t}$ : satisfies the ODE by the Cell-D verification pattern.

Cell J — Exam twist (build the ODE from a solution)

Worked example Cell J: reverse-engineer

A second-order constant-coefficient ODE has general solution $y = e^{- 2 x} (c_{1} cos x + c_{2} sin x)$ . Find the ODE. Forecast: the solution shape tells you the roots. What are $α$ and $β$ ?

Read off roots: envelope $e^{- 2 x}$ → $α = - 2$ ; frequency of $cos x, sin x$ → $β = 1$ . So $r = - 2 \pm i$ . Why this step? Complex-case form $e^{α x} (c_{1} cos β x + c_{2} sin β x)$ encodes $α, β$ directly.
Rebuild the polynomial: $r = - 2 \pm i$ means factors $(r + 2 - i) (r + 2 + i)$ . $(r + 2)^{2} - i^{2} = (r + 2)^{2} + 1 = r^{2} + 4 r + 5.$ Why this step? Multiplying conjugate factors clears the imaginary part, giving real coefficients.
Char. eq. → ODE: $r^{2} + 4 r + 5 = 0$ corresponds to $y^{''} + 4 y^{'} + 5 y = 0.$ Why this step? Reverse the substitution $r^{k} \leftrightarrow y^{(k)}$ .

Verify: solve forward: $Δ = 16 - 20 = - 4$ , $r = \frac{- 4 \pm 2 i}{2} = - 2 \pm i$ ✓ — matches the given solution.

Recall Quick self-test across the matrix

Root pair $- 1 \pm 4 i$ → which cell and what solution? ::: Cell F, $y = e^{- x} (c_{1} cos 4 x + c_{2} sin 4 x)$ (decaying oscillation). $c = 0$ in $a y^{''} + b y^{'} + cy = 0$ forces which root? ::: $r = 0$ , giving a constant solution $c_{1}$ (Cell C). $Δ = 0$ physical meaning in a damped system? ::: Critical damping — fastest return with no oscillation (Cell D/I). Solution $(c_{1} + c_{2} x) e^{5 x}$ came from which discriminant sign? ::: $Δ = 0$ , repeated root $r = 5$ . Both roots real and positive ( $b < 0, c > 0$ ) → behaviour? ::: Pure growth, no oscillation (Cell A2).

Connections

Homogeneous with constant coefficients — characteristic equation
Linear Differential Operators and Superposition
Euler's Formula and Complex Exponentials
Wronskian and Linear Independence of Solutions
Method of Undetermined Coefficients
Damped Harmonic Oscillator
Second Order Linear ODE — Reduction of Order