Case 3 - complex conjugate roots — Euler's formula connection: Worked examples

Child of Case 3 — complex conjugate roots — Euler's formula connection. Here we hunt down every kind of complex-root problem an exam can throw at you and solve each one fully. If the parent note taught you the recipe, this page is the drill hall where we run the recipe through every sign, every degenerate input, and every real-world dressing.

Before we start: recall the one sentence that powers everything. For $a y^{''} + b y^{'} + cy = 0$ with characteristic roots $m = α \pm i β$ (a complex conjugate pair, meaning $β \neq = 0$ ), the real general solution is $y (x) = e^{α x} (C_{1} cos β x + C_{2} sin β x) .$ Here $α = - \frac{b}{2 a}$ is the envelope exponent (real part of the root) and $β = \frac{4 a c - b ^{2}}{2 a} > 0$ is the angular frequency (size of the imaginary part). See Characteristic Equation of Linear ODEs for where $m$ comes from.

The scenario matrix

Every Case-3 problem lives in exactly one cell below. Our job is to hit all of them.

Cell	What makes it distinct	Sign of $α$	Example
A	Pure oscillation, no first-derivative term ( $b = 0$ )	$α = 0$	Ex 1
B	Decaying oscillation ( $α < 0$ )	$α < 0$	Ex 2
C	Growing oscillation ( $α > 0$ )	$α > 0$	Ex 3
D	IVP — pin down $C_{1}, C_{2}$ from data	any	Ex 4
E	Non-monic leading coeff $a \neq = 1$ (fractions in $α, β$ )	$α < 0$	Ex 5
F	Degenerate boundary: $b^{2} = 4 a c$ almost — checking we did NOT fall into Case 2	$α < 0$	Ex 6
G	Real-world word problem (spring–mass with damping)	$α < 0$	Ex 7
H	Exam twist: convert answer to single-cosine amplitude–phase form	$α = 0$	Ex 8

Figure — Case 3 - complex conjugate roots — Euler's formula connection

Cell A — pure oscillation ( $b = 0$ , $α = 0$ )

Cell B — decaying oscillation ( $α < 0$ )

Worked example Example 2 —

$y^{''} + 4 y^{'} + 13 y = 0$ Forecast: the middle term $+ 4 y^{'}$ is friction. Will the wiggle grow or shrink?

Step 1. Characteristic: $m^{2} + 4 m + 13 = 0$ . Why this step? $y^{''} \to m^{2}, y^{'} \to m, y \to 1$ .

Step 2. Quadratic formula: $m = \frac{- 4 \pm 16 - 52}{2} = \frac{- 4 \pm - 36}{2}$ . Why this step? Discriminant $Δ = 16 - 52 = - 36 < 0$ confirms Case 3 (complex roots).

Step 3. $- 36 = 6 i$ , so $m = \frac{- 4 \pm 6 i}{2} = - 2 \pm 3 i$ . Thus $α = - 2, β = 3$ . Why this step? Divide both parts by 2. The real part $- 2$ is the envelope exponent; the imaginary part $3$ is the frequency.

Step 4. $y = e^{- 2 x} (C_{1} cos 3 x + C_{2} sin 3 x) .$ Why this step? $α = - 2 < 0$ → amplitude shrinks like $e^{- 2 x}$ (a swing with friction). Connects to Damped Harmonic Oscillator (RLC / spring-mass).

Verify: with $y_{1} = e^{- 2 x} cos 3 x$ , compute $y_{1}^{'} = e^{- 2 x} (- 2 cos 3 x - 3 sin 3 x)$ and $y_{1}^{''} = e^{- 2 x} (- 5 cos 3 x + 12 sin 3 x)$ . Then $y_{1}^{''} + 4 y_{1}^{'} + 13 y_{1} = e^{- 2 x} [(- 5 - 8 + 13) cos 3 x + (12 - 12) sin 3 x] = 0$ . ✔

Cell C — growing oscillation ( $α > 0$ )

Cell D — initial value problem (pin the constants)

Worked example Example 4 —

$y^{''} + 6 y^{'} + 13 y = 0$ , $y (0) = 1$ , $y^{'} (0) = - 3$ Forecast: two conditions → two equations → unique $C_{1}, C_{2}$ . Guess whether it decays.

Step 1. $m^{2} + 6 m + 13 = 0 \Rightarrow m = \frac{- 6 \pm 36 - 52}{2} = \frac{- 6 \pm 4 i}{2} = - 3 \pm 2 i$ . Why this step? $Δ = - 16 < 0$ ; $- 16 = 4 i$ . So $α = - 3, β = 2$ .

Step 2. General solution $y = e^{- 3 x} (C_{1} cos 2 x + C_{2} sin 2 x)$ . Why this step? Apply the Case-3 template.

Step 3. Use $y (0) = 1$ : at $x = 0$ , $e^{0} = 1$ , $cos 0 = 1$ , $sin 0 = 0$ , so $y (0) = C_{1} = 1$ . Why this step? Substituting $x = 0$ collapses every trig term to its value at zero, isolating $C_{1}$ .

Step 4. Differentiate (product rule): $y^{'} = e^{- 3 x} [(- 3 C_{1} + 2 C_{2}) cos 2 x + (- 3 C_{2} - 2 C_{1}) sin 2 x] .$ At $x = 0$ : $y^{'} (0) = - 3 C_{1} + 2 C_{2}$ . Set equal to $- 3$ : $- 3 (1) + 2 C_{2} = - 3 \Rightarrow C_{2} = 0$ . Why this step? The derivative must match the given slope at the start; only the $cos$ term survives at $x = 0$ .

Step 5. $y = e^{- 3 x} cos 2 x .$

Verify: $y (0) = e^{0} cos 0 = 1$ ✔. $y^{'} = e^{- 3 x} (- 3 cos 2 x - 2 sin 2 x)$ , so $y^{'} (0) = - 3$ ✔. And $y^{''} + 6 y^{'} + 13 y$ : with $y^{''} = e^{- 3 x} (5 cos 2 x + 12 sin 2 x)$ , sum $= e^{- 3 x} [(5 - 18 + 13) cos 2 x + (12 - 12) sin 2 x] = 0$ ✔.

Cell E — non-monic leading coefficient (fractions appear)

Cell F — degenerate boundary (make sure it's Case 3, not Case 2)

Worked example Example 6 —

$y^{''} + 2 y^{'} + 2 y = 0$ vs. its dangerous neighbour Forecast: what if $c$ were slightly smaller and $Δ$ hit exactly zero? Would our formula still apply?

Step 1. $m^{2} + 2 m + 2 = 0$ , $Δ = 4 - 8 = - 4 < 0$ . Why this step? We must check the sign of $Δ$ first. Only $Δ < 0$ is Case 3; $Δ = 0$ would be Case 2 — repeated real roots and $Δ > 0$ would be Case 1 — real distinct roots.

Step 2. $m = \frac{- 2 \pm - 4}{2} = \frac{- 2 \pm 2 i}{2} = - 1 \pm i$ . So $α = - 1, β = 1$ . Why this step? $- 4 = 2 i$ ; divide by $2$ . Note $β = 1$ — small but nonzero, so genuinely oscillatory.

Step 3. $y = e^{- x} (C_{1} cos x + C_{2} sin x) .$ Why this step? Envelope $e^{- x}$ , slowest possible whole oscillation.

Step 4 (the boundary lesson). Had the equation been $y^{''} + 2 y^{'} + 1 y = 0$ , then $Δ = 4 - 4 = 0$ : not Case 3 — a repeated root $m = - 1$ , giving $y = (C_{1} + C_{2} x) e^{- x}$ with no wiggle. The moment $β \to 0$ the cosine flattens to $1$ and Case 3 hands over to Case 2. Why this step? This is the limiting behaviour of Case 3: as $β \to 0$ the oscillation period $2 π / β \to \infty$ , i.e. it stops oscillating entirely.

Verify: for $y = e^{- x} cos x$ : $y^{'} = e^{- x} (- cos x - sin x)$ , $y^{''} = e^{- x} (2 sin x)$ . Then $y^{''} + 2 y^{'} + 2 y = e^{- x} [(0 - 2 + 2) cos x + (2 - 2 + 0) sin x] = 0$ ✔.

Cell G — real-world word problem (damped spring)

Worked example Example 7 — spring–mass with damping

A mass $m = 1 kg$ hangs on a spring of stiffness $k = 25 N/m$ with a damper of coefficient $c = 6 N\cdots/m$ . Newton's law gives $x^{''} + 6 x^{'} + 25 x = 0$ where $x (t)$ is displacement in metres. Start from $x (0) = 0.1 m$ , released from rest $x^{'} (0) = 0$ . Find $x (t)$ .

Forecast: underdamped (it should wiggle while dying) or overdamped (crawl back without wiggle)?

Step 1. Characteristic: $m^{2} + 6 m + 25 = 0$ , $Δ = 36 - 100 = - 64 < 0$ . Why this step? $Δ < 0$ means underdamped — the physical name for Case 3. Real damped motion of an Damped Harmonic Oscillator (RLC / spring-mass).

Step 2. $m = \frac{- 6 \pm - 64}{2} = \frac{- 6 \pm 8 i}{2} = - 3 \pm 4 i$ . So $α = - 3 s^{- 1}, β = 4 rad/s$ . Why this step? $α$ has units of inverse seconds (a decay rate); $β$ is angular frequency in rad/s. Units confirm we set it up right.

Step 3. $x (t) = e^{- 3 t} (C_{1} cos 4 t + C_{2} sin 4 t)$ .

Step 4. $x (0) = C_{1} = 0.1$ . Why? trig collapses at $t = 0$ .

Step 5. $x^{'} = e^{- 3 t} [(- 3 C_{1} + 4 C_{2}) cos 4 t + (- 3 C_{2} - 4 C_{1}) sin 4 t]$ ; $x^{'} (0) = - 3 C_{1} + 4 C_{2} = 0 \Rightarrow C_{2} = \frac{3 C _{1}}{4} = 0.075$ . Why? "released from rest" means zero starting velocity.

Step 6. $x (t) = e^{- 3 t} (0.1 cos 4 t + 0.075 sin 4 t) m .$ Why this step? The mass oscillates at $4$ rad/s inside a shrinking envelope $e^{- 3 t}$ : it wobbles a few times and settles. See the figure for the damped trace.

Verify: $x (0) = 0.1$ ✔; $x^{'} (0) = - 3 (0.1) + 4 (0.075) = - 0.3 + 0.3 = 0$ ✔ (rest). Units: metres throughout ✔.

Cell H — exam twist: amplitude–phase form

Worked example Example 8 — rewrite

$y = 3 cos 2 x + 4 sin 2 x$ as one cosine This is the answer to $y^{''} + 4 y = 0$ with $y (0) = 3, y^{'} (0) = 8$ . Exams often want the single amplitude–phase form $R cos (2 x - φ)$ .

Forecast: what is the peak height $R$ , and does the phase $φ$ land in quadrant I, II, III or IV?

Step 1. Want $R cos (2 x - φ) = R cos φ cos 2 x + R sin φ sin 2 x$ . Why this step? The cosine subtraction identity is the bridge — it turns "two waves" into "one shifted wave". Match coefficients to $3 cos 2 x + 4 sin 2 x$ .

Step 2. Match: $R cos φ = 3$ , $R sin φ = 4$ . Why this step? Two equations, two unknowns $R, φ$ .

Step 3. $R = 3^{2} + 4^{2} = 25 = 5$ . Why this step? Square and add: $R^{2} (cos^{2} φ + sin^{2} φ) = 9 + 16$ , and $cos^{2} + sin^{2} = 1$ (unit-circle identity from Euler's Formula and the Unit Circle).

Step 4. $φ = arctan (\frac{R s i n φ}{R c o s φ}) = arctan \frac{4}{3} \approx 0.9273 rad$ . Why this step? Both $R cos φ = 3 > 0$ and $R sin φ = 4 > 0$ , so $φ$ is in quadrant I — the plain $arctan$ is safe here (no quadrant fix needed). If either had been negative we'd adjust by $\pm π$ .

Step 5. $y = 5 cos (2 x - 0.9273) .$ Why this step? One clean wave: amplitude $5$ , phase shift $0.9273$ rad.

Verify: at $x = 0$ : $5 cos (- 0.9273) = 5 cos (0.9273) = 5 \cdot \frac{3}{5} = 3 = y (0)$ ✔. And $y^{'} = - 10 sin (2 x - 0.9273)$ ; $y^{'} (0) = - 10 sin (- 0.9273) = 10 sin (0.9273) = 10 \cdot \frac{4}{5} = 8 = y^{'} (0)$ ✔.

Recall checkpoint

Recall Which cell am I in? (self-test)

Given $3 y^{''} + 0 \cdot y^{'} + 12 y = 0$ , which cell? ::: Cell A — $b = 0$ so $α = 0$ , pure oscillation ( $m = \pm 2 i$ ). Given roots $m = 2 \pm i$ , does the amplitude grow or decay? ::: Grows — $α = + 2 > 0$ , envelope $e^{2 x}$ (Cell C). As $β \to 0$ , Case 3 turns into which case? ::: Case 2 — repeated real roots — the oscillation period blows up and the wiggle disappears. What single number decides Case 1 vs 2 vs 3? ::: The discriminant $Δ = b^{2} - 4 a c$ : $> 0, = 0, < 0$ respectively. In a spring problem, what physical name does Case 3 carry? ::: Underdamped oscillation.

Back to Case 3 — complex conjugate roots — Euler's formula connection · foundations in Superposition Principle for Linear ODEs and Wronskian and Linear Independence.