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

These exercises drill the whole machine of Case 3: spotting a negative discriminant, reading off $α$ and $β$ , and writing the real solution $y = e^{α x} (C_{1} cos β x + C_{2} sin β x)$ . Every problem has a collapsible full solution — try first, then reveal. Difficulty climbs from L1 (recognise) to L5 (mastery).

Before we start, one shared reminder of the machine, so no symbol is used unexplained.

The figure below is the mental picture behind every answer: an envelope squeezing a wiggle. (Description: a blue curve oscillates left-to-right; two dashed yellow curves — one above, one mirrored below — form a shrinking funnel that traps the blue wiggle, showing amplitude fading as $x$ grows because $α < 0$ .)

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

Level 1 — Recognition

Goal: given a quadratic, decide the case and read off $α, β$ . No solving yet.

L1.1 For $y^{''} + 16 y = 0$ , find the roots $m$ and state $α, β$ .

Recall Solution L1.1

Step 1 (WHAT + WHY). Form the characteristic equation by replacing $y^{''} \to m^{2}$ , $y \to 1$ , giving $m^{2} + 16 = 0$ . Why this replacement: substituting the trial solution $y = e^{m x}$ turns each derivative into a power of $m$ ( $y^{''} \to m^{2} e^{m x}$ , $y \to e^{m x}$ ), and dividing off the never-zero factor $e^{m x}$ leaves exactly $m^{2} + 16 = 0$ . Step 2 (WHY complex). Here $a = 1, b = 0, c = 16$ , so $Δ = b^{2} - 4 a c = 0 - 64 = - 64 < 0$ → a negative discriminant means the roots are complex (Case 3). Step 3. $m^{2} = - 16 \Rightarrow m = \pm - 16 = \pm 4 i$ . So $α = 0$ , $β = 4$ . Answer: $m = \pm 4 i$ , $α = 0$ , $β = 4$ .

L1.2 For $y^{''} + 6 y^{'} + 13 y = 0$ , compute $Δ$ and confirm it is Case 3; then read off $α, β$ .

Recall Solution L1.2

Step 1. $a = 1, b = 6, c = 13$ . $Δ = 6^{2} - 4 (1) (13) = 36 - 52 = - 16 < 0$ ✔ Case 3. Step 2 (WHY the formula). The quadratic formula splits into $α = - \frac{b}{2 a} = - \frac{6}{2} = - 3$ (real part). Step 3. $β = \frac{4 a c - b ^{2}}{2 a} = \frac{52 - 36}{2} = \frac{16}{2} = \frac{4}{2} = 2$ . Answer: $α = - 3$ , $β = 2$ , so $m = - 3 \pm 2 i$ .

L1.3 Which of these are Case 3 (complex roots)? (a) $m^{2} - 5 m + 6 = 0$ (b) $m^{2} + 4 = 0$ (c) $m^{2} + 2 m + 1 = 0$ .

Recall Solution L1.3

Compute each discriminant $Δ = b^{2} - 4 a c$ :

(a) $Δ = 25 - 24 = 1 > 0$ → real distinct, Case 1.
(b) $Δ = 0 - 16 = - 16 < 0$ → complex, Case 3 ✔
(c) $Δ = 4 - 4 = 0$ → repeated real, Case 2. Answer: only (b) is Case 3.

Level 2 — Application

Goal: go from quadratic all the way to the real general solution.

L2.1 Solve $y^{''} + 25 y = 0$ .

Recall Solution L2.1

Step 1. $m^{2} + 25 = 0 \Rightarrow m = \pm 5 i$ . So $α = 0, β = 5$ . Step 2 (WHY the general solution form). Complex roots $α \pm i β$ give real independent solutions $e^{α x} cos β x$ and $e^{α x} sin β x$ (Euler + superposition, built in the parent note); here $α = 0$ so $e^{α x} = e^{0} = 1$ and the envelope is flat — undamped. Answer: $y = C_{1} cos 5 x + C_{2} sin 5 x$ .

L2.2 Solve $y^{''} + 4 y^{'} + 8 y = 0$ .

Recall Solution L2.2

Step 1 (WHAT + WHY). Characteristic equation $m^{2} + 4 m + 8 = 0$ (trial $y = e^{m x}$ , divide off $e^{m x}$ ). Apply the quadratic formula because it solves any $a m^{2} + bm + c = 0$ exactly: $m = \frac{- 4 \pm 16 - 32}{2} = \frac{- 4 \pm - 16}{2} = \frac{- 4 \pm 4 i}{2} = - 2 \pm 2 i$ . Step 2. $α = - 2, β = 2$ . Step 3 (WHY this final form). Two real independent solutions $e^{- 2 x} cos 2 x$ , $e^{- 2 x} sin 2 x$ combine by superposition into the general solution. Answer: $y = e^{- 2 x} (C_{1} cos 2 x + C_{2} sin 2 x)$ . The $e^{- 2 x}$ is the decaying envelope (damped oscillation).

L2.3 Solve $4 y^{''} + y = 0$ (careful — $a \neq = 1$ ).

Recall Solution L2.3

Step 1. $4 m^{2} + 1 = 0 \Rightarrow m^{2} = - \frac{1}{4} \Rightarrow m = \pm \frac{1}{2} i$ . Step 2. $α = 0, β = \frac{1}{2}$ . Answer: $y = C_{1} cos \frac{x}{2} + C_{2} sin \frac{x}{2}$ . Note $β = \frac{1}{2}$ is small → slow oscillation, period $2 π / β = 4 π$ .

Level 3 — Analysis

Goal: initial value problems and reading physical meaning off the answer.

L3.1 Solve the IVP $y^{''} + 16 y = 0$ , $y (0) = 3$ , $y^{'} (0) = 8$ .

Recall Solution L3.1

Step 1 (WHAT + WHY). From L1.1 the roots are $m = \pm 4 i$ , so by the Case 3 form the general solution is $y = C_{1} cos 4 x + C_{2} sin 4 x$ (envelope $e^{0 x} = 1$ ). Why two constants: a second-order ODE needs two independent solutions, and the two initial data $y (0), y^{'} (0)$ will pin down $C_{1}, C_{2}$ . Step 2 (apply $y (0)$ ). $y (0) = C_{1} cos 0 + C_{2} sin 0 = C_{1} = 3$ . Why: $cos 0 = 1, sin 0 = 0$ . Step 3 (differentiate then apply $y^{'} (0)$ ). $y^{'} = - 4 C_{1} sin 4 x + 4 C_{2} cos 4 x$ , so $y^{'} (0) = 4 C_{2} = 8 \Rightarrow C_{2} = 2$ . Answer: $y = 3 cos 4 x + 2 sin 4 x$ .

L3.2 Solve the IVP $y^{''} + 2 y^{'} + 2 y = 0$ , $y (0) = 1$ , $y^{'} (0) = 0$ .

Recall Solution L3.2

Step 1. $m = \frac{- 2 \pm 4 - 8}{2} = \frac{- 2 \pm 2 i}{2} = - 1 \pm i$ , so $α = - 1, β = 1$ . Step 2. $y = e^{- x} (C_{1} cos x + C_{2} sin x)$ . Step 3 ( $y (0)$ ). $y (0) = e^{0} (C_{1} \cdot 1 + C_{2} \cdot 0) = C_{1} = 1$ . Step 4 ( $y^{'}$ ). Use product rule: $y^{'} = - e^{- x} (C_{1} cos x + C_{2} sin x) + e^{- x} (- C_{1} sin x + C_{2} cos x)$ . At $x = 0$ : $y^{'} (0) = - (C_{1}) + (C_{2}) = - C_{1} + C_{2} = - 1 + C_{2}$ . Set $= 0 \Rightarrow C_{2} = 1$ . Answer: $y = e^{- x} (cos x + sin x)$ .

L3.3 A damped oscillator gives $y = e^{- 0.5 x} (C_{1} cos 3 x + C_{2} sin 3 x)$ . What is (a) the oscillation period, (b) the factor by which the envelope shrinks over one period?

Recall Solution L3.3

(a) Period. $β = 3$ , so period $T = \frac{2 π}{β} = \frac{2 π}{3} \approx 2.094$ . (b) Envelope over one period. Envelope is $e^{- 0.5 x}$ . Over $Δ x = T = \frac{2 π}{3}$ it multiplies by $e^{- 0.5 \cdot 2 π /3} = e^{- π /3} \approx 0.351$ . Answer: period $\frac{2 π}{3} \approx 2.094$ ; amplitude drops to $\approx 0.351$ (about $35%$ ) each period.

Level 4 — Synthesis

Goal: run the machine backwards, or under a twist.

L4.1 Construct a real second-order ODE whose general solution is $y = e^{2 x} (C_{1} cos 5 x + C_{2} sin 5 x)$ .

Recall Solution L4.1

Step 1 (read off roots + WHY this reverses the machine). The form $e^{α x} (C_{1} cos β x + C_{2} sin β x)$ is produced only by roots $m = α \pm i β$ ; so matching it against the given solution forces $α = 2, β = 5$ , i.e. $m = 2 \pm 5 i$ . Step 2 (build the quadratic from its roots). A monic quadratic with roots $m_{1}, m_{2}$ is $(m - m_{1}) (m - m_{2}) = m^{2} - (m_{1} + m_{2}) m + m_{1} m_{2}$ .

Sum $m_{1} + m_{2} = (2 + 5 i) + (2 - 5 i) = 4$ .
Product $m_{1} m_{2} = (2 + 5 i) (2 - 5 i) = 4 - 25 i^{2} = 4 + 25 = 29$ . Step 3. Characteristic equation $m^{2} - 4 m + 29 = 0$ , so the ODE is $y^{''} - 4 y^{'} + 29 y = 0$ . Check: $Δ = 16 - 116 = - 100 < 0$ ✔, $α = - b /2 a = 4/2 = 2$ ✔, $β = 100 /2 = 5$ ✔. Answer: $y^{''} - 4 y^{'} + 29 y = 0$ .

L4.2 Find $a, b, c$ (with $a = 1$ ) so that $a y^{''} + b y^{'} + cy = 0$ has solutions that oscillate with period $π$ and whose amplitude halves every period. (Take the envelope $e^{α x}$ ; "halves every period" means $e^{α T} = \frac{1}{2}$ .)

Recall Solution L4.2

Step 1 (frequency from period). Period $T = \frac{2 π}{β} = π \Rightarrow β = 2$ . Step 2 (envelope condition). $e^{α T} = \frac{1}{2}$ with $T = π$ : $α π = ln \frac{1}{2} = - ln 2 \Rightarrow α = - \frac{ln 2}{π} \approx - 0.2206$ . Step 3 (build coefficients). With $a = 1$ : $b = - 2 α = \frac{2 ln 2}{π} \approx 0.4413$ , and $c = α^{2} + β^{2}$ (since $α^{2} + β^{2} = ∣ m ∣^{2} =$ product of the conjugate roots $= c / a$ ). $c \approx 0.220 6^{2} + 4 \approx 4.0487$ . Answer: $y^{''} + \frac{2 ln 2}{π} y^{'} + (\frac{ln ^{2} 2}{π ^{2}} + 4) y = 0$ , numerically $y^{''} + 0.4413 y^{'} + 4.0487 y = 0$ .

Level 5 — Mastery

Goal: everything at once — conjugate structure, Euler, independence, limits.

L5.1 Show directly (using Euler's formula) that recombining $y_{1} = e^{(α + i β) x}$ and $y_{2} = e^{(α - i β) x}$ as $u = \frac{1}{2} (y_{1} + y_{2})$ gives a real function, and compute it.

Recall Solution L5.1

Step 1 (Euler on each). $y_{1} = e^{α x} (cos β x + i sin β x)$ , $y_{2} = e^{α x} (cos β x - i sin β x)$ . Step 2 (add). $y_{1} + y_{2} = e^{α x} [(cos β x + cos β x) + i (sin β x - sin β x)] = e^{α x} \cdot 2 cos β x$ . The $i$ -parts cancel — that's the whole point of pairing conjugates. Step 3 (halve). $u = \frac{1}{2} (y_{1} + y_{2}) = e^{α x} cos β x$ — purely real ✔. By superposition, $u$ is again a solution. (Similarly $v = \frac{1}{2 i} (y_{1} - y_{2}) = e^{α x} sin β x$ .)

L5.2 Verify that $u = e^{α x} cos β x$ and $v = e^{α x} sin β x$ are linearly independent by computing their Wronskian, and state where it could ever fail.

Recall Solution L5.2

Step 0 (WHAT the Wronskian is). For two functions $u, v$ , the Wronskian is the determinant $W = u u^{'} v v^{'} = u v^{'} - v u^{'} .$ Why it matters: if $W \neq = 0$ at even one point, $u$ and $v$ cannot be constant multiples of each other → they are independent → they form a valid basis for the solution space. Step 1 (derivatives, product rule). $u^{'} = e^{α x} (α cos β x - β sin β x)$ , $v^{'} = e^{α x} (α sin β x + β cos β x)$ . Step 2 (determinant). $W = u v^{'} - v u^{'}$ . Factor out $e^{2 α x}$ : $W = e^{2 α x} [cos β x (α sin β x + β cos β x) - sin β x (α cos β x - β sin β x)] .$ The $α$ -terms cancel; the $β$ -terms give $β (cos^{2} β x + sin^{2} β x) = β$ . Step 3. $W = β e^{2 α x}$ . When could it fail? $W = 0$ only if $β = 0$ . But Case 3 requires $Δ < 0 \Rightarrow β > 0$ , and $e^{2 α x} > 0$ always. So $W \neq = 0$ everywhere → independent ✔.

L5.3 As $c \to$ its boundary value, Case 3 degenerates into Case 2. For $y^{''} + 2 y^{'} + cy = 0$ , find the value of $c$ where the roots stop being complex, and describe what happens to $β$ and to the solution shape as $c$ approaches it from above.

Recall Solution L5.3

Step 1 (discriminant). $a = 1, b = 2$ : $Δ = 4 - 4 c$ . Complex requires $Δ < 0 \Rightarrow c > 1$ . Step 2 (boundary). At $c = 1$ , $Δ = 0$ → repeated real root Case 2, $m = - 1$ (double). Step 3 (limit of $β$ ). $β = \frac{4 c - 4}{2} = c - 1$ . As $c \to 1^{+}$ , $β \to 0^{+}$ : the oscillation slows to an infinitely long period ( $T = 2 π / β \to \infty$ ). The wiggle straightens out into the non-oscillating $e^{- x} (C_{1} + C_{2} x)$ of Case 2. Answer: boundary at $c = 1$ ; as $c \to 1^{+}$ , $β \to 0$ and the oscillation disappears (period $\to \infty$ ), matching Case 2 continuously.

The figure below makes L5.3's limit visible. (Description: three coloured curves show $e^{α x} sin β x / β$ for shrinking $β$ = 1.5 (red), 0.6 (yellow), 0.15 (green). As $β$ falls the wiggle stretches out and its first hump widens; the dashed blue curve $x e^{α x}$ is the $β \to 0$ limit that all three approach — the Case 2 solution emerging from the vanishing oscillation.)

Read the picture together with the algebra: because $\frac{sin β x}{β} \to x$ as $β \to 0$ , the green ( $β = 0.15$ ) curve already hugs the dashed blue $x e^{α x}$ — the oscillation has almost entirely turned into the Case 2 factor $x e^{α x}$ . This is the continuous bridge between Case 3 and Case 2 that the L5 mistake box warns about.

Recall One-line self-test recap

Case 3 in three moves ::: (1) $Δ < 0$ ; (2) read $α = - b /2 a$ , $β = 4 a c - b^{2} /2 a$ ; (3) write $y = e^{α x} (C_{1} cos β x + C_{2} sin β x)$ . Reverse-engineering an ODE from roots $α \pm i β$ ::: sum $= 2 α$ , product $= α^{2} + β^{2}$ , so $m^{2} - 2 α m + (α^{2} + β^{2}) = 0$ .