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Lesson 95 — 2nd-order linear ODEs with constant coefficients

ay'' + by' + cy = 0. Characteristic equation and three regimes: distinct real roots, repeated real root, complex conjugate roots.

Used in: Spécialité Maths (France, Terminale) · Math III japonês (avançado) · Leistungskurs alemão Klasse 12

ay'' + by' + cy = 0 \;\Longrightarrow\; a\lambda^2 + b\lambda + c = 0

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Rigorous notation, full derivation, hypotheses

Characteristic equation — three cases

Ansatz and characteristic equation

Substitute $y = e^{\lambda x}$ : $y' = \lambda e^{\lambda x}$ , $y'' = \lambda^2 e^{\lambda x}$ .

$a\lambda^2 e^{\lambda x} + b\lambda e^{\lambda x} + c e^{\lambda x} = 0 \;\Rightarrow\; a\lambda^2 + b\lambda + c = 0$

Since $e^{\lambda x} \neq 0$ , everything reduces to the quadratic algebraic equation.

"If $b^2 - 4ac > 0$ , the characteristic equation has two distinct real roots $r_1, r_2$ , and the general solution of [the ODE] is $y = c_1 e^{r_1 x} + c_2 e^{r_2 x}$ ." — Lebl, Notes on Diffy Qs, §2.2

Qualitative diagram — behavior by case

Qualitative profiles: case 1 (pure exponential), case 2 (critical boundary), case 3 (oscillatory).

Non-homogeneous equations

For $ay'' + by' + cy = q(x)$ : general solution $y = y_h + y_p$ .

$y_h$ : solution to the homogeneous equation (two free parameters).

$y_p$ : any particular solution — obtained by undetermined coefficients (when $q$ is a combination of polynomials, exponentials, sines/cosines) or variation of parameters (general).

Solved examples

Example— 1· Case 1 — distinct real roots

Problem. Solve $y'' - 5y' + 6y = 0$ , $y(0) = 2$ , $y'(0) = 3$ .

Strategy. $\Delta = 25 - 24 = 1 > 0$ : two distinct real roots.

Resolution.

Characteristic equation: $\lambda^2 - 5\lambda + 6 = (\lambda-2)(\lambda-3) = 0$ .

$\lambda_1 = 2$ , $\lambda_2 = 3$ . General solution: $y = C_1 e^{2x} + C_2 e^{3x}$ .

$y(0) = C_1 + C_2 = 2$ .

$y'(0) = 2C_1 + 3C_2 = 3$ .

System: $C_1 = 3$ , $C_2 = -1$ .

$y = 3e^{2x} - e^{3x}$

Verification. $y'' - 5y' + 6y = (12-15+6\cdot 3)e^{2x} + (-9+15-6)e^{3x} = 12e^{2x} \cdot 0 + 0 = 0$ . $y(0) = 3 - 1 = 2$ , $y'(0) = 6 - 3 = 3$ . Correct.

Source. Lebl, Notes on Diffy Qs, §2.2, Example 2.2.1 — CC-BY-SA.

Example— 2· Case 2 — repeated root

Problem. Solve $y'' - 4y' + 4y = 0$ , $y(0) = 1$ , $y'(0) = 0$ .

Strategy. $\Delta = 16 - 16 = 0$ : repeated root.

Resolution.

$\lambda^2 - 4\lambda + 4 = (\lambda-2)^2 = 0$ . Repeated root $\lambda = 2$ .

$y = (C_1 + C_2 x)e^{2x}$ .

$y(0) = C_1 = 1$ .

$y'(x) = C_2 e^{2x} + 2(C_1 + C_2 x)e^{2x}$ .

$y'(0) = C_2 + 2C_1 = 0 \Rightarrow C_2 = -2$ .

$y = (1 - 2x)e^{2x}$

Verification. $y' = -2e^{2x} + 2(1-2x)e^{2x} = (2-4x-2)e^{2x} = -4xe^{2x}$ . $y'' = -4e^{2x} - 8xe^{2x}$ . $y''-4y'+4y = (-4-8x)e^{2x} - 4(-4x)e^{2x} + 4(1-2x)e^{2x} = (-4-8x+16x+4-8x)e^{2x} = 0$ . Correct.

Source. Lebl, Notes on Diffy Qs, §2.2, Example 2.2.2 — CC-BY-SA.

Example— 3· Case 3 — complex conjugate roots (damped oscillation)

Problem. Solve $y'' + 2y' + 5y = 0$ , $y(0) = 1$ , $y'(0) = 2$ .

Strategy. $\Delta = 4 - 20 = -16 < 0$ : complex. $\alpha = -1$ , $\beta = 2$ .

Resolution.

$\lambda = (-2 \pm \sqrt{-16})/2 = -1 \pm 2i$ .

$y = e^{-x}(C_1\cos 2x + C_2\sin 2x)$ .

$y(0) = C_1 = 1$ .

$y'(x) = -e^{-x}(C_1\cos 2x + C_2\sin 2x) + e^{-x}(-2C_1\sin 2x + 2C_2\cos 2x)$ .

$y'(0) = -C_1 + 2C_2 = 2 \Rightarrow 2C_2 = 3 \Rightarrow C_2 = 3/2$ .

$y = e^{-x}\!\left(\cos 2x + \frac{3}{2}\sin 2x\right)$

Verification. The characteristic equation has roots $-1 \pm 2i$ , both with real part $-1 < 0$ : solution decays exponentially. $y(0) = 1$ , $y'(0) = -1 + 3 = 2$ . Correct.

Source. Lebl, Notes on Diffy Qs, §2.2, Example 2.2.3 — CC-BY-SA.

Example— 4· Non-homogeneous by undetermined coefficients

Problem. Solve $y'' - y' - 2y = e^{3x}$ .

Strategy. $y = y_h + y_p$ . For $q = e^{3x}$ and $3$ is not a characteristic root: guess $y_p = Ae^{3x}$ .

Resolution.

$\lambda^2 - \lambda - 2 = (\lambda-2)(\lambda+1) = 0$ . $y_h = C_1 e^{2x} + C_2 e^{-x}$ .

Guess: $y_p = Ae^{3x}$ . Substitute:

$9Ae^{3x} - 3Ae^{3x} - 2Ae^{3x} = e^{3x} \Rightarrow 4A = 1 \Rightarrow A = 1/4$ .

$y = C_1 e^{2x} + C_2 e^{-x} + \frac{1}{4}e^{3x}$

Verification. $y_p'' - y_p' - 2y_p = (9/4 - 3/4 - 2/4)e^{3x} = (9-3-2)/4\,e^{3x} = e^{3x}$ . Correct.

Source. Lebl, Notes on Diffy Qs, §2.3, Example 2.3.1 — CC-BY-SA.

Example— 5· Non-homogeneous with resonance (modification rule)

Problem. Solve $y'' - 3y' + 2y = e^{2x}$ .

Strategy. Try $y_p = Ae^{2x}$ — but $\lambda = 2$ is a characteristic root! Modification rule: multiply by $x$ .

Resolution.

$\lambda^2 - 3\lambda + 2 = (\lambda-1)(\lambda-2) = 0$ . $y_h = C_1 e^x + C_2 e^{2x}$ .

$\lambda = 2$ is a simple root: try $y_p = Axe^{2x}$ .

$y_p' = Ae^{2x} + 2Axe^{2x}$ , $y_p'' = 4Ae^{2x} + 4Axe^{2x}$ .

$y_p'' - 3y_p' + 2y_p = (4A + 4Ax)e^{2x} - (3A + 6Ax)e^{2x} + 2Axe^{2x} = Ae^{2x}$ .

$A = 1$ . Thus $y_p = xe^{2x}$ .

$y = C_1 e^x + C_2 e^{2x} + xe^{2x}$

Verification. $y_p = xe^{2x}$ : $y_p'' - 3y_p' + 2y_p = (4+4x)e^{2x} - 3(1+2x)e^{2x} + 2xe^{2x} = (4+4x-3-6x+2x)e^{2x} = e^{2x}$ . Correct.

Source. Lebl, Notes on Diffy Qs, §2.3, Example 2.3.2 — CC-BY-SA.

Exercise list

30 exercises · 7 with worked solution (25%)

Application 18Understanding 3Modeling 4Challenge 3Proof 2

Ex. 95.1Application
Solve $y'' - 3y' + 2y = 0$ .
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Ex. 95.2Application
Solve $y'' + 4y' + 4y = 0$ .
Solve online
Ex. 95.3ApplicationAnswer key
Solve $y'' + 9y = 0$ .
Solve online
Ex. 95.4Application
Solve $y'' + 2y' + 10y = 0$ .
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Ex. 95.5Application
Solve $y'' + 4y' + 3y = 0$ , $y(0) = 1$ , $y'(0) = 0$ .
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Ex. 95.6Application
Solve $y'' - 2y' + y = 0$ , $y(0) = 2$ , $y'(0) = -1$ .
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Ex. 95.7Application
Solve $y'' + 4y = 0$ , $y(0) = 0$ , $y'(0) = 1$ .
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Ex. 95.8Application
Solve $y'' - y' - 6y = 0$ .
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Ex. 95.9Application
Solve $y'' + 2y' + 5y = 0$ .
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Ex. 95.10ApplicationAnswer key
Solve $y'' + 6y' + 9y = 0$ .
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Ex. 95.11UnderstandingAnswer key
What is the correct general solution form for $y'' - 2y' + 2y = 0$ ?
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Ex. 95.12Understanding
$y'' + by' + cy = 0$ with $a = 1$ . Under what condition(s) on $b, c$ does the solution decay to zero?
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Ex. 95.13Application
Solve $y'' - y = e^{2x}$ .
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Ex. 95.14Application
Solve $y'' + 4y = 3\sin x$ .
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Ex. 95.15ApplicationAnswer key
Solve $y'' + 2y' = 4x^2$ .
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Ex. 95.16ApplicationAnswer key
Solve $y'' + 4y = \sin 2x$ (resonance case).
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Ex. 95.17Application
Calculate the Wronskian of $y_1 = e^x$ and $y_2 = e^{2x}$ and confirm they form a fundamental set of solutions for $y'' - 3y' + 2y = 0$ .
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Ex. 95.18Application
LC circuit: $L\ddot q + q/C = 0$ , $q(0) = Q_0$ , $\dot q(0) = 0$ . Find $q(t)$ and $\omega_0$ .
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Ex. 95.19Understanding
Is $\{\cos 3x, \sin 3x\}$ a fundamental set of solutions for $y'' + 9y = 0$ ?
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Ex. 95.20Modeling
Mass-spring without friction: $m = 2\,\text{kg}$ , $k = 500\,\text{N/m}$ , $y(0) = 0{,}1\,\text{m}$ , $y'(0) = 0$ . Find $\omega_0$ , period, and displacement $y(t)$ .
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Ex. 95.21Modeling
Damped oscillator: $y'' + 5y' + 6y = 0$ , $y(0) = 1$ , $y'(0) = 0$ . Classify (under/over/critically) and solve.
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Ex. 95.22ModelingAnswer key
Critical damping: $2y'' + 4y' + 2y = 0$ , $y(0) = 1$ , $y'(0) = -1$ . Solve and classify.
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Ex. 95.23Modeling
Underdamped system: $y'' + 4y' + 13y = 0$ , $y(0) = 1$ , $y'(0) = 0$ . Solve and identify damped frequency $\omega_d$ .
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Ex. 95.24Application
Solve $y'' + 4y' + 3y = 5e^{-x}$ (modification rule required).
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Ex. 95.25Application
Solve $y'' - 4y' + 4y = e^{2x}$ (repeated root — guess increases to $x^2$ ).
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Ex. 95.26Challenge
Apply variation of parameters to $y'' - y = \sec x$ . Does the result have a closed form?
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Ex. 95.27Challenge
Cauchy-Euler equation: $x^2 y'' + xy' + y = 0$ . Try $y = x^m$ and solve for $m$ .
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Ex. 95.28Challenge
Use reduction of order with $y_1 = e^{3x}$ to find the second solution of $y'' - y' - 6y = 0$ .
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Ex. 95.29Proof
Demonstrate that, when $\lambda$ is a repeated root of $a\lambda^2 + b\lambda + c = 0$ , the function $y_2 = xe^{\lambda x}$ satisfies $ay'' + by' + cy = 0$ .
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Ex. 95.30ProofAnswer key
State and justify the existence-uniqueness theorem for $ay'' + by' + cy = q(x)$ , $y(x_0) = y_0$ , $y'(x_0) = y_1$ .
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Sources

Notes on Diffy Qs — Jiří Lebl · v6.6 · §2.2–2.3 · EN · CC-BY-SA. Primary source.
Calculus Volume 2 — OpenStax · §7.1–7.2 · EN · CC-BY-NC-SA.
Elementary Differential Equations — William F. Trench · §5.1–5.2, §5.6–5.7 · EN · open.

Lesson 95 — 2nd-order linear ODEs with constant coefficients

Characteristic equation — three cases

General problem

Ansatz and characteristic equation

The repeated root case

Qualitative diagram — behavior by case

Non-homogeneous equations

Solved examples

Exercise list

Sources