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Lesson 2: Characteristic Equation -- Three Cases

Estimated time: 40-50 minutes

Learning Objectives

By the end of this lesson, you will be able to:

Write the general solution for two distinct real roots
Handle the repeated root case with the te^rt trick
Use Euler's formula to convert complex roots to real-valued solutions
Identify oscillatory, decaying, and growing behavior from the roots
Solve IVPs for all three cases

Case 1: Two Distinct Real Roots (D > 0)

When b² - 4ac > 0, the characteristic equation ar² + br + c = 0 has two distinct real roots r₁ ≠ r₂.

Case 1 General Solution: y(t) = c₁e^r₁t + c₂e^r₂t

Example 1: Distinct Real Roots

Solve y'' + y' - 6y = 0.

Characteristic equation: r² + r - 6 = 0 ⇒ (r+3)(r-2) = 0 ⇒ r = -3, 2.

General solution: y(t) = c₁e^-3t + c₂e^2t.

Behavior: The e^-3t term decays; the e^2t term grows. Unless c₂ = 0, solutions grow unboundedly.

Case 2: Repeated Root (D = 0)

When b² - 4ac = 0, there is one repeated root r = -b/(2a). We get only one exponential solution e^rt. The second linearly independent solution is te^rt.

Case 2 General Solution: y(t) = c₁e^rt + c₂te^rt = (c₁ + c₂t)e^rt

Example 2: Repeated Root

Solve y'' - 4y' + 4y = 0.

Characteristic equation: r² - 4r + 4 = 0 ⇒ (r-2)² = 0 ⇒ r = 2 (repeated).

General solution: y(t) = c₁e^2t + c₂te^2t = (c₁ + c₂t)e^2t.

Example 3: Repeated Root IVP

Solve y'' + 6y' + 9y = 0, y(0) = 2, y'(0) = -1.

Step 1: r² + 6r + 9 = (r+3)² = 0 ⇒ r = -3 (repeated).

Step 2: y = (c₁ + c₂t)e^-3t.

Step 3: y(0) = c₁ = 2.

Step 4: y' = c₂e^-3t - 3(c₁ + c₂t)e^-3t. y'(0) = c₂ - 3c₁ = c₂ - 6 = -1, so c₂ = 5.

Answer: y(t) = (2 + 5t)e^-3t.

Case 3: Complex Conjugate Roots (D < 0)

When b² - 4ac < 0, the roots are complex conjugates: r = α ± βi where α = -b/(2a) and β = √(4ac - b²)/(2a).

Euler's Formula: e^iβt = cos(βt) + i sin(βt). This lets us convert complex exponentials into real-valued functions.

Case 3 General Solution: y(t) = e^αt[c₁cos(βt) + c₂sin(βt)]

The factor e^αt controls growth/decay. The trig functions create oscillations with frequency β.

Example 4: Complex Roots

Solve y'' + 4y' + 13y = 0.

Step 1: r² + 4r + 13 = 0. D = 16 - 52 = -36.

Step 2: r = (-4 ± √(-36))/2 = (-4 ± 6i)/2 = -2 ± 3i.

So α = -2, β = 3.

General solution: y(t) = e^-2t[c₁cos(3t) + c₂sin(3t)].

Behavior: Decaying oscillation (since α = -2 < 0).

Example 5: Pure Oscillation

Solve y'' + 9y = 0.

Characteristic equation: r² + 9 = 0 ⇒ r = ±3i. Here α = 0, β = 3.

General solution: y(t) = c₁cos(3t) + c₂sin(3t).

Behavior: Pure oscillation with no decay or growth (since α = 0). This is simple harmonic motion.

Summary of All Three Cases

Discriminant	Roots	General Solution
D > 0	r₁, r₂ real distinct	c₁e^r₁t + c₂e^r₂t
D = 0	r repeated	(c₁ + c₂t)e^rt
D < 0	α ± βi	e^αt[c₁cos(βt) + c₂sin(βt)]

Check Your Understanding

1. Solve y'' - 2y' - 8y = 0.

Answer: r² - 2r - 8 = (r-4)(r+2) = 0. Roots: r = 4, -2. y = c₁e^4t + c₂e^-2t.

2. Solve y'' + 2y' + y = 0.

Answer: r² + 2r + 1 = (r+1)² = 0. Repeated root r = -1. y = (c₁ + c₂t)e^-t.

3. Solve y'' + 2y' + 5y = 0.

Answer: r² + 2r + 5 = 0. r = (-2 ± √(4-20))/2 = -1 ± 2i. α = -1, β = 2. y = e^-t[c₁cos(2t) + c₂sin(2t)].

4. For complex roots α ± βi, under what condition do solutions decay?

Answer: Solutions decay when α < 0. The exponential envelope e^αt shrinks to zero. When α = 0, solutions oscillate forever. When α > 0, oscillations grow.

Key Takeaways

Distinct real roots r₁, r₂: y = c₁e^r₁t + c₂e^r₂t.
Repeated root r: y = (c₁ + c₂t)e^rt. The extra factor of t provides the second independent solution.
Complex roots α ± βi: y = e^αt[c₁cos βt + c₂sin βt]. Euler's formula bridges complex and real forms.
α < 0 means damped oscillation, α = 0 means pure oscillation, α > 0 means growing oscillation.

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