📄 Save or print this lesson:

← Module 4 Home Lesson 1 of 4 Next Lesson →

Lesson 1: Homogeneous with Constant Coefficients

Estimated time: 35-45 minutes

Learning Objectives

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

Recognize the standard form ay'' + by' + cy = 0
Explain the superposition principle for linear homogeneous ODEs
Derive the characteristic equation from the exponential guess y = e^rt
State the general solution structure using two linearly independent solutions
Verify linear independence using the Wronskian

The Standard Form

A second-order linear ODE with constant coefficients has the form:

ay'' + by' + cy = g(t)

where a, b, c are constants and g(t) is the forcing function. When g(t) = 0, the equation is homogeneous.

Homogeneous Second-Order Linear ODE: ay'' + by' + cy = 0 with a ≠ 0. This is the equation we solve in this lesson.

The Superposition Principle

Superposition Principle: If y₁ and y₂ are solutions to the homogeneous equation ay'' + by' + cy = 0, then c₁y₁ + c₂y₂ is also a solution for any constants c₁, c₂.

This means the set of all solutions forms a two-dimensional vector space. We need exactly two linearly independent solutions to span it.

The Characteristic Equation

We guess y = e^rt for some constant r. Substituting into ay'' + by' + cy = 0:

a r² e^rt + b r e^rt + c e^rt = 0

e^rt(ar² + br + c) = 0

Since e^rt ≠ 0, we need:

Characteristic Equation: ar² + br + c = 0. The roots r₁ and r₂ determine the form of the general solution.

The discriminant D = b² - 4ac determines three cases:

D > 0: Two distinct real roots (covered in Example 1 below; detailed in Lesson 2)
D = 0: One repeated real root (detailed in Lesson 2)
D < 0: Complex conjugate roots (detailed in Lesson 2)

Example 1: Distinct Real Roots

Solve y'' - 5y' + 6y = 0.

Step 1: Characteristic equation: r² - 5r + 6 = 0.

Step 2: Factor: (r - 2)(r - 3) = 0, so r₁ = 2, r₂ = 3.

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

Example 2: Finding Particular Solutions from IVPs

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

Step 1: General solution: y = c₁e^2t + c₂e^3t (from Example 1).

Step 2: Apply y(0) = 2: c₁ + c₂ = 2.

Step 3: y' = 2c₁e^2t + 3c₂e^3t. Apply y'(0) = 5: 2c₁ + 3c₂ = 5.

Step 4: Solve the system: c₁ = 1, c₂ = 1.

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

Linear Independence and the Wronskian

Wronskian: For two functions y₁ and y₂, the Wronskian is W(y₁, y₂) = y₁y₂' - y₂y₁'. If W ≠ 0, the functions are linearly independent.

Example 3: Checking Linear Independence

Verify that y₁ = e^2t and y₂ = e^3t are linearly independent.

W = e^2t · 3e^3t - e^3t · 2e^2t = 3e^5t - 2e^5t = e^5t ≠ 0.

Since W ≠ 0, the solutions are linearly independent, confirming they form a fundamental set.

Structure of the General Solution

General Solution (Homogeneous): If y₁ and y₂ are linearly independent solutions, then the general solution is y = c₁y₁ + c₂y₂, where c₁ and c₂ are arbitrary constants. An IVP with y(t₀) = y₀, y'(t₀) = y₀' determines c₁ and c₂ uniquely.

Check Your Understanding

1. Write the characteristic equation for y'' + 4y' + 3y = 0 and find its roots.

Answer: r² + 4r + 3 = 0 ⇒ (r+1)(r+3) = 0 ⇒ r = -1, r = -3. General solution: y = c₁e^-t + c₂e^-3t.

2. Solve y'' - y = 0.

Answer: r² - 1 = 0 ⇒ r = 1, r = -1. General solution: y = c₁e^t + c₂e^-t.

3. Compute the Wronskian of y₁ = e^-t and y₂ = e^-3t.

Answer: W = e^-t(-3e^-3t) - e^-3t(-e^-t) = -3e^-4t + e^-4t = -2e^-4t ≠ 0. Linearly independent.

4. Why does a second-order ODE need two initial conditions to determine a unique solution?

Answer: The general solution has two arbitrary constants (c₁ and c₂). Two equations (from y(t₀) and y'(t₀)) are needed to determine these two unknowns.

Key Takeaways

For ay'' + by' + cy = 0, guess y = e^rt to get the characteristic equation ar² + br + c = 0.
The superposition principle says any linear combination of solutions is also a solution.
Two linearly independent solutions form a fundamental set; the general solution is y = c₁y₁ + c₂y₂.
The Wronskian W = y₁y₂' - y₂y₁' tests linear independence (W ≠ 0 means independent).
An IVP needs two initial conditions (y and y') to determine the two constants.

Ready for More?

Next Lesson

In Lesson 2, you will master all three cases of the characteristic equation in detail.

Start Lesson 2

Module Progress

You have completed Lesson 1 of 4.

Back to Module Home