Second-order linear differential equations with constant coefficients appear throughout engineering and applied mathematics, including mechanical vibrations, electrical circuits (RLC systems), structural analysis, and control systems.

The central tool for solving these equations is the characteristic equation, which transforms the differential equation into an algebraic problem.

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This guide explains the method clearly and systematically.

How to Solve These Equations (Quick Reference)

If you are given:

  ay″ + by′ + cy = 0

1. Form the characteristic equation:
     am² + bm + c = 0

2. Compute the discriminant:
     Δ = b² − 4ac

3. Use Δ to determine the solution form:

Once the roots are classified correctly, the structure of the solution follows immediately.

What Is a Second-Order Homogeneous Linear Differential Equation?

A differential equation of the form

  ay″ + by′ + cy = 0
  where a, b, c ∈ ℝ and a ≠ 0

is called:

• Second order because the highest derivative is y″

• Linear because y, y′, and y″ appear only to the first power

• Homogeneous because the right-hand side equals zero

• With constant coefficients because a, b, and c are constants

Why the Characteristic Equation Works

Suppose we try a solution of the form:

  y = e^(mx)

Then:

  y′ = me^(mx)
  y″ = m²e^(mx)

Substituting into

  ay″ + by′ + cy = 0

gives:

  (am² + bm + c)e^(mx) = 0

Since e^(mx) ≠ 0, the equation reduces to:

  am² + bm + c = 0

This quadratic equation is called the characteristic equation.
Its roots determine the form of the solution.

Step 1 — Form the Characteristic Equation

Assume:

  y = e^(mx)

Substitute into the differential equation and simplify to obtain:

  am² + bm + c = 0

This algebraic equation forms the foundation of the solution method.

Step 2 — Classify the Roots and Write the General Solution

Use the quadratic formula:

  m = (-b ± √(b² − 4ac)) / (2a)

Let Δ = b² − 4ac.

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

If the roots are m₁ and m₂, then the general solution is:

  y = C₁e^(m₁x) + C₂e^(m₂x)

Example

y″ − y′ − 6y = 0

Characteristic equation:

  m² − m − 6 = 0
  (m + 2)(m − 3) = 0

Roots:

  m = −2, 3

Solution:

  y = C₁e^(−2x) + C₂e^(3x)

When Δ > 0, the solution consists entirely of exponential growth or decay terms.

Case 2: One Repeated Root (Δ = 0)

If the characteristic equation has a repeated real root m, the solution becomes:

  y = C₁e^(mx) + C₂xe^(mx)

The factor of x ensures the two solutions are linearly independent.

Example

y″ − 10y′ + 25y = 0

Characteristic equation:

  (m − 5)² = 0

Root:

  m = 5

Solution:

  y = C₁e^(5x) + C₂xe^(5x)

Case 3: Complex Conjugate Roots (Δ < 0)

If the roots are:

  m = α ± βi

the real-valued solution is:

  y = e^(αx)(C₁cos βx + C₂sin βx)

Example

y″ − 6y′ + 13y = 0

Characteristic equation:

  m² − 6m + 13 = 0

Roots:

  m = 3 ± 2i

Solution:

  y = e^(3x)(C₁cos 2x + C₂sin 2x)

When Δ < 0, the solution exhibits oscillatory behavior combined with exponential growth or decay.

Solving Initial Value Problems (IVPs)

An initial value problem specifies conditions such as:

  y(0) = value
  y′(0) = value

These determine the constants C₁ and C₂.

Example

y″ + 16y = 0
y(0) = 2
y′(0) = −1

Characteristic equation:

  m² + 16 = 0
  m = ±4i

General solution:

  y = C₁cos 4x + C₂sin 4x

Applying the initial conditions:

  C₁ = 2
  C₂ = −1/4

Final solution:

  y = 2cos 4x − (1/4)sin 4x

Solving Boundary Value Problems (BVPs)

A boundary value problem specifies conditions at two different points.

Example

y″ + 8y′ + 16y = 0
y(0) = 1
y(1) = 0

Characteristic equation:

  (m + 4)² = 0

Root:

  m = −4

General solution:

  y = C₁e^(−4x) + C₂xe^(−4x)

Applying the boundary conditions:

  C₁ = 1
  C₂ = −1

Final solution:

  y = (1 − x)e^(−4x)

Non-Homogeneous Equations (Overview)

If the equation has the form

  ay″ + by′ + cy = g(x)

the solution is written as:

  y = yₕ + yₚ

where:

• yₕ is the homogeneous solution

• yₚ is a particular solution that depends on g(x)

Two common methods are:

1. Method of Undetermined Coefficients, used when g(x) is a polynomial, exponential, sine/cosine function, or combinations of these.

2. Variation of Parameters, a general method applicable to broader forcing functions.

Common Mistakes

• Forgetting the extra x in the repeated root case

• Misclassifying the discriminant

• Overlapping the trial solution in undetermined coefficients

• Dropping constants or misapplying signs

• Failing to verify independence of solutions

Extending to Higher-Order Equations

For equations such as

  aₙy^(n) + aₙ₋₁y^(n−1) + … + a₁y′ + a₀y = 0

form the characteristic polynomial, find its roots (including multiplicities), and build the solution using:

• exponentials for real roots

• x-multipliers for repeated roots

• sine and cosine for complex roots

Example

y‴ − 4y″ − 5y′ = 0

Characteristic equation:

  m(m − 5)(m + 1) = 0

Solution:

  y = C₁ + C₂e^(5x) + C₃e^(−x)

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