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MathidiaMathematics Academy
Differential equations
Lesson 53 of 75
Standard practice

Second-order equations

Objective. Solve constant-coefficient homogeneous equations.

  1. 1. Learn
  2. 2. Worked example
  3. 3. Practice
  4. 4. Feedback
  5. 5. Continue

Learn: the key idea

A second-order equation here means a constant-coefficient homogeneous ODE ay'' + by' + cy = 0, solved via the characteristic equation aλ² + bλ + c = 0. Real distinct roots λ₁, λ₂ give y = Ae^(λ₁x) + Be^(λ₂x); a repeated root λ gives y = (A + Bx)e^(λx); complex roots p ± qi give y = e^(px)(Acos(qx) + Bsin(qx)). For y'' − 3y' + 2y = 0 the characteristic equation λ² − 3λ + 2 = 0 factors as (λ−1)(λ−2), giving y = Ae^x + Be^(2x). Two conditions are needed to fix both constants.

Worked example

Solve the second-order equation y'' − 5y' + 6y = 0, giving the general solution.

5 … 6
Coordinate grid with the line 5 … 6
  1. 1Characteristic: λ² − 5λ + 6 = 0
  2. 2Factor: (λ−2)(λ−3) = 0
  3. 3Roots: λ = 2, λ = 3
  4. 4General solution: y = Ae^(2x) + Be^(3x)

Answer: y = Ae^(2x) + Be^(3x)

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Common mistake. Treating a repeated root λ like two independent roots and writing y = Ae^(λx) + Be^(λx); this collapses to one constant. The correct second solution is xe^(λx), not another copy of e^(λx).

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