The Characteristic Equation (2nd order)
Ch 6 — Part I — Explicit & Qualitative Methods
The characteristic equation is the polynomial whose roots generate exponential solutions of a constant-coefficient linear ODE. It converts calculus into algebra.
General rule (any order)
For \(a_n y^{(n)} + a_{n-1} y^{(n-1)} + \cdots + a_0 y = 0\), the characteristic polynomial is $$p(\lambda) = a_n \lambda^n + a_{n-1} \lambda^{n-1} + \cdots + a_0.$$ Each root \(\lambda\) of multiplicity \(m\) contributes \(m\) solutions $$e^{\lambda t}, t e^{\lambda t}, \ldots, t^{m-1} e^{\lambda t}.$$
Complex roots in conjugate pairs
A conjugate pair \(r \pm i s\) with multiplicity \(m\) contributes \(2m\) real solutions $$e^{rt}\cos(st), e^{rt}\sin(st), t e^{rt}\cos(st), t e^{rt}\sin(st), \ldots$$
Why it works
Differential operators with constant coefficients commute and factor like polynomials. Solving \(p(D) y = 0\) reduces to solving \(p(\lambda) = 0\) and combining kernels of \((D - \lambda)^m\).
Practice Problems
Hint
Solution
\(\lambda^3 - \lambda^2 - \lambda + 1 = \lambda^2(\lambda - 1) - (\lambda - 1) = (\lambda-1)(\lambda^2-1) = (\lambda-1)^2(\lambda+1)\). Roots \(1\) (double) and \(-1\).
Answer: \(y = (c_1 + c_2 t) e^t + c_3 e^{-t}\).
Hint
Solution
Roots \(\pm i\) each with multiplicity 2. Solutions: \(\cos t, \sin t, t\cos t, t\sin t\).
Answer: \(y = c_1\cos t + c_2 \sin t + c_3 t\cos t + c_4 t\sin t.\)
Hint
Solution
\(a^2 - 4 < 0\Rightarrow |a| < 2\).
Answer: \(-2 < a < 2.\)