Second-Order Homogeneous Linear ODEs
Ch 6 — Part I — Explicit & Qualitative Methods
A second-order linear homogeneous ODE with constant coefficients has the form $$y'' + a y' + b y = 0, \quad a, b \in \mathbb{R}.$$ It has a two-dimensional solution space. Finding a basis of solutions reduces to algebra on the characteristic polynomial.
Ansatz: exponential
Guess \(y = e^{\lambda t}\). Substitution gives \((\lambda^2 + a\lambda + b) e^{\lambda t} = 0\). So \(\lambda\) must solve the characteristic equation $$\lambda^2 + a \lambda + b = 0.$$
Three cases for roots
Let \(\Delta = a^2 - 4b\).
1. \(\Delta > 0\), distinct real roots \(\lambda_1 \ne \lambda_2\): $$y(t) = c_1 e^{\lambda_1 t} + c_2 e^{\lambda_2 t}.$$ 2. \(\Delta < 0\), complex conjugate \(\lambda = r_0 \pm i s_0\): $$y(t) = e^{r_0 t} (c_1 \cos s_0 t + c_2 \sin s_0 t).$$ 3. \(\Delta = 0\), repeated real \(\lambda\): $$y(t) = (c_1 + c_2 t) e^{\lambda t}.$$
Superposition & basis
The two base functions are linearly independent (check the Wronskian), so every solution is a unique linear combination.
Practice Problems
Hint
Solution
Roots \(\lambda = 2, 3\). General solution \(c_1 e^{2t} + c_2 e^{3t}\). \(y(0) = c_1 + c_2 = 1,\; y'(0) = 2c_1 + 3c_2 = 0\Rightarrow c_1 = 3,\; c_2 = -2\).
Answer: \(y(t) = 3 e^{2t} - 2 e^{3t}.\)
Hint
Solution
Roots \(\lambda = -1 \pm 2i\). So \(y = e^{-t}(c_1 \cos 2t + c_2 \sin 2t)\). From \(y(0)=1\): \(c_1 = 1\). \(y'(t) = -e^{-t}(c_1\cos 2t + c_2\sin 2t) + e^{-t}(-2c_1\sin 2t + 2c_2\cos 2t)\). \(y'(0) = -c_1 + 2c_2 = -1 + 2c_2 = -1\Rightarrow c_2 = 0\).
Answer: \(y(t) = e^{-t}\cos 2t.\)
Hint
Solution
\(\lambda = 2\) (double). \(y = (c_1 + c_2 t) e^{2t}\). \(y(0) = c_1 = 2\). \(y' = c_2 e^{2t} + 2(c_1 + c_2 t) e^{2t}\), so \(y'(0) = c_2 + 2c_1 = c_2 + 4 = 1\Rightarrow c_2 = -3\).
Answer: \(y(t) = (2 - 3t) e^{2t}.\)
Hint
Solution
\(\lambda = -1\) double. Solutions \((c_1 + c_2 t)e^{-t}\) — critically damped, decays to zero without oscillation.
Answer: Critically damped decay.