From Scalar to Vector

Differential Equations · Axiom Academy

How a second-order equation becomes a first-order system, and the companion matrix that carries it 1. Starting Point: One Function, One Equation Consider the general second-order linear equation with constant coefficients: This is a scalar equation — a single unknown function y(t) appears together with its first and second derivatives. Solving it means finding one curve y(t) that makes the whole expression balance to zero at every instant t . 2. The Key Substitution: Naming the Velocity Introduce two new variables that track the function and its own rate of change as separate coordinates: Differentiate both definitions. The first is immediate; the second calls on the original equation itself, solved for y'' : Collecting both derivative relations gives a system of two first-order equations — every term is a first derivative; y'' has vanished entirely: One equation, involves y'' . Needs y(0) and y'(0) to pin down a solution. Two equations, only first derivatives x_1' , x_2' . Needs x_1(0) and x_2(0) — the same two numbers. 4. Matrix Form: The Companion Matrix The two-equation system is exactly a matrix-vector product. Watch the coefficients of x_1 and x_2 in each equation slot directly into a matrix A : Once the equation is , the entire machinery of linear algebra applies. For our running case y'' - 5y' + 6y = 0 , the companion matrix is: A 's eigenvalues are exactly the scalar equation's characteristic roots — here , matching r^2 - 5r + 6 = (r-2)(r-3) = 0 .

This is the written version of the interactive lesson above. See the full Differential Equations course.