Systems of First-Order Linear ODEs

The state-vector reduction is the starting point of control theory, where any linear time-invariant plant description, however high its order, is rewritten as $\dot{\mathbf x}=A\mathbf x+B\mathbf u$ so that stability and controllability can be studied via the eigenvalues of $A$ — see the foreshadowed Eigenvalue Method for Linear Systems. In numerical analysis, virtually every ODE solver (Runge–Kutta, multistep methods) is coded to accept only first-order systems, so higher-order equations are always silently converted this way before numerical integration. In physics, the reduction to phase space (position and momentum, or a full state vector) underlies Hamiltonian and Lagrangian mechanics, as well as normal-mode analysis of coupled oscillators. In electrical engineering, circuits with multiple inductors and capacitors give rise naturally to first-order vector systems via Kirchhoff's laws, with the companion form appearing when a single-loop circuit's governing equation is high order. In statistics and signal processing, state-space models (Kalman filters) are literally the discrete-time analogue of $\mathbf{x}'=A\mathbf{x}+\mathbf{g}$, and the matrix exponential of $A$ (the natural next topic, the Matrix Exponential) gives the exact transition operator between states. Finally, in pure mathematics, this reduction is the entry point to the general theory of linear systems with variable coefficients, fundamental matrices, and Floquet theory for periodic $A(t)$.

Ask the tutor: How do I choose a particular solution ansatz when the forcing term resonates with an eigenvalue of $A$?

Think of a scalar $n$th-order equation as describing a single particle whose position obeys a complicated rule involving its own velocity, acceleration, and so on up through the $(n-1)$th derivative. The trick of the companion reduction is to stop thinking of "position, velocity, acceleration, …" as a hierarchy of derived quantities and instead treat them as independent coordinates of a single point moving in $n$-dimensional space. The rule that used to be one complicated equation of order $n$ becomes $n$ simple equations of order 1: the first $n-1$ just say "the next coordinate is the derivative of the previous one" ($x_1' = x_2$, $x_2' = x_3$, etc.), and only the last equation carries the actual physics, restating the original ODE for $y^{(n)}$ in terms of the coordinates. This is exactly the passage from configuration space to phase space familiar from mechanics: instead of tracking $y(t)$ alone, we track the point $(y, y', \dots, y^{(n-1)})(t)$ moving through $\mathbb{R}^n$, and the entire dynamics is now a velocity field $A(t)\mathbf{x} + \mathbf{g}(t)$ prescribing, at every point and time, exactly how the state vector should move. Once in this form, matrix algebra — eigenvalues, determinants, fundamental matrices — becomes available in a way it never was for the scalar higher-order picture.

Example 1 (Homogeneous, order 3, converted and solved as a system). Convert

to a first-order system, then find the general solution. Reduction. Let $x_1 = y,\ x_2 = y',\ x_3 = y''$. Then $x_1' = x_2$, $x_2' = x_3$, and from the ODE, $x_3' = -2x_1 + x_2 + 2x_3$. In matrix form,

Solving. Because this is the companion matrix of the scalar ODE, its characteristic polynomial equals the ODE's characteristic polynomial (up to sign convention), $\lambda^3 - 2\lambda^2 - \lambda + 2 = 0$. Factor:

Roots $\lambda = 2, 1, -1$, all distinct. The scalar solutions are $y = e^{2t}, e^{t}, e^{-t}$, giving state vectors $(y,y',y'')$:

Check these are eigenvectors: for $\lambda=2$: $A(1,2,4)^T = (2, 4, -2+2+8)^T=(2,4,8)^T = 2(1,2,4)^T$. ✓. For $\lambda=1$: $A(1,1,1)^T = (1,1,-2+1+2)^T = (1,1,1)^T$. ✓. For $\lambda=-1$: $A(1,-1,1)^T = (-1,1,-2-1+2)^T=(-1,1,-1)^T = -1\cdot(1,-1,1)^T$. ✓. Since $\{\boldsymbol\phi_1,\boldsymbol\phi_2,\boldsymbol\phi_3\}$ are eigenvector solutions for three distinct eigenvalues, they are linearly independent (standard fact from linear algebra: eigenvectors for distinct eigenvalues are independent), hence form a basis of the 3-dimensional solution space guaranteed by the theorem. General solution:

and reading off the first component, $y(t) = c_1 e^{2t} + c_2 e^t + c_3 e^{-t}$ — matching what characteristic-root theory for the scalar ODE would predict directly, confirming the reduction is faithful.

Example 2 (Non-symmetric first-order system with forcing, solved by direct elimination). Solve the initial value problem

Setup. Here $A = \begin{pmatrix} 4 & 1 \\ -2 & 1\end{pmatrix}$, $\mathbf{g}(t) = (t, 1)^T$ — continuous everywhere, so by the theorem a unique solution exists on all of $\mathbb{R}$. Homogeneous part. Characteristic polynomial: $\det(A-\lambda I) = (4-\lambda)(1-\lambda) - (1)(-2) = \lambda^2 - 5\lambda + 4 + 2 = \lambda^2 - 5\lambda + 6 = (\lambda-2)(\lambda-3)$. For $\lambda=2$: $(A-2I)\mathbf{v}=0 \Rightarrow \begin{pmatrix}2&1\\-2&-1\end{pmatrix}\mathbf{v}=0 \Rightarrow v_2 = -2v_1$. Eigenvector $(1,-2)^T$. For $\lambda=3$: $(A-3I)\mathbf{v}=0 \Rightarrow \begin{pmatrix}1&1\\-2&-2\end{pmatrix}\mathbf{v}=0 \Rightarrow v_2=-v_1$. Eigenvector $(1,-1)^T$. Homogeneous general solution:

Particular solution. Try $\mathbf{x}_p = \mathbf{a}t + \mathbf{b}$ with $\mathbf{a}=(a_1,a_2)^T,\ \mathbf{b}=(b_1,b_2)^T$ (undetermined coefficients, since $\mathbf{g}$ is linear in $t$ and no resonance with $\lambda=0$ occurs as $0$ is not an eigenvalue). Then $\mathbf{x}_p' = \mathbf{a}$, and we need

where $\mathbf{e}_1=(1,0)^T,\ \mathbf{e}_2=(0,1)^T$. Matching coefficients of $t$: $A\mathbf{a} + \mathbf{e}_1 = 0 \Rightarrow A\mathbf{a} = -\mathbf{e}_1$. Matching constants: $\mathbf{a} = A\mathbf{b}+\mathbf{e}_2$. Solve $A\mathbf{a}=-\mathbf{e}_1$: $4a_1+a_2=-1$, $-2a_1+a_2=0 \Rightarrow a_2 = 2a_1$. Substituting: $4a_1+2a_1=-1 \Rightarrow a_1 = -1/6$, $a_2=-1/3$. So $\mathbf{a} = (-1/6, -1/3)^T$. Solve $A\mathbf{b} = \mathbf{a}-\mathbf{e}_2 = (-1/6, -1/3-1)^T = (-1/6,-4/3)^T$:

Subtract: $6b_1 = -\tfrac16+\tfrac43 = \tfrac{-1+8}{6}=\tfrac{7}{6} \Rightarrow b_1 = \tfrac{7}{36}$. Then $b_2 = -\tfrac43+2b_1 = -\tfrac{48}{36}+\tfrac{14}{36} = -\tfrac{34}{36}=-\tfrac{17}{18}$. So

General solution and initial conditions.

At $t=0$: $x_1(0) = c_1+c_2+\tfrac{7}{36} = 1 \Rightarrow c_1+c_2 = \tfrac{29}{36}$. $x_2(0) = -2c_1-c_2-\tfrac{17}{18} = 0 \Rightarrow -2c_1-c_2 = \tfrac{17}{18}=\tfrac{34}{36}$. Add the two equations: $-c_1 = \tfrac{29}{36}+\tfrac{34}{36}=\tfrac{63}{36}=\tfrac{7}{4} \Rightarrow c_1 = -\tfrac{7}{4}$. Then $c_2 = \tfrac{29}{36}-c_1 = \tfrac{29}{36}+\tfrac{63}{36} = \tfrac{92}{36}=\tfrac{23}{9}$. Verification at $t=0$: $x_1(0) = -\tfrac74+\tfrac{23}{9}+\tfrac{7}{36}$. Common denominator 36: $-\tfrac{63}{36}+\tfrac{92}{36}+\tfrac{7}{36} = \tfrac{36}{36}=1$. ✓ matches given $x_1(0)=1$. $x_2(0) = -2(-\tfrac74)-\tfrac{23}{9}-\tfrac{17}{18} = \tfrac{7}{2}-\tfrac{23}{9}-\tfrac{17}{18}$. Common denominator 18: $\tfrac{63}{18}-\tfrac{46}{18}-\tfrac{17}{18}=0$. ✓ matches given $x_2(0)=0$. Final answer: