First-Order Linear System Form: Matrix Form for ODE Systems

By Vegard Gjerde Based on Masterful Learning 11 min read Published
first-order-linear-system-form differential-equations math learning-strategies

First-Order Linear System Form says a system of first-order equations can be written as x=A(t)x+f(t)\mathbf{x}^{\prime}=A(t)\mathbf{x}+\mathbf{f}(t), where a coefficient matrix acts on the unknown vector and a forcing vector is added. It applies when the matrix and forcing vector are defined on the working interval. Use it to recognize a coupled linear system before checking the extra requirements for constant-coefficient, homogeneous, or nonhomogeneous solution methods.

Unisium hero image titled First-Order Linear System Form showing the principle equation and a conditions card.
The first-order linear system template x=A(t)x+f(t)\mathbf{x}^{\prime}=A(t)\mathbf{x}+\mathbf{f}(t) with the canonical condition that the matrix and forcing vector are defined on the working interval.

On this page: The Principle | Conditions | Misconceptions | EE Questions | Retrieval Practice | Worked Example | Solve a Problem | Related Principles | FAQ


The Principle

Statement

A first-order linear system has first-order linear system form when the unknowns are collected into a vector x(t)\mathbf{x}(t) and the derivatives are collected into x(t)\mathbf{x}^{\prime}(t). The system is linear when the right-hand side is a coefficient matrix times the unknown vector, plus a known forcing vector:

x=A(t)x+f(t)\mathbf{x}^{\prime}=A(t)\mathbf{x}+\mathbf{f}(t)

The matrix A(t)A(t) can couple the components of x\mathbf{x}. That coupling is still linear when each right-hand side is a linear combination of the unknown components with coefficients depending only on tt, plus known terms depending only on tt.

Linear here means the unknown functions do not multiply each other, appear inside nonlinear functions, or appear with powers other than one.

Mathematical Form

x=A(t)x+f(t)\mathbf{x}^{\prime}=A(t)\mathbf{x}+\mathbf{f}(t)

Where:

  • tt = independent variable
  • x(t)\mathbf{x}(t) = vector of unknown functions
  • x(t)\mathbf{x}^{\prime}(t) = vector of first derivatives
  • A(t)A(t) = known coefficient matrix on the working interval
  • f(t)\mathbf{f}(t) = known forcing vector on the working interval

What the form tells you

This form is a recognition and organization step. It tells you that a coupled first-order system belongs to the linear system family. It does not solve the system by itself, and it does not require A(t)A(t) to be constant.

If f(t)=0\mathbf{f}(t)=\mathbf{0}, the system is homogeneous. If f(t)0\mathbf{f}(t)\ne\mathbf{0}, the system is nonhomogeneous. Both cases still use the same representational form.


Conditions of Applicability

Condition: matrix and forcing vector defined on working interval

Practical modeling notes

  • The entries of A(t)A(t) and f(t)\mathbf{f}(t) must be known functions on the interval being used.
  • Coupling between variables is allowed. For example, x1=2x1x2x_1^{\prime}=2x_1-x_2 is linear because the right-hand side is a linear combination of the unknown components with known coefficients.
  • Extra solution-method requirements, such as constant coefficients or diagonalizability, belong to later principles. They are not part of this principle’s canonical condition.

When It Doesn’t Apply

This principle does not cover:

  • Nonlinear coupling: x1=x1x2x_1^{\prime}=x_1x_2 is a first-order system term, but it is not linear in the unknown vector.
  • Undefined matrix entries: if A(t)A(t) has an entry such as 1/(t2)1/(t-2), the form is not valid on an interval that crosses t=2t=2; choose a working interval that avoids the singular point.
  • Unknown coefficients: if an entry of A(t)A(t) is another unknown function to solve for, the system is not yet a closed linear system in x\mathbf{x} alone.

Want the complete framework behind this guide? Read Masterful Learning.


Common Misconceptions

Misconception 1: “Coupled means nonlinear”

The truth: coupling means one equation may involve several unknown components. The system is still linear if those components enter through a matrix multiplying x\mathbf{x}.

Why this matters: a coupled system can still be handled by linear system tools once it is written in matrix form.

Misconception 2: “The matrix must be constant”

The truth: first-order linear system form allows A(t)A(t) and f(t)\mathbf{f}(t) to depend on tt.

Why this matters: constant-coefficient methods are a special case. Do not reject the system form just because the entries vary with time.

Misconception 3: “Known input terms can stay mixed into the equations”

The truth: known terms that do not multiply unknown components belong in the forcing vector f(t)\mathbf{f}(t).

Why this matters: mixing forcing terms into the coefficient matrix hides whether the system is homogeneous or nonhomogeneous, which affects later solution choices.


Elaborative Encoding

Use these questions to build deep understanding. (See Elaborative Encoding for the full method.)

Within the Principle

  • In x=A(t)x+f(t)\mathbf{x}^{\prime}=A(t)\mathbf{x}+\mathbf{f}(t), which pieces are known before solving, and which piece is unknown?
  • How can a matrix entry create coupling between two unknown functions while keeping the system linear?

For the Principle

  • Before using the system form on an interval, what must you check about the entries of A(t)A(t) and f(t)\mathbf{f}(t)?
  • If a term does not multiply an unknown component, why does it belong in f(t)\mathbf{f}(t) instead of A(t)xA(t)\mathbf{x}?

Between Principles

Generate an Example

  • Write a two-variable system that fits the matrix form, then write one near miss that fails because it contains a product of unknowns.

Retrieval Practice

Answer from memory, then click to reveal and check. (See Retrieval Practice for the full method.)

State the principle in words: _____A first-order linear system can be written with the derivative vector equal to a known matrix times the unknown vector plus a known forcing vector.
Write the canonical equation: _____x=A(t)x+f(t)\mathbf{x}^{\prime}=A(t)\mathbf{x}+\mathbf{f}(t)
State the canonical condition: _____matrix and forcing vector defined on working interval

Worked Example

Use this worked example to practice Self-Explanation.

Problem

For the system

x1=2x1x2+sint,x2=3x1+4x2+t,x_1^{\prime}=2x_1-x_2+\sin t,\qquad x_2^{\prime}=3x_1+4x_2+t,

write the system in first-order linear system form and identify A(t)A(t) and f(t)\mathbf{f}(t) on the interval (,)(-\infty,\infty).

Step 1: Verbal Decoding

Target: A(t)A(t), f(t)\mathbf{f}(t), first-order linear system form
Given: tt, x1x_1, x2x_2
Constraints: unknown components appear linearly; coefficient matrix and forcing vector must be defined on the working interval

Step 2: Visual Decoding

Draw two columns labeled unknown-vector terms and forcing terms. Place terms containing x1x_1 or x2x_2 in the first column and terms depending only on tt in the second. (The split separates A(t)xA(t)\mathbf{x} from f(t)\mathbf{f}(t).)

Step 3: Mathematical Modeling

  1. [x1x2]=A(t)[x1x2]+f(t)\begin{bmatrix}x_1^{\prime}\\ x_2^{\prime}\end{bmatrix}=A(t)\begin{bmatrix}x_1\\ x_2\end{bmatrix}+\mathbf{f}(t)

Step 4: Mathematical Procedures

  1. A(t)=[2134]A(t)=\begin{bmatrix}2&-1\\ 3&4\end{bmatrix}
  2. f(t)=[sintt]\mathbf{f}(t)=\begin{bmatrix}\sin t\\ t\end{bmatrix}
  3. [x1x2]=[2134][x1x2]+[sintt]\underline{\begin{bmatrix}x_1^{\prime}\\ x_2^{\prime}\end{bmatrix}=\begin{bmatrix}2&-1\\ 3&4\end{bmatrix}\begin{bmatrix}x_1\\ x_2\end{bmatrix}+\begin{bmatrix}\sin t\\ t\end{bmatrix}}

Step 5: Reflection

  • Verification: multiplying the matrix by [x1x2]\begin{bmatrix}x_1\\ x_2\end{bmatrix} reproduces the unknown-vector terms in both equations.
  • Domain check: all entries of A(t)A(t) and f(t)\mathbf{f}(t) are defined for every real tt.
  • Connection to concept: the matrix form organizes the coupled system before any solving method is chosen.

Before moving on: self-explain the model

Try explaining Step 3 out loud (or in writing): why the unknown terms belong in the matrix product, why sint\sin t and tt belong in the forcing vector, and why coupling does not break linearity.

Mathematical model with explanation

Principle: First-Order Linear System Form - x=A(t)x+f(t)\mathbf{x}^{\prime}=A(t)\mathbf{x}+\mathbf{f}(t).

Conditions: the coefficient matrix and forcing vector are defined on the working interval (,)(-\infty,\infty).

Relevance: the problem asks for the system representation, so the useful move is to collect the unknown functions into a vector and separate known forcing terms.

Description: The coefficients 22, 1-1, 33, and 44 form the matrix that multiplies the unknown vector. The terms sint\sin t and tt do not multiply unknown functions, so they form the forcing vector.

Goal: express the two scalar equations as one vector equation in the canonical linear system form.


Solve a Problem

Apply what you’ve learned with Problem Solving.

Problem

For the system

x1=tx1+x2+1,x2=etx2t2,x_1^{\prime}=t x_1+x_2+1,\qquad x_2^{\prime}=e^t x_2-t^2,

write the system in first-order linear system form and identify the working-interval condition.

Hint (if needed): separate the terms that multiply unknown components from terms depending only on tt.

Show Solution

Step 1: Verbal Decoding

Target: A(t)A(t), f(t)\mathbf{f}(t), condition check
Given: tt, x1x_1, x2x_2
Constraints: unknown components appear linearly; matrix entries and forcing entries must be defined on the interval

Step 2: Visual Decoding

Draw a two-row matrix slot next to the vector [x1x2]\begin{bmatrix}x_1\\ x_2\end{bmatrix}, then reserve a separate two-row forcing vector for terms with no unknown component. (The coefficient of any missing unknown component is zero.)

Step 3: Mathematical Modeling

  1. [x1x2]=A(t)[x1x2]+f(t)\begin{bmatrix}x_1^{\prime}\\ x_2^{\prime}\end{bmatrix}=A(t)\begin{bmatrix}x_1\\ x_2\end{bmatrix}+\mathbf{f}(t)

Step 4: Mathematical Procedures

  1. A(t)=[t10et]A(t)=\begin{bmatrix}t&1\\ 0&e^t\end{bmatrix}
  2. f(t)=[1t2]\mathbf{f}(t)=\begin{bmatrix}1\\ -t^2\end{bmatrix}
  3. [x1x2]=[t10et][x1x2]+[1t2], valid on (,)\underline{\begin{bmatrix}x_1^{\prime}\\ x_2^{\prime}\end{bmatrix}=\begin{bmatrix}t&1\\ 0&e^t\end{bmatrix}\begin{bmatrix}x_1\\ x_2\end{bmatrix}+\begin{bmatrix}1\\ -t^2\end{bmatrix},\ \text{valid on }(-\infty,\infty)}

Step 5: Reflection

  • Verification: the zero in the lower-left matrix entry correctly removes any x1x_1 term from the second equation.
  • Domain check: tt, ete^t, 11, and t2-t^2 are defined for every real tt.
  • Connection to concept: variable coefficients are allowed because the form only requires known entries on the interval.

PrincipleRelationship to First-Order Linear System Form
First-Order Linear Standard FormThe scalar linear form is the one-unknown analogue of this vector system form.
Eigenvalue-Eigenvector Solution ModeFor constant homogeneous systems, eigenpairs later give exponential solution modes.
Matrix Exponential Solution FormMatrix exponentials solve constant-coefficient homogeneous systems after the system form is identified.

See Differential Equations Subdomain for the full systems lane, and Principle Structures for keeping equations, conditions, and neighboring principles separate in memory.


FAQ

What is First-Order Linear System Form?

It is the matrix representation x=A(t)x+f(t)\mathbf{x}^{\prime}=A(t)\mathbf{x}+\mathbf{f}(t) for a system of first-order differential equations. The unknown functions are collected in x\mathbf{x}, the coefficient functions are collected in A(t)A(t), and the known input terms are collected in f(t)\mathbf{f}(t).

When does First-Order Linear System Form apply?

It applies when the coefficient matrix and forcing vector are defined on the working interval. The unknown components must enter as a linear combination with coefficients depending only on the independent variable; the coefficient entries may still vary with tt.

Does a coupled system have to be nonlinear?

No. Coupling means equations share unknown components. A coupled system is still linear when those components appear through a matrix times the unknown vector.

What is the difference between homogeneous and nonhomogeneous system form?

The form is homogeneous when f(t)=0\mathbf{f}(t)=\mathbf{0} and nonhomogeneous when f(t)0\mathbf{f}(t)\ne\mathbf{0}. Both use the same coefficient matrix structure.

Does matrix form solve the system?

No. Matrix form organizes the system and identifies the linear family. Later principles choose solution methods for special cases, such as constant homogeneous systems.



How This Fits in Unisium

Within the differential equations subdomain, Unisium treats First-Order Linear System Form as the organization step before system-specific solution methods. Pair this guide with elaborative encoding, retrieval practice, and self-explanation so you can separate matrix terms from forcing terms before choosing a method.

Ready to study differential equations with structure? Check access and join the Unisium waitlist or explore the full framework in Masterful Learning.

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