Linear System of ODEs: Complete Guide with Step-by-Step Solutions

Quick Navigation

• What is a Linear System of ODEs?
• The General Solution Method
• Classification of Equilibrium Points
• Stability Analysis
• Phase Portraits: Visualizing Solutions
• Real-World Applications
• Step-by-Step Examples
• Special Cases
• Frequently Asked Questions
• Practice Problems
• Related Topics

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What is a Linear System of ODEs?

Definition

A linear system of ordinary differential equations (ODEs) with constant coefficients is a system of the form:

$$ \frac{d\mathbf{x}}{dt} = A\mathbf{x} $$

Where:

• $\mathbf{x}(t)$ is a vector of $n$ unknown functions $[x_1(t), x_2(t), \dots, x_n(t)]^T$.
• $A$ is an $n \times n$ constant matrix.
• The system is homogeneous (no external forcing term).

Component Form

For a 2D system ($n=2$), this expands to:

$$ \begin{aligned} \frac{dx_1}{dt} &= a_{11}x_1 + a_{12}x_2 \ \frac{dx_2}{dt} &= a_{21}x_1 + a_{22}x_2 \end{aligned} $$

The Matrix Exponential Solution

The general solution to this system is given by the matrix exponential:

$$ \mathbf{x}(t) = e^{At}\mathbf{x}(0) $$

However, computing $e^{At}$ directly is difficult. Instead, we use eigenvalues and eigenvectors to construct the solution explicitly.

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The General Solution Method

The form of the solution depends on the eigenvalues $\lambda$ of the matrix $A$.

Case 1: Distinct Real Eigenvalues

If $A$ has distinct real eigenvalues $\lambda_1, \dots, \lambda_n$ with corresponding eigenvectors $\mathbf{v}_1, \dots, \mathbf{v}_n$, the general solution is:

$$ \mathbf{x}(t) = c_1 \mathbf{v}_1 e^{\lambda_1 t} + c_2 \mathbf{v}_2 e^{\lambda_2 t} + \dots + c_n \mathbf{v}_n e^{\lambda_n t} $$

Case 2: Complex Conjugate Eigenvalues

If $A$ has complex eigenvalues $\lambda = \alpha \pm i\beta$ with eigenvector $\mathbf{v} = \mathbf{a} \pm i\mathbf{b}$, the real-valued general solution involves sines and cosines:

$$ \mathbf{x}(t) = c_1 e^{\alpha t} (\mathbf{a} \cos(\beta t) - \mathbf{b} \sin(\beta t)) + c_2 e^{\alpha t} (\mathbf{a} \sin(\beta t) + \mathbf{b} \cos(\beta t)) $$

Case 3: Repeated Eigenvalues (Defective Matrix)

If an eigenvalue $\lambda$ has algebraic multiplicity $k$ but fewer than $k$ linearly independent eigenvectors, we must find generalized eigenvectors. For a defect of 1 (one missing eigenvector):

$$ \mathbf{x}(t) = c_1 \mathbf{v} e^{\lambda t} + c_2 (\mathbf{v}t + \mathbf{w}) e^{\lambda t} $$

Where $\mathbf{w}$ is a generalized eigenvector satisfying $(A - \lambda I)\mathbf{w} = \mathbf{v}$.

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Classification of Equilibrium Points

The origin $\mathbf{x}=\mathbf{0}$ is always an equilibrium point. Its type is determined by the eigenvalues of $A$ (for 2D systems).

Eigenvalues ($\lambda_1, \lambda_2$)	Type	Stability	Phase Portrait Shape
Real, $\lambda_1 < \lambda_2 < 0$	Stable Node	Asymptotically Stable	Trajectories approach origin along straight lines
Real, $0 < \lambda_1 < \lambda_2$	Unstable Node	Unstable	Trajectories diverge from origin
Real, $\lambda_1 < 0 < \lambda_2$	Saddle Point	Unstable	Approaches along one axis, diverges along other
Complex, $\text{Re}(\lambda) < 0$	Stable Spiral	Asymptotically Stable	Spirals into origin
Complex, $\text{Re}(\lambda) > 0$	Unstable Spiral	Unstable	Spirals away from origin
Purely Imaginary, $\text{Re}(\lambda) = 0$	Center	Lyapunov Stable	Closed elliptical orbits
Repeated, Defective	Improper Node	Depends on sign	Parabolic trajectories

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Stability Analysis

Stability Definitions

Term	Meaning	Mathematical Definition
Asymptotically stable	Solutions approach origin as $t \to \infty$	$\lim_{t \to \infty} |\mathbf{x}(t)| = 0$
Lyapunov stable	Solutions stay near origin if started near it	$|\mathbf{x}(0)| < \delta \implies |\mathbf{x}(t)| < \epsilon$
Unstable	At least one solution moves away from origin	Not stable

Stability Criterion (General n)

Condition	Stability
All eigenvalues have $\text{Re}(\lambda) < 0$	Asymptotically stable
All eigenvalues have $\text{Re}(\lambda) \leq 0$, and those with $\text{Re}=0$ are non-defective	Lyapunov stable
Any eigenvalue has $\text{Re}(\lambda) > 0$	Unstable

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Phase Portraits: Visualizing Solutions

What is a Phase Portrait?

A phase portrait shows solution curves (trajectories) in the state space (e.g., the $x_1$-$x_2$ plane for 2D systems). It provides a geometric understanding of the system's behavior without solving for specific time values.

Reading Phase Portraits

Direction of time: Arrows indicate increasing $t$.

• Arrows point toward origin → stable
• Arrows point away from origin → unstable

Trajectory shapes:

• Straight lines → Motion along eigenvectors
• Curves → Linear combinations of modes
• Closed loops → Periodic solutions (centers)
• Spirals → Complex eigenvalues with non-zero real part

How to Sketch by Hand (2D)

1. Find eigenvalues $\lambda_1, \lambda_2$ and eigenvectors $\mathbf{v}_1, \mathbf{v}_2$.
2. Draw eigenvector lines through the origin.
3. Determine flow direction:
   • If $\lambda > 0$, arrows point away from origin along that eigenvector.
   • If $\lambda < 0$, arrows point toward origin.
4. Sketch sample trajectories that are tangent to the dominant eigenvector (largest $|\lambda|$) near the origin.

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Real-World Applications

🏗️ Engineering: Spring-Mass Systems

A two-mass spring system can be modeled as a second-order system, which converts to a 4-dimensional first-order linear system:

$$ \frac{d}{dt}\begin{bmatrix} x_1 \ x_2 \ v_1 \ v_2 \end{bmatrix} = \left[\begin{array}{cccc} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ -\frac{k_1+k_2}{m_1} & \frac{k_2}{m_1} & 0 & 0 \\ \frac{k_2}{m_2} & -\frac{k_2}{m_2} & 0 & 0 \end{array}\right] \begin{bmatrix} x_1 \ x_2 \ v_1 \ v_2 \end{bmatrix} $$

Eigenvalues give natural frequencies: $\omega = \sqrt{|\lambda|}$ (for imaginary eigenvalues).

🧬 Biology: Predator-Prey Models (Linearized)

Near an equilibrium point, the non-linear Lotka-Volterra equations can be linearized:

$$ \frac{d}{dt}\begin{bmatrix} P \ Q \end{bmatrix} = \left[\begin{array}{cc} \alpha & -\beta \\ -\delta & \gamma \end{array}\right] \begin{bmatrix} P \ Q \end{bmatrix} $$

Eigenvalues determine: Oscillation frequencies and stability of the ecosystem.

🔌 Electrical Engineering: RLC Circuits

A series RLC circuit is governed by a second-order ODE, which converts to a 2D linear system:

$$ \frac{d}{dt}\begin{bmatrix} q \ i \end{bmatrix} = \left[\begin{array}{cc} 0 & 1 \\ -\frac{1}{LC} & -\frac{R}{L} \end{array}\right] \begin{bmatrix} q \ i \end{bmatrix} $$

Eigenvalues tell you:

• Complex $\lambda$: Underdamped (oscillates)
• Real distinct $\lambda$: Overdamped
• Real repeated $\lambda$: Critically damped

🚀 Control Theory: State-Space Models

Any linear control system is written as: $$ \dot{\mathbf{x}} = A\mathbf{x} + B\mathbf{u} $$ The homogeneous part $\dot{\mathbf{x}} = A\mathbf{x}$ determines the natural stability of the system. If $A$ has any eigenvalue with positive real part, the open-loop system is unstable.

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Step-by-Step Examples

Example 1: Stable Node (Both $\lambda$ Negative)

System: $$ \frac{d\mathbf{x}}{dt} = \left[\begin{array}{cc} -2 & 0 \\ 0 & -1 \end{array}\right] \mathbf{x} $$

Step 1: Find eigenvalues Since the matrix is diagonal, $\lambda_1 = -2,\quad \lambda_2 = -1$.

Step 2: Find eigenvectors $$ \mathbf{v}_1 = \left[\begin{array}{c} 1 \\ 0 \end{array}\right],\quad \mathbf{v}_2 = \left[\begin{array}{c} 0 \\ 1 \end{array}\right] $$

Step 3: General solution $$ \mathbf{x}(t) = c_1 \left[\begin{array}{c} 1 \\ 0 \end{array}\right] e^{-2t} + c_2 \left[\begin{array}{c} 0 \\ 1 \end{array}\right] e^{-t} $$

Step 4: Classification

• Type: Stable node (sink)
• Stability: Asymptotically stable
• Behavior: All trajectories approach the origin. The $x_1$ direction decays faster ($\lambda = -2$) than the $x_2$ direction ($\lambda = -1$).

Example 2: Saddle Point (Mixed Signs)

System: $$ \frac{d\mathbf{x}}{dt} = \left[\begin{array}{cc} 1 & 0 \\ 0 & -2 \end{array}\right] \mathbf{x} $$

Step 1: Find eigenvalues $$ \lambda_1 = 1,\quad \lambda_2 = -2 $$

Step 2: General solution $$ \mathbf{x}(t) = c_1 \left[\begin{array}{c} 1 \\ 0 \end{array}\right] e^{t} + c_2 \left[\begin{array}{c} 0 \\ 1 \end{array}\right] e^{-2t} $$

Step 3: Classification

• Type: Saddle point
• Stability: Unstable
• Behavior: Trajectories approach the origin along the $x_2$-axis (stable manifold, $\lambda_2 < 0$) but diverge along the $x_1$-axis (unstable manifold, $\lambda_1 > 0$).

Example 3: Stable Spiral (Complex $\lambda$ with $\alpha < 0$)

System: $$ \frac{d\mathbf{x}}{dt} = \left[\begin{array}{cc} -1 & -2 \\ 2 & -1 \end{array}\right] \mathbf{x} $$

Step 1: Find eigenvalues Characteristic equation: $(\lambda+1)^2 + 4 = 0 \Rightarrow \lambda = -1 \pm 2i$. Here $\alpha = -1, \beta = 2$.

Step 2: Find eigenvector for $\lambda = -1 + 2i$ Solving $(A - \lambda I)\mathbf{v} = 0$ yields $\mathbf{v} = \left[\begin{array}{c} 1 \\ -i \end{array}\right] = \left[\begin{array}{c} 1 \\ 0 \end{array}\right] + i\left[\begin{array}{c} 0 \\ -1 \end{array}\right]$. So $\mathbf{a} = [1, 0]^T$ and $\mathbf{b} = [0, -1]^T$.

Step 3: General real solution $$ \mathbf{x}(t) = c_1 e^{-t} \left( \left[\begin{array}{c} 1 \\ 0 \end{array}\right] \cos(2t) - \left[\begin{array}{c} 0 \\ -1 \end{array}\right] \sin(2t) \right) + c_2 e^{-t} \left( \left[\begin{array}{c} 1 \\ 0 \end{array}\right] \sin(2t) + \left[\begin{array}{c} 0 \\ -1 \end{array}\right] \cos(2t) \right) $$

Step 4: Classification

• Type: Stable spiral
• Stability: Asymptotically stable
• Behavior: Solutions spiral into the origin with frequency 2 rad/s.

Example 4: Defective Matrix (Repeated $\lambda$, One Eigenvector)

System: $$ A = \left[\begin{array}{cc} 2 & 1 \\ 0 & 2 \end{array}\right] $$

Step 1: Find eigenvalues $\det(A-\lambda I) = (2-\lambda)^2 = 0 \Rightarrow \lambda = 2$ (multiplicity 2).

Step 2: Find eigenvectors $(A - 2I)\mathbf{v} = \left[\begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array}\right]\mathbf{v} = \mathbf{0} \Rightarrow \mathbf{v}_1 = \left[\begin{array}{c} 1 \\ 0 \end{array}\right]$. Only one independent eigenvector exists.

Step 3: Find generalized eigenvector $\mathbf{w}$ Solve $(A - 2I)\mathbf{w} = \mathbf{v}_1$: $$ \left[\begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array}\right] \left[\begin{array}{c} w_1 \\ w_2 \end{array}\right] = \left[\begin{array}{c} 1 \\ 0 \end{array}\right] \implies w_2 = 1 $$ We can choose $w_1 = 0$, so $\mathbf{w} = \left[\begin{array}{c} 0 \\ 1 \end{array}\right]$.

Step 4: General solution $$ \mathbf{x}(t) = c_1 \left[\begin{array}{c} 1 \\ 0 \end{array}\right] e^{2t} + c_2 \left( \left[\begin{array}{c} 1 \\ 0 \end{array}\right]t + \left[\begin{array}{c} 0 \\ 1 \end{array}\right] \right) e^{2t} $$

Step 5: Classification

• Type: Improper node (source)
• Stability: Unstable ($\lambda > 0$)

Example 5: Initial Value Problem

System: Same as Example 1 with IC: $\mathbf{x}(0) = [1, 2]^T$.

Step 1: General solution $$ x_1(t) = c_1 e^{-2t},\quad x_2(t) = c_2 e^{-t} $$

Step 2: Apply initial conditions ($t = 0$) $$ x_1(0) = c_1 = 1 \implies c_1 = 1 $$ $$ x_2(0) = c_2 = 2 \implies c_2 = 2 $$

Step 3: Particular solution $$ x_1(t) = e^{-2t},\quad x_2(t) = 2e^{-t} $$

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Special Cases

Case: Zero Matrix ($A = 0$)

System: $\frac{d\mathbf{x}}{dt} = \mathbf{0}$ Solution: $\mathbf{x}(t) = \mathbf{x}(0)$ (Constant) Classification: Every point is an equilibrium (degenerate).

Case: Nilpotent Matrix

Example: $$ A = \left[\begin{array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{array}\right] $$ Property: $A^3 = 0$. The matrix exponential series terminates: $$ e^{At} = I + At + \frac{1}{2}A^2t^2 $$

Case: Skew-Symmetric Matrix

Example: $$ A = \left[\begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array}\right] $$ Property: $A^T = -A$. Eigenvalues are purely imaginary ($\pm i$). Result: Center (stable periodic orbits). Energy is conserved.

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Frequently Asked Questions

Q: Why do we need eigenvalues for ODEs?

A: Eigenvalues determine the growth rate and oscillation frequency of the system. They tell you if the system will explode, decay, or oscillate without solving the entire time history.

Q: What if $A$ has complex eigenvalues?

A: The solution will involve sine and cosine terms, representing oscillation. The real part of the eigenvalue determines if the oscillation grows (unstable) or decays (stable).

Q: What is the difference between algebraic and geometric multiplicity?

A:

• Algebraic: How many times $\lambda$ appears as a root of the characteristic polynomial.
• Geometric: The number of linearly independent eigenvectors for $\lambda$.
• If Geometric < Algebraic, the matrix is defective, and you need generalized eigenvectors.

Q: Can I solve non-homogeneous systems ($\dot{\mathbf{x}} = A\mathbf{x} + \mathbf{f}(t)$)?

A: Yes, using variation of parameters: $$ \mathbf{x}(t) = e^{At}\mathbf{x}(0) + \int_0^t e^{A(t-s)} \mathbf{f}(s) ds $$

Q: How do I know if my system is stable?

A: Check the real parts of all eigenvalues:

• All $\text{Re}(\lambda) < 0$ → Asymptotically stable
• Any $\text{Re}(\lambda) > 0$ → Unstable
• $\text{Re}(\lambda) = 0$ → Marginally stable (if non-defective)

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Practice Problems

Beginner Level

1. Find eigenvalues: For $A = \left[\begin{array}{cc} -3 & 0 \\ 0 & -5 \end{array}\right]$, classify the equilibrium.
2. Write system: Write $\frac{d}{dt}[x_1, x_2]^T = \left[\begin{array}{cc} 1 & 2 \\ 3 & 4 \end{array}\right][x_1, x_2]^T$ as two separate equations.

Intermediate Level

3. Solve completely: Find the general solution for $\frac{d\mathbf{x}}{dt} = \left[\begin{array}{cc} 1 & 1 \\ 1 & 1 \end{array}\right] \mathbf{x}$.
4. Complex eigenvalues: Solve $\frac{d\mathbf{x}}{dt} = \left[\begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array}\right] \mathbf{x}$ and classify the equilibrium.

Advanced Level

5. Defective matrix: Find the general solution for $A = \left[\begin{array}{cc} 2 & 1 \\ 0 & 2 \end{array}\right]$.
6. Initial Value Problem: Solve $\frac{d\mathbf{x}}{dt} = \left[\begin{array}{cc} -2 & 1 \\ 0 & -2 \end{array}\right] \mathbf{x}$ with $\mathbf{x}(0) = [1, 1]^T$.

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Summary

Key Takeaways

Concept	Summary
System form	$\frac{d\mathbf{x}}{dt} = A\mathbf{x}$
General solution	Linear combination of $e^{\lambda t}$ times eigenvectors
Stability	Determined by signs of $\text{Re}(\lambda)$
Phase portrait	Geometric picture of trajectories in state space
Fundamental matrix	$\Phi(t) = e^{At}$ (columns are basis solutions)

Algorithm for Solving

1. Compute eigenvalues $\lambda$ of matrix $A$.
2. Compute eigenvectors $\mathbf{v}$ (and generalized if needed).
3. Build solution components:
   • Real distinct: $c \mathbf{v} e^{\lambda t}$
   • Complex: $e^{\alpha t}(\cos(\beta t)\mathbf{a} - \sin(\beta t)\mathbf{b})$
   • Repeated/defective: Include polynomials $t^k e^{\lambda t}$
4. Combine with constants for general solution.
5. Apply Initial Conditions (if given) to find particular solution.

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Related Topics

Explore these related concepts in our Linear Algebra Toolbox:

• Eigenvalue/Eigenvector Calculator - Core of ODE solution method
• Matrix Exponential Calculator - Direct computation of $e^{At}$
• Diagonalization Calculator - Simplifies ODEs when $A = PDP^{-1}$
• Phase Portrait Visualizer - See the geometry of solutions

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Try It Yourself!

Use the calculator above to explore linear ODE systems:

1. Enter your matrix A (2×2 to 4×4)
2. Optional: Add initial condition $\mathbf{x}(0)$
3. Click "Calculate" to see:
   • General solution (symbolic)
   • Particular solution (if IC given)
   • Eigenvalues and eigenvectors
   • Equilibrium classification
   • Stability analysis

Test these examples:

Stable node (sink): $$ A = \left[\begin{array}{cc} -2 & 0 \\ 0 & -1 \end{array}\right] $$

Saddle point: $$ A = \left[\begin{array}{cc} 1 & 0 \\ 0 & -2 \end{array}\right] $$

Stable spiral: $$ A = \left[\begin{array}{cc} -1 & -2 \\ 2 & -1 \end{array}\right] $$

Center (periodic): $$ A = \left[\begin{array}{cc} 0 & -2 \\ 2 & 0 \end{array}\right] $$

Defective (Jordan block): $$ A = \left[\begin{array}{cc} 2 & 1 \\ 0 & 2 \end{array}\right] $$

Check initial condition: Try $\mathbf{x}(0) = [1, 0]^T$ with the stable node example to see exponential decay!

Pro tip: The eigenvalues tell you everything about system behavior:

• Negative real parts → decay to equilibrium
• Positive real parts → growth/instability
• Imaginary parts → oscillations
• Zero real parts → marginal stability