Chapter 4 · Systems of DEs

Solving Homogeneous Linear Systems

x' = Ax — The Eigenvalue Method

Dr. Mohamed Mabrok · Qatar University

The Problem: Homogeneous Linear Systems

Standard Form

      $$\mathbf{x}' = A\mathbf{x}$$
    
      where $\mathbf{x}(t)$ is a vector and $A$ is a constant $2 \times 2$ (or $n \times n$) matrix.

The Key Insight

      If $\lambda$ is an eigenvalue of $A$ and $\mathbf{v}$ is a corresponding eigenvector, then:
    
      $$\mathbf{x}(t) = e^{\lambda t}\mathbf{v}$$
    
      is a solution to $\mathbf{x}' = A\mathbf{x}$. This bridges linear algebra and differential equations!

Fundamental Theorem: Eigenvalue → Solution

Theorem

      If $A\mathbf{v} = \lambda\mathbf{v}$, then $\mathbf{x}(t) = e^{\lambda t}\mathbf{v}$ satisfies $\mathbf{x}' = A\mathbf{x}$.

Proof

      Left side: $\mathbf{x}' = \frac{d}{dt}(e^{\lambda t}\mathbf{v}) = \lambda e^{\lambda t}\mathbf{v}$
    
      Right side: $A\mathbf{x} = A(e^{\lambda t}\mathbf{v}) = e^{\lambda t}(A\mathbf{v}) = e^{\lambda t}(\lambda\mathbf{v}) = \lambda e^{\lambda t}\mathbf{v}$
    
      Both sides equal! ✓

Superposition

      If $\mathbf{x}_1(t), \mathbf{x}_2(t), \ldots, \mathbf{x}_n(t)$ are solutions, then any linear combination is also a solution:
    
      $$\mathbf{x}(t) = c_1\mathbf{x}_1(t) + c_2\mathbf{x}_2(t) + \cdots + c_n\mathbf{x}_n(t)$$

The Eigenvalue Method: 5 Steps

1 Find eigenvalues: solve $\det(A - \lambda I) = 0$

2 For each eigenvalue, find eigenvectors: solve $(A - \lambda I)\mathbf{v} = \mathbf{0}$

3 Form fundamental solutions: $\mathbf{x}_i(t) = e^{\lambda_i t}\mathbf{v}_i$

4 Write the general solution: $\mathbf{x}(t) = c_1\mathbf{x}_1(t) + c_2\mathbf{x}_2(t) + \cdots$

5 Apply initial conditions to find $c_1, c_2, \ldots$

Key: The Characteristic Polynomial

      $$\det(A - \lambda I) = 0$$
    
      This polynomial has degree $n$ (the size of $A$), so it has $n$ roots (counting multiplicity).

Case 1: Distinct Real Eigenvalues

When It Occurs

      When the characteristic polynomial has $n$ distinct real roots: $\lambda_1 \neq \lambda_2 \neq \cdots \neq \lambda_n$

General Solution

      $$\mathbf{x}(t) = c_1e^{\lambda_1 t}\mathbf{v}_1 + c_2e^{\lambda_2 t}\mathbf{v}_2 + \cdots$$
    
      The eigenvectors $\mathbf{v}_1, \mathbf{v}_2, \ldots$ are linearly independent.

Behavior

      The solution is a superposition of exponential modes. Each mode grows (if $\lambda_i > 0$) or decays (if $\lambda_i < 0$) independently. The most dominant eigenvalue determines long-term behavior.

Worked Example

Solve: $\mathbf{x}' = \begin{pmatrix} 1 & 2 \\ 3 & 2 \end{pmatrix}\mathbf{x}$

STEP 1 Characteristic equation: $\det(A - \lambda I) = 0$

\det\begin{pmatrix} 1-\lambda & 2 \\ 3 & 2-\lambda \end{pmatrix} = (1-\lambda)(2-\lambda) - 6 = \lambda^2 - 3\lambda - 4 = 0

STEP 2 Factor and find eigenvalues

(\lambda - 4)(\lambda + 1) = 0 \quad \Rightarrow \quad \lambda_1 = 4, \lambda_2 = -1

STEP 3 Find eigenvectors (for $\lambda_1 = 4$)

(A - 4I)\mathbf{v}_1 = 0 \Rightarrow \begin{pmatrix} -3 & 2 \\ 3 & -2 \end{pmatrix}\begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = 0 \Rightarrow \mathbf{v}_1 = \begin{pmatrix} 2 \\ 3 \end{pmatrix}

STEP 4 Find eigenvectors (for $\lambda_2 = -1$)

(A + I)\mathbf{v}_2 = 0 \Rightarrow \begin{pmatrix} 2 & 2 \\ 3 & 3 \end{pmatrix}\begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = 0 \Rightarrow \mathbf{v}_2 = \begin{pmatrix} 1 \\ -1 \end{pmatrix}

STEP 5 Write general solution

\mathbf{x}(t) = c_1e^{4t}\begin{pmatrix} 2 \\ 3 \end{pmatrix} + c_2e^{-t}\begin{pmatrix} 1 \\ -1 \end{pmatrix}

Worked Example with Initial Condition

Same system with $\mathbf{x}(0) = \begin{pmatrix} 5 \\ 2 \end{pmatrix}$

STEP 1 Set up the initial condition equation

\mathbf{x}(0) = c_1\begin{pmatrix} 2 \\ 3 \end{pmatrix} + c_2\begin{pmatrix} 1 \\ -1 \end{pmatrix} = \begin{pmatrix} 5 \\ 2 \end{pmatrix}

STEP 2 Solve the linear system

2c_1 + c_2 = 5, \quad 3c_1 - c_2 = 2 \quad \Rightarrow \quad c_1 = \frac{7}{5}, c_2 = \frac{11}{5}

STEP 3 Write particular solution

\mathbf{x}(t) = \frac{7}{5}e^{4t}\begin{pmatrix} 2 \\ 3 \end{pmatrix} + \frac{11}{5}e^{-t}\begin{pmatrix} 1 \\ -1 \end{pmatrix}

Case 2: Complex Conjugate Eigenvalues

When It Occurs

      When characteristic polynomial has complex roots: $\lambda = \alpha \pm \beta i$ (conjugate pairs)

Real-Valued General Solution

      For a $2 \times 2$ system with $\lambda = \alpha \pm \beta i$, if $\mathbf{v} = \mathbf{u} + i\mathbf{w}$ is a complex eigenvector:
    
      $$\mathbf{x}(t) = c_1e^{\alpha t}[\cos(\beta t)\mathbf{u} - \sin(\beta t)\mathbf{w}] + c_2e^{\alpha t}[\sin(\beta t)\mathbf{u} + \cos(\beta t)\mathbf{w}]$$

Behavior

      The real part $\alpha$ controls exponential growth/decay. The imaginary part $\beta$ controls oscillation frequency. If $\alpha < 0$, oscillations decay. If $\alpha > 0$, they grow.

Worked Example

Solve: $\mathbf{x}' = \begin{pmatrix} 1 & -2 \\ 1 & 3 \end{pmatrix}\mathbf{x}$

STEP 1 Characteristic equation

\det\begin{pmatrix} 1-\lambda & -2 \\ 1 & 3-\lambda \end{pmatrix} = \lambda^2 - 4\lambda + 5 = 0

STEP 2 Solve using quadratic formula

\lambda = \frac{4 \pm \sqrt{16-20}}{2} = \frac{4 \pm 2i}{2} = 2 \pm i

STEP 3 Identify $\alpha = 2$, $\beta = 1$

STEP 4 Find complex eigenvector for $\lambda = 2 + i$

(A - (2+i)I)\mathbf{v} = 0 \Rightarrow \mathbf{v} = \begin{pmatrix} 2 \\ -1-i \end{pmatrix} = \begin{pmatrix} 2 \\ -1 \end{pmatrix} + i\begin{pmatrix} 0 \\ -1 \end{pmatrix}

STEP 5 Write general solution (real form)

\mathbf{x}(t) = c_1e^{2t}\left[\cos(t)\begin{pmatrix} 2 \\ -1 \end{pmatrix} - \sin(t)\begin{pmatrix} 0 \\ -1 \end{pmatrix}\right] + c_2e^{2t}\left[\sin(t)\begin{pmatrix} 2 \\ -1 \end{pmatrix} + \cos(t)\begin{pmatrix} 0 \\ -1 \end{pmatrix}\right]

Case 3: Repeated Eigenvalue

The Challenge

      When $\lambda$ is a repeated root of the characteristic polynomial, we may have fewer than $n$ linearly independent eigenvectors!

Sub-Case A: Full Set of Eigenvectors

      If $A = \lambda I$ (the matrix is already diagonal), we have $n$ independent eigenvectors.
    
      $$\mathbf{x}(t) = c_1e^{\lambda t}\mathbf{v}_1 + c_2e^{\lambda t}\mathbf{v}_2 + \cdots$$

Sub-Case B: Generalized Eigenvector Needed

      If we have only one independent eigenvector, we need a generalized eigenvector $\mathbf{w}$ satisfying:
    
      $$(A - \lambda I)\mathbf{w} = \mathbf{v}$$
    
      Then the second solution is: $\mathbf{x}_2(t) = e^{\lambda t}(t\mathbf{v} + \mathbf{w})$

Worked Example

Solve: $\mathbf{x}' = \begin{pmatrix} 2 & -1 \\ 1 & 0 \end{pmatrix}\mathbf{x}$

STEP 1 Characteristic equation

\det\begin{pmatrix} 2-\lambda & -1 \\ 1 & -\lambda \end{pmatrix} = \lambda^2 - 2\lambda + 1 = (\lambda - 1)^2 = 0

STEP 2 Eigenvalue is $\lambda = 1$ (repeated)

STEP 3 Find eigenvector

(A - I)\mathbf{v} = 0 \Rightarrow \begin{pmatrix} 1 & -1 \\ 1 & -1 \end{pmatrix}\mathbf{v} = 0 \Rightarrow \mathbf{v} = \begin{pmatrix} 1 \\ 1 \end{pmatrix}

STEP 4 Find generalized eigenvector

(A - I)\mathbf{w} = \mathbf{v} \Rightarrow \begin{pmatrix} 1 & -1 \\ 1 & -1 \end{pmatrix}\mathbf{w} = \begin{pmatrix} 1 \\ 1 \end{pmatrix} \Rightarrow \mathbf{w} = \begin{pmatrix} 1 \\ 0 \end{pmatrix}

STEP 5 Write general solution

\mathbf{x}(t) = c_1e^t\begin{pmatrix} 1 \\ 1 \end{pmatrix} + c_2e^t\left(t\begin{pmatrix} 1 \\ 1 \end{pmatrix} + \begin{pmatrix} 1 \\ 0 \end{pmatrix}\right) = e^t\begin{pmatrix} c_1 + c_2 + c_2t \\ c_1 + c_2t \end{pmatrix}

Converting nth-Order ODEs to Systems

The Method

      Any $n$-th order ODE can be converted to a system of $n$ first-order ODEs.

Example: Second-Order ODE

      $$y'' + 3y' + 2y = 0$$
    
      Let $x_1 = y$ and $x_2 = y'$. Then:
    
      $$x_1' = x_2, \quad x_2' = -2x_1 - 3x_2$$
    
      In matrix form: $\mathbf{x}' = \begin{pmatrix} 0 & 1 \\ -2 & -3 \end{pmatrix}\mathbf{x}$ (companion matrix)

Worked Example

Solve ODE via System: $y'' - y' - 6y = 0$, $y(0)=1, y'(0)=0$

STEP 1 Convert to system form

\mathbf{x}' = \begin{pmatrix} 0 & 1 \\ 6 & 1 \end{pmatrix}\mathbf{x}, \quad \mathbf{x}(0) = \begin{pmatrix} 1 \\ 0 \end{pmatrix}

STEP 2 Find eigenvalues

\det\begin{pmatrix} -\lambda & 1 \\ 6 & 1-\lambda \end{pmatrix} = \lambda^2 - \lambda - 6 = (\lambda - 3)(\lambda + 2) = 0

STEP 3-4 Find eigenvectors and general solution

\mathbf{v}_1 = \begin{pmatrix} 1 \\ 3 \end{pmatrix}, \quad \mathbf{v}_2 = \begin{pmatrix} 1 \\ -2 \end{pmatrix}

STEP 5 Apply initial conditions to find $c_1, c_2$

c_1\begin{pmatrix} 1 \\ 3 \end{pmatrix} + c_2\begin{pmatrix} 1 \\ -2 \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \end{pmatrix} \Rightarrow c_1 = \frac{2}{5}, c_2 = \frac{3}{5}

y(t) = \frac{2}{5}e^{3t} + \frac{3}{5}e^{-2t}

Summary: The Three Cases

Case 1: Distinct Real

$\lambda_1 \neq \lambda_2$ (real)

\mathbf{x} = c_1e^{\lambda_1 t}\mathbf{v}_1 + c_2e^{\lambda_2 t}\mathbf{v}_2

Most common. Exponential growth/decay modes.

Case 2: Complex Pair

$\lambda = \alpha \pm \beta i$

\mathbf{x} = e^{\alpha t}[c_1\cos(\beta t)\mathbf{u} + c_2\sin(\beta t)\mathbf{u}]

Oscillatory with exponential envelope.

Case 3: Repeated Root

$\lambda$ (multiplicity $> 1$)

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

May need generalized eigenvectors.

Complete Worked Example

Solve IVP: $\mathbf{x}' = \begin{pmatrix} -3 & 1 \\ 0 & -2 \end{pmatrix}\mathbf{x}$, $\mathbf{x}(0) = \begin{pmatrix} 1 \\ 4 \end{pmatrix}$

STEP 1 Find eigenvalues (upper triangular = diagonal values)

\lambda_1 = -3, \quad \lambda_2 = -2

STEP 2-3 Find eigenvectors

\mathbf{v}_1 = \begin{pmatrix} 1 \\ 0 \end{pmatrix}, \quad \mathbf{v}_2 = \begin{pmatrix} 1 \\ -1 \end{pmatrix}

STEP 4 General solution

\mathbf{x}(t) = c_1e^{-3t}\begin{pmatrix} 1 \\ 0 \end{pmatrix} + c_2e^{-2t}\begin{pmatrix} 1 \\ -1 \end{pmatrix}

STEP 5 Apply initial condition

c_1\begin{pmatrix} 1 \\ 0 \end{pmatrix} + c_2\begin{pmatrix} 1 \\ -1 \end{pmatrix} = \begin{pmatrix} 1 \\ 4 \end{pmatrix} \Rightarrow c_1 = -3, c_2 = 4

\mathbf{x}(t) = -3e^{-3t}\begin{pmatrix} 1 \\ 0 \end{pmatrix} + 4e^{-2t}\begin{pmatrix} 1 \\ -1 \end{pmatrix}

Common Mistakes to Avoid

Characteristic Equation & Eigenvalues

Forgetting to compute $\det(A - \lambda I)$ correctly. Check your algebra!
Using $\det(A)$ instead of $\det(A - \lambda I)$.
Sign errors when expanding the determinant.

Eigenvectors

Forgetting to substitute $\lambda$ back into $(A - \lambda I)\mathbf{v} = \mathbf{0}$ and solve!
Treating the zero vector as an eigenvector (it's not).
Scaling eigenvectors inconsistently (OK to scale, just be clear).

Repeated & Complex Cases

Forgetting to use generalized eigenvectors for repeated eigenvalues with insufficient eigenvectors.
Using complex solution directly instead of converting to real form.
Confusing the real and imaginary parts of eigenvectors.

Initial Conditions & Final Form

Forgetting to apply the initial condition $\mathbf{x}(0) = \mathbf{x}_0$.
Solving the system incorrectly to find $c_1, c_2$.
Not verifying your solution by checking $\mathbf{x}(0)$ and $\mathbf{x}'(t) = A\mathbf{x}(t)$.

Summary: Eigenvalue Method for x' = Ax

Distinct Real

Find eigenvalues and eigenvectors. Solution is sum of exponential modes: $e^{\lambda_i t}\mathbf{v}_i$.

Complex Conjugates

Real part = exponential envelope, imaginary part = oscillation frequency. Convert to real form.

Repeated Root

Check if you have enough independent eigenvectors. If not, use generalized eigenvectors.

Key Takeaway

      The eigenvalues of $A$ completely determine the solution structure and long-term behavior of the system. Distinct real eigenvalues give pure exponential behavior, complex conjugates add oscillation, and repeated eigenvalues require special care.

Next Up

      Phase Portraits — visualize system behavior WITHOUT solving! Use eigenvalues and eigenvectors to classify stability and predict trajectories in the phase plane.