4.2 Solving Homogeneous Linear Systems $\mathbf{x}' = A\mathbf{x}$

Eigenvalues, eigenvectors, and complete solution methods

Learning Objectives

By the end of this section, you will be able to:

  1. Find eigenvalues and eigenvectors of a matrix
  2. Classify solutions based on eigenvalue types (real distinct, complex, repeated)
  3. Write the general solution for all three cases
  4. Solve initial value problems for linear systems
  5. Visualize solution behavior in the phase plane

The Homogeneous Linear System

We focus on systems of the form:

$$\mathbf{x}' = A\mathbf{x}$$

where $\mathbf{x}(t) = \begin{pmatrix} x_1(t) \\ x_2(t) \end{pmatrix}$ is a vector of unknown functions and $A$ is a constant $2 \times 2$ matrix. The solution method depends entirely on the eigenvalues and eigenvectors of $A$.

Key Insight

The exponential matrix function $e^{At}$ forms the fundamental matrix, and eigenvalues tell us the growth rate along eigenvector directions. This is one of the most elegant connections between linear algebra and differential equations.

Theory: The Eigenvalue Method

Fundamental Theorem

Theorem

If $\lambda$ is an eigenvalue of $A$ and $\mathbf{v}$ is a corresponding eigenvector (so $A\mathbf{v} = \lambda\mathbf{v}$), then

$$\mathbf{x}(t) = e^{\lambda t}\mathbf{v}$$

is a solution to $\mathbf{x}' = A\mathbf{x}$.

Proof

Compute the derivative:

$$\mathbf{x}'(t) = \lambda e^{\lambda t}\mathbf{v}$$

Since $A\mathbf{v} = \lambda\mathbf{v}$, we have:

$$A\mathbf{x}(t) = A(e^{\lambda t}\mathbf{v}) = e^{\lambda t}(A\mathbf{v}) = e^{\lambda t}(\lambda\mathbf{v}) = \lambda e^{\lambda t}\mathbf{v}$$

Therefore $\mathbf{x}' = A\mathbf{x}$. ✓

Case 1: Distinct Real Eigenvalues

When $A$ has two distinct real eigenvalues $\lambda_1 \neq \lambda_2$ with corresponding eigenvectors $\mathbf{v}_1$ and $\mathbf{v}_2$, both $e^{\lambda_1 t}\mathbf{v}_1$ and $e^{\lambda_2 t}\mathbf{v}_2$ are solutions.

General Solution (Case 1)
$$\mathbf{x}(t) = C_1 e^{\lambda_1 t}\mathbf{v}_1 + C_2 e^{\lambda_2 t}\mathbf{v}_2$$

where $C_1, C_2$ are arbitrary constants determined by initial conditions.

Superposition Principle: Any linear combination of solutions is also a solution. This follows from the linearity of the system.

Case 2: Complex Eigenvalues

When $A$ has complex eigenvalues $\lambda = \alpha \pm \beta i$ (with $\beta \neq 0$) and corresponding eigenvector $\mathbf{v} = \mathbf{a} + i\mathbf{b}$ (where $\mathbf{a}, \mathbf{b}$ are real vectors), the complex exponential $e^{(\alpha + \beta i)t}\mathbf{v}$ is a solution. However, we need real-valued solutions.

Using Euler's formula $e^{i\theta} = \cos\theta + i\sin\theta$:

$$e^{(\alpha+\beta i)t}(\mathbf{a}+i\mathbf{b}) = e^{\alpha t}(\cos(\beta t) + i\sin(\beta t))(\mathbf{a}+i\mathbf{b})$$

Separating real and imaginary parts:

$$\mathbf{x}_1(t) = e^{\alpha t}[\mathbf{a}\cos(\beta t) - \mathbf{b}\sin(\beta t)]$$
$$\mathbf{x}_2(t) = e^{\alpha t}[\mathbf{a}\sin(\beta t) + \mathbf{b}\cos(\beta t)]$$
General Solution (Case 2: Complex)
$$\mathbf{x}(t) = C_1\mathbf{x}_1(t) + C_2\mathbf{x}_2(t)$$

where $\mathbf{x}_1(t)$ and $\mathbf{x}_2(t)$ are the real and imaginary parts as shown above.

Interpretation: The factor $e^{\alpha t}$ controls exponential growth/decay. If $\alpha > 0$, solutions spiral outward (unstable); if $\alpha < 0$, they spiral inward (stable). The frequency $\beta$ determines the rotation speed.

Case 3: Repeated Eigenvalue

When the characteristic polynomial has a repeated eigenvalue $\lambda$ (eigenvalue of multiplicity 2), there are two subcases:

Case 3a: Two Independent Eigenvectors

If there are two linearly independent eigenvectors $\mathbf{v}_1, \mathbf{v}_2$, then:

$$\mathbf{x}(t) = C_1 e^{\lambda t}\mathbf{v}_1 + C_2 e^{\lambda t}\mathbf{v}_2$$

Case 3b: One Eigenvector (Defective Matrix)

If there is only one linearly independent eigenvector $\mathbf{v}$, we need a generalized eigenvector $\mathbf{w}$ satisfying:

$$(A - \lambda I)\mathbf{w} = \mathbf{v}$$

Then the two linearly independent solutions are:

$$\mathbf{x}_1(t) = e^{\lambda t}\mathbf{v}, \quad \mathbf{x}_2(t) = e^{\lambda t}(t\mathbf{v} + \mathbf{w})$$
General Solution (Case 3: Repeated, Defective)
$$\mathbf{x}(t) = C_1 e^{\lambda t}\mathbf{v} + C_2 e^{\lambda t}(t\mathbf{v} + \mathbf{w})$$

where $\mathbf{v}$ is the eigenvector and $\mathbf{w}$ is the generalized eigenvector.

Important Note

A repeated eigenvalue does not always require a generalized eigenvector. If the eigenspace has dimension 2 (two independent eigenvectors exist), use Case 3a. Always check the rank of $(A - \lambda I)$ to determine the number of independent eigenvectors.

Step-by-Step Solution Method

  1. Find the characteristic equation: Compute $\det(A - \lambda I) = 0$ to find eigenvalues.
    For $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$, the characteristic equation is $\lambda^2 - (a+d)\lambda + (ad-bc) = 0$.
  2. Find eigenvectors: For each eigenvalue $\lambda$, solve $(A - \lambda I)\mathbf{v} = \mathbf{0}$ to find the eigenvector(s).
  3. Determine the case: Check if eigenvalues are real/distinct, real/repeated, or complex.
  4. Write the general solution: Apply the appropriate formula from the three cases above.
  5. Apply initial conditions: If $\mathbf{x}(0) = \mathbf{x}_0$ is given, solve for constants $C_1, C_2$.

Worked Examples

Problem

Solve the system:

$$\mathbf{x}' = \begin{pmatrix} 1 & 3 \\ 5 & 3 \end{pmatrix}\mathbf{x}$$

Solution

Step 1: Find eigenvalues

$$\det(A - \lambda I) = \det\begin{pmatrix} 1-\lambda & 3 \\ 5 & 3-\lambda \end{pmatrix} = 0$$

Expand:

$$(1-\lambda)(3-\lambda) - 15 = 0$$
$$\lambda^2 - 4\lambda + 3 - 15 = 0$$
$$\lambda^2 - 4\lambda - 12 = 0$$

Factor or use the quadratic formula:

$$(\lambda - 6)(\lambda + 2) = 0$$

So $\lambda_1 = 6$ and $\lambda_2 = -2$ (distinct real eigenvalues).

Step 2: Find eigenvectors

For $\lambda_1 = 6$: Solve $(A - 6I)\mathbf{v}_1 = \mathbf{0}$

$$\begin{pmatrix} -5 & 3 \\ 5 & -3 \end{pmatrix}\begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = \mathbf{0}$$

From the first row: $-5v_1 + 3v_2 = 0 \Rightarrow v_2 = \frac{5}{3}v_1$. Choose $v_1 = 3$: $\mathbf{v}_1 = \begin{pmatrix} 3 \\ 5 \end{pmatrix}$

For $\lambda_2 = -2$: Solve $(A + 2I)\mathbf{v}_2 = \mathbf{0}$

$$\begin{pmatrix} 3 & 3 \\ 5 & 5 \end{pmatrix}\begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = \mathbf{0}$$

From the first row: $3v_1 + 3v_2 = 0 \Rightarrow v_1 = -v_2$. Choose $v_2 = -1$: $\mathbf{v}_2 = \begin{pmatrix} 1 \\ -1 \end{pmatrix}$

Step 3: Write the general solution

$$\mathbf{x}(t) = C_1 e^{6t}\begin{pmatrix} 3 \\ 5 \end{pmatrix} + C_2 e^{-2t}\begin{pmatrix} 1 \\ -1 \end{pmatrix}$$

In component form:

$$x_1(t) = 3C_1 e^{6t} + C_2 e^{-2t}$$
$$x_2(t) = 5C_1 e^{6t} - C_2 e^{-2t}$$
Problem

Solve the system:

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

Solution

Step 1: Find eigenvalues

$$\det(A - \lambda I) = \det\begin{pmatrix} -1-\lambda & 2 \\ -2 & -1-\lambda \end{pmatrix} = 0$$

Expand:

$$(-1-\lambda)^2 + 4 = 0$$
$$\lambda^2 + 2\lambda + 1 + 4 = 0$$
$$\lambda^2 + 2\lambda + 5 = 0$$

By the quadratic formula:

$$\lambda = \frac{-2 \pm \sqrt{4-20}}{2} = \frac{-2 \pm \sqrt{-16}}{2} = \frac{-2 \pm 4i}{2} = -1 \pm 2i$$

So $\alpha = -1$ and $\beta = 2$.

Step 2: Find the eigenvector for $\lambda = -1 + 2i$

Solve $(A - (-1+2i)I)\mathbf{v} = \mathbf{0}$:

$$\begin{pmatrix} -2i & 2 \\ -2 & -2i \end{pmatrix}\begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = \mathbf{0}$$

From the first row: $-2iv_1 + 2v_2 = 0 \Rightarrow v_2 = iv_1$. Choose $v_1 = 1$: $\mathbf{v} = \begin{pmatrix} 1 \\ i \end{pmatrix}$

So $\mathbf{a} = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix} 0 \\ 1 \end{pmatrix}$

Step 3: Write the real-valued solutions

$$\mathbf{x}_1(t) = e^{-t}\left[\begin{pmatrix} 1 \\ 0 \end{pmatrix}\cos(2t) - \begin{pmatrix} 0 \\ 1 \end{pmatrix}\sin(2t)\right] = e^{-t}\begin{pmatrix} \cos(2t) \\ -\sin(2t) \end{pmatrix}$$
$$\mathbf{x}_2(t) = e^{-t}\left[\begin{pmatrix} 1 \\ 0 \end{pmatrix}\sin(2t) + \begin{pmatrix} 0 \\ 1 \end{pmatrix}\cos(2t)\right] = e^{-t}\begin{pmatrix} \sin(2t) \\ \cos(2t) \end{pmatrix}$$

Step 4: General solution

$$\mathbf{x}(t) = C_1 e^{-t}\begin{pmatrix} \cos(2t) \\ -\sin(2t) \end{pmatrix} + C_2 e^{-t}\begin{pmatrix} \sin(2t) \\ \cos(2t) \end{pmatrix}$$

This represents a spiral converging to the origin (stable focus) because $\alpha = -1 < 0$.

Problem

Solve the system:

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

Solution

Step 1: Find eigenvalues

$$\det(A - \lambda I) = \det\begin{pmatrix} 3-\lambda & -1 \\ 1 & 1-\lambda \end{pmatrix} = 0$$

Expand:

$$(3-\lambda)(1-\lambda) + 1 = 0$$
$$3 - 3\lambda - \lambda + \lambda^2 + 1 = 0$$
$$\lambda^2 - 4\lambda + 4 = 0$$
$$(\lambda - 2)^2 = 0$$

So $\lambda = 2$ (repeated, multiplicity 2).

Step 2: Find eigenvector(s)

Solve $(A - 2I)\mathbf{v} = \mathbf{0}$:

$$\begin{pmatrix} 1 & -1 \\ 1 & -1 \end{pmatrix}\begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = \mathbf{0}$$

From the first row: $v_1 - v_2 = 0 \Rightarrow v_1 = v_2$. Choose $v_1 = 1$: $\mathbf{v} = \begin{pmatrix} 1 \\ 1 \end{pmatrix}$

Only one eigenvector exists (rank is 1), so this is a defective matrix. We need a generalized eigenvector.

Step 3: Find generalized eigenvector

Solve $(A - 2I)\mathbf{w} = \mathbf{v}$:

$$\begin{pmatrix} 1 & -1 \\ 1 & -1 \end{pmatrix}\begin{pmatrix} w_1 \\ w_2 \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \end{pmatrix}$$

This gives $w_1 - w_2 = 1$. Choose $w_1 = 1$: $w_2 = 0$, so $\mathbf{w} = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$

Step 4: General solution

$$\mathbf{x}(t) = C_1 e^{2t}\begin{pmatrix} 1 \\ 1 \end{pmatrix} + C_2 e^{2t}\left(t\begin{pmatrix} 1 \\ 1 \end{pmatrix} + \begin{pmatrix} 1 \\ 0 \end{pmatrix}\right)$$

Simplifying:

$$\mathbf{x}(t) = e^{2t}\begin{pmatrix} C_1 + C_2(t+1) \\ C_1 + C_2 t \end{pmatrix}$$
Problem

Solve the IVP:

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

Solution

Step 1: Find eigenvalues

$$\det(A - \lambda I) = \det\begin{pmatrix} 2-\lambda & 3 \\ 4 & 3-\lambda \end{pmatrix} = 0$$

Expand:

$$(2-\lambda)(3-\lambda) - 12 = 0$$
$$\lambda^2 - 5\lambda + 6 - 12 = 0$$
$$\lambda^2 - 5\lambda - 6 = 0$$
$$(\lambda - 6)(\lambda + 1) = 0$$

So $\lambda_1 = 6$ and $\lambda_2 = -1$.

Step 2: Find eigenvectors

For $\lambda_1 = 6$: $(A - 6I)\mathbf{v}_1 = \mathbf{0}$

$$\begin{pmatrix} -4 & 3 \\ 4 & -3 \end{pmatrix}\begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = \mathbf{0}$$

From the first row: $-4v_1 + 3v_2 = 0 \Rightarrow v_2 = \frac{4}{3}v_1$. Choose $v_1 = 3$: $\mathbf{v}_1 = \begin{pmatrix} 3 \\ 4 \end{pmatrix}$

For $\lambda_2 = -1$: $(A + I)\mathbf{v}_2 = \mathbf{0}$

$$\begin{pmatrix} 3 & 3 \\ 4 & 4 \end{pmatrix}\begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = \mathbf{0}$$

From the first row: $3v_1 + 3v_2 = 0 \Rightarrow v_1 = -v_2$. Choose $v_2 = -1$: $\mathbf{v}_2 = \begin{pmatrix} 1 \\ -1 \end{pmatrix}$

Step 3: General solution

$$\mathbf{x}(t) = C_1 e^{6t}\begin{pmatrix} 3 \\ 4 \end{pmatrix} + C_2 e^{-t}\begin{pmatrix} 1 \\ -1 \end{pmatrix}$$

Step 4: Apply initial conditions

At $t=0$:

$$\begin{pmatrix} 1 \\ 2 \end{pmatrix} = C_1\begin{pmatrix} 3 \\ 4 \end{pmatrix} + C_2\begin{pmatrix} 1 \\ -1 \end{pmatrix}$$

This gives the system:

$$3C_1 + C_2 = 1$$
$$4C_1 - C_2 = 2$$

Adding the equations: $7C_1 = 3 \Rightarrow C_1 = \frac{3}{7}$

From the first: $C_2 = 1 - 3C_1 = 1 - \frac{9}{7} = -\frac{2}{7}$

Particular solution:

$$\mathbf{x}(t) = \frac{3}{7}e^{6t}\begin{pmatrix} 3 \\ 4 \end{pmatrix} - \frac{2}{7}e^{-t}\begin{pmatrix} 1 \\ -1 \end{pmatrix}$$

Or in component form:

$$x_1(t) = \frac{9}{7}e^{6t} - \frac{2}{7}e^{-t}$$
$$x_2(t) = \frac{12}{7}e^{6t} + \frac{2}{7}e^{-t}$$

Converting Higher-Order DEs to Systems

Any $n$th-order ODE can be rewritten as a system of $n$ first-order equations by introducing new variables for each derivative. This lets us apply the eigenvalue method to equations that are not originally written as systems.

The Conversion Recipe

Given an $n$th-order ODE in $y$, define new state variables:

$$x_1 = y, \quad x_2 = y', \quad x_3 = y'', \quad \ldots, \quad x_n = y^{(n-1)}$$

Then the system becomes $x_1' = x_2$, $x_2' = x_3$, $\ldots$, and the last equation comes from the original ODE solved for the highest derivative.

Problem

Convert the following to a first-order system and solve:

$$y'' + y' - 6y = 0$$

Step 1: Convert to a System

Let $x_1 = y$ and $x_2 = y'$. Then:

$$x_1' = y' = x_2$$

From the original ODE: $y'' = -y' + 6y = -x_2 + 6x_1$, so:

$$x_2' = 6x_1 - x_2$$

In matrix form:

$$\begin{pmatrix} x_1 \\ x_2 \end{pmatrix}' = \begin{pmatrix} 0 & 1 \\ 6 & -1 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \end{pmatrix}$$

Step 2: Find Eigenvalues

$$\det(A - \lambda I) = \det\begin{pmatrix} -\lambda & 1 \\ 6 & -1-\lambda \end{pmatrix} = 0$$
$$-\lambda(-1-\lambda) - 6 = \lambda^2 + \lambda - 6 = 0$$
$$(\lambda + 3)(\lambda - 2) = 0 \quad \Longrightarrow \quad \lambda_1 = 2, \quad \lambda_2 = -3$$

Step 3: Find Eigenvectors

For $\lambda_1 = 2$: solve $(A - 2I)\mathbf{v} = \mathbf{0}$

$$\begin{pmatrix} -2 & 1 \\ 6 & -3 \end{pmatrix}\begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = \mathbf{0} \quad \Longrightarrow \quad -2v_1 + v_2 = 0 \quad \Longrightarrow \quad v_2 = 2v_1$$

Choose $v_1 = 1$: $\quad \mathbf{v}_1 = \begin{pmatrix} 1 \\ 2 \end{pmatrix}$

For $\lambda_2 = -3$: solve $(A + 3I)\mathbf{v} = \mathbf{0}$

$$\begin{pmatrix} 3 & 1 \\ 6 & 2 \end{pmatrix}\begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = \mathbf{0} \quad \Longrightarrow \quad 3v_1 + v_2 = 0 \quad \Longrightarrow \quad v_2 = -3v_1$$

Choose $v_1 = 1$: $\quad \mathbf{v}_2 = \begin{pmatrix} 1 \\ -3 \end{pmatrix}$

Step 4: Write the General Solution

The system solution is:

$$\mathbf{x}(t) = C_1 e^{2t}\begin{pmatrix} 1 \\ 2 \end{pmatrix} + C_2 e^{-3t}\begin{pmatrix} 1 \\ -3 \end{pmatrix}$$

Since $y = x_1$, the solution to the original ODE is:

$$\boxed{y(t) = C_1 e^{2t} + C_2 e^{-3t}}$$
Verification

The characteristic equation of $y'' + y' - 6y = 0$ is $r^2 + r - 6 = (r+3)(r-2) = 0$, giving $r = 2, -3$. This matches our eigenvalues exactly — the eigenvalues of the companion matrix are always the roots of the characteristic equation!

Problem

Convert to a first-order system and solve:

$$y'' + 2y' + 5y = 0, \qquad y(0) = 1, \quad y'(0) = -1$$

Step 1: Convert to a System

Let $x_1 = y$ and $x_2 = y'$. Then $x_1' = x_2$ and from the ODE: $y'' = -2y' - 5y$, so $x_2' = -5x_1 - 2x_2$.

$$\mathbf{x}' = \begin{pmatrix} 0 & 1 \\ -5 & -2 \end{pmatrix}\mathbf{x}, \qquad \mathbf{x}(0) = \begin{pmatrix} 1 \\ -1 \end{pmatrix}$$

Step 2: Find Eigenvalues

$$\det(A - \lambda I) = \lambda^2 + 2\lambda + 5 = 0$$
$$\lambda = \frac{-2 \pm \sqrt{4 - 20}}{2} = \frac{-2 \pm 4i}{2} = -1 \pm 2i$$

We have complex eigenvalues with $\alpha = -1$ and $\beta = 2$.

Step 3: Find an Eigenvector

For $\lambda = -1 + 2i$: solve $(A - (-1+2i)I)\mathbf{v} = \mathbf{0}$

$$\begin{pmatrix} 1-2i & 1 \\ -5 & -1-2i \end{pmatrix}\begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = \mathbf{0}$$

From the first row: $(1-2i)v_1 + v_2 = 0$, so $v_2 = -(1-2i)v_1 = (-1+2i)v_1$.

Choose $v_1 = 1$:

$$\mathbf{v} = \begin{pmatrix} 1 \\ -1 + 2i \end{pmatrix} = \underbrace{\begin{pmatrix} 1 \\ -1 \end{pmatrix}}_{\mathbf{a}} + i\underbrace{\begin{pmatrix} 0 \\ 2 \end{pmatrix}}_{\mathbf{b}}$$

Step 4: Write the General Solution

Using the formula $\mathbf{x}(t) = e^{\alpha t}\big[C_1(\mathbf{a}\cos\beta t - \mathbf{b}\sin\beta t) + C_2(\mathbf{a}\sin\beta t + \mathbf{b}\cos\beta t)\big]$:

$$\mathbf{x}(t) = e^{-t}\left[C_1\left(\begin{pmatrix}1\\-1\end{pmatrix}\cos 2t - \begin{pmatrix}0\\2\end{pmatrix}\sin 2t\right) + C_2\left(\begin{pmatrix}1\\-1\end{pmatrix}\sin 2t + \begin{pmatrix}0\\2\end{pmatrix}\cos 2t\right)\right]$$

So $x_1(t) = y(t) = e^{-t}(C_1 \cos 2t + C_2 \sin 2t)$.

Step 5: Apply Initial Conditions

$y(0) = 1$: $\quad C_1 \cdot 1 + C_2 \cdot 0 = 1 \quad \Longrightarrow \quad C_1 = 1$

For $y'(0) = -1$, compute $y' = e^{-t}[(-C_1 + 2C_2)\cos 2t + (-C_2 - 2C_1)\sin 2t]$:

$y'(0) = -C_1 + 2C_2 = -1 + 2C_2 = -1 \quad \Longrightarrow \quad C_2 = 0$

$$\boxed{y(t) = e^{-t}\cos 2t}$$
Physical Interpretation

This describes a damped oscillation: the $\cos 2t$ factor produces oscillation with frequency $\omega = 2$, while the $e^{-t}$ envelope causes the amplitude to decay over time. The negative real part $\alpha = -1$ guarantees stability — the system spirals toward the origin.

Problem

Convert the following third-order ODE to a first-order system and solve:

$$y''' - 6y'' + 11y' - 6y = 0$$

Step 1: Convert to a System

A third-order ODE needs three state variables:

$$x_1 = y, \qquad x_2 = y', \qquad x_3 = y''$$

Then:

  • $x_1' = x_2$
  • $x_2' = x_3$
  • $x_3' = y''' = 6y'' - 11y' + 6y = 6x_3 - 11x_2 + 6x_1$ (from the original ODE)

In matrix form:

$$\begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix}' = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 6 & -11 & 6 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix}$$
Companion Matrix

This $3 \times 3$ matrix is called a companion matrix. It always has 1's on the superdiagonal, and the coefficients of the ODE (with appropriate signs) in the last row. The eigenvalues of this matrix are exactly the roots of the characteristic equation of the original ODE.

Step 2: Find Eigenvalues

The characteristic equation of the companion matrix equals the characteristic equation of the ODE:

$$r^3 - 6r^2 + 11r - 6 = 0$$

Try $r = 1$: $1 - 6 + 11 - 6 = 0$ ✓. So $(r-1)$ is a factor. Divide:

$$r^3 - 6r^2 + 11r - 6 = (r-1)(r^2 - 5r + 6) = (r-1)(r-2)(r-3)$$

Eigenvalues: $\lambda_1 = 1, \quad \lambda_2 = 2, \quad \lambda_3 = 3$.

Step 3: Find Eigenvectors

For a companion matrix, the eigenvector for eigenvalue $\lambda$ is always:

$$\mathbf{v}_\lambda = \begin{pmatrix} 1 \\ \lambda \\ \lambda^2 \end{pmatrix}$$

This gives us:

$$\mathbf{v}_1 = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}, \qquad \mathbf{v}_2 = \begin{pmatrix} 1 \\ 2 \\ 4 \end{pmatrix}, \qquad \mathbf{v}_3 = \begin{pmatrix} 1 \\ 3 \\ 9 \end{pmatrix}$$
Why This Pattern?

Since $x_1 = y$, $x_2 = y'$, and $x_3 = y''$, and if $y = e^{\lambda t}$, then $y' = \lambda e^{\lambda t}$ and $y'' = \lambda^2 e^{\lambda t}$. The ratios $1 : \lambda : \lambda^2$ form the eigenvector — each component is one more derivative of $e^{\lambda t}$.

Step 4: Write the General Solution

$$\mathbf{x}(t) = C_1 e^{t}\begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} + C_2 e^{2t}\begin{pmatrix} 1 \\ 2 \\ 4 \end{pmatrix} + C_3 e^{3t}\begin{pmatrix} 1 \\ 3 \\ 9 \end{pmatrix}$$

Since $y = x_1$:

$$\boxed{y(t) = C_1 e^{t} + C_2 e^{2t} + C_3 e^{3t}}$$

And we can verify: $y' = x_2 = C_1 e^t + 2C_2 e^{2t} + 3C_3 e^{3t}$ and $y'' = x_3 = C_1 e^t + 4C_2 e^{2t} + 9C_3 e^{3t}$, which are consistent with differentiating $y(t)$.

Key Takeaway

Converting a higher-order ODE to a system is not just an exercise — it reveals deeper structure. The eigenvalues of the companion matrix are exactly the roots of the characteristic equation, and the eigenvector structure encodes the derivative relationships. For numerical computation, the system form is essential since all ODE solvers (like MATLAB's ode45) work with first-order systems.

Interactive System Solver

Enter a 2×2 matrix and visualize the solution trajectories in the phase plane.

Exam-Style Practice Problems

Work through these problems to test your understanding. Click to reveal solutions.

Solution

Characteristic equation:

$$\det\begin{pmatrix} 4-\lambda & -2 \\ 5 & -3-\lambda \end{pmatrix} = (4-\lambda)(-3-\lambda) + 10 = 0$$
$$\lambda^2 - \lambda - 12 + 10 = 0$$
$$\lambda^2 - \lambda - 2 = 0$$
$$(\lambda - 2)(\lambda + 1) = 0$$

Eigenvalues: $\lambda_1 = 2, \lambda_2 = -1$ (distinct real)

Eigenvectors:

For $\lambda_1 = 2$: $(A-2I)\mathbf{v}_1 = \mathbf{0}$ gives $2v_1 - 2v_2 = 0$, so $\mathbf{v}_1 = \begin{pmatrix} 1 \\ 1 \end{pmatrix}$

For $\lambda_2 = -1$: $(A+I)\mathbf{v}_2 = \mathbf{0}$ gives $5v_1 - 2v_2 = 0$, so $\mathbf{v}_2 = \begin{pmatrix} 2 \\ 5 \end{pmatrix}$

General solution:

$$\mathbf{x}(t) = C_1 e^{2t}\begin{pmatrix} 1 \\ 1 \end{pmatrix} + C_2 e^{-t}\begin{pmatrix} 2 \\ 5 \end{pmatrix}$$

Solution

Characteristic equation:

$$\det\begin{pmatrix} 3-\lambda & -4 \\ 2 & -1-\lambda \end{pmatrix} = (3-\lambda)(-1-\lambda) + 8 = 0$$
$$\lambda^2 - 2\lambda - 3 + 8 = 0$$
$$\lambda^2 - 2\lambda + 5 = 0$$
$$\lambda = \frac{2 \pm \sqrt{4-20}}{2} = \frac{2 \pm 4i}{2} = 1 \pm 2i$$

Eigenvalues: $\lambda = 1 \pm 2i$ (complex with $\alpha = 1, \beta = 2$)

Eigenvector for $\lambda = 1 + 2i$:

Solve $(A - (1+2i)I)\mathbf{v} = \mathbf{0}$:

$$\begin{pmatrix} 2-2i & -4 \\ 2 & -2-2i \end{pmatrix}\begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = \mathbf{0}$$

From the first row: $(2-2i)v_1 - 4v_2 = 0 \Rightarrow v_1 = \frac{4v_2}{2-2i} = \frac{4v_2(2+2i)}{8} = v_2(1+i)$

Choose $v_2 = 1$: $\mathbf{v} = \begin{pmatrix} 1+i \\ 1 \end{pmatrix}$, so $\mathbf{a} = \begin{pmatrix} 1 \\ 1 \end{pmatrix}$, $\mathbf{b} = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$

Real-valued solutions:

$$\mathbf{x}_1(t) = e^{t}\begin{pmatrix} \cos(2t) - \sin(2t) \\ \cos(2t) \end{pmatrix}$$
$$\mathbf{x}_2(t) = e^{t}\begin{pmatrix} \sin(2t) + \cos(2t) \\ \sin(2t) \end{pmatrix}$$

General solution:

$$\mathbf{x}(t) = C_1 e^{t}\begin{pmatrix} \cos(2t) - \sin(2t) \\ \cos(2t) \end{pmatrix} + C_2 e^{t}\begin{pmatrix} \sin(2t) + \cos(2t) \\ \sin(2t) \end{pmatrix}$$

Note: System is unstable (spiraling outward) since $\alpha = 1 > 0$.

Solution

Characteristic equation:

$$\det\begin{pmatrix} 5-\lambda & -4 \\ 4 & -3-\lambda \end{pmatrix} = (5-\lambda)(-3-\lambda) + 16 = 0$$
$$\lambda^2 - 2\lambda - 15 + 16 = 0$$
$$\lambda^2 - 2\lambda + 1 = 0$$
$$(\lambda - 1)^2 = 0$$

Repeated eigenvalue: $\lambda = 1$ (multiplicity 2)

Eigenspace: $(A - I)\mathbf{v} = \mathbf{0}$ gives

$$\begin{pmatrix} 4 & -4 \\ 4 & -4 \end{pmatrix}\begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = \mathbf{0}$$

So $v_1 = v_2$, giving one eigenvector $\mathbf{v} = \begin{pmatrix} 1 \\ 1 \end{pmatrix}$

Generalized eigenvector: Solve $(A-I)\mathbf{w} = \mathbf{v}$:

$$\begin{pmatrix} 4 & -4 \\ 4 & -4 \end{pmatrix}\begin{pmatrix} w_1 \\ w_2 \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \end{pmatrix}$$

From the first row: $4w_1 - 4w_2 = 1 \Rightarrow w_1 = w_2 + \frac{1}{4}$. Choose $w_2 = 0$: $\mathbf{w} = \begin{pmatrix} 1/4 \\ 0 \end{pmatrix}$

General solution:

$$\mathbf{x}(t) = C_1 e^{t}\begin{pmatrix} 1 \\ 1 \end{pmatrix} + C_2 e^{t}\left(t\begin{pmatrix} 1 \\ 1 \end{pmatrix} + \begin{pmatrix} 1/4 \\ 0 \end{pmatrix}\right)$$

Apply initial conditions: At $t=0$:

$$\begin{pmatrix} 3 \\ 4 \end{pmatrix} = C_1\begin{pmatrix} 1 \\ 1 \end{pmatrix} + C_2\begin{pmatrix} 1/4 \\ 0 \end{pmatrix}$$

This gives: $C_1 + \frac{C_2}{4} = 3$ and $C_1 = 4$

From the second: $C_1 = 4$, so $4 + \frac{C_2}{4} = 3 \Rightarrow C_2 = -4$

Particular solution:

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