6.2 Systems of Linear Differential Equations

The reason eigenvalues matter outside linear algebra. A coupled system of differential equations is impossible to attack directly — but in the eigenvector basis it decouples into

n

independent copies of

y^{'} = λ y

, each with the solution you already know.

A first-order system $y_{i}^{'} = a_{i 1} y_{1} + \dots + a_{in} y_{n}$ packages into $Y^{'} = A Y$ , where $Y$ and $Y^{'}$ are vector-valued functions of $t$ . For $n = 1$ this is $y^{'} = a y$ with solution $y = c e^{a t}$ — the whole method is the guess that the general case looks the same.

Try $Y = e^{λ t} x$ . Then $Y^{'} = λ e^{λ t} x$ while $A Y = e^{λ t} A x$ . These match for all $t$ exactly when $A x = λ x$ — so every eigenpair gives a solution.

Structure of the solution set

Linear combinations of solutions are solutions, and the solution set of an

n \times n

system

Y^{'} = A Y

is an $n$ -dimensional subspace of the space of continuous vector-valued functions. So

n

independent eigenvectors give the general solution

Y = c_{1} e^{λ_{1} t} x_{1} + c_{2} e^{λ_{2} t} x_{2} + \dots + c_{n} e^{λ_{n} t} x_{n} .

Adding $Y (0) = Y_{0}$ makes it an initial value problem with a unique particular solution.

Worked example

An initial value problem end to end

0/6 steps

Solve

y_{1}^{'} = 3 y_{1} + 4 y_{2}

y_{2}^{'} = 3 y_{1} + 2 y_{2}

with

y_{1} (0) = 6

y_{2} (0) = 1

Complex eigenvalues

If $λ = a + bi$ with eigenvector $x = Re x + i Im x$ , then $e^{λ t} x$ and $e^{λ ˉ t} x ˉ$ are both solutions. Their average and their difference over $2 i$ are real, giving two real solutions:

Y_{1} Y_{2} = Re (e^{λ t} x) = e^{a t} [(cos b t) Re x - (sin b t) Im x] = Im (e^{λ t} x) = e^{a t} [(cos b t) Im x + (sin b t) Re x]

★Key idea

Read the dynamics straight off the eigenvalue: the real part $a$ controls growth (

a > 0

) or decay (

a < 0

); the imaginary part $b$ sets the oscillation frequency. Purely imaginary eigenvalues ⟹ closed orbits.

Higher-order systems

A second-order system $Y^{''} = A_{1} Y + A_{2} Y^{'}$ becomes first order by introducing $Y_{1} = Y$ and $Y_{2} = Y^{'}$ :

(Y_{1}^{'} Y_{2}^{'}) = (O A_{1} I A_{2}) (Y_{1} Y_{2})

The price is size: an $m$ th-order system in $n$ unknowns becomes a first-order system of size $mn$ , requiring $mn$ initial conditions. Special case: if the system is simply $Y^{(m)} = A Y$ , no new variables are needed — take $σ$ to be an $m$ th root of an eigenvalue $λ$ , and $Y = e^{σ t} x$ is a solution.

Check your work

Eigenvalues, eigenspaces & diagonalizability

Builds the characteristic polynomial exactly, factors out rational roots, then solves the null space of A − λI.

Square matrix

characteristic polynomial

p(λ) = λ³ − 2λ² + λ = 0

trace = Σλ

det = Πλ

diagonalizable

yes

λ = 0algebraic mult. 1 · geometric mult. 1

eigenspace basis:

λ = 1algebraic mult. 2 · geometric mult. 2

eigenspace basis:

-1

Get the eigenvalues and eigenvectors you need for the general solution.

Check yourself

For

Y^{'} = A Y

with

A

n \times n

real matrix, a solution of the form

e^{λ t} x

works precisely when:

T / F

To write the general solution of

Y^{'} = A Y

c_{1} e^{λ_{1} t} x_{1} + \dots + c_{n} e^{λ_{n} t} x_{n}

, you need

n

linearly independent eigenvectors.

A

is real with complex eigenvalue

λ = a + bi

and eigenvector

x

. Two real-valued solutions of

Y^{'} = A Y

are:

T / F

A second-order system

Y^{''} = A_{1} Y + A_{2} Y^{'}

with

Y \in R^{n}

converts into a first-order system of size

2 n

6.2Systems of Linear Differential Equations

What you must be able to do

An initial value problem end to end

Complex eigenvalues

Higher-order systems

Eigenvalues, eigenspaces & diagonalizability

Check yourself