System of Differential Equations¶

Definition¶

A system of differential equations consists of $n$ first-order differential equations for $n$ unknown functions $x_1(t),\dots,x_n(t)$ . For $i=1,\dots,n$ $1, \dots, n$ ,

x_i'(t)=f_i\big(t,x_1(t),\dots,x_n(t)\big),

where $x_i'(t)=\dfrac{\mathrm{d}x_i}{\mathrm{d}t}(t)$ $\frac{d x _{i}}{d t} (t)$ . With the state vector $\boldsymbol{x}(t)=\big(x_1(t),\dots,x_n(t)\big)^{\top}$ it reads $\boldsymbol{x}'(t)=\boldsymbol{f}\big(t,\boldsymbol{x}(t)\big)$ $f (t, x (t))$ . The system is linear if

\boldsymbol{x}'(t)=A(t)\,\boldsymbol{x}(t)+\boldsymbol{b}(t)

for an $n\times n$ matrix function $A(t)$ and a vector function $\boldsymbol{b}(t)$ ; it is coupled if some $x_i'$ depends on some $x_j$ with $j\neq i$ $i$ . An initial value problem prescribes the whole vector $\boldsymbol{x}(t_0)\in\mathbb{R}^n$ $R^{n}$ .

Interpretation¶

A system describes several interacting quantities, such as two competing populations or prices in linked markets. In the linear case the matrix product $A\boldsymbol{x}$ packs all couplings into one coefficient matrix; the entry $a_{ij}$ measures how $x_j$ affects the rate of change of $x_i$ .

In the phase plane of the coupled example, solutions move out along the line through (1,1) and in along the line through (1,−1), and the green curve passes through x(0) = (3,1).

The figure shows the coupled example below in the $(x_1,x_2)$ plane. Along the eigenvector $(1,1)^{\top}$ (blue, $\lambda=3$ $3$ ) solutions move outwards like $\mathrm{e}^{3t}$ , along $(1,-1)^{\top}$ (red, $\lambda=-1$ $- 1$ ) they move inwards like $\mathrm{e}^{-t}$ , and every other solution (grey) combines the two. The green curve is the solution with $\boldsymbol{x}(0)=(3,1)^{\top}$ $(3, 1)^{⊤}$ : it comes in near the $(1,-1)^{\top}$ line and then leaves in the $(1,1)^{\top}$ direction. Neither coordinate can be solved on its own, but in eigenvector coordinates the two motions separate.

Drag $\boldsymbol{x}_0$ and move $t$ : the component along $(1,1)^T$ grows like $e^{3t}$ and the component along $(1,-1)^T$ decays like $e^{-t}$ , so every orbit off the line of $(1,-1)^T$ eventually leaves along $(1,1)^T$ .

Every scalar equation of order $n$ is a first-order system in disguise: with $y_1=x$ $x$ , $y_2=x'$ $x^{'}$ , ..., $y_n=x^{(n-1)}$ the equation $x^{(n)}=f(t,x,\dots,x^{(n-1)})$ becomes $y_1'=y_2,\ \dots,\ y_{n-1}'=y_n,\ y_n'=f(t,y_1,\dots,y_n)$ $f (t, y_{1}, \dots, y_{n})$ . For a second-order differential equation this gives a two-dimensional system. So first-order systems are the general framework, and one theory covers all orders.

Coupling is what makes systems harder than $n$ separate scalar equations. For constant, diagonalizable $A$ the eigenvalues and eigenvectors remove it: a coordinate change along eigenvectors turns the system into decoupled scalar equations. A common error is to solve each row as if the other variables were constants. The discrete-time counterpart is the system of linear difference equations $\boldsymbol{x}_{t+1}=A\boldsymbol{x}_t$ $A x_{t}$ .

Key Results¶

For a linear system with continuous $A(t),\boldsymbol{b}(t)$ on an interval $I$ , each initial value problem has exactly one solution on $I$ .
If $\boldsymbol{b}=\boldsymbol{0}$ and $A$ is constant and diagonalizable, $A=PDP^{-1}$ (diagonalization), then $\boldsymbol{y}=P^{-1}\boldsymbol{x}$ satisfies $\boldsymbol{y}'=D\boldsymbol{y}$ $D y$ , i.e. $y_i'=\lambda_iy_i$ $λ_{i} y_{i}$ , so $y_i(t)=y_i(0)\mathrm{e}^{\lambda_it}$ $y_{i} (0) e^{λ_{i} t}$ .
In that case $\displaystyle \boldsymbol{x}(t)=\sum_i c_i\mathrm{e}^{\lambda_it}\boldsymbol{v}_i$ $i \sum c_{i} e^{λ_{i} t} v_{i}$ , with $\boldsymbol{v}_i$ the columns of $P$ ; for any constant $A$ , $\boldsymbol{x}(t)=\mathrm{e}^{At}\boldsymbol{x}(0)$ with the matrix exponential.
The companion matrix of $x''+a_1x'+a_0x=0$ is $\begin{pmatrix}0&1\\-a_0&-a_1\end{pmatrix}$ , with characteristic polynomial $\lambda^2+a_1\lambda+a_0$ $a_{0}$ .

Examples¶

Coupled linear system. $x_1'=x_1+2x_2$ $2 x_{2}$ , $x_2'=2x_1+x_2$ $x_{2}$ , so $A=\begin{pmatrix}1&2\\2&1\end{pmatrix}$ $(1221)$ . Eigenpairs: $A(1,1)^{\top}=3(1,1)^{\top}$ and $A(1,-1)^{\top}=-(1,-1)^{\top}$ $- (1, - 1)^{⊤}$ . Hence

\boldsymbol{x}(t)=c_1\mathrm{e}^{3t}\begin{pmatrix}1\\1\end{pmatrix}+c_2\mathrm{e}^{-t}\begin{pmatrix}1\\-1\end{pmatrix}.

With $\boldsymbol{x}(0)=(3,1)^{\top}$ $(3, 1)^{⊤}$ : $c_1+c_2=3$ $3$ , $c_1-c_2=1$ $1$ , so $c_1=2$ $2$ , $c_2=1$ and $\boldsymbol{x}(t)=\big(2\mathrm{e}^{3t}+\mathrm{e}^{-t},\ 2\mathrm{e}^{3t}-\mathrm{e}^{-t}\big)^{\top}$ $e^{- t})^{⊤}$ .

Reduction of order. $x''+3x'+2x=0$ becomes, with $y_1=x$ $x$ , $y_2=x'$ $x^{'}$ , the system $\boldsymbol{y}'=\begin{pmatrix}0&1\\-2&-3\end{pmatrix}\boldsymbol{y}$ $(0 - 2 1 - 3) y$ , whose eigenvalues $-1,-2$ are the roots of $r^2+3r+2$ $2$ .

Decoupled system. $x_1'=-x_1$ $- x_{1}$ , $x_2'=4x_2$ has diagonal $A$ and solution $\big(x_1(0)\mathrm{e}^{-t},\ x_2(0)\mathrm{e}^{4t}\big)^{\top}$ .

Prerequisites¶

Sources¶

Math V: Advanced Linear Algebra · Week 4

Aliases: system of ODEs, coupled differential equations, system of ordinary differential equations, ODE system, vector differential equation Keywords: x'(t)=f(t,x(t)), first-order system, companion matrix, decoupling, state vector Symbols: element of, lambda, real numbers, sum

Last updated: 17 October 2026 21:53