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$,

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

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$$i$. An initial value problem prescribes the whole vector $\boldsymbol{x}(t_0)\in\mathbb{R}^n$${\mathbb{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 $x_i$.

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{)}^{\top }$: 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.

Every scalar equation of order $n$ is a first-order system in disguise: with $y_1=x$$x$, $y_2=x'$$x^{\prime }$, ..., $y_n=x^{(n-1)}$ 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{\mathbf{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}$$D\mathbf{y}$, i.e. $y_i'=\lambda_iy_i$${\lambda }_i{y}_i$, so $y_i(t)=y_i(0)\mathrm{e}^{\lambda_it}$$y_i(0){\mathrm{e}}^{{\lambda }_{i}t}$.
• In that case $\displaystyle \boldsymbol{x}(t)=\sum_i c_i\mathrm{e}^{\lambda_it}\boldsymbol{v}_i$$\sum _i{c}_i{\mathrm{e}}^{{\lambda }_{i}t}{\mathbf{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$$2x_2$, $x_2'=2x_1+x_2$$x_2$, so $A=\begin{pmatrix}1&2\\2&1\end{pmatrix}$$\left(\begin{array}{cc}1& 2\\ 2& 1\end{array}\right)$. Eigenpairs: $A(1,1)^{\top}=3(1,1)^{\top}$ and $A(1,-1)^{\top}=-(1,-1)^{\top}$$-(1,-1{)}^{\top }$. Hence

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

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

• First-Order Differential Equation
• Matrix Product
• Eigenvalue and Eigenvector
• Diagonalization

Related Concepts¶

• Linear Vector ODE
• Matrix Exponential
• Second-Order Differential Equation
• System of Linear Difference Equations

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