Differential Equation¶

Definition¶

Let $x$ be an unknown real-valued function of time $t$ , and write $x'(t)=\dfrac{\mathrm{d}x}{\mathrm{d}t}(t)$ for its derivative and $x^{(k)}(t)$ for its $k$ -th higher-order derivative. An ordinary differential equation (ODE) of order $n$ is a relation

F\big(t,\,x(t),\,x'(t),\,\dots,\,x^{(n)}(t)\big)=0,

in which $x^{(n)}$ actually occurs. A solution on an interval $I$ is an $n$ times differentiable function $x\colon I\to\mathbb{R}$ that satisfies the relation for every $t\in I$ $I$ .

The equation is linear if it has the form $x^{(n)}+a_{n-1}(t)x^{(n-1)}+\dots+a_0(t)x=g(t)$ $g (t)$ , homogeneous (linear case) if $g\equiv0$ $0$ , and autonomous if $t$ does not appear explicitly, as in $x'=f(x)$ $f (x)$ . An initial value problem adds conditions $x(t_0)=x_0,\ x'(t_0)=x_1,\dots,x^{(n-1)}(t_0)=x_{n-1}$ $x_{n - 1}$ .

Interpretation¶

A differential equation describes continuous-time change: it states how the rate of change of a quantity depends on time and on the current state. It is the continuous-time counterpart of a difference equation, where the state jumps from $x_t$ to $x_{t+1}$ .

Short grey slope segments of x' = 3x with several solution curves Ce^(3t) following them, and the curve 2e^(3t) through the initial point x(0) = 2 highlighted in blue.

The figure draws the exponential-growth example $x'(t)=3x(t)$ $3 x (t)$ . Each short grey segment has slope $3x$ , the rate of change the equation prescribes at state $x$ . The grey curves are the solutions $C\mathrm{e}^{3t}$ for $C=-2,-1,0,1,3$ $- 2, - 1, 0, 1, 3$ ; each follows the segments everywhere. The initial condition $x(0)=2$ picks out one member of this family, the blue curve $2\mathrm{e}^{3t}$ .

Move $x(0)$ to pick one curve $Ce^{at}$ from the family that follows the slope segments, and change $a$ to switch between growth ( $a >$ $a>0$ $0$ ) and decay ( $a <$ $a<0$ $0$ ).

The unknown is a whole function, not a number. An equation without initial conditions therefore has a family of solutions, usually indexed by $n$ free constants; the $n$ initial conditions select one member of that family.

A common misunderstanding is to treat "solution" as a formula valid for all $t$ . A solution lives on an interval, and for nonlinear equations it can cease to exist after finite time (see the last example).

Key Results¶

The order is the highest derivative that occurs; $x''+(x')^3=0$ has order 2, not 3.
The simplest case $x'(t)=g(t)$ with $g$ continuous is solved by any antiderivative: $x(t)=x(t_0)+\int_{t_0}^t g(s)\,\mathrm{d}s$ $\int_{t_{0}}^{t} g (s) d s$ .
For a linear equation with continuous coefficients $a_k(t)$ and $g(t)$ on $I$ , every initial value problem has exactly one solution, defined on all of $I$ .
For a homogeneous linear equation, sums and scalar multiples of solutions are solutions, so the solutions form a vector space.
A nonlinear equation can have solutions that exist only on a bounded interval.

Examples¶

Exponential growth. For $x'(t)=3x(t)$ $3 x (t)$ , every function $x(t)=C\mathrm{e}^{3t}$ is a solution, since $x'(t)=3C\mathrm{e}^{3t}$ $3 C e^{3 t}$ . The condition $x(0)=2$ gives $C=2$ $2$ , so $x(t)=2\mathrm{e}^{3t}$ $2 e^{3 t}$ .

Second order. For $x''(t)=6t$ $6 t$ , integrating twice gives $x'(t)=3t^2+c_1$ and $x(t)=t^3+c_1t+c_2$ $c_{2}$ : two free constants for an equation of order 2. With $x(0)=1$ $1$ , $x'(0)=0$ we get $x(t)=t^3+1$ $1$ .

Finite-time blow-up. For the nonlinear equation $x'(t)=x(t)^2$ with $x(0)=1$ $1$ , the function $x(t)=\dfrac{1}{1-t}$ satisfies $x'(t)=\dfrac{1}{(1-t)^2}=x(t)^2$ $x (t)^{2}$ . It is a solution only on $(-\infty,1)$ and tends to $+\infty$ as $t\to1^-$ $1^{-}$ .

Prerequisites¶

Sources¶

Math V: Advanced Linear Algebra · Week 3

Aliases: ordinary differential equation, ODE Keywords: x'(t), order of a differential equation, initial value problem, autonomous differential equation, linear differential equation Symbols: element of, infinity, integral, real numbers

Last updated: 14 October 2026 08:21