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Differential Equation


Definition

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

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

in which x(n)x^{(n)} actually occurs. A solution on an interval II is an nn times differentiable function x ⁣:I→Rx\colon I\to\mathbb{R} that satisfies the relation for every t∈It\in I.

The equation is linear if it has the form x(n)+an−1(t)x(n−1)+⋯+a0(t)x=g(t)x^{(n)}+a_{n-1}(t)x^{(n-1)}+\dots+a_0(t)x=g(t), homogeneous (linear case) if g≡0g\equiv0, and autonomous if tt does not appear explicitly, as in x′=f(x)x'=f(x). An initial value problem adds conditions x(t0)=x0, x′(t0)=x1,…,x(n−1)(t0)=xn−1x(t_0)=x_0,\ x'(t_0)=x_1,\dots,x^{(n-1)}(t_0)=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 xtx_t to xt+1x_{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)x'(t)=3x(t). Each short grey segment has slope 3x3x, the rate of change the equation prescribes at state xx. The grey curves are the solutions Ce3tC\mathrm{e}^{3t} for C=−2,−1,0,1,3C=-2,-1,0,1,3; each follows the segments everywhere. The initial condition x(0)=2x(0)=2 picks out one member of this family, the blue curve 2e3t2\mathrm{e}^{3t}.

The unknown is a whole function, not a number. For the linear equations with continuous coefficients considered below, the general solution has nn free constants, and nn initial values select exactly one solution. For a general nonlinear equation, the order alone guarantees neither existence nor uniqueness of an initial value problem.

A common misunderstanding is to treat "solution" as a formula valid for all tt. 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=0x''+(x')^3=0 has order 2, not 3.
  • The simplest case x′(t)=g(t)x'(t)=g(t) with gg continuous is solved by any antiderivative: x(t)=x(t0)+∫t0tg(s) dsx(t)=x(t_0)+\int_{t_0}^t g(s)\,\mathrm{d}s.
  • For a linear equation with continuous coefficients ak(t)a_k(t) and g(t)g(t) on II, every initial value problem has exactly one solution, defined on all of II.
  • 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)x'(t)=3x(t), every function x(t)=Ce3tx(t)=C\mathrm{e}^{3t} is a solution, since x′(t)=3Ce3tx'(t)=3C\mathrm{e}^{3t}. The condition x(0)=2x(0)=2 gives C=2C=2, so x(t)=2e3tx(t)=2\mathrm{e}^{3t}.

Second order. For x′′(t)=6tx''(t)=6t, integrating twice gives x′(t)=3t2+c1x'(t)=3t^2+c_1 and x(t)=t3+c1t+c2x(t)=t^3+c_1t+c_2: two free constants for an equation of order 2. With x(0)=1x(0)=1, x′(0)=0x'(0)=0 we get x(t)=t3+1x(t)=t^3+1.

Finite-time blow-up. For the nonlinear equation x′(t)=x(t)2x'(t)=x(t)^2 with x(0)=1x(0)=1, the function x(t)=11−tx(t)=\dfrac{1}{1-t} satisfies x′(t)=1(1−t)2=x(t)2x'(t)=\dfrac{1}{(1-t)^2}=x(t)^2. It is a solution only on (−∞,1)(-\infty,1) and tends to +∞+\infty as t→1−t\to1^-.


Prerequisites



Sources

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: 27 October 2026 11:52