# Integro-differential equation

In mathematics, an integro-differential equation is an equation that involves both integrals and derivatives of a function.

## General first order linear equations

The general first-order, linear integro-differential equation is of the form

$\frac{d}{dx}u(x) + \int_{x_0}^x f(t,u(t))\,dt = g(x,u(x)), \qquad u(x_0) = u_0, \qquad x_0 \ge 0.$

As is typical with differential equations, obtaining a closed-form solution can often be difficult. In the relatively few cases where a solution can be found, it is often by some kind of integral transform, where the problem is first transformed into an algebraic setting. In such situations, the solution of the problem may be derived by applying the inverse transform to the solution of this algebraic equation.

### Example

Consider the following first-order problem,

$u'(x) + 2u(x) + 5\int_{0}^{x}u(t)\,dt = \left\{ \begin{array}{ll} 1, \qquad x \geq 0\\ 0, \qquad x < 0 \end{array} \right. \qquad \text{with} \qquad u(0)=0.$

The Laplace transform is defined by,

$U(s) = \mathcal{L} \left\{u(x)\right\}=\int_0^{\infty} e^{-sx} u(x) \,dx.$

Upon taking term-by-term Laplace transforms, and utilising the rules for derivatives and integrals, the integro-differential equation is converted into the following algebraic equation,

$s U(s) - u(0) + 2U(s) + \frac{5}{s}U(s) = \frac{1}{s}.$

Thus,

$U(s) = \frac{1}{s^2 + 2s + 5}$.

Inverting the Laplace transform using contour integral methods then gives

$u(x) = \frac{1}{2} e^{-x} \sin(2x)$.

## Applications

Integro-differential equations model many situations from science and engineering. A particularly rich source is electrical circuit analysis.[citation needed]

The activity of interacting inhibitory and excitatory neurons can be described by a system of integro-differential equations, see for example the Wilson-Cowan model.