📂Odinary Differential Equations

Ordinary Differential Equation

Definition¹

For the univariate function $u(t)$, the following form is called an ordinary differential equation (ODE).

$$F(t, u(t), u^{\prime}(t), \dots, u^{(n)}(t)) = 0 \tag{1}$$

Here, $u^{\prime}$ is the derivative of $u$, and $u^{(n)}$ is the $n$-th order derivative of $u$, or simply referred to as $y = u(t)$,

$$F(t, y, y^{\prime}, \dots, y^{(n)}) = 0$$

Explanation

In $(1)$, $n$ is referred to as the order of the equation. In undergraduate studies of ordinary differential equations, the focus is primarily on first-order ordinary differential equations and second-order ordinary differential equations.

A function $u$ that satisfies $(1)$ is called a solution of the differential equation, and the phrase ‘solving the differential equation’ is synonymous with ‘finding the solution of the differential equation’.

Ordinary differential equations refer to differential equations with a single independent variable. The independent variable is usually denoted by $t$ or $x$. Remember that if denoted by $t$, it signifies time. Differentiation with respect to time is often expressed simply by a dot above the character.

$$\dfrac{dx}{dt} = \dot{x} \qquad \dfrac{d^{2}x}{dt^{2}} = \ddot{x}$$

Initial Value Problem²

Suppose we are given the following ordinary differential equation.

$$F(t, u(t), u^{\prime}(t), \dots, u^{(n)}(t)) = 0 \tag{1}$$ $$\begin{aligned} u(t_{0}) &= u_{0} \\ u^{\prime}(t_{0}) &= u_{1} \\ &\vdots \\ u^{(n-1)}(t_{0}) &= u_{n-1} \end{aligned} \tag{2}$$

In this case, $(2)$ is called the initial condition, and $(1)$ combined with $(2)$ forms what is known as the initial value problem. To find the solution of a $n$-th order differential equation, $n$ initial values are required.

Boundary Value Problem³

Suppose we have a function $y(x)$ defined in the interval $[a, b]$, with the following second-order ordinary differential equation given.

$$y^{\prime \prime}(x) + p(x)y^{\prime}(x) + q(x)y(x) + r(x) = 0 \tag{4}$$ $$y(a) = y_{0}, \quad y(b) = y_{1} \tag{5}$$

Here, $(5)$ is referred to as the boundary condition, and $(4)$ combined with $(5)$ is called the boundary value problem. In boundary value problems, the independent variable often signifies space.