# Ordinary differential equation (ODE)

An ODE is an equation where the unknowns are one or several functions that depend on an independent variable. Also, to evaluate the equation at a given point we only need to know the value of the unknown functions and its derivatives at this point. Other type of differential equations will not be called ordinary.

An example would be the equation: $$y'=y$$ where $$y=y(x)$$, is a function of one variable and to evaluate the equation we only need to know $$y$$ and $$y'$$ at a point.

The order of an ODE is the order of the derivative of higher order that appears in the equation. For example, in the previous case, the order of the ODE is $$1$$.

An example would be the equation: $$y'=y$$ where $$y=y(x)$$, is a function of one variable and to evaluate the equation we only need to know $$y$$ and $$y'$$ at a point.

The order of an ODE is the order of the derivative of higher order that appears in the equation.

For example, in the previous case, the order of the ODE is 1.

A result exists that says that we can write any ODE as a system of ODEs of order $$1$$.

So whenever we shall refer to an ODE, we will write it as $$y'=f(x,y)$$ understanding that perhaps $$y$$ is a vector. For example,

If we have the ODE: $$\displaystyle 2 \cdot \frac{y'}{x^2}+5x^3=\frac{3y^2+5}{8x}$$$we can re-write it as: $$\displaystyle 2\cdot \frac{y'}{x^2}=\frac{3y^2+5}{8x}-5x^3 \Longrightarrow y'=\frac{3y^2x+5x}{16}-\frac{5}{2}x^5$$$ so that $$\displaystyle f(x,y)=\frac{3y^2x+5x}{16}-{5}{2}x^5$$$We define a problem of initial values (PVI) as the system: $$\left\{\begin{matrix}y' = f(x,y) \\ y(x_0) = y_0 \end{matrix}\right.$$$ that is, we look for a function that satisfies the differential equation and, also, takes the value $$y_0$$ at $$x_0$$.

We may wonder whether a solution will always exist. Well, there is a result that guarantees the existence and uniqueness of solution of a PVI if f is sufficiently good, i.e. if it is differentiable.