Ordinary differential equation

An ordinary differential equation (ODE) relates a function and its derivatives. We usually use $y$ to denote the function and $t$ to denote the variable.

Ordinary means that the equation has one variable, as opposed to partial differential.

There is no general method to solve ODEs. We separate them by classes and solve them individually.

Example

An example of an ODE is the following

$y'=y$

The general solution of the above is

$y(t)=ce^{t}$

Notably, the solution is homogeneous, meaning that $y(t)=0$ is a solution. This will probably be covered later.

Classification

An ODE is linear if all terms are proportional to $y,y',y''.\ldots$ or are given functions of $t$. This distinction is especially useful since linear combination can be used to construct solutions.

The order of an ODE is the order of its highest derivative.

In a scalar, there is only one unknown function $y(t)$. In a system, there are several, and you have to solve them simultaneously.

Here is a list of ODEs we study, from simple to complex:

• Linear First Order ODE

Applications

Since the derivative can be described as the rate of change, and the function itself is the state, ODEs arises as mathematical models of systems whose rate of change depends on the state of the system.

The following are brief descriptions of some applications of ODEs.

1. Radioactive decay, where the function is the (large) number of atoms.
   • Atoms decay at an average constant rate $r$
   • ${\frac {dN}{dt}}=-rN$
2. Object falling under gravity, where the function is the velocity of the object
   • ${\frac {dv}{dt}}=g-{\frac {\gamma v}{m}}$

Here are some more general questions that are solved by ODEs

• Initial Value Problem: Use an initial value of a function and an ODE to generate a function.
  • Equilibrium Solution: Find the equilibrium solution of an initial value problem

Dimensional Analysis

The two sides of the equation must match in dimensions (aka. units). This also applies in ODEs.

Consider radioactive decay.

${\begin{cases}{\frac {dN}{dt}}=-rN\\N(0)=N_{0}\end{cases}}$

The solution comes to

$N(t)=N_{0}e^{-rt}$

We use time constant $\tau$ to get a sense of how fast it is decaying. Its units is time.

$\tau ={\frac {1}{r}}$