Ordinary Differential Equations

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1 Introduction
2 Linear First Order ODEs
3 Separable First Order ODEs

[edit] Introduction

An ordinary differential equations (ODE) is an equation that contains a function and its derivatives, but no partial derivatives. If an equation contains partial derivatives it is called a partial differential equation (PDE).

A simple example of an ODE with a function of time $y(t)\,\!$ is

$2 y(t) + \frac{dy}{dt} = 4 t$

or written without Leibniz notation and using Langrange notation instead for the derivative

$2 y(t) + y'(t) = 4 t\,\!$

The order of an ODE is given by the highest derivative that occurs in the equation. For the example given above the order is 1, since only the first derivative occurs.

[edit] Linear First Order ODEs

An important class of ODEs are linear ODEs. They are important because they are usually easy to solve, as we shall see in a bit. An ODE is called linear if $y(t)\,\!$ and its derivatives only occur as linear terms, i.e. either by themselves or at most multiplied with a function that doesn't depend on $y(t)\,\!$ but only on time. The example given above is a linear first order ODE, the ODE $2 y(t) + y'(t)^2 = 4 t\,\!$ is not because of the occurrence of $y'(t)^2\,\!$ .

The general form of a linear first order ODE is given by

$a(t) y(t) + y'(t) = b(t)\,\!$ (or similarly $y'(t) = a(t) y(t) + b(t)\,\!$ , which is sometimes used, but we'll stick to the former)

where $y(t),a(t),b(t)\,\!$ are continuous functions over their domain. In this form, the ODE is often also called inhomogeneous because there's a function $b(t)\,\!$ on the right hand side of the equation.

In the example given in the introduction $a(t) = 2, b(t) = 4 t\,\!$ .

[edit] Homogeneous Form

In order to solve this ODE, let us first solve the homogeneous form of this equation, which we get by setting $b(t) = 0\,\!$ .

$\begin{array}{ll} & a(t) y(t) + y'(t) = 0 \\ \Leftrightarrow & y'(t) = -a(t) y(t) \\ \Leftrightarrow & \frac{y'(t)}{y(t)} = -a(t) \\ \Leftrightarrow & \int\frac{y'(t)}{y(t)}dt = -\int a(t)dt \\ \Leftrightarrow & ln \left|y(t)\right| = -\int a(t)dt \\ \Leftrightarrow & y(t) = exp \left( -\int a(t)dt \right) \end{array}$

[edit] Inhomogeneous Form

Now that we have the homogeneous solution, let us return to the inhomogeneous, linear first order ODE and see if we can use the homogeneous solution to solve the inhomogeneous equation. The "trick" is to find a function $\mu (t)\,\!$ , the so-called integrating factor, which the equation is multiplied by on both sides.

$a(t) y(t) \mu(t) + y'(t) \mu(t) = b(t) \mu(t)\,\!$

This hasn't really simplified anything yet. However, if we choose $\mu (t)\,\!$ so that we can simplify the above equation to

$\frac{d}{dt} \left( something \right) = b(t) \mu(t)$

we can solve this equation by integrating on both sides. An obvious choice for the integrating factor is a function that satisfies

$\frac{d}{dt} \mu(t) = a(t) \mu(t)$

because then with the product rule

$\frac{d}{dt} \left( \mu(t) y(t) \right) = \frac{d}{dt} \mu(t) \, y(t) + \mu(t) \, \frac{d}{dt}y(t) = \mu'(t) y(t) + \mu(t) y'(t) = a(t) y(t) \mu(t) + y'(t) \mu(t)$

Now all we need to do is determine $\mu (t)\,\!$ , which we can do by solving the homogeneous, linear first order ODE

$\frac{d}{dt} \mu(t) = a(t) \mu(t) \Leftrightarrow \mu'(t) - a(t) \mu(t) = 0$

for which we already have the solution

$\mu(t) = exp \left( \int a(t)dt \right)$

We can now use this to solve the inhomogeneous equation:

$\begin{array}{ll} & a(t) y(t) + y'(t) = b(t) \\ \Leftrightarrow & a(t) y(t) \mu(t) + y'(t) \mu(t) = b(t) \mu(t) \\ \Leftrightarrow & \frac{d}{dt} \left( \mu(t) y(t) \right) = b(t) \mu(t) \\ \Leftrightarrow & \int\frac{d}{dt} \left( \mu(t) y(t) \right) dt = \int b(t) \mu(t) dt \\ \Leftrightarrow & \mu(t) y(t) = \int b(t) \mu(t) dt \\ \Leftrightarrow & y(t) = \frac{1}{\mu(t)} \int b(t) \mu(t) dt \\ \Leftrightarrow & y(t) = exp\left( - \int a(t) dt \right) \int b(t) \mu(t) dt \\ \Leftrightarrow & y(t) = exp\left( - \int a(t) dt \right) \int b(t) exp\left( \int a(t) dt \right) dt \end{array}$

[edit] Vector ODEs

For a vector-valued function y the linear first order ODE looks like this:

$y_i'(x) = \sum_{j=1}^n a_{i,j}(x) y_j + b_i(x) \, \mathrm{,} \quad i = 1,\ldots,n$

or in vector notation

$\mathbf{y}^'(x) = \mathbf{A}(x) \mathbf{y}(x) + \mathbf{b}(x)$

[edit] Separable First Order ODEs

A class of non-linear ODEs that can easily be solved are separable ODEs. A first order ODE is called separable if it can be brought into this form:

$f(y) \frac{dy}{dt} = g(t)\,\! \Leftrightarrow f(y(t)) y'(t) = g(t)$

where f is a continuous function depending only on y and g is a continuous function depending only on t. In order to solve it, we first note that

$\frac{d}{dt}F(y(t)) = f(y(t)) y'(t) = g(t)$

where F is any anti-derivative of f, i.e.

$F(y) = \int f(y) dy$

We then integrate on both sides of the equation to get

$F(y(t)) = \int g(t) dt + c$

where c is an arbitrary constant. This is an implicit solution of the differential equation. Sometimes it is possible to find an explicit solution by solving for y(t). If this is not the case, we can still use numeric methods to compute a solution to arbitrary precision.

To solve the initial value problem with $y(t_0) = y_0\,\!$ we can either solve for the constant in the above equation or we can simplify the equation using the initial values as integration bounds:

$\begin{array}{ll} & F(y(t)) - F(y(t_0)) = \int_{t_0}^{t} g(s) ds \\ \Leftrightarrow & \int_{y_0}^{y} f(z) dz = \int_{t_0}^{t} g(s) ds \\ \end{array}$