An **ordinary differential equation** or **ODE** (as opposed to a partial differential equation) is a type of differential equation that involves a function of only one independent variable. It can most simply be defined, for a layman, as any equation that involves any combination of the following:

- An independent variable ($ x $)
- Functions of the independent variable (or dependent variables) ($ g(x) $)
- A primary dependent variable (the function in question) ($ y(x) $)
- And, necessarily, any number of and degrees of derivatives of the primary function ($ \frac{dy}{dx}, \frac{d^2 y}{dx^2} $)

An example differential equation is as follows:

$ x^2 y + 3 \frac{d^2 y}{dx^2} - xy\frac{d^2 y}{dx^2} = 3 $

## Methods of solving

### First order

#### Separable equations

Separable equations are the easiest to solve. For any ODE in the form

- $ \frac{dy}{dx} = f(x) g(y) $

the solution is

- $ \int \frac{dy}{g(y)} = \int f(x) dx $

#### General linear equations

For linear equations in the form

- $ \frac{dy}{dx} + f(x)y = g(x) $,

the solution can be found with the formula

- $ y = \frac{1}{\mu} \int \mu g(x) dx, \mu = e^{\int f(x)dx} $

#### Bernoulli equations

Bernoulli differential equations are those in the form

- $ \frac{dy}{dx} + f(x)y = g(x)y^n $,

They can be solved by transforming them into linear differential equations by substituting.

#### Exact equations

Exact differential equations are those in the form

- $ m(x,y) dx + n(x,y) dy = 0 \, \, \text{if} \, \, \frac{\partial }{\partial y} m(x,y) = \frac{\partial}{\partial x} n(x,y) $

and will have the solution

- $ \int m(x,y)dx + \int n(x,y)dy = C $

#### Systems of differential equations

A system of ODEs in the form

- $ \begin{bmatrix} x'_1 \\ \vdots \\ x'_n \end{bmatrix} = \begin{bmatrix} C_{11} & \cdots & C_{1n} \\ \vdots & \ddots & \vdots\\ C_{n1} & \cdots & C_{nn} \end{bmatrix} \begin{bmatrix} x_1 \\ \vdots \\ x_n \end{bmatrix} $

or, in more compact notation,

- $ \vec{x}' = A \vec{x} $

the general solution will be

- $ \vec{x} = C_1 \vec{v}_1 e^{\lambda_1 t} + \cdots + C_n \vec{v}_n e^{\lambda_n t} $

where $ \vec{v} $ and $ \lambda $ represent the eigenvectors and eigenvalues of *A*, respectively.
Higher order differential equations can be converted into such a system by making the substitution

- $ x_n = y^{(n)} (t) $

differentiating, and substituting variables from the original equation for derivatives of *y* to yield a system of first order ODEs. For example:

- $ y'' - 2y' + y = 0 $

We can make the substition

- $ \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} y \\ y' \end{bmatrix} $
- $ \begin{bmatrix} x_1 \\ x_2 \end{bmatrix}' = \begin{bmatrix} y' \\ y'' \end{bmatrix} = \begin{bmatrix} x_2 \\ 2x_2 - x_1 \end{bmatrix} $
- $ \vec{x}' = \begin{bmatrix} 0 & 1 \\ 2 & -1 \end{bmatrix} \vec{x} $

### Higher order equations

In general, if *f(x)* and *g(x)* are both solutions to a differential equation, *C _{1}f(x) + C_{2}g(x)* is also a solution.

#### Homogenous equations

- $ a\frac{d^2 y}{dx^2} + b\frac{dy}{dx}+ cy = 0 $

Differential equations in this form can be solved by first finding the roots of the auxiliary or characteristic equation, which is equal to

- $ ar^2 + br + c = 0 $

The roots may be real or complex. If the roots are $ r_1, r_2 $ then the general solution will be:

- $ y=C_1 e^{r_1 x} + C_2 e^{r_2 x} $

If the characteristic equation has only one solution, the general solution of the ODE will be

- $ y=C_1 e^{r x} + C_2 x e^{r x} $

If the roots are complex and equal to $ a \pm b i $, by Euler's formula the solution simplifies to

- $ y=e^{a x} (C_1 \cos (b x) + C_2 \sin (b x)) $

The constant of integration is not restricted to real numbers in this case.

#### Method of undetermined coefficients

This method is used to find a particular solution for non-homogeneous equations by using a similar function as a potential particular solution *Y _{P}* and adjusting the coefficients. It only works when the extra term is a polynomial, trigonometric function, or exponential function. For example:

- $ y'' + 3y' - 4y = 3x^2 $
- $ Y_p = Ax^2 + Bx + C, \;\; Y'_p = 2Ax + B, \;\; Y''_p = 2A $
- $ 2A + 3(2Ax + B) - 4(Ax^2 + Bx + C) = 3x^2 $

This equality now creates a linear system of equations.

- $ -4A = 3 $
- $ 6A - 4B = 0 $
- $ 2A + 3B - 4C = 0 $
- $ A = -\frac{3}{4}, \;\; B = -\frac{9}{8}, \;\; C = -\frac{39}{32} $
- $ Y_p = -\frac{3}{4} x^2 -\frac{9}{8} x -\frac{39}{32} $

This particular solution, combined with the general solution of the associated homogeneous equation, gives a final general solution of

- $ y = C_1 e^{-4x} + C_2 e^{x} -\frac{3}{4} x^2 -\frac{9}{8} x -\frac{39}{32} $

## Real world examples

### Murder case

Detectives find a murder victim in room with the thermostat set to 20 °C at 4:30 AM. The detectives take a measurement as soon as they arrive and find the body to be 27 °C. An hour later the body is at 25 °C. What time was the victim killed?

This problem can be solved by using Newton's Law of Cooling, which states

- $ \frac{dT}{dt}=-k(T-a) $

$ T $ is temperature, the dependent variable, $ t $ is time, the independent variable, $ a $ is the ambient temperature and $ k $ is a constant. Since the variables can be separated, we can find the solution using that method.

- $ \frac{dT}{dt}=-k(T-a) $
- $ \frac{dT}{(T-a)}=-kdt $
- $ \int \frac{dT}{(T-a)}=\int -kdt $
- $ \ln(T-a)=-kt + C $
- $ T=Ce^{-kt} + a $
- $ T=Ce^{-kt} + 20 $

Firstly, we must find $ C $. We can do this with the initial value $ T(0) = 36.8 $ (human body temperature).

- $ 36.8=Ce^0 + 20 $
- $ 36.8=C + 20 $
- $ C=16.8 $

We can now solve for $ x $ and $ k $. We will assume that 4:30 is $ x $ hours after death.

- $ 27=16.8e^{-kx} + 20, 25=16.8e^{-k(x+1)} + 20 $
- $ k=-\frac{\ln(\frac{7}{16.8})}{x}=\frac{0.875}{x}, k=-\frac{\ln(\frac{5}{16.8})}{x+1} = \frac{1.212}{x+1} $
- $ \frac{0.875}{x} = \frac{1.212}{x+1} $
- $ x = 2.597 $

The murder took place 2.597 hours ago, or at 1:56 AM. We can also find $ k $ and find our final solution, although this is unnecessary as far as the detectives are concerned since we already know the time of death.

- $ 27=16.8e^{-2.597k} + 20 $
- $ k=\frac{0.875}{x}=\frac{0.875}{2.597}=0.337 $
- $ T=16.8e^{-0.337t} + 20 $

### Radioactive decay

Polunium-208 has a half life of 2.898 years. If the original sample was 10 grams, how much will remain after 1 year?

We can set this up as a differential equation, with $ A $ as the amount, $ t $ as time, and $ k $ as a constant.

- $ \frac{dA}{dt} = kA $

This is a separable differential equation, and can be solved to give

- $ \ln(A) = kt + C $
- $ A = Ce^{kt} $

Since we have the initial value $ A(0) = 10 $, we can find C.

- $ 10 = Ce^{k0} = Ce^0 = C $

- $ A = 10e^{kt} $

$ k $ can be found since we know that after 2.898 years there will be 5 grams.

- $ 5 = 10e^{2.898k} $
- $ k = \frac{\ln(\frac{1}{2})}{2.898} = -0.239 $

We know have a complete formula, which we can use to calculate out answer.

- $ A = 10e^{-0.239t} $
- $ A(1) = 10e^{-0.239} = 10(0.787) = 7.87 $

Our amount of Polunium-208 after 1 year is 7.87 grams.

### Simple harmonic motion

A system which obeys Hooke's law, or

- $ F = -kx $

can be rewritten as a second-order differential equation.

- $ F = - kx = m \frac{d^2 x}{dt^2} $
- $ \frac{d^2 x}{dt^2} + \frac{k}{m} x = 0 $

This equation will have the characteristic equation

- $ r^2 + \frac{k}{m} = 0 $
- $ r = \pm \sqrt{-\tfrac{k}{m}} = \pm \sqrt{\tfrac{k}{m}} i = \pm \omega i $

The solution to the differential equation is

- $ x(t) = C_1 e^{r_1 t} + C_2 e^{r_2 t} = C_1 e^{\omega i t} + C_2 e^{-\omega i t} $

By using Euler's formula this can be written in the form

- $ x(t) = C_1 (\cos (\omega t) + i \sin(\omega t)) + C_2 (\cos (\omega t) - i \sin(\omega t)) $
- $ x(t) = (C_1 + C_2) \cos (\omega t) + (C_1 - C_2) \sin (\omega t) $
- $ x(t) = C_3 \cos (\omega t) + C_4 \sin (\omega t) $

which, by trigonometric identities, can be written as

- $ x(t) = A \cos (\omega t - \varphi) $

where $ A = \sqrt{C_3^{ \ 2} + C_4^{ \ 2}} $ and $ \tan \varphi = \tfrac{C_4}{C_3} $. Systems which follow this equation are said to exhibit simple harmonic motion.