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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

First order first degree differential equations in the form

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

Can be solved with direct integration.

#### Separable equations

Separable equations are the one of 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, C1f(x) + C2g(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 YP and adjusting the coefficients. It is especially useful when the forcing term is a polynomial, trigonometric function, or exponential function, however if there is information known about the particular integral required different guesses can be tried.

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.

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

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.

Using Newton's 2nd Law,

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

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.

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