Slope Field

Possible solutions to the differential equation \frac{dy}{dx}=x^2-x-2 represented by a slope field.

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

Van der pols equation phase portrait

By viewing the matrix A as a linear transformation, one can create a vector field called the phase portrait based on the mapping from x (the vector representing a given point) to x', similarly to a slope field. This particular phase portrait is based on the equation \tfrac{d^2y}{dt^2}+(y^2-1)\tfrac{dy}{dt}+y=0.

A system of ODEs in the form

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


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.

\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

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
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.