Where Differential Equations Come From
Chapter Eighty-Two
Syllabus topic Module 2, "2.1 Equation of the first order and of the first degree"
Pages 180 to 181 of 303
In one line
Write down what you know about how something changes, and you have written a differential equation.
The pattern
Every model in this chapter is built the same way.
- Name the quantity you care about and the variable it depends on.
- Say, in words, what its rate of change is proportional to.
- Write that sentence as an equation, with a constant of proportionality.
- Fix the constant and any others from what you are told.
The mathematics is in step three, and it is one line. The thinking is in step two.
Radioactive decay, and why it applies to a cache
The number of atoms that decay in the next second is proportional to how many there are. Nothing else affects it.
dN/dt = -k N
The minus sign is because N is falling. Separating the variables, which the next section does properly, gives the solution.
dN/dt = -3N
N = C e^(-3t)
So the amount decays exponentially. The half life is the time for N to halve, and setting the exponential to one half gives it.
log(2)/3 = log(2)/3
The same equation, with a different name on the letter, describes a capacitor discharging, a hot object cooling towards room temperature, a drug leaving the bloodstream, and the number of entries left in a cache that is evicted at a rate proportional to its size. One equation, many subjects, which is the real reason this module is worth learning.
Newton's law of cooling
The rate at which something cools is proportional to how much hotter it is than its surroundings.
dT/dt = -k(T - R)
R is the room temperature. Notice that it is not the temperature that matters but the excess over the room, which is why a cup of tea cools quickly at first and slowly later.
The solution, which the substitution u = T minus R turns into the decay equation above:
dT/dt = -2(T - 20)
T = 20 + C e^(-2t)
As t grows the exponential dies and T approaches 20, which is the room temperature. Any model whose answer does not do that is wrong, and checking the long-term behaviour is a good way to catch a sign error.
A charging capacitor
A capacitor C charged through a resistor R from a supply of E volts. The current is proportional to the voltage still missing.
dV/dt = (E - V)/(R C)
dV/dt = (5 - V)/2
V = 5 + C1 e^(-t/2)
With V(0) = 0 the constant is minus 5, and the voltage rises from 0 towards 5, quickly at first and then slowly. The product RC is the time constant, and the voltage gets to about 63 per cent of the way in one time constant, a fact every electronics student learns as a number and which comes from here.
Where Differential Equations Come From
A population with a limit
A population grows in proportion to its own size, which gives exponential growth, but it is also limited by the space available. The simplest model multiplies the two.
dP/dt = k P (M - P)
M is the maximum the environment supports. When P is small, the second bracket is nearly M and the growth is nearly exponential. When P approaches M, the bracket approaches zero and the growth stops.
This is the logistic equation, and its solution is the S-shaped curve that describes the adoption of a new technology, the spread of a rumour through a network, and the growth of a bacterial culture. It is non-linear, because P appears squared, so none of Module 2's linear techniques applies to it; it is separable, and the first section of this module solves it.
A falling body with air resistance
Acceleration is gravity minus a drag proportional to speed.
dv/dt = g - k v
dv/dt = 10 - v
v = 10 + C e^(-t)
With v(0) = 0 the constant is minus 10, and the speed rises from 0 towards 10, which is the terminal velocity: the speed at which the drag exactly balances gravity, so the acceleration is zero. Setting dv/dt to zero in the equation gives v = g over k directly, without solving anything, which is the quickest way to find a terminal value of any of these models.
What they have in common
Four of the five above are of the form below, with a and b constants.
dy/dt = a - b y
That equation is linear, first order, first degree, separable, and exact after an integrating factor, so four of the techniques in section 2.1 will solve it. Meeting it early is useful, because you will recognise it repeatedly.
Its solution always has the same shape: a constant that it settles at, which is a over b, plus a decaying exponential.
dy/dt = 6 - 3y
y = 2 + C e^(-3t)
Check yourself
Write the equation for each, then check the solution offered.
| In words | The equation |
|---|---|
| the rate of growth is proportional to the amount | dy/dt = ky |
| the rate of decay is proportional to the amount | dy/dt = -ky |
| the rate of change is proportional to the difference from 100 | dy/dt = k(100 - y) |
| the rate is proportional to the product of y and 50 minus y | dy/dt = ky(50 - y) |
dy/dt = 4y
y = C e^(4t)
dy/dt = 5(100 - y)
y = 100 + C e^(-5t)
The rest of this subject
These notes are cut from the University's printed syllabus. Open the syllabus itself for the same subject.