The equation y′ = k·y says a quantity changes at a rate proportional to its size, and separation solves it: y = y₀·e^(kt). Euler's method on the same equation multiplies by the same factor 1 + kh at every step, so its output is the geometric sequence y₀·(1 + kh)ⁿ, an exact fraction at every step. The lab prints that fraction beside e^(kt), which is rounded, and the gap between them shrinks as the step does.
Core ideas
Three ideas: the equation and its exact solution, the method's solution, and the distance between the two.
y′ = k·y with k a constant. The bigger the quantity, the faster it changes, in the same proportion. For k > 0 that is growth, for k < 0 decay, and the number k has the units of one over time. Nothing in the equation says the growth is fast; a small k is slow, and the solution is still exponential.
Writing the equation as (1/y)·y′ = k and integrating gives ln(|y|) = kt + C, and with y(0) = y₀ the solution is y = y₀·e^(kt). The value e^(kt) is irrational for rational kt ≠ 0, so every figure from the closed form on this page is rounded and carries ≈.
One Euler step is yₙ₊₁ = yₙ + h·k·yₙ = (1 + kh)·yₙ. The factor does not depend on n. So yₙ = y₀·(1 + kh)ⁿ, which is a geometric sequence with ratio 1 + kh, and for rational k, h and y₀ every term is an exact fraction.
The material
Separation first, then Euler's steps, then the two side by side, with the lab's numbers for step sizes one quarter and one sixteenth.
A quantity y(t) obeys exponential growth when y′ = k·y for a constant k > 0, and exponential decay when k < 0. The number k is the rate constant.
The name describes the form of the solution, not its speed: y₀·e^(kt) is exponential for every k ≠ 0, however small.
The solution of y′ = k·y with y(0) = y₀ is y = y₀·e^(kt).
Take y₀ > 0; for y₀ < 0 the same lines run with |y|, and y₀ = 0 gives the constant solution y = 0, which the formula covers as well. While y > 0, the equation separates as (1/y)·y′ = k, so ln y = kt + C by the method of “Separable Equations”, with 1/y integrated to ln y as in “The Integral of 1/t”.
At t = 0 that says C = ln y₀, so ln y − ln y₀ = kt, which is y = y₀·e^(kt). The check is the rate of e^(kt) from “The Exponential and Its Rate”: the rate of y₀·e^(kt) is k·y₀·e^(kt), which is k·y.
□
That is a solution in closed form, and it is what every figure on this page will be compared with. It is also the only kind of statement here that is not an exact fraction: e^(kt) has no finite fraction, so the lab prints it rounded to six figures and marks it ≈.
Take y′ = y, y(0) = 1, h = 1/4. The first step is 1 + (1/4)·1 = 5/4. The second is 5/4 + (1/4)·(5/4) = 25/16, which is (5/4)². Every step multiplies by 5/4, because the rate is the value itself, and so after four steps, at t = 1, the lab prints 625/256, which is (5/4)⁴ exactly.
This is the geometric sequence of “Geometric Sequences and Series” in Algebra's Sequences and Series: each term is the previous one times a fixed ratio, here 1 + kh. Euler's method on a growth equation is that sequence under another name. The ratio is a rational number, so nothing in it is rounded.
The value 625/256 is about 2.44141; the closed form at t = 1 is e ≈ 2.71828, rounded. The lab's error tile prints the difference as ≈ 0.276876. That is not because any step was computed wrongly: each step assumed the rate stayed at its starting value across the step, while the real rate kept climbing. The exact sequence is always below the curve for growth.
As the step shrinks the sequence climbs towards e, which is the sequence met in “The Number e” in Algebra's Exponential and Logarithmic Functions, built here directly from a differential equation. That the values approach e is a claim the table suggests and does not prove; “Euler's Error and the Step Size” in the previous course measured how fast, and found the error roughly halving each time the step does.
Interactive
A geometric sequence beside an exponential
Method
Do both, and compare them. Either one alone leaves you with a number and no idea how far to trust it.
Write the equation as y′ = k·y and identify k with its sign. If the right side is −3y, then k = −3, and the quantity decays.
(1/y)·y′ = k, ln(|y|) = kt + C, and y = y₀·e^(kt) from the initial value. Check by differentiating: the rate of e^(kt) is k·e^(kt).
1 + kh, a fraction when k and h are. Write it down before any steps; the whole table is its powers.
yₙ = y₀·(1 + kh)ⁿ against y₀·e^(k·nh). The difference is the method's error at this step size, and it is of the sign the factor dictates.
Halve h, double n, and watch the error fall. That it falls is the reason Euler's method is useful, and how fast is the reason the later lessons exist.
The factor is 1 + 1·(1/4) = 5/4 and the end time is t = 4·(1/4) = 1. Every number below is exact until the last line, which is the closed form and says so.
Every step is the one before multiplied by 5/4, and there is no other arithmetic in the column. The sequence is geometric, and the error is the whole of what is wrong with it.
Now run the third preset: the step is 1/16, there are 16 steps, and the end value is (17/16)¹⁶, about 2.63793. The end time is still t = 1, and the sequence has moved most of the way to e ≈ 2.71828. The second preset has a different rate and start (k = 1/2, y₀ = 2), and its factor is 9/8.
You should be able to write the exact solution y₀·e^(kt), compute Euler's factor 1 + kh as a fraction, write yₙ as a power of it, recognise the sequence as geometric, and report the rounded error with the step size beside it.
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