When both a and b are constants the equation has one constant solution, the steady state b/a, and every other solution is that constant plus a gap that is multiplied by the same factor in equal stretches of time. The gap shrinks to nothing without ever reaching it when a is positive and grows away when a is negative. Euler's method multiplies the gap by 1 − a·h at each step, an exact geometric sequence beside a curve that is not.
Core ideas
Three facts, each one line of algebra, and the second one is the solution. The third is what the lab shows beside it.
From y′ + a·y = b the rate is y′ = b − a·y, and that is zero when y = b/a. So y = b/a is a constant solution, the only one when a is not zero. It is the equilibrium from Equilibria, Stability and Phase Lines, and here it can be written down without drawing a phase line.
Let u = y − b/a, the gap. Then u′ = y′ = b − a·y = −a·(y − b/a) = −a·u. The gap satisfies the equation from Separable Equations, Growth and Decay, so u = u₀·e^(−a·t), and y = b/a + (y₀ − b/a)·e^(−a·t). The starting value enters only as the starting gap.
One Euler step is yₙ₊₁ = yₙ + h·(b − a·yₙ). Subtracting b/a from both sides gives yₙ₊₁ − b/a = (1 − a·h)·(yₙ − b/a). So the gap after n steps is the starting gap times (1 − a·h)ⁿ, a geometric sequence of fractions, to be compared with the true factor e^(−a·h) per step, which is irrational.
The material
The steady state, the shifted exponential and why it solves the equation, what never reaching the steady state means, the repelling case, and Euler's geometric sequence beside the closed form.
Take y′ + 2y = 6 with y(0) = 0. Before solving anything, ask where the rate is zero: y′ = 6 − 2y, which is 0 at y = 3. If y ever sat at 3 it would stay there for ever. The question the rest of the lesson answers is what a solution does when it starts somewhere else.
For y′ + a·y = b with a ≠ 0, the steady state is the constant y* = b/a. It is the one constant solution, and it is attracting when a > 0, repelling when a < 0.
The solution of y′ + a·y = b with y(0) = y₀ is y = b/a + (y₀ − b/a)·e^(−a·t).
Let u = y − b/a. Then u′ = y′, and from the equation y′ = b − a·y = −a·(y − b/a) = −a·u.
So u′ = −a·u, which is the decay equation and has the solution u = u₀·e^(−a·t), a claim from Separable Equations, Growth and Decay that is not reproved here. The starting gap is u₀ = y₀ − b/a, and adding b/a back gives the formula.
□
The gap 3·e^(−2t) is positive for every t, because an exponential is never zero. So y comes closer to 3 and does not reach it at any finite time, however long you wait. A reader who sees the curve flatten in the drawing is looking at the gap getting smaller than a pixel, which is not the same as the gap being zero. The claim the lesson makes is about every finite t, and the formula proves it.
Nor is the steady state the starting value. The solution starts at y(0) = 0 and the steady state is 3; the starting value fixes the gap and the steady state fixes where the gap is measured from. Start at 5 instead and the gap is +2, giving y = 3 + 2·e^(−2t), which falls toward 3 from above and never crosses it, since the gap never changes sign.
Take y′ − y = 2, so a = −1, b = 2 and y* = −2. The solution from y(0) = 0 is y = −2 + 2·e^t: the gap is multiplied by e^t, so it grows. The lab prints the steady state as −2 (repelling). Starting exactly at −2 stays there; starting anywhere else, however close, leaves.
Run Euler on y′ + 2y = 6 from y(0) = 0 with h = 1/4. The factor 1 − a·h = 1 − 1/2 = 1/2 is an exact fraction, and so is every value: yₙ = 3 − 3·(1/2)ⁿ. The closed form at the same times is irrational, so the lab prints it rounded, to six figures, with ≈, in a separate column.
The fraction 45/16 is exactly 3 − 3/16, and it is 2.8125. The true value is 3 − 3·e^(−2) ≈ 2.59399, so Euler's polygon ends about 0.218506 above the curve. Every fraction in the column is right; none of them is the solution. Euler's gap shrinks by 1/2 per step, and the true gap shrinks by e^(−1/2) ≈ 0.606531 per step, which is why the polygon runs ahead.
From y(0) = 5 the gap is +2 and Euler gives yₙ = 3 + 2·(1/2)ⁿ, so y₄ = 3 + 2/16 = 25/8: above the steady state at every step, as the true curve is.
For y′ − y = 2 from y(0) = 0 the factor is 1 − (−1)·(1/4) = 5/4, so y₄ = −2 + 2·(5/4)⁴ = 369/128. The gap is multiplied by a number bigger than one, and the polygon leaves the steady state exactly as the curve does.
One warning for later. The factor 1 − a·h is positive and below one for a small step, but it goes negative when h > 1/a, and its size passes one when h > 2/a. At h = 1/2 exactly, here, the factor is 0: the polygon lands on 3 after one step and stays there, which the curve never does. A polygon that follows the curve for h = 1/4 can run away for a larger step on the same equation. “Stiffness: When the Step Is Too Big” takes that up.
Interactive
μ, y_h, y_p and the constant, exactly
Method
Find the constant, measure the gap, let the gap decay or grow, then check. Euler is one more line.
Set the rate to zero: b − a·y = 0, so y* = b/a. Say whether it attracts (a > 0) or repels (a < 0).
y₀ − y*, with its sign. The sign says which side of the steady state the solution lives on for ever.
y = y* + (y₀ − y*)·e^(−a·t). The exponent is −a·t: negative for an attracting steady state, positive for a repelling one.
At t = 0 the exponential is 1, so y = y* + (y₀ − y*) = y₀. Then substitute into y′ + a·y and read b.
The gap after n steps is (y₀ − y*)·(1 − a·h)ⁿ. Add y* and the whole column follows, as exact fractions.
The equation is y′ + 2y = 6 with y(0) = 0 and step h = 1/4. The steady state comes first, then the gap, then the formula, then Euler as one factor.
The two numbers at the foot are different quantities and the lab prints both: 45/16 is exact arithmetic on Euler's approximation, and 2.59399 is a rounded value of the curve. The distance between them, about 0.218506, is the error of this method at this step, and it is the only part of the table that is not exact. Halve the step and it falls, as Differential Equations and Euler's Method measured.
For a rehearsal, change the start to y(0) = 5 without running the lab. The steady state is unchanged, the gap is now +2, and the factor is still 1/2. Predict y₄ as a fraction, then pick the second preset and compare.
You should be able to find b/a, write the starting gap with its sign, put the gap on e^(−a·t), check the result at t = 0 and by substitution, say why the solution never reaches the steady state in finite time, and predict Euler's value after n steps as an exact fraction from the factor 1 − a·h.
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