Approximating Solutions Using Euler’s Method: Study Guide (Calculus BC 7.5)

BC-only material. Euler’s method is part of Calculus BC only. Calculus AB students do not need this topic.

Before you start: prerequisites

You need these earlier ideas. If one is shaky, revisit it from the Calculus BC roadmap first.

Prerequisite	Topic	What you use it for here
Tangent line approximation	4.6	Every Euler step is one tangent line
Differential equations	7.1	The equation gives the slope at any point
Verifying solutions	7.2	Checking an exact solution used for comparison
Slope fields and particular solutions	7.3, 7.4	Euler’s method follows the field from a starting point
Concavity from the second derivative	Unit 5	Deciding if an estimate is too high or too low

A note on notation

The differential equation is written dy/dx = f(x, y): a rule that gives the slope at any point (x, y). The step size is h (some books write Δx). The starting point is (x₀, y₀), the next point is (x₁, y₁), and so on. The point (xₙ, yₙ) is the estimate after n steps. In context the letters change (for example dA/dt), but the method is the same.

The idea: follow tangent lines in short steps

Suppose y = f(x) is the particular solution through (x₀, y₀). You cannot always find a formula for it. But the differential equation tells you its slope at (x₀, y₀). From Topic 4.6, the tangent line there gives a local linear approximation:

y(x₀ + h) ≈ y₀ + h · (slope at (x₀, y₀))

This works well for small h and badly for large h, because the curve bends away from its tangent line.

Euler’s method fixes this by taking several short steps instead of one long one. After each step you are at a new point. The differential equation gives you a new slope there, so you draw a new short tangent segment and carry on. You are walking through the slope field, re-reading the direction at every stop.

The procedure

For dy/dx = f(x, y), starting at (x₀, y₀), with step size h:

xₙ₊₁ = xₙ + h yₙ₊₁ = yₙ + h · f(xₙ, yₙ)

In words: new y = old y + step size × slope at the old point.

To keep your work clear, use a table with one row per step:

Step	(xₙ, yₙ)	Slope f(xₙ, yₙ)	Change h · slope	New point

The number of steps is (target x − starting x) ÷ h. To get from x = 0 to x = 1 with h = 0.5 you need 2 steps; with h = 0.25 you need 4.

Worked example 1: two steps, and a comparison with one

Question (no calculator). Let y = f(x) be the particular solution of dy/dx = x² − y with f(0) = 1.

(a) Use the tangent line at x = 0 to approximate f(1). (b) Use Euler’s method with two steps of equal size to approximate f(1).

(a) At (0, 1) the slope is 0² − 1 = −1. The tangent line is y = 1 − x, so f(1) ≈ 0. This is the same as one Euler step with h = 1.

(b) Two equal steps from 0 to 1 means h = 0.5.

Step	(xₙ, yₙ)	Slope xₙ² − yₙ	Change 0.5 × slope	New point
1	(0, 1)	0 − 1 = −1	−0.5	(0.5, 0.5)
2	(0.5, 0.5)	0.25 − 0.5 = −0.25	−0.125	(1, 0.375)

So f(1) ≈ 0.375.

Check against the exact solution. In Topics 7.6 and later you learn to solve some equations exactly. This one cannot be solved by separating the variables, so the solution is simply given here: y = x² − 2x + 2 − e^(−x). You can confirm it with Topic 7.2: y′ = 2x − 2 + e^(−x), and x² − y = x² − x² + 2x − 2 + e^(−x) = 2x − 2 + e^(−x). ✓ Also y(0) = 2 − 1 = 1. ✓ So f(1) = 1 − e⁻¹ ≈ 0.632.

Both estimates are too low, but two steps did much better than one. Figure 1 shows why.

Interpretation. The first Euler step is exactly the tangent line. The second step corrects the direction using the slope at (0.5, 0.5). The estimate is still low, because the true curve bends upward away from each straight step.

How step size affects accuracy

Repeat Worked example 1 with smaller steps (the longer runs were done with a calculator or a short program):

Step size h	Steps	Estimate of f(1)	Error (exact − estimate)
1	1	0	0.632
0.5	2	0.375	0.257
0.25	4	0.513	0.119
0.1	10	0.586	0.046

The exact value is 1 − e⁻¹ ≈ 0.632.

• Smaller steps usually give better estimates, because each tangent segment is shorter and has less room to drift from the curve.
• The error does not reach zero. Euler’s method is always an approximation.
• Background, not examined: for well-behaved equations the error is roughly proportional to h. In the table, halving h from 0.5 to 0.25 roughly halves the error. You will not be asked to prove this.

Overestimate or underestimate? Use concavity

A tangent line lies below a curve that is concave up and above a curve that is concave down. Each Euler step is a tangent segment, so:

• If d²y/dx² > 0 (concave up) in the region the steps pass through, Euler’s method gives an underestimate.
• If d²y/dx² < 0 (concave down) there, it gives an overestimate.

Find d²y/dx² by differentiating the differential equation implicitly (Topic 7.4), remembering that y is a function of x.

For Worked example 1: d²y/dx² = d/dx (x² − y) = 2x − dy/dx = 2x − (x² − y) = 2x − x² + y. For 0 ≤ x ≤ 1, 2x − x² ≥ 0, and y > 0 along the steps, so d²y/dx² > 0. At the two points used, the values are 1 and 1.25. The solutions are concave up, so 0.375 is an underestimate. This matches the exact value 0.632.

Careful. Whether y is increasing or decreasing tells you nothing here. Only concavity decides. And if the concavity changes sign along the path, you cannot conclude either way without more information.

Worked example 2: Euler’s method in context

Question (calculator allowed). The area A, in square metres, of an algae mat on a pond (fictional data) grows according to dA/dt = 0.4√A, where t is in days. At t = 0 the area is 25 m².

(a) Find the rate of growth at t = 0, with units. (b) Use Euler’s method with step size 2 days to approximate A(4). (c) Is your answer an overestimate or an underestimate? Justify.

(a) dA/dt at A = 25 is 0.4 × √25 = 2 m² per day.

(b) From t = 0 to t = 4 with h = 2 needs 2 steps.

Step	(tₙ, Aₙ)	Slope 0.4√Aₙ (m²/day)	Change 2 × slope (m²)	New point
1	(0, 25)	2	4	(2, 29)
2	(2, 29)	0.4√29 ≈ 2.154066	≈ 4.308132	(4, 33.308132)

A(4) ≈ 33.308 m². (Keep the unrounded slope in step 2; round only at the end.)

(c) Differentiate dA/dt = 0.4√A with respect to t, using the chain rule:

d²A/dt² = 0.4 · (1 / (2√A)) · dA/dt = (0.2/√A) · 0.4√A = 0.08.

This is positive for every A > 0, so every solution is concave up. Each Euler step lies below the curve, so 33.308 m² is an underestimate.

Checks.

• Units: (m² per day) × days = m². ✓
• Exact comparison: the solution here is A = (5 + 0.2t)². Check: dA/dt = 2(5 + 0.2t)(0.2) = 0.4(5 + 0.2t) = 0.4√A, and A(0) = 25. ✓ So A(4) = 5.8² = 33.64 m², which is above 33.308, as predicted.
• Smaller steps: with h = 1 (4 steps) the estimate is about 33.471 m², closer to 33.64.

Interpretation. After 4 days the algae mat covers about 33.3 m²; the true area is a little larger because growth speeds up as the mat grows.

Stepping backwards

To approximate a value to the left of the starting point, use a negative step. The formula does not change.

For dy/dx = x² − y with f(0) = 1, approximate f(−0.5) with one step, h = −0.5:

f(−0.5) ≈ 1 + (−0.5)(−1) = 1.5.

The exact value is 3.25 − e^0.5 ≈ 1.601. Here the estimate is below the curve again: at (0, 1), d²y/dx² = 1 > 0, so the tangent line lies below the curve on both sides.

Common misconceptions

• Using the slope at the new point. Each step uses the slope at the point you are leaving, (xₙ, yₙ).
• Forgetting to multiply by h. The change in y is h × slope, not the slope itself.
• Updating x but not y (or the reverse) before finding the next slope. Both coordinates change every step.
• Reporting the change instead of the new value. yₙ₊₁ = yₙ + change; add the change on.
• Rounding at each step. Small rounding errors grow across steps. Keep full values and round once at the end.
• “Increasing means underestimate.” Over or under depends on concavity (the sign of d²y/dx²), not on whether y is increasing.
• “With small enough steps, Euler is exact.” The error shrinks but does not vanish.
• Using the wrong number of steps. Number of steps = (distance in x) ÷ h. Two steps of 0.5 reach x = 1; two steps of 0.25 reach only x = 0.5.

Where this leads

Euler’s method gives numbers when you cannot, or do not yet, find a formula. Next, Topic 7.6 shows how to find exact solutions by separating the variables, and Topic 7.7 uses an initial condition to pick the particular solution, so you can compare your Euler estimates with exact values. The same “follow the slope” idea appears again in the BC logistic model (Topic 7.9).

Next topic: Finding General Solutions Using Separation of Variables. Previous topic: Reasoning Using Slope Fields. Return to the Calculus BC roadmap to see the order.

Try the practice questions now, then use the revision notes and the checklist to consolidate.