AP Calculus AB/BC · Unit 7 · Topic 7.5 · BC Only
Approximating Solutions Using Euler’s Method
Advance along tangent-line steps from an initial condition.
1. Topic Focus
Model rates with differential equations, read slope fields, approximate solutions, solve separable equations, and interpret exponential or logistic models.
This topic: Advance along tangent-line steps from an initial condition.
2. Key Relationship
Read every symbol with its domain, direction, units, and hypotheses before applying the relationship.
3. Visual Connection
4. Worked Example
Starting at (0,1) with y′=x+y and h=0.1 gives y₁=1.1.
Write the governing relationship first, carry out the algebra cleanly, and finish with a sentence that answers the mathematical question.
5. Concept Development
Euler's Method approximates the solution of an initial-value problem by repeatedly following a tangent line for one short step. For
the tangent-line approximation at \((x_n,y_n)\) predicts
Here \(h\) is the step size. The method replaces a smooth solution curve with a chain of tangent-line steps, recalculating the slope at every newly approximated point.
Why the update formula works
Local linearity gives
Euler's Method substitutes the known approximation \(y_n\) for \(y(x_n)\) and uses the differential equation to estimate \(y'(x_n)\) by \(F(x_n,y_n)\). Thus each step follows the slope field from the current approximate point.
The five-column workflow
| Column | What to record |
|---|---|
| \(n\) | step number |
| \(x_n\) | current input |
| \(y_n\) | current approximate output |
| \(F(x_n,y_n)\) | new slope evaluated at the current point |
| \(hF(x_n,y_n)\) | predicted change added to obtain \(y_{n+1}\) |
- Compute the number of steps: \(N=(x_{\text{target}}-x_0)/h\).
- Begin with the exact initial pair \((x_0,y_0)\).
- Evaluate the slope using both current coordinates.
- Multiply the slope by \(h\) and add it to the current \(y\)-value.
- Advance \(x\) by \(h\), then repeat with the new approximate pair.
Step size and direction
A positive \(h\) moves to the right; a negative \(h\) approximates values to the left of the initial point. The target must lie on the Euler grid \(x_n=x_0+nh\), so choose \(h\) and the number of steps consistently. A smaller \(|h|\) usually gives a better approximation because the tangent line is followed over a shorter interval, but it requires more steps.
Overestimates and underestimates
On an interval where the exact solution is concave up, its tangent lines lie below the curve, so forward Euler steps generally underestimate. For a concave-down solution, tangent lines lie above the curve, so the method generally overestimates.
| Exact solution shape | Forward Euler tendency |
|---|---|
| \(y''>0\) throughout the steps | underestimate |
| \(y''<0\) throughout the steps | overestimate |
| concavity changes | no single error direction follows without further analysis |
This comparison requires the concavity of the exact solution, not merely the sign of \(y'\). A solution can be decreasing and concave up.
Error and numerical care
Each tangent step introduces local error, and those errors accumulate. Under standard smoothness assumptions, Euler's local truncation error is proportional to \(h^2\), while the accumulated error over a fixed interval is proportional to \(|h|\). This explains why halving the step often roughly halves the overall error, although it is not an AP error-bound formula.
- Keep several guard digits and round only the final answer unless instructed otherwise.
- Do not replace an approximate \(y_n\) with an exact value midway through the table.
- Check units: \(h\) has input units, \(F\) has output per input, and \(hF\) has output units.
- A numerical approximation is not a general solution formula.
6. Detailed Worked Example and Error Check
Example 1: Two standard steps
Approximate \(y(1)\) for \(y'=x-y\), \(y(0)=1\), with \(h=0.5\).
| \(n\) | \(x_n\) | \(y_n\) | \(F(x_n,y_n)\) | \(y_{n+1}\) |
|---|---|---|---|---|
| 0 | 0 | 1 | \(-1\) | \(1+0.5(-1)=0.5\) |
| 1 | 0.5 | 0.5 | \(0\) | \(0.5+0.5(0)=0.5\) |
Therefore \(y(1)\approx0.5\). The second slope must be evaluated at the new pair \((0.5,0.5)\).
Example 2: A nonlinear table
Use \(h=0.2\) to approximate \(y(0.6)\) for \(y'=y-x^2\), \(y(0)=1\).
| \(n\) | \(x_n\) | \(y_n\) | \(y_n-x_n^2\) | next \(y\) |
|---|---|---|---|---|
| 0 | 0 | 1 | 1 | 1.2 |
| 1 | 0.2 | 1.2 | 1.16 | 1.432 |
| 2 | 0.4 | 1.432 | 1.272 | 1.6864 |
Thus \(y(0.6)\approx1.6864\).
Example 3: Working backward
Approximate \(y(0.5)\) for \(y'=x+y\), \(y(1)=2\), using \(h=-0.25\).
The corresponding inputs are \(0.75\) and \(0.5\), so \(y(0.5)\approx0.75\). The negative step changes the sign of each predicted change.
Example 4: Using tabulated slopes
Suppose \(y(0)=2\), \(h=0.2\), and a table gives \(F(0,2)=1.5\) and \(F(0.2,2.3)=1.1\). Then
Therefore \(y(0.4)\approx2.52\). The second table lookup uses the approximated output \(2.3\).
Example 5: Population context and units
A population satisfies \(P'=0.4P(1-P/1000)\), where time is in years and \(P(0)=100\). With \(h=0.5\),
The slope has units people per year; multiplying by \(0.5\) year produces a change in people.
Example 6: Error direction from concavity
For \(y'=y\), \(y(0)=1\), the exact solution is concave up because \(y''=y'=y>0\). With \(h=0.5\), Euler's Method gives \(y(1)\approx2.25\), while the exact value is \(e\approx2.718\). The underestimate agrees with the tangent-line geometry.
Example 7: Smaller steps
For the same problem \(y'=y\), \(y(0)=1\), using \(h=0.25\) gives four updates:
This is closer to \(e\) than \(2.25\), though it remains an underestimate because the solution stays concave up.
Example 8: Diagnosing a stale-slope error
For \(y'=x+y\), \(y(0)=1\), and \(h=0.25\), the first slope is \(1\), giving \(y_1=1.25\). The next slope is not \(1\); it is
Therefore \(y_2=1.25+0.25(1.5)=1.625\). Reusing the initial slope would turn Euler's Method into one long tangent line.
Common calculation errors
- Evaluating every slope at the original point.
- Updating \(y\) but forgetting to update \(x\).
- Using \(F(x_{n+1},y_n)\) instead of \(F(x_n,y_n)\).
- Taking the wrong number of steps or stepping past the target.
- Losing the sign of a negative step when working backward.
- Rounding each row so aggressively that the errors compound.
7. AP Reasoning Routine
Translate the context into a rate equation, verify candidate solutions by substitution, carry constants through integration, and apply initial conditions last.
- Identify the representation and requested quantity.
- State the rule or theorem and verify its conditions.
- Keep exact values until the final requested approximation.
- Interpret sign, units, interval, and context.
Use Euler's Method and show the updates or a labeled table.
(a) For \(y'=F(x,y)\), \(y(2)=5\), and \(F(2,5)=-3\), use one step of size \(0.1\) to approximate \(y(2.1)\).
(b) Approximate \(y(0.5)\) for \(y'=x+y\), \(y(0)=1\), with \(h=0.25\).
(c) Approximate \(y(0.4)\) for \(y'=y-x^2\), \(y(0)=1\), with \(h=0.2\).
(d) Approximate \(y(0.5)\) for \(y'=x-y\), \(y(1)=2\), with \(h=-0.25\).
(e) How many steps are required to move from \(x=-1\) to \(x=1\) with \(h=0.25\)?
(f) If \(t\) is measured in hours and \(Q'=F(t,Q)\) is measured in liters per hour, explain the units of \(hF(t_n,Q_n)\).
(g) The exact solution is concave down throughout a forward Euler interval. Is the Euler estimate expected to be above or below the exact value?
(h) Let \(y(0)=3\), \(h=0.5\), \(F(0,3)=-2\), and \(F(0.5,2)=-1.4\). Approximate \(y(1)\).
(i) A population satisfies \(P'=0.2P(1-P/500)\), \(P(0)=50\). Use \(h=1\) to approximate \(P(2)\).
(j) For \(y'=2x\), \(y(0)=0\), compare Euler approximations of \(y(1)\) using \(h=0.5\) and \(h=0.25\). The exact value is \(1\).
Check the solution
(a) \(y(2.1)\approx5+0.1(-3)=4.7\).
(b) \(y_1=1+0.25(1)=1.25\), then \(y_2=1.25+0.25(0.25+1.25)=1.625\).
(c) First \(y_1=1+0.2(1)=1.2\). Then \(y_2=1.2+0.2(1.2-0.2^2)=1.432\).
(d) At \((1,2)\), the slope is \(-1\), so \(y_1=2+(-0.25)(-1)=2.25\) at \(x=0.75\). The new slope is \(0.75-2.25=-1.5\), so \(y_2=2.25+(-0.25)(-1.5)=2.625\). Thus \(y(0.5)\approx2.625\).
(e) \(N=[1-(-1)]/0.25=8\) steps.
(f) \(h\) has units hours, so \(hF\) has units \((\text{hours})(\text{liters/hour})=\text{liters}\), matching the units of the change in \(Q\).
(g) Above. Tangent lines to a concave-down curve lie above the curve, so forward Euler generally overestimates.
(h) \(y_1=3+0.5(-2)=2\), then \(y_2=2+0.5(-1.4)=1.3\).
(i) \(P_1=50+0.2(50)(0.9)=59\). Then \(P_2=59+0.2(59)(1-59/500)=69.4076\).
(j) With \(h=0.5\), the updates use slopes \(0\) and \(1\), giving \(0.5\). With \(h=0.25\), the slopes are \(0,0.5,1,1.5\), giving \(0.75\). Both underestimate the exact value \(1\), and the smaller step is closer.