Approximating Solutions Using Euler’s Method

3. Visual Connection

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

1. Compute the number of steps: \(N=(x_{\text{target}}-x_0)/h\).
2. Begin with the exact initial pair \((x_0,y_0)\).
3. Evaluate the slope using both current coordinates.
4. Multiply the slope by \(h\) and add it to the current \(y\)-value.
5. 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.

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 \(h=0.5\).

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 \(y'=y-x^2\), \(y(0)=1\).

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.