Why Build an Euler Table?
A differential equation gives a rule for how a quantity changes. When an exact solution is unavailable or not needed, Euler’s method uses that rule to calculate a sequence of approximate values. A table makes the repeated calculations visible: each row records the current estimate, the slope there, the change over one step, and the next estimate.
The focus here is carrying out that process accurately by hand. We will use autonomous differential equations, in which the slope depends on the dependent variable \(y\), and a fixed step size \(h\). The same bookkeeping is useful whenever you apply the Forward Euler Update Rule.
How to Read Each Row
A useful table separates the slope from the change it produces. They are not interchangeable: the slope is a rate of change, while the increment is that rate multiplied by the step size. For a constant step size, the columns are:
| Column | Meaning |
|---|---|
| \(n\) | Step number; \(n=0\) is the initial condition |
| \(t_n\) | Independent-variable value at the start of the step |
| \(y_n\) | Current approximate solution value |
| \(f(y_n)\) | Slope evaluated using the current value |
| \(h f(y_n)\) | Euler increment, or estimated change over the step |
| \(y_{n+1}\) | Updated estimate, found by adding the increment to \(y_n\) |
After a row is calculated, its last value becomes the next row’s current value. In particular, the slope for the next step must be evaluated using the updated estimate, not the earlier one. If \(f(y_n)\) is positive, the increment is positive; if it is negative, the increment is negative. This sign check is a quick way to catch some arithmetic or transcription errors.
Why the Update Uses the Current Slope
At a point \((t_n,y_n)\), the differential equation specifies the slope of the solution as \(f(y_n)\). The line through that point with this slope is the tangent-line approximation \(y \approx y_n+f(y_n)(t-t_n)\). At the next grid point, \(t_{n+1}=t_n+h\), this line gives \(y_n+h f(y_n)\). That is exactly the Euler update.
This explains both parts of the table: use the slope at the start of the step, and multiply it by the horizontal step length. The tangent line is an approximation to the solution curve over the step, so a table value generally should not be presented as an exact solution value.
A Reliable Hand-Calculation Process
Copy \(t_0\) and \(y_0\) from the initial condition. Do not begin by evaluating a slope at a value that has not been recorded.
Substitute the current estimate \(y_n\) into the right-hand side \(f(y)\), and show the substitution.
Multiply the slope by \(h\), keeping its sign and units.
Set \(t_{n+1}=t_n+h\) and \(y_{n+1}=y_n+h f(y_n)\). Use this new estimate in the next slope evaluation.
Substitute the recorded current value into \(f\), confirm the multiplication by \(h\), and confirm that current estimate plus increment equals next estimate.
Worked Example: A Squared-Value Slope
Worked Example: A Squared-Value Slope
Construct an Euler table for \(y'=y^2\), \(y(0)=1\), using \(h=0.1\), through the estimate at \(t=0.4\).
Start at \(t_0=0\), \(y_0=1\). At step \(n=0\), the slope is \(f(y_0)=1^2=1\). The increment is \(0.1(1)=0.1\), so \(y_1=1+0.1=1.1\), at \(t_1=0+0.1=0.1\).
At step \(n=1\), use the current estimate \(y_1=1.1\): the slope is \(1.1^2=1.21\). The increment is \(0.1(1.21)=0.121\), so \(y_2=1.1+0.121=1.221\), at \(t_2=0.1+0.1=0.2\).
At step \(n=2\), the slope is \(1.221^2=1.490841\). The increment is \(0.1(1.490841)=0.1490841\), so \(y_3=1.221+0.1490841=1.3700841\), at \(t_3=0.2+0.1=0.3\).
At step \(n=3\), square the listed estimate \(1.3700841\): \(1.3700841^2=1.87713044107281\). The increment is \(0.1(1.87713044107281)=0.187713044107281\). Thus \(y_4=1.3700841+0.187713044107281=1.557797144107281\), at \(t_4=0.3+0.1=0.4\).
| \(n\) | \(t_n\) | \(y_n\) | \(f(y_n)=y_n^2\) | \(h f(y_n)\) | \(y_{n+1}\) |
|---|---|---|---|---|---|
| 0 | 0 | 1 | 1 | 0.1 | 1.1 |
| 1 | 0.1 | 1.1 | 1.21 | 0.121 | 1.221 |
| 2 | 0.2 | 1.221 | 1.490841 | 0.1490841 | 1.3700841 |
| 3 | 0.3 | 1.3700841 | 1.87713044107281 | 0.187713044107281 | 1.557797144107281 |
Check the last row directly: \(1.3700841^2=1.87713044107281\), \(0.1(1.87713044107281)=0.187713044107281\), and \(1.3700841+0.187713044107281=1.557797144107281\). Each row’s next estimate is therefore consistent with the displayed slope and step size.
Worked Example: A Slope That Decreases Toward Zero
Worked Example: A Slope That Decreases Toward Zero
Use Euler’s method for \(y'=3-y\), \(y(0)=1\), with \(h=0.25\), for three steps.
Initially, \(t_0=0\) and \(y_0=1\). The slope is \(3-1=2\), so the increment is \(0.25(2)=0.5\). Therefore \(t_1=0.25\) and \(y_1=1+0.5=1.5\).
At \(y_1=1.5\), the slope is \(3-1.5=1.5\). The increment is \(0.25(1.5)=0.375\), giving \(t_2=0.5\) and \(y_2=1.5+0.375=1.875\).
At \(y_2=1.875\), the slope is \(3-1.875=1.125\). The increment is \(0.25(1.125)=0.28125\), giving \(t_3=0.75\) and \(y_3=1.875+0.28125=2.15625\).
| \(n\) | \(t_n\) | \(y_n\) | \(3-y_n\) | \(0.25(3-y_n)\) | \(y_{n+1}\) |
|---|---|---|---|---|---|
| 0 | 0 | 1 | 2 | 0.5 | 1.5 |
| 1 | 0.25 | 1.5 | 1.5 | 0.375 | 1.875 |
| 2 | 0.5 | 1.875 | 1.125 | 0.28125 | 2.15625 |
The calculations check: \(1+0.5=1.5\), \(1.5+0.375=1.875\), and \(1.875+0.28125=2.15625\). Also, each displayed slope equals \(3-y_n\) for that row. The slopes remain positive but decrease as the estimate approaches \(3\).
Worked Example: Negative Slopes and Decreasing Estimates
Worked Example: Negative Slopes and Decreasing Estimates
Construct three Euler steps for \(y'=-0.4y\), \(y(0)=100\), using \(h=1\).
At \(t_0=0\), \(y_0=100\), so the slope is \(-0.4(100)=-40\). The increment is \(1(-40)=-40\), and the next estimate is \(100+(-40)=60\), at \(t_1=1\).
At \(y_1=60\), the slope is \(-0.4(60)=-24\). The increment is \(1(-24)=-24\), so \(y_2=60+(-24)=36\), at \(t_2=2\).
At \(y_2=36\), the slope is \(-0.4(36)=-14.4\). The increment is \(1(-14.4)=-14.4\), so \(y_3=36+(-14.4)=21.6\), at \(t_3=3\).
| \(n\) | \(t_n\) | \(y_n\) | \(-0.4y_n\) | \(1(-0.4y_n)\) | \(y_{n+1}\) |
|---|---|---|---|---|---|
| 0 | 0 | 100 | -40 | -40 | 60 |
| 1 | 1 | 60 | -24 | -24 | 36 |
| 2 | 2 | 36 | -14.4 | -14.4 | 21.6 |
For a check, the updates are \(100-40=60\), \(60-24=36\), and \(36-14.4=21.6\). The negative slope correctly produces negative increments and decreasing estimates.
Common Mistake: Confusing Slope and Increment
Suppose the current estimate is \(y_n=2\), the equation gives \(y'=y^2\), and \(h=0.1\). The slope is \(2^2=4\), but the increment is \(0.1(4)=0.4\). The correct next estimate is \(2+0.4=2.4\).
A wrong calculation would write \(y_{n+1}=2+4=6\). This adds the slope directly and forgets that the step lasts only \(0.1\) units. The correction is to keep the two calculations distinct: first evaluate \(f(y_n)\), then multiply by \(h\), then add that product to \(y_n\).
A related error is to evaluate the next slope using the old estimate again. After the correct update to \(2.4\), the next slope for \(y'=y^2\) is \(2.4^2=5.76\), not \(2^2=4\). Each row begins with the preceding row’s updated estimate.
What a Completed Table Tells You
A completed table is a chain of approximate solution values at equally spaced independent-variable values. Its usefulness depends on accurate, transparent bookkeeping. Before accepting a row, verify the slope substitution, the multiplication by \(h\), and the addition that gives the next estimate. Keep sufficient digits through intermediate calculations; if rounding is necessary, state the rounding and apply it consistently.
The table gives estimates, not a guarantee that the exact solution passes through every listed point. It also does not, by itself, quantify the error. Its immediate purpose is to follow the differential equation’s prescribed slope one step at a time and record exactly how each approximation was produced.
Check Your Understanding
For each question, show the slope, the increment, and the updated estimate rather than recording only the final value.
- For \(y'=2y\), \(y(0)=3\), and \(h=0.1\), calculate the first Euler step.
- For \(y'=4-y\), \(y(0)=2\), and \(h=0.5\), calculate two Euler steps.
- If \(f(y_n)=-3\) and \(h=0.2\), what is the increment? Explain its sign.
- A student uses the slope itself as the increment. What factor is missing when the step size is \(h\)?
- For \(y'=y^2\), \(y_n=1.5\), and \(h=0.1\), calculate the slope, increment, and next estimate.
