From an Initial-Value Problem to a Table
An initial-value problem gives a differential equation and a starting value. For example, an autonomous first-order problem has the form \(y'=f(y)\), with \(y(t_0)=y_0\). The equation specifies the slope at a given solution value, while the initial condition supplies the first value from which to begin computing.
Euler's method turns that information into a sequence of approximate solution values. A table makes each step visible: it records the time, the current estimate, the slope evaluated at that estimate, and the next estimate. The essential habit is to calculate each row from the current estimate in that same row—not from a value that has not yet been reached.
The product \(h f(y_n)\) is the change in the estimate during that step. It is useful to calculate and record this increment explicitly, even if a table format combines it with the next-estimate calculation. The next row begins with the previous row's next estimate, and its time is one step later.
For a table with \(N\) steps, there are \(N\) updates but \(N+1\) time-value pairs: the initial pair \((t_0,y_0)\), followed by the estimates through \((t_N,y_N)\). The last row used to perform an update is row \(N-1\), since it produces \(y_N\).
A Reliable Row-by-Row Procedure
Before filling in any rows, write down the initial time, initial estimate, step size, and number of steps. For an autonomous equation, the slope depends on the current estimate through \(f(y_n)\); the time column still advances by \(h\), but time is not an input to this particular slope function.
Set \(t_0\) and \(y_0\) from the initial condition. These are given values, not Euler updates.
In row \(n\), substitute the current estimate \(y_n\) into the right-hand side to find \(f(y_n)\).
Multiply the evaluated slope by the step size: \(\Delta y_n=h f(y_n)\).
Set \(y_{n+1}=y_n+\Delta y_n\) and \(t_{n+1}=t_n+h\). Copy those values into the next row as its current time and estimate.
Use the same step size in every row. Keep enough digits to avoid unnecessary rounding during intermediate calculations, and check that the increment's sign agrees with the evaluated slope: because \(h>0\), a positive slope gives a positive increment and a negative slope gives a negative increment.
Worked Tables
Worked Example: A Slope That Decreases with the Estimate
Complete four Euler steps for \(y'=2-y\), with \(y(0)=0\) and \(h=0.5\). Here \(f(y)=2-y\). In each row, evaluate \(2-y_n\), multiply by \(0.5\), and add the result to \(y_n\).
The time entries follow \(t_{n+1}=t_n+0.5\): \(0\), \(0.5\), \(1\), \(1.5\), and finally \(2\). Thus the fourth update gives the estimate \(y_4=1.875\) at time \(t_4=2\). To check the final update, use the last row's current value: \(f(1.75)=0.25\), so \(1.75+0.5(0.25)=1.875\), as recorded.
Worked Example: A Slope Equal to the Current Estimate
Complete three steps for \(y'=y\), with \(y(0)=1\) and \(h=0.5\). Since \(f(y_n)=y_n\), each slope is the current estimate itself.
The final estimate is \(y_3=3.375\) at \(t_3=1.5\). Check the last row directly: the slope at \(y_2=2.25\) is \(2.25\), the increment is \(0.5(2.25)=1.125\), and the next value is \(2.25+1.125=3.375\). Notice that the slope in a row is based on that row's current estimate, not on its next estimate.
Worked Example: Keeping Track of Negative Slopes
Complete four steps for \(y'=-2y\), with \(y(0)=4\) and \(h=0.25\). The slope is negative for these positive estimates, so each increment should decrease the estimate.
The four updates reach \(t_4=1\) and give \(y_4=0.25\). The last update checks as \(0.5+0.25(-1)=0.5-0.25=0.25\). The negative sign belongs to the increment; writing \(0.5+0.25\) would contradict the slope \(-1\).
Why the Table Recurrence Is Consistent
A completed row must describe one Euler update from the pair \((t_n,y_n)\) to the next pair. The time changes by \(h\), so \(t_{n+1}=t_n+h\). The equation supplies the slope \(f(y_n)\) at the current estimate. Over a step of length \(h\), Euler's rule uses that slope to calculate a change of \(h f(y_n)\), giving \(y_{n+1}=y_n+h f(y_n)\).
This argument explains why the columns have their particular order. The slope column is calculated from the current estimate; the increment column is calculated from the slope; the next-estimate column is calculated from the current estimate and increment. Rearranging these dependencies—for example, using a new estimate to calculate the slope that was supposed to produce it—would no longer be the stated Euler update.
Indexing, Checks, and Common Errors
Table labels can be a source of errors even when the arithmetic is simple. Row \(n\) starts with \(t_n\) and \(y_n\), and its update produces \(t_{n+1}\) and \(y_{n+1}\). The next row then begins with those same values. A table may display the next estimate in the last column of one row or as the current estimate in the next row; either layout is valid as long as the values agree.
Other useful checks include counting the rows, verifying the time sequence, and checking signs:
- For \(N\) steps, there should be \(N\) slope evaluations and \(N+1\) time-value pairs, including the initial pair.
- Every successive time should differ by the same amount \(h\).
- The difference between consecutive estimates should equal the recorded increment.
- For \(h>0\), the increment should have the same sign as the slope.
- The next row's current estimate should equal the preceding row's next estimate.
An Euler table records the numerical procedure, not an exact solution of the differential equation. Its final entry is an approximation produced by repeated updates. The table is valuable because it makes those updates auditable: one can locate whether a discrepancy comes from evaluating a slope, multiplying by the step size, adding the increment, or advancing the time.
Check Your Understanding
Use the definitions and row-by-row procedure above to answer these questions.
- For \(y'=3-y\), \(y(0)=1\), and \(h=0.2\), what slope, increment, and next estimate belong in the first row?
- If a table has six Euler updates, how many time-value pairs should it contain, and what is the index of the last update row?
- For \(y'=-y\) with a positive current estimate and \(h>0\), what sign should the increment have? Explain using the update rule.
- A row records \(y_n=2\), \(f(y_n)=3\), \(h=0.25\), and \(y_{n+1}=2.75\). Is the update consistent? Show the calculation.
- Why must the slope in a row be evaluated using \(y_n\), rather than the next estimate \(y_{n+1}\)?
