Why Euler’s Method Uses a Table
A differential equation such as \(y'=f(x,y)\) specifies the slope of a solution curve at each point. If we know one point on the curve, the differential equation tells us the slope there. Euler’s method uses that slope to estimate a nearby point, then repeats the process from the new estimate.
This is useful when a solution is difficult to find exactly or when a question asks for a numerical estimate at a specified input. In a free-response solution, the table should make the calculation traceable: show the current point, the slope there, the step size, and the resulting next estimate.
The value \(y_n\) is an approximation to the actual solution value \(y(x_n)\). Unless the differential equation and the steps happen to make the numerical updates exact, the table does not give the exact solution.
Why the Update Formula Works
At a point \((x_n,y_n)\), the differential equation assigns slope \(f(x_n,y_n)\). A short move of size \(h\) along a line with that slope changes the vertical coordinate by slope multiplied by horizontal change. Thus the tangent-line estimate at the next input is the current value plus \(h f(x_n,y_n)\).
Adding this estimated change to \(y_n\) gives Euler’s update. The approximation comes from treating the slope as though it stayed constant throughout the step. In reality, the slope may change as \(x\) and \(y\) change.
How to Complete an Euler Table
For a constant positive step size, the input values increase by the same amount each row. At every row, calculate the slope from the differential equation using that row’s current \(x\) and \(y\). Then multiply by \(h\) and add the result to the current \(y\)-value.
Begin with the given initial values \((x_0,y_0)\), or use the values in the latest completed row.
Substitute the current \(x\) and \(y\) into \(f(x,y)\) to find \(f(x_n,y_n)\).
Calculate the change \(h f(x_n,y_n)\).
Set \(x_{n+1}=x_n+h\) and \(y_{n+1}=y_n+h f(x_n,y_n)\). Use this new point for the next slope.
Worked Example: A Three-Step Euler Table
Worked Example: A Three-Step Euler Table
Use Euler’s method with step size \(h=0.2\) to estimate \(y(0.6)\) for \(y'=x-y\), given \(y(0)=1\).
Start with \(x_0=0\), \(y_0=1\). At this point the slope is \(0-1=-1\). The change in \(y\) over the first step is \((-1)(0.2)=-0.2\), so \(y_1=1-0.2=0.8\), at \(x_1=0.2\).
At \((0.2,0.8)\), the slope is \(0.2-0.8=-0.6\). The next change in \(y\) is \((-0.6)(0.2)=-0.12\), so \(y_2=0.8-0.12=0.68\), at \(x_2=0.4\).
At \((0.4,0.68)\), the slope is \(0.4-0.68=-0.28\). The next change is \((-0.28)(0.2)=-0.056\), so \(y_3=0.68-0.056=0.624\), at \(x_3=0.6\).
The updates are \(1+(-0.2)=0.8\), \(0.8+(-0.12)=0.68\), and \(0.68+(-0.056)=0.624\). The table therefore gives \(y(0.6)\approx 0.624\). As a check, the last row’s input is \(0.4\), so one more step of \(0.2\) reaches the requested input \(0.6\); the last update uses exactly that row’s slope, \(-0.28\).
Step Size and Approximation
The step size \(h\) is the horizontal distance advanced in one update. If the input represents time in seconds, \(h\) is measured in seconds, and the slope is measured in units of \(y\) per second. The product \(h f(x_n,y_n)\) then has the same units as \(y\), as required for the update.
A smaller step means more updates are needed to reach the same final input. It also means each tangent-line estimate covers a shorter interval, so the slope has less distance over which to change before the next update. For a smooth solution, this generally improves the approximation over a fixed interval. However, it is not correct to claim that every individual smaller-step estimate must be closer for every differential equation and every endpoint.
A Justification for the Step-Size Effect
Suppose an exact solution \(y(x)\) has a continuous second derivative over the interval of interest. Taylor’s theorem says that, for a step \(h\), there is a point \(\xi\) between \(x\) and \(x+h\) such that
Since \(y'(x)=f(x,y(x))\), Euler’s update starting from the exact value \(y(x)\) would give \(y(x)+h y'(x)\). The difference between the true next value and this one-step Euler estimate is therefore
If \(|y''(x)|\le M\) throughout the interval, the magnitude of this one-step difference is at most \(Mh^2/2\). This is a local error estimate: it compares one Euler step started at an exact solution value with the exact value one step later.
In a full table, later steps also use earlier approximations, so local errors can accumulate. For example, if \(f\) is Lipschitz in \(y\) with constant \(L\), meaning \(|f(x,u)-f(x,v)|\le L|u-v|\), then an existing error can affect the next slope. If \(e_n\) is the absolute error at step \(n\), the local bound and this slope condition give
On a fixed interval, repeated steps under these conditions lead to an accumulated error that is proportional to \(h\) as \(h\) becomes small. This explains why using more, shorter steps usually improves the approximation for a well-behaved problem. It does not say that the error must decrease monotonically at every particular endpoint when \(h\) changes.
Worked Example: Comparing Two Step Sizes
Worked Example: Comparing Two Step Sizes
Approximate \(y(1)\) for \(y'=y\), \(y(0)=1\), using first \(h=0.5\) and then \(h=0.25\). Compare the estimates with the exact solution \(y=e^x\), so \(y(1)=e\approx2.71828\).
With \(h=0.5\), the first slope is \(f(0,1)=1\). The change is \(0.5(1)=0.5\), giving \(y_1=1.5\) at \(x_1=0.5\). The next slope is \(f(0.5,1.5)=1.5\). The change is \(0.5(1.5)=0.75\), giving \(y_2=1.5+0.75=2.25\) at \(x_2=1\).
With \(h=0.25\), the successive updates are:
The input advances by \(0.25\) each time, so these four updates reach \(x=1\). The estimates are \(2.25\) and \(2.44140625\), while the exact value is approximately \(2.71828\). Their absolute errors are approximately \(2.71828-2.25=0.46828\) and \(2.71828-2.44140625=0.27687\), respectively. In this example, the smaller step gives a closer estimate. This comparison illustrates the usual effect under favorable conditions; it is not a universal guarantee for every problem.
Changing Step Sizes
Not every table uses a constant step. If consecutive input values are specified directly, find each step from the difference between the next and current inputs. For a step from \(x_n\) to \(x_{n+1}\), use \(h_n=x_{n+1}-x_n\), then update \(y\) with that same \(h_n\).
Worked Example: Using Unequal Steps
Use Euler’s method for \(y'=x-y\), starting from \(y(1)=2\), to estimate the solution at \(x=1.5\), using the specified intermediate input \(x=1.2\).
For the first step, \(h_0=1.2-1=0.2\). At \((1,2)\), the slope is \(1-2=-1\). Thus the change in \(y\) is \(0.2(-1)=-0.2\), and the new estimate is \(y=2-0.2=1.8\).
For the second step, \(h_1=1.5-1.2=0.3\). At \((1.2,1.8)\), the slope is \(1.2-1.8=-0.6\). The change is \(0.3(-0.6)=-0.18\), so the estimate at \(x=1.5\) is \(1.8-0.18=1.62\).
Check the input increments: \(1+0.2=1.2\) and \(1.2+0.3=1.5\). Each vertical update also equals its own step multiplied by the slope at that step’s starting point. Therefore the Euler estimate is \(y(1.5)\approx1.62\).
Common Mistake: Reusing an Old Slope
Each Euler slope belongs to the current estimated point. Reusing a slope from an earlier row ignores the fact that the differential equation may assign a different slope at the new point.
Worked Example: Correcting a Reused-Slope Error
For \(y'=x+y\), \(y(0)=1\), use \(h=0.1\) to estimate \(y(0.2)\).
The first slope is \(0+1=1\), so the first update gives \(y_1=1+0.1(1)=1.1\) at \(x_1=0.1\). The new slope is \(0.1+1.1=1.2\), so the second update gives \(y_2=1.1+0.1(1.2)=1.22\) at \(x_2=0.2\).
A common incorrect calculation reuses the first slope and writes \(y_2=1.1+0.1(1)=1.2\). That is wrong because the second slope must be evaluated at \((0.1,1.1)\), where \(x+y=1.2\). The corrected estimate is \(1.22\). Checking the table’s final input, \(0.1+0.1=0.2\), confirms that the two correct updates reach the requested input.
Writing a Clear Free-Response Explanation
A complete numerical response should show enough work to make the method verifiable. State the step size, identify the slope used for each update, and show the product of slope and step before giving the new \(y\)-value. If the question asks about changing step size, describe the comparison as a general tendency under suitable conditions rather than promising that every smaller step gives a closer answer.
- Use the current estimated \(y\)-value when evaluating the slope.
- Keep the step size and units consistent with the input variable.
- Show the arithmetic for each requested update.
- Check that the input values reach the requested endpoint.
- Distinguish a numerical estimate from an exact solution value.
Check Your Understanding
For each question, show the slope-times-step calculation in every Euler update.
- For \(y'=2x-y\), \(y(0)=1\), use \(h=0.25\) to estimate \(y(0.5)\).
- For \(y'=y\), \(y(0)=2\), use \(h=0.5\) to estimate \(y(1)\). What changes in the calculation if the step size is \(0.25\)?
- In an Euler table, the current point is \((2,3)\), the differential equation is \(y'=x-y\), and the next input is \(2.4\). What slope, step, and next \(y\)-estimate should be used?
- Explain why the slope for an update must be evaluated at the current estimated point rather than at the next input.
- Why does a smaller step usually improve Euler’s approximation for a smooth solution, and why is that not an unconditional guarantee about every endpoint estimate?
