[provisional cross-reference: WORK-IN-PROGRESS]
\begin{equation*}
y' = y + t, \quad y(0) = 1
\end{equation*}
Section Eulerโs Method: The Full Process
In the previous section, we zoomed in on the single โstepโ at the heart of Eulerโs method: move forward by a small amount \(h\) in \(t\text{,}\) and adjust \(y\) by \(h\) times the slope. Now weโll zoom out. By repeating that step again and again, we can march across an interval, building an approximate solution point by point.
Subsection A Concrete Step Through Eulerโs Method
Before writing any formulas, it helps to see what Eulerโs Method feels like. Think of Eulerโs method like crossing a landscape with only a compass. You know where you start. At each step, you check your direction (the slope from the differential equation), take a small stride that way, then check again. That back-and-forth of โstep, check, adjustโ is the rhythm of the method.
The Goal.
Suppose we have the initial value problem
\begin{equation}
y'(t) = 6t + y(t), \quad y(0) = -2, \quad t \text{ in } [0, 1.5]\text{.}\tag{27}
\end{equation}
Letโs approximate the solution at the equally-spaced \(t\)-values: \(0, 0.5, 1, 1.5\text{.}\) This tells us that our goal is to approximate the \(y\)-values at the following points:
\begin{equation*}
\text{Approximation} = \big\{\ (\ 0\ ,\ ?\ ),\ (\ 0.5\ ,\ ?\ ),\ (\ 1\ ,\ ?\ ),\ (\ 1.5\ ,\ ?\ )\ \big\}\text{.}
\end{equation*}
Luckily, the first \(y\) value is just the initial condition, \(y(0) = -2\text{,}\) so we add it to the list:
\begin{equation*}
\text{Approximation} = \left\{\ (0,-2)\ ,\ (0.5,\ ?\ )\ ,\ (1,\ ?\ )\ ,\ (1.5,\ ?\ )\ \right\}\text{.}
\end{equation*}
Starting at \((0,-2)\text{,}\) Eulerโs method finds the rest of the points, one-by-one.
Approximating the Next Point.
From \((0, -2)\text{,}\) we need to get to \((0.5,?)\text{.}\) Clearly, the run is \(0.5\text{,}\) but we also need the slope for our movement direction. This comes directly from the differential equation:
\begin{equation*}
\text{slope at}\ (0,-2)\ \os{\large t=0}{\quad\longrightarrow\quad} y'(0) = 6(0) + y(0) = -2\text{.}
\end{equation*}
Moving from \((0, -2)\) with a slope of \(-2\) and a run of \(0.5\) is given by:
\begin{equation*}
(0, -2) \rightarrow (0 + \text{run}, -2 + \text{slope} \times \text{run}) = (0.5, -2 + -2 \cdot 0.5) = (0.5, -3)\text{.}
\end{equation*}
Therefore, the next point is \((0.5, -3)\) and our approximation list becomes:
\begin{equation*}
\text{Approximation} = \left\{\ (0,-2)\ ,\ (0.5,-3)\ ,\ (1,\ ?\ )\ ,\ (1.5,\ ?\ )\ \right\}\text{.}
\end{equation*}
Approximating the Rest of the Points.
To get the third and fourth point, we just repeat what we just did.
At \((0.5, -3)\text{,}\) we get the slope from:
\begin{equation*}
\text{slope at}\ (0.5, -3)\ \os{\large t=0.5}{\quad\longrightarrow\quad} y'(0.5) = 6(0.5) + y(0.5) = 3 + (-3) = 0\text{.}
\end{equation*}
Using this slope and the same run, we get the third point:
\begin{equation*}
(0.5 + \text{run}, -3 + \text{slope} \times \text{run}) = (1, -3 + 0 \cdot 0.5) = (1, -3)\text{.}
\end{equation*}
Now we are at \((1, -3)\) and the slope we move this time is:
\begin{equation*}
\text{slope at}\ (1, -3)\ \os{\large t=0}{\quad\longrightarrow\quad} y'(1) = 6(1) + y(1) = 6 + (-3) = 3\text{.}
\end{equation*}
and our final point is given by:
\begin{equation*}
(1 + \text{run}, -3 + \text{slope} \times \text{run}) = (1.5, -3 + 3 \cdot 0.5) = (1.5, -1.5)\text{.}
\end{equation*}
With this, our approximation is now complete:
\begin{equation*}
\text{Approximation} = \left\{\ (0,-2)\ ,\ (0.5,-3)\ ,\ (1,-3)\ ,\ (1.5,-1.5)\ \right\}\text{.}
\end{equation*}
A plot of the process is given in Figureย 122.
๐ฎ Interactive 123. [provisional cross-reference: WORK-IN-PROGRESS].
[provisional cross-reference: WORK-IN-PROGRESS].
Subsection Eulerโs Method
Now letโs generalize the concrete example. As we do, check the footnotes for the concrete version of each equation and step.
Consider an initial value problem:
โ2โ
\begin{equation*}
y'(t) = 6t + y(t), \quad y(0) = -2, \quad t\ \text{in}\ [0, 1.5]
\end{equation*}
\begin{equation*}
y'(t) = f(t, y), \quad y(t_0) = y_0, \quad t\ \text{in}\ [t_0, t_N]\text{.}
\end{equation*}
We divide \([t_0, t_N]\) into equally spaced values (a distance \(h\) apart):
โ3โ
\begin{equation*}
\us{\ds t_0}{\us{\uparrow}{0}},\
\us{\ds t_1}{\us{\uparrow}{0.5}},\
\us{\ds t_2}{\us{\uparrow}{1}},\
\us{\ds t_3}{\us{\uparrow}{1.5}}\qquad h = 0.5
\end{equation*}
\begin{equation*}
t_0,\ t_1,\ t_2,\ \ldots,\ t_N, \quad\text{where}\quad t_{k+1} = t_k + h
\end{equation*}
Using the shorthand \(y_k = y(t_k)\text{,}\) our goal is to approximate the points:
โ4โ
\begin{equation*}
\text{Approximation} = \{\ (0,?),\ (0.5,?),\ (1,?),\ (1.5,?)\ \big\}
\end{equation*}
\begin{equation*}
(t_0, y_0), (t_1, y_1), \ldots, (t_N, y_N).
\end{equation*}
The first point, \((t_0, y_0)\text{,}\) is known since it is the initial condition. The remaining points are approximated using the movement rule, where โcurโ is \(k\) and โnewโ is \(k+1\text{:}\)
\begin{equation*}
(t_{k+1},\ y_{k+1}) = (t_k + h,\ y_k + h\cdot\text{slope})\text{.}
\end{equation*}
The slope is found from \(f(t_k, y_k)\) in the differential equation, so the rule becomes:
โ5โ
\begin{equation*}
6t + y(t)
\end{equation*}
\begin{equation*}
(t_{k+1},\ y_{k+1}) = (t_k + h,\ y_k + h\cdot f(t_k, y_k)).
\end{equation*}
Extracting just the \(y\)-coordinate, gives the update rule for Eulerโs method:
\begin{equation*}
y_{k+1} = y_k + h\cdot f(t_k, y_k).
\end{equation*}
Thatโs it. All of the \(y\) values in the table of Figureย 122 can be found with this formula.
Checkpoint 124. Match Eulerโs Method Parts to their Meaning.
โ๐ป Method 4. Eulerโs Method.
Given an initial value problem
\begin{equation*}
y' = f(t, y), \quad y(t_0) = y_0, \quad t_0 \le t \le t_N
\end{equation*}
approximate the solution with the following steps:
- Select step size
- Choosing a step size \(h\) determines \(t_0,\ t_1,\ t_2,\ ...,\ t_{N}\) where \(t_k = t_0 + kh\) for \(k = 0, 1, 2, ..., N\text{.}\)
- Apply Eulerโs update rule
- For \(k = 0\) to \(N - 1\text{,}\) compute\begin{equation*} y_{k+1} = y_k + h \cdot f(t_k, y_k). \end{equation*}
Desipte being based on relatively simple idea, Eulerโs method reveals something profound: with nothing more than a starting point and slope formula, you can approximate a solution that no closed-form equation could describe.
Subsection Examples
๐ Example 125. Eulerโs Method Example.
Use Eulerโs method to approximate the solution to the differential equation
\begin{equation*}
y' = t + y, \quad y(0) = -\frac{7}{8}
\end{equation*}
over the interval \([0, 1.5]\) with a step size of \(h = 0.5\text{.}\)
Solution.
Select step size. The step size is \(h = 0.5\text{,}\) so the approximation locations are \(t_0 = 0, t_1 = 0.5, t_2 = 1.0, t_3 = 1.5\text{.}\)
Apply Eulerโs update rule. The initial condition gives the first value in our approximation, \(y_0 = -\frac{7}{8} = -0.875\text{.}\)
Now, we can use the formula
\begin{equation*}
y_{k+1} = y_k + h\left(t_k + y_k\right)
\end{equation*}
to find \(y_1\text{,}\) \(y_2\) and \(y_3\) as follows:
\begin{align*}
\text{G} \amp\text{iven} \\
k \amp = 0 \\
k \amp = 1 \\
k \amp = 2
\end{align*}
\begin{align*}
y_0 \amp = -0.875 \\
y_1 \amp = y_0 + h\left(t_0 + y_0\right) = -0.875 + 0.5\left(0 - 0.875\right) \\
y_2 \amp = y_1 + h\left(t_1 + y_1\right) = -1.3125 + 0.5\left(0.5 - 1.3125\right) \\
y_3 \amp = y_2 + h\left(t_2 + y_2\right) = -1.5625 + 0.5\left(1.0 - 1.5625\right)
\end{align*}
The simplified calculations are summarized in the following table:
| \(k\) | \(t_k\) | \(y_k\) |
| \(0\) | \(0.0\) | \(-0.875\) |
| \(1\) | \(0.5\) | \(-0.75\) |
| \(2\) | \(1.0\) | \(-0.5\) |
| \(3\) | \(1.5\) | \(-0.125\) |
Thus, the approximate solution to the initial-value problem is
\begin{equation*}
y(0) \approx -\frac{7}{8}, \quad y(0.5) \approx -\frac{3}{4}, \quad y(1.0) \approx -\frac{1}{2}, \quad y(1.5) \approx -\frac{1}{8}.
\end{equation*}
We can visualize the approximation by plotting the points \((t_k, y_k)\) and connecting them with line segments. The exact solution to the initial-value problem is given by
\begin{equation*}
y(t) = \frac{1}{8}e^t - t - 1.
\end{equation*}
The following plot shows the approximation and the exact solution.
The red curve is the exact solution, while the green points are the approximated values at \(t = 0, 0.5, 1.0, 1.5\text{.}\) The blue line segments connect the points in the approximation, showing how Eulerโs method steps through the solution.
The following plot shows the approximation and the exact solution, with a grid for better visualization.
\begin{align*}
y'(t_k) \amp = (y(t_k))^2 - t_k \\
\frac{y(t_{k+1}) - y(t_k)}{h} \amp = (y(t_k))^2 - t_k \\
\frac{y_{k+1} - y_k}{h} \amp = (y_k)^2 - t_k \\
y(t_{k+1}) - y(t_k) \amp = h\left((y(t_k))^2 - t_k\right) \\
y(t_{k+1}) \amp = y(t_k) + h\left((y(t_k))^2 - t_k\right)
\end{align*}
๐ Example 126. Analytical and Numerical Solutions.
Consider the differential equation
\begin{equation*}
y' + 4ty = 0, \, y(0) = 1\text{.}
\end{equation*}
-
Find the analytical solution to this equation using the method of separation of variables.
-
Use Eulerโs method with a step size of \(h = 0.1\) to approximate the solution at \(t = 0.5\text{.}\)
-
Compare the analytical solution with the numerical approximation obtained from Eulerโs method.
Solution.
To solve the differential equation
\begin{equation*}
y' + 4ty = 0
\end{equation*}
using the method of separation of variables, we separate the variables as follows:
\begin{equation*}
\frac{dy}{y} = -4t \, dt
\end{equation*}
Integrating both sides gives:
\begin{equation*}
\ln |y| = -2t^2 + C
\end{equation*}
Exponentiating both sides results in:
\begin{equation*}
y = Ce^{-2t^2}
\end{equation*}
Since the initial condition, \(y(0) = 1\text{,}\) gives \(C = 1\text{,}\) the analytical solution is:
\begin{equation*}
y(t) = e^{-2t^2}
\end{equation*}
Next, we apply Eulerโs method to approximate the solution at \(t = 0.5\text{.}\) The general formula for Eulerโs method is:
\begin{equation*}
y_{k+1} = y_k + h f(t_k, y_k)
\end{equation*}
where \(f(t, y) = -4ty\) and \(h=0.1\text{.}\) So, we can update the formula as:
\begin{equation*}
y_{k+1} = y_k - 0.4 t_k y_k \quad \text{where}\ y_0 = 1.
\end{equation*}
For \(k = 0, 1, \ldots, 4\text{,}\) we contruct the following table:
| \(k\) | \(t_k\) | \(y_k\) | \(y_{k+1}\) |
| \(0\) | \(0.0\) | \(y_0 = {\DLBa 1}\) | \(y_1 = {\DLBa 1} + 0.1(-4 \cdot 0.0 \cdot {\DLBa 1}) = {\DLGa 1}\) |
| \(1\) | \(0.1\) | \(y_1 = {\DLBa 1}\) | \(y_2 = {\DLBa 1} + 0.1(-4 \cdot 0.1 \cdot {\DLBa 1}) = {\DLGa 0.96}\) |
| \(2\) | \(0.2\) | \(y_2 = {\DLBa 0.96}\) | \(y_3 = {\DLBa 0.96} + 0.1(-4 \cdot 0.2 \cdot {\DLBa 0.96}) = {\DLGa 0.8832}\) |
| \(3\) | \(0.3\) | \(y_3 = {\DLBa 0.8832}\) | \(y_4 = {\DLBa 0.8832} + 0.1(-4 \cdot 0.3 \cdot {\DLBa 0.8832}) \approx {\DLGa 0.7772}\) |
| \(4\) | \(0.4\) | \(y_4 = {\DLBa 0.7772}\) | \(y_5 = {\DLBa 0.7772} + 0.1(-4 \cdot 0.4 \cdot {\DLBa 0.7772}) \approx {\DLGa 0.6529}\) |
Since \(t_5=0.5\text{,}\) the approximation we are looking for is \(y_5 \approx 0.6529\text{.}\)
Comparing this with the analytical solution at \(t = 0.5\text{,}\) we see the true value is:
\begin{equation*}
y(0.5) = e^{-2(0.5)^2} = e^{-0.5} \approx 0.6065.
\end{equation*}
Although there is some error between the two values, it is more due to the step size we chose, rather than the method itself. A smaller step size would yield a more accurate approximation. For example, using \(h = 0.001\) gives the approximation: \(y(0.5) \approx 0.6069\text{.}\)
๐ Example 127. Example: Eulerโs Method with \(h = 0.5\).
Use Eulerโs method to approximate the solution to
\begin{equation*}
y'(t) = 6t + y(t), \quad y(0) = -2
\end{equation*}
over \(t = 0\) to \(1.5\) using step size \(h = 0.5\text{.}\)
Solution.
The formula is:
\begin{equation*}
y_{k+1} = y_k + 0.5 \cdot (6t_k + y_k)
\end{equation*}
| \(k\) | \(t_k\) | \(y_k\) | \(y_{k+1} = y_k + 0.5(6t_k + y_k)\) |
| 0 | 0.0 | -2.0 | -3.0 |
| 1 | 0.5 | -3.0 | -3.0 |
| 2 | 1.0 | -3.0 | -1.5 |
Thus, the approximate values are:
\begin{equation*}
y(0) \approx -2,\quad y(0.5) \approx -3,\quad y(1.0) \approx -3,\quad y(1.5) \approx -1.5
\end{equation*}
Eulerโs method runs on an intuitive idea: use the slope at each point to โpredictโ the next point. Itโs simple, but that simplicity is its strength. This method is often the first stepping stone into the world of numerical methods and the same pattern of โpredict, step, repeatโ forms the backbone of many other techniques.
๐ค Wrap-Up.
๐๏ธ Key Takeaways...
-
Eulerโs Method turns slope information into a sequence of small steps forward.
-
Each step uses the simple rule: \(y_{k+1} = y_k + h f(t_k, y_k)\text{.}\)
-
Smaller step sizes usually mean better accuracy โ but more steps.
-
This method is the foundation for more advanced numerical techniques.
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