Slope fields and Euler's method

What you are looking at

A differential equation $\dfrac{dy}{dx} = f(x, y)$ gives the gradient of the solution curve at every point without giving the curve itself. The slope field draws a short segment with that gradient at each point of a grid spaced $0.5$ apart. Every solution curve runs along the segments, so the field shows the shape of the whole family of solutions at once; the green starting point $(x_0, y_0)$ picks out one of them.

Euler's method replaces the curve with straight steps. From $(x_n, y_n)$ it moves a distance $h$ to the right along the gradient at that point: $x_{n+1} = x_n + h$ and $y_{n+1} = y_n + h\,f(x_n, y_n)$. Because the gradient is only updated at the start of each step, the yellow Euler path drifts away from the exact solution, drawn in blue. The working shows the first step, the point reached after $n$ steps and the error there.

Try this

1. 1With the default $\dfrac{dy}{dx} = x + y$, $y(0) = 0$, $h = 0.5$ and $n = 4$, Euler's method reaches $y_4 = 2.0625$ at $x = 2$, while the exact solution $y = e^x - x - 1$ gives $e^2 - 3 \approx 4.39$.
2. 2Halve the step to $h = 0.25$ and double $n$ to $8$ so that you still finish at $x = 2$. The error falls from about $2.33$ to about $1.43$: smaller steps follow the curve more closely, but the error does not vanish.
3. 3Press Play: $n$ goes back to $0$ and one step is added at a time until the path reaches the right-hand edge. Watch whether each new point lands above or below the blue curve.
4. 4Choose $\dfrac{dy}{dx} = x - y$ and drag the green point to different starting heights. Every exact solution, $y = x - 1 + Ce^{-x}$, approaches the line $y = x - 1$, and the Euler path is drawn towards it too.

Mistakes students make

• Using the wrong gradient in a step. Euler's method uses $f(x_n, y_n)$, the gradient at the point you are leaving, not at the new point and not at the new $x$ with the old $y$.
• Expecting Euler's method to be exact. Each step follows a tangent, so it tends to fall below a solution that curves upwards (as for $\dfrac{dy}{dx} = x + y$ from $(0, 0)$) and to overshoot one that curves downwards.
• Swapping $x$ and $y$ when working out a segment. For $\dfrac{dy}{dx} = x - y$ at the point $(2, 1)$ the gradient is $2 - 1 = 1$, not $-1$.

Where it is taught

• IB Maths AA HL 5.18 · Slope fields and Euler's method
• AP Calculus AB/BC Unit 7 · Sketching and reasoning with slope fields (7.3, 7.4)
• AP Calculus BC Unit 7 · Approximating solutions with Euler's method (7.5)