Euler's method

Three quick questions from earlier lessons.

$y' = y$ , $y(0) = 1$ . At $(0, 1)$ the slope is 1. Follow the tangent line for a step of $0.5$ . Where do you land?

Walk along short tangent lines.

Euler's method for

y' = f(x, y)

,

y(x_0) = y_0

, step

h

:

x_{n+1} = x_n + h

,

y_{n+1} = y_n + h\,f(x_n, y_n)

. Smaller steps give better estimates.

$y' = x + y$ , $y(0) = 1$ , $h = 0.5$ . Estimate $y(1)$ .

y_1 = 1 + 0.5 \cdot 1 = 1.5

Slope at (0, 1) is 1.

y_2 = 1.5 + 0.5 \cdot 2 = 2.5

Slope at (0.5, 1.5) is 2.

y(1) \approx 2.5

Two steps reach x = 1.

Your turn: $y' = 2y$ , $y(0) = 3$ , one step with $h = 0.1$ .

y(0.1) \approx 3.6

3 + 0.1 × 6.

$\dfrac{dy}{dx} = 1x + y$ and $y(1) = 3$ . Use one step of Euler's method with $h = 0.1$ to estimate $y(1.1)$ .
$\dfrac{dy}{dx} = 4x + y$ and $y(1) = 8$ . Use one step of Euler's method with $h = 0.1$ to estimate $y(1.1)$ .
$\dfrac{dy}{dx} = 9x + y$ and $y(1) = 4$ . Use one step of Euler's method with $h = 0.1$ to estimate $y(1.1)$ .

$\dfrac{dy}{dx} = y$ and $y(0) = 8$ . Use Euler's method with two steps of size $h = \tfrac{1}{2}$ to estimate $y(1)$ .
$\dfrac{dy}{dx} = y$ and $y(0) = 4$ . Use Euler's method with two steps of size $h = \tfrac{1}{2}$ to estimate $y(1)$ .

The solution of $\dfrac{dy}{dx} = y$ , $y(0) = 1$ , is concave up. Does Euler's method with $h > 0$ over- or underestimate $y(1)$ ?
1. Underestimate: each step follows a tangent line, which lies below a concave-up curve
2. Overestimate
3. It is exact

Three questions to check you have got it.

$\dfrac{dy}{dx} = 1x + y$ and $y(1) = 3$ . Use one step of Euler's method with $h = 0.1$ to estimate $y(1.1)$ .
$\dfrac{dy}{dx} = 4x + y$ and $y(1) = 8$ . Use one step of Euler's method with $h = 0.1$ to estimate $y(1.1)$ .
$\dfrac{dy}{dx} = 9x + y$ and $y(1) = 4$ . Use one step of Euler's method with $h = 0.1$ to estimate $y(1.1)$ .

Euler's method

$y_{n+1} = y_n + h\,f(x_n, y_n)$ : slope at the current point × step.

Watch out: update the slope at each new point.

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