Räknare för Eulers metod

Bestäm $$$y{\left(1 \right)}$$$ för $$$y^{\prime }\left(t\right) = t y$$$, givet $$$y{\left(0 \right)} = 3$$$, $$$h = \frac{1}{5}$$$, med Eulers metod.

Eulers metod anger att $$$y_{n+1} = y_{n} + h\cdot f{\left(t_{n},y_{n} \right)}$$$, där $$$t_{n+1} = t_{n} + h$$$.

Vi har att $$$h = \frac{1}{5}$$$, $$$t_{0} = 0$$$, $$$y_{0} = 3$$$ och $$$f{\left(t,y \right)} = t y$$$.

Steg 1

$$$t_{1} = t_{0} + h = 0 + \frac{1}{5} = \frac{1}{5}$$$

$$$y_{1} = y{\left(t_{1} \right)} = y{\left(\frac{1}{5} \right)} = y_{0} + h\cdot f{\left(t_{0},y_{0} \right)} = 3 + h\cdot f{\left(0,3 \right)} = 3 + \frac{1}{5} \cdot 0 = 3$$$

Steg 2

$$$t_{2} = t_{1} + h = \frac{1}{5} + \frac{1}{5} = \frac{2}{5}$$$

$$$y_{2} = y{\left(t_{2} \right)} = y{\left(\frac{2}{5} \right)} = y_{1} + h\cdot f{\left(t_{1},y_{1} \right)} = 3 + h\cdot f{\left(\frac{1}{5},3 \right)} = 3 + \frac{1}{5} \cdot 0.6 = 3.12$$$

Steg 3

$$$t_{3} = t_{2} + h = \frac{2}{5} + \frac{1}{5} = \frac{3}{5}$$$

$$$y_{3} = y{\left(t_{3} \right)} = y{\left(\frac{3}{5} \right)} = y_{2} + h\cdot f{\left(t_{2},y_{2} \right)} = 3.12 + h\cdot f{\left(\frac{2}{5},3.12 \right)} = 3.12 + \frac{1}{5} \cdot 1.248 = 3.3696$$$

Steg 4

$$$t_{4} = t_{3} + h = \frac{3}{5} + \frac{1}{5} = \frac{4}{5}$$$

$$$y_{4} = y{\left(t_{4} \right)} = y{\left(\frac{4}{5} \right)} = y_{3} + h\cdot f{\left(t_{3},y_{3} \right)} = 3.3696 + h\cdot f{\left(\frac{3}{5},3.3696 \right)} = 3.3696 + \frac{1}{5} \cdot 2.02176 = 3.773952$$$

Steg 5

$$$t_{5} = t_{4} + h = \frac{4}{5} + \frac{1}{5} = 1$$$

$$$y_{5} = y{\left(t_{5} \right)} = y{\left(1 \right)} = y_{4} + h\cdot f{\left(t_{4},y_{4} \right)} = 3.773952 + h\cdot f{\left(\frac{4}{5},3.773952 \right)} = 3.773952 + \frac{1}{5} \cdot 3.0191616 = 4.37778432$$$

$$$y{\left(1 \right)}\approx 4.37778432$$$A