オイラー法計算機

あなたの入力

オイラー法を使用して、 $y{\left(0 \right)} = 3$$h = \frac{1}{5}$ときに$y^{\prime } = t y$ $y{\left(1 \right)}$を見つけます。

解決

オイラー法は、 $y_{n+1} = y_{n} + h\cdot f{\left(t_{n},y_{n} \right)}$ 、ここで$t_{n+1} = t_{n} + h$ます。

$h = \frac{1}{5}$, $t_{0} = 0$$y_{0} = 3$$f{\left(t,y \right)} = t y$あります。

ステップ1

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

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

ステップ2

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

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

ステップ3

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

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

ステップ4

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

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

ステップ5

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

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

答え

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