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I tried to do euler's method with this problem: $\frac{dw}{dt} = (3-w)(w+1),\quad w(0) = 0,\quad \text{and }t\in [0,5].\quad \Delta t = 0.5.$

But am getting weird results like for $y_2 = 5.25$, $y_3= -15.84375$; but checked against the book but this is not correct. Can you please work out some iterations?

$y_2 = 1.5 + (3-1.5)(1.5+1)(1) = 5.25$, for example

1 Answers 1

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Euler's method in this case is $ y_{k+1} = y_k + \Delta t \cdot (3-y_k)(y_k+1) $, $y_0=0$. This gives $y_2 = 3.375$ and $y_3=2.555$. Note that $\Delta t = 0.5$.