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yields
: <math>y_{k+1}=y_k-\Delta t y_k^2 \quad \quad \quad(3)\, </math>
for each <math>k=0, 1, \dots, n.</math>
With the backward Euler method
:<math>\frac{y_{k+1}-y_k}{\Delta t} = - y_{k+1}^2</math>
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