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I am attempting to compute an approximation of the solution with the forward Euler method in $[0,1]$ with step lengths $h_{1}= 0.2$, $h_{2}= 0.1$ given the initial value problem below

$$\frac{dy}{dz}=\frac{1}{1+z}-y(z)\quad y(0)=1$$

I am not sure what to do when I am given two step sizes instead of one. I know how to compute it if it was given with a step size. Am I supposed to find out the approximation for two different step sizes? Or is there anything I am missing?

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The problem asks for solving the differential equation twice. Once for the step size of $h=.1 $ and once for the step size of $h= .2$ and compare the results. As you know different step sizes give you different results with the smaller step size smaller error is made .

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  • $\begingroup$ Ah ok I understood completely another thing! $\endgroup$ – gimusi Oct 21 '18 at 15:09
  • $\begingroup$ Thanks a lot for the explanation $\endgroup$ – enes Oct 21 '18 at 17:29
  • $\begingroup$ Thanks for your attention and understanding $\endgroup$ – Mohammad Riazi-Kermani Oct 21 '18 at 18:19
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We can apply the Euler’s method as usual using $h_1$ for the first solution that is

$$y_{i+1}=y_i+h_1F(z_i,y_i)$$

and $h_2$ for the second one that is

$$y_{i+1}=y_i+h_2F(z_i,y_i)$$

in order to compare the results since smaller isbthe step more accurate is the solution.

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