Euler's method initial value problem — how should I plot the solution?












2












$begingroup$


We are given the following initial value problem:



$qquad y' = t,e^{3,t} -2,y, quad 0 le t le 1, quad y(0)=0$



Use Euler`s method to approximate the solution to this initial value problem. In this case, the step size is $h=0.5$.



Plot the approximating solution $y_k$ along with the exact solution $y(t)$.










share|improve this question











$endgroup$












  • $begingroup$
    Are you sure that you are in the right forum? This one is about the Mathematica software. Perhaps this might fit better in math.stackexchange.com?
    $endgroup$
    – MarcoB
    Mar 24 at 1:36










  • $begingroup$
    Yes, plot the calculated $y$ vs $t$ at the computed points. Below is an example computed in Mathematica. $n=20$ steps.
    $endgroup$
    – mjw
    Mar 24 at 2:18






  • 2




    $begingroup$
    A related question.
    $endgroup$
    – J. M. is slightly pensive
    Mar 24 at 2:41










  • $begingroup$
    @J.M., Yes, your method there (at the linked question) is a nice way to go about it!
    $endgroup$
    – mjw
    Mar 24 at 5:27
















2












$begingroup$


We are given the following initial value problem:



$qquad y' = t,e^{3,t} -2,y, quad 0 le t le 1, quad y(0)=0$



Use Euler`s method to approximate the solution to this initial value problem. In this case, the step size is $h=0.5$.



Plot the approximating solution $y_k$ along with the exact solution $y(t)$.










share|improve this question











$endgroup$












  • $begingroup$
    Are you sure that you are in the right forum? This one is about the Mathematica software. Perhaps this might fit better in math.stackexchange.com?
    $endgroup$
    – MarcoB
    Mar 24 at 1:36










  • $begingroup$
    Yes, plot the calculated $y$ vs $t$ at the computed points. Below is an example computed in Mathematica. $n=20$ steps.
    $endgroup$
    – mjw
    Mar 24 at 2:18






  • 2




    $begingroup$
    A related question.
    $endgroup$
    – J. M. is slightly pensive
    Mar 24 at 2:41










  • $begingroup$
    @J.M., Yes, your method there (at the linked question) is a nice way to go about it!
    $endgroup$
    – mjw
    Mar 24 at 5:27














2












2








2


1



$begingroup$


We are given the following initial value problem:



$qquad y' = t,e^{3,t} -2,y, quad 0 le t le 1, quad y(0)=0$



Use Euler`s method to approximate the solution to this initial value problem. In this case, the step size is $h=0.5$.



Plot the approximating solution $y_k$ along with the exact solution $y(t)$.










share|improve this question











$endgroup$




We are given the following initial value problem:



$qquad y' = t,e^{3,t} -2,y, quad 0 le t le 1, quad y(0)=0$



Use Euler`s method to approximate the solution to this initial value problem. In this case, the step size is $h=0.5$.



Plot the approximating solution $y_k$ along with the exact solution $y(t)$.







plotting differential-equations homework






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Mar 28 at 16:51









mjw

1,24810




1,24810










asked Mar 24 at 0:38









KevinKevin

111




111












  • $begingroup$
    Are you sure that you are in the right forum? This one is about the Mathematica software. Perhaps this might fit better in math.stackexchange.com?
    $endgroup$
    – MarcoB
    Mar 24 at 1:36










  • $begingroup$
    Yes, plot the calculated $y$ vs $t$ at the computed points. Below is an example computed in Mathematica. $n=20$ steps.
    $endgroup$
    – mjw
    Mar 24 at 2:18






  • 2




    $begingroup$
    A related question.
    $endgroup$
    – J. M. is slightly pensive
    Mar 24 at 2:41










  • $begingroup$
    @J.M., Yes, your method there (at the linked question) is a nice way to go about it!
    $endgroup$
    – mjw
    Mar 24 at 5:27


















  • $begingroup$
    Are you sure that you are in the right forum? This one is about the Mathematica software. Perhaps this might fit better in math.stackexchange.com?
    $endgroup$
    – MarcoB
    Mar 24 at 1:36










  • $begingroup$
    Yes, plot the calculated $y$ vs $t$ at the computed points. Below is an example computed in Mathematica. $n=20$ steps.
    $endgroup$
    – mjw
    Mar 24 at 2:18






  • 2




    $begingroup$
    A related question.
    $endgroup$
    – J. M. is slightly pensive
    Mar 24 at 2:41










  • $begingroup$
    @J.M., Yes, your method there (at the linked question) is a nice way to go about it!
    $endgroup$
    – mjw
    Mar 24 at 5:27
















$begingroup$
Are you sure that you are in the right forum? This one is about the Mathematica software. Perhaps this might fit better in math.stackexchange.com?
$endgroup$
– MarcoB
Mar 24 at 1:36




$begingroup$
Are you sure that you are in the right forum? This one is about the Mathematica software. Perhaps this might fit better in math.stackexchange.com?
$endgroup$
– MarcoB
Mar 24 at 1:36












$begingroup$
Yes, plot the calculated $y$ vs $t$ at the computed points. Below is an example computed in Mathematica. $n=20$ steps.
$endgroup$
– mjw
Mar 24 at 2:18




$begingroup$
Yes, plot the calculated $y$ vs $t$ at the computed points. Below is an example computed in Mathematica. $n=20$ steps.
$endgroup$
– mjw
Mar 24 at 2:18




2




2




$begingroup$
A related question.
$endgroup$
– J. M. is slightly pensive
Mar 24 at 2:41




$begingroup$
A related question.
$endgroup$
– J. M. is slightly pensive
Mar 24 at 2:41












$begingroup$
@J.M., Yes, your method there (at the linked question) is a nice way to go about it!
$endgroup$
– mjw
Mar 24 at 5:27




$begingroup$
@J.M., Yes, your method there (at the linked question) is a nice way to go about it!
$endgroup$
– mjw
Mar 24 at 5:27










2 Answers
2






active

oldest

votes


















6












$begingroup$

Euler's method to solve, plot, and compare to known solution:



$y^prime = te^{3t}-2y, quad y(0)=0, quad 0 le t le 1. $



y = 0; t = 0.0;
n = 20; h = 1/n;
f[y_, t_] := t Exp[3 t] - 2 y;
ξ = {y};
Do[(
y = y + f[y, t] h;
t = t + h;
ξ = Join[ξ, {y}]
), n
]

p = Transpose[{Range[0, n]/n, ξ}];

Clear[y, t];
DSolve[{y'[t] == t Exp[3 t] - 2 y[t], y[0] == 0}, y[t], t]

q = Plot[Evaluate[y[t] /. %], {t, 0, 1}, PlotStyle -> Gray];

Show[q, ListPlot[p, PlotStyle -> Blue]]


Here is the output from DSolve:



{{y[t] -> 1/25 E^(-2 t) (1 - E^(5 t) + 5 E^(5 t) t)}}


Or in plain English, the solution is:



$y(t) = frac{1}{25} (e^{-2 t} - e^{3t}+5t e^{3t})$



Here is the plot of the computed points along with the known solution (computed with DSolve above).



enter image description here



With $h=0.5$, Euler's method does not do so well:



enter image description here



Again, for $h=1/20$, we can compute y[t] at each approximation point and store it in the list $eta$:



m = DSolve[{y'[t] == t Exp[3 t] - 2 y[t], y[0] == 0}, y[t], t];
[Eta] = y[t] /. m /. t -> Range[0.0, n]/n;


Then we can tabulated $t, xi, eta$ (The list $xi$ contains the Euler approximations).



TableForm[Transpose@Join[Transpose@N@p, [Eta]]]


$begin{array}{lll}
0.00 &0.00000000 &0.00000000\
0.05 &0.00000000 &0.00133847\
0.10 &0.00290459 &0.00575205\
0.15 &0.00936342 &0.01394960\
0.20 &0.02018940 &0.02681280\
0.25 &0.03639170 &0.04543120\
0.30 &0.05921500 &0.07114450\
0.35 &0.09018750 &0.10559300\
0.40 &0.13117800 &0.15077800\
0.45 &0.18446200 &0.20913400\
0.50 &0.25280800 &0.28361700\
0.55 &0.33957000 &0.37780300\
0.60 &0.44880500 &0.49602000\
0.65 &0.58541300 &0.64348300\
0.70 &0.75530400 &0.82648100\
0.75 &0.96559000 &1.05258000\
0.80 &1.22482000 &1.33086000\
0.85 &1.54327000 &1.67223000\
0.90 &1.93324000 &2.08977000\
0.95 &2.40951000 &2.59915000\
1.00 &2.98972000 &3.21910000\
end{array}$



We can also plot the error $ = eta - xi$:



ListPlot[Transpose@{Range[0, n]/n, Flatten@[Eta] - [Xi]}, 
Axes -> False, Frame -> True,
FrameLabel -> {t, "Error=y[t]-Euler Approx."}]


enter image description here






share|improve this answer











$endgroup$





















    2












    $begingroup$

    forwardEuler[{t_, y_}] := {t + h, y + h f[t, y]} // N

    f[t_, y_] = t*Exp[3 t] - 2 y;
    h = 0.5;
    steps = Floor[(1 - 0)/h]; (*0 <= t <=1 *)
    ics = {0, 0};


    With h=0.5 you get only 3 points,



    NestList[forwardEuler, ics, steps]
    {{0, 0}, {0.5, 0.}, {1., 1.12042}}


    You have to choose more steps!



    h = 0.05;
    steps = Floor[(1 - 0)/h];
    euler = NestList[forwardEuler, ics, steps];
    ListLinePlot[euler, GridLines -> Automatic]


    enter image description here






    share|improve this answer









    $endgroup$













    • $begingroup$
      Looking at your solution, I realized that I updated $t_{n+1} = t_n +h$ too early. I've made the correction in my answer. Thank you very much!
      $endgroup$
      – mjw
      Mar 24 at 13:40












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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    6












    $begingroup$

    Euler's method to solve, plot, and compare to known solution:



    $y^prime = te^{3t}-2y, quad y(0)=0, quad 0 le t le 1. $



    y = 0; t = 0.0;
    n = 20; h = 1/n;
    f[y_, t_] := t Exp[3 t] - 2 y;
    ξ = {y};
    Do[(
    y = y + f[y, t] h;
    t = t + h;
    ξ = Join[ξ, {y}]
    ), n
    ]

    p = Transpose[{Range[0, n]/n, ξ}];

    Clear[y, t];
    DSolve[{y'[t] == t Exp[3 t] - 2 y[t], y[0] == 0}, y[t], t]

    q = Plot[Evaluate[y[t] /. %], {t, 0, 1}, PlotStyle -> Gray];

    Show[q, ListPlot[p, PlotStyle -> Blue]]


    Here is the output from DSolve:



    {{y[t] -> 1/25 E^(-2 t) (1 - E^(5 t) + 5 E^(5 t) t)}}


    Or in plain English, the solution is:



    $y(t) = frac{1}{25} (e^{-2 t} - e^{3t}+5t e^{3t})$



    Here is the plot of the computed points along with the known solution (computed with DSolve above).



    enter image description here



    With $h=0.5$, Euler's method does not do so well:



    enter image description here



    Again, for $h=1/20$, we can compute y[t] at each approximation point and store it in the list $eta$:



    m = DSolve[{y'[t] == t Exp[3 t] - 2 y[t], y[0] == 0}, y[t], t];
    [Eta] = y[t] /. m /. t -> Range[0.0, n]/n;


    Then we can tabulated $t, xi, eta$ (The list $xi$ contains the Euler approximations).



    TableForm[Transpose@Join[Transpose@N@p, [Eta]]]


    $begin{array}{lll}
    0.00 &0.00000000 &0.00000000\
    0.05 &0.00000000 &0.00133847\
    0.10 &0.00290459 &0.00575205\
    0.15 &0.00936342 &0.01394960\
    0.20 &0.02018940 &0.02681280\
    0.25 &0.03639170 &0.04543120\
    0.30 &0.05921500 &0.07114450\
    0.35 &0.09018750 &0.10559300\
    0.40 &0.13117800 &0.15077800\
    0.45 &0.18446200 &0.20913400\
    0.50 &0.25280800 &0.28361700\
    0.55 &0.33957000 &0.37780300\
    0.60 &0.44880500 &0.49602000\
    0.65 &0.58541300 &0.64348300\
    0.70 &0.75530400 &0.82648100\
    0.75 &0.96559000 &1.05258000\
    0.80 &1.22482000 &1.33086000\
    0.85 &1.54327000 &1.67223000\
    0.90 &1.93324000 &2.08977000\
    0.95 &2.40951000 &2.59915000\
    1.00 &2.98972000 &3.21910000\
    end{array}$



    We can also plot the error $ = eta - xi$:



    ListPlot[Transpose@{Range[0, n]/n, Flatten@[Eta] - [Xi]}, 
    Axes -> False, Frame -> True,
    FrameLabel -> {t, "Error=y[t]-Euler Approx."}]


    enter image description here






    share|improve this answer











    $endgroup$


















      6












      $begingroup$

      Euler's method to solve, plot, and compare to known solution:



      $y^prime = te^{3t}-2y, quad y(0)=0, quad 0 le t le 1. $



      y = 0; t = 0.0;
      n = 20; h = 1/n;
      f[y_, t_] := t Exp[3 t] - 2 y;
      ξ = {y};
      Do[(
      y = y + f[y, t] h;
      t = t + h;
      ξ = Join[ξ, {y}]
      ), n
      ]

      p = Transpose[{Range[0, n]/n, ξ}];

      Clear[y, t];
      DSolve[{y'[t] == t Exp[3 t] - 2 y[t], y[0] == 0}, y[t], t]

      q = Plot[Evaluate[y[t] /. %], {t, 0, 1}, PlotStyle -> Gray];

      Show[q, ListPlot[p, PlotStyle -> Blue]]


      Here is the output from DSolve:



      {{y[t] -> 1/25 E^(-2 t) (1 - E^(5 t) + 5 E^(5 t) t)}}


      Or in plain English, the solution is:



      $y(t) = frac{1}{25} (e^{-2 t} - e^{3t}+5t e^{3t})$



      Here is the plot of the computed points along with the known solution (computed with DSolve above).



      enter image description here



      With $h=0.5$, Euler's method does not do so well:



      enter image description here



      Again, for $h=1/20$, we can compute y[t] at each approximation point and store it in the list $eta$:



      m = DSolve[{y'[t] == t Exp[3 t] - 2 y[t], y[0] == 0}, y[t], t];
      [Eta] = y[t] /. m /. t -> Range[0.0, n]/n;


      Then we can tabulated $t, xi, eta$ (The list $xi$ contains the Euler approximations).



      TableForm[Transpose@Join[Transpose@N@p, [Eta]]]


      $begin{array}{lll}
      0.00 &0.00000000 &0.00000000\
      0.05 &0.00000000 &0.00133847\
      0.10 &0.00290459 &0.00575205\
      0.15 &0.00936342 &0.01394960\
      0.20 &0.02018940 &0.02681280\
      0.25 &0.03639170 &0.04543120\
      0.30 &0.05921500 &0.07114450\
      0.35 &0.09018750 &0.10559300\
      0.40 &0.13117800 &0.15077800\
      0.45 &0.18446200 &0.20913400\
      0.50 &0.25280800 &0.28361700\
      0.55 &0.33957000 &0.37780300\
      0.60 &0.44880500 &0.49602000\
      0.65 &0.58541300 &0.64348300\
      0.70 &0.75530400 &0.82648100\
      0.75 &0.96559000 &1.05258000\
      0.80 &1.22482000 &1.33086000\
      0.85 &1.54327000 &1.67223000\
      0.90 &1.93324000 &2.08977000\
      0.95 &2.40951000 &2.59915000\
      1.00 &2.98972000 &3.21910000\
      end{array}$



      We can also plot the error $ = eta - xi$:



      ListPlot[Transpose@{Range[0, n]/n, Flatten@[Eta] - [Xi]}, 
      Axes -> False, Frame -> True,
      FrameLabel -> {t, "Error=y[t]-Euler Approx."}]


      enter image description here






      share|improve this answer











      $endgroup$
















        6












        6








        6





        $begingroup$

        Euler's method to solve, plot, and compare to known solution:



        $y^prime = te^{3t}-2y, quad y(0)=0, quad 0 le t le 1. $



        y = 0; t = 0.0;
        n = 20; h = 1/n;
        f[y_, t_] := t Exp[3 t] - 2 y;
        ξ = {y};
        Do[(
        y = y + f[y, t] h;
        t = t + h;
        ξ = Join[ξ, {y}]
        ), n
        ]

        p = Transpose[{Range[0, n]/n, ξ}];

        Clear[y, t];
        DSolve[{y'[t] == t Exp[3 t] - 2 y[t], y[0] == 0}, y[t], t]

        q = Plot[Evaluate[y[t] /. %], {t, 0, 1}, PlotStyle -> Gray];

        Show[q, ListPlot[p, PlotStyle -> Blue]]


        Here is the output from DSolve:



        {{y[t] -> 1/25 E^(-2 t) (1 - E^(5 t) + 5 E^(5 t) t)}}


        Or in plain English, the solution is:



        $y(t) = frac{1}{25} (e^{-2 t} - e^{3t}+5t e^{3t})$



        Here is the plot of the computed points along with the known solution (computed with DSolve above).



        enter image description here



        With $h=0.5$, Euler's method does not do so well:



        enter image description here



        Again, for $h=1/20$, we can compute y[t] at each approximation point and store it in the list $eta$:



        m = DSolve[{y'[t] == t Exp[3 t] - 2 y[t], y[0] == 0}, y[t], t];
        [Eta] = y[t] /. m /. t -> Range[0.0, n]/n;


        Then we can tabulated $t, xi, eta$ (The list $xi$ contains the Euler approximations).



        TableForm[Transpose@Join[Transpose@N@p, [Eta]]]


        $begin{array}{lll}
        0.00 &0.00000000 &0.00000000\
        0.05 &0.00000000 &0.00133847\
        0.10 &0.00290459 &0.00575205\
        0.15 &0.00936342 &0.01394960\
        0.20 &0.02018940 &0.02681280\
        0.25 &0.03639170 &0.04543120\
        0.30 &0.05921500 &0.07114450\
        0.35 &0.09018750 &0.10559300\
        0.40 &0.13117800 &0.15077800\
        0.45 &0.18446200 &0.20913400\
        0.50 &0.25280800 &0.28361700\
        0.55 &0.33957000 &0.37780300\
        0.60 &0.44880500 &0.49602000\
        0.65 &0.58541300 &0.64348300\
        0.70 &0.75530400 &0.82648100\
        0.75 &0.96559000 &1.05258000\
        0.80 &1.22482000 &1.33086000\
        0.85 &1.54327000 &1.67223000\
        0.90 &1.93324000 &2.08977000\
        0.95 &2.40951000 &2.59915000\
        1.00 &2.98972000 &3.21910000\
        end{array}$



        We can also plot the error $ = eta - xi$:



        ListPlot[Transpose@{Range[0, n]/n, Flatten@[Eta] - [Xi]}, 
        Axes -> False, Frame -> True,
        FrameLabel -> {t, "Error=y[t]-Euler Approx."}]


        enter image description here






        share|improve this answer











        $endgroup$



        Euler's method to solve, plot, and compare to known solution:



        $y^prime = te^{3t}-2y, quad y(0)=0, quad 0 le t le 1. $



        y = 0; t = 0.0;
        n = 20; h = 1/n;
        f[y_, t_] := t Exp[3 t] - 2 y;
        ξ = {y};
        Do[(
        y = y + f[y, t] h;
        t = t + h;
        ξ = Join[ξ, {y}]
        ), n
        ]

        p = Transpose[{Range[0, n]/n, ξ}];

        Clear[y, t];
        DSolve[{y'[t] == t Exp[3 t] - 2 y[t], y[0] == 0}, y[t], t]

        q = Plot[Evaluate[y[t] /. %], {t, 0, 1}, PlotStyle -> Gray];

        Show[q, ListPlot[p, PlotStyle -> Blue]]


        Here is the output from DSolve:



        {{y[t] -> 1/25 E^(-2 t) (1 - E^(5 t) + 5 E^(5 t) t)}}


        Or in plain English, the solution is:



        $y(t) = frac{1}{25} (e^{-2 t} - e^{3t}+5t e^{3t})$



        Here is the plot of the computed points along with the known solution (computed with DSolve above).



        enter image description here



        With $h=0.5$, Euler's method does not do so well:



        enter image description here



        Again, for $h=1/20$, we can compute y[t] at each approximation point and store it in the list $eta$:



        m = DSolve[{y'[t] == t Exp[3 t] - 2 y[t], y[0] == 0}, y[t], t];
        [Eta] = y[t] /. m /. t -> Range[0.0, n]/n;


        Then we can tabulated $t, xi, eta$ (The list $xi$ contains the Euler approximations).



        TableForm[Transpose@Join[Transpose@N@p, [Eta]]]


        $begin{array}{lll}
        0.00 &0.00000000 &0.00000000\
        0.05 &0.00000000 &0.00133847\
        0.10 &0.00290459 &0.00575205\
        0.15 &0.00936342 &0.01394960\
        0.20 &0.02018940 &0.02681280\
        0.25 &0.03639170 &0.04543120\
        0.30 &0.05921500 &0.07114450\
        0.35 &0.09018750 &0.10559300\
        0.40 &0.13117800 &0.15077800\
        0.45 &0.18446200 &0.20913400\
        0.50 &0.25280800 &0.28361700\
        0.55 &0.33957000 &0.37780300\
        0.60 &0.44880500 &0.49602000\
        0.65 &0.58541300 &0.64348300\
        0.70 &0.75530400 &0.82648100\
        0.75 &0.96559000 &1.05258000\
        0.80 &1.22482000 &1.33086000\
        0.85 &1.54327000 &1.67223000\
        0.90 &1.93324000 &2.08977000\
        0.95 &2.40951000 &2.59915000\
        1.00 &2.98972000 &3.21910000\
        end{array}$



        We can also plot the error $ = eta - xi$:



        ListPlot[Transpose@{Range[0, n]/n, Flatten@[Eta] - [Xi]}, 
        Axes -> False, Frame -> True,
        FrameLabel -> {t, "Error=y[t]-Euler Approx."}]


        enter image description here







        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited Mar 24 at 19:21

























        answered Mar 24 at 2:17









        mjwmjw

        1,24810




        1,24810























            2












            $begingroup$

            forwardEuler[{t_, y_}] := {t + h, y + h f[t, y]} // N

            f[t_, y_] = t*Exp[3 t] - 2 y;
            h = 0.5;
            steps = Floor[(1 - 0)/h]; (*0 <= t <=1 *)
            ics = {0, 0};


            With h=0.5 you get only 3 points,



            NestList[forwardEuler, ics, steps]
            {{0, 0}, {0.5, 0.}, {1., 1.12042}}


            You have to choose more steps!



            h = 0.05;
            steps = Floor[(1 - 0)/h];
            euler = NestList[forwardEuler, ics, steps];
            ListLinePlot[euler, GridLines -> Automatic]


            enter image description here






            share|improve this answer









            $endgroup$













            • $begingroup$
              Looking at your solution, I realized that I updated $t_{n+1} = t_n +h$ too early. I've made the correction in my answer. Thank you very much!
              $endgroup$
              – mjw
              Mar 24 at 13:40
















            2












            $begingroup$

            forwardEuler[{t_, y_}] := {t + h, y + h f[t, y]} // N

            f[t_, y_] = t*Exp[3 t] - 2 y;
            h = 0.5;
            steps = Floor[(1 - 0)/h]; (*0 <= t <=1 *)
            ics = {0, 0};


            With h=0.5 you get only 3 points,



            NestList[forwardEuler, ics, steps]
            {{0, 0}, {0.5, 0.}, {1., 1.12042}}


            You have to choose more steps!



            h = 0.05;
            steps = Floor[(1 - 0)/h];
            euler = NestList[forwardEuler, ics, steps];
            ListLinePlot[euler, GridLines -> Automatic]


            enter image description here






            share|improve this answer









            $endgroup$













            • $begingroup$
              Looking at your solution, I realized that I updated $t_{n+1} = t_n +h$ too early. I've made the correction in my answer. Thank you very much!
              $endgroup$
              – mjw
              Mar 24 at 13:40














            2












            2








            2





            $begingroup$

            forwardEuler[{t_, y_}] := {t + h, y + h f[t, y]} // N

            f[t_, y_] = t*Exp[3 t] - 2 y;
            h = 0.5;
            steps = Floor[(1 - 0)/h]; (*0 <= t <=1 *)
            ics = {0, 0};


            With h=0.5 you get only 3 points,



            NestList[forwardEuler, ics, steps]
            {{0, 0}, {0.5, 0.}, {1., 1.12042}}


            You have to choose more steps!



            h = 0.05;
            steps = Floor[(1 - 0)/h];
            euler = NestList[forwardEuler, ics, steps];
            ListLinePlot[euler, GridLines -> Automatic]


            enter image description here






            share|improve this answer









            $endgroup$



            forwardEuler[{t_, y_}] := {t + h, y + h f[t, y]} // N

            f[t_, y_] = t*Exp[3 t] - 2 y;
            h = 0.5;
            steps = Floor[(1 - 0)/h]; (*0 <= t <=1 *)
            ics = {0, 0};


            With h=0.5 you get only 3 points,



            NestList[forwardEuler, ics, steps]
            {{0, 0}, {0.5, 0.}, {1., 1.12042}}


            You have to choose more steps!



            h = 0.05;
            steps = Floor[(1 - 0)/h];
            euler = NestList[forwardEuler, ics, steps];
            ListLinePlot[euler, GridLines -> Automatic]


            enter image description here







            share|improve this answer












            share|improve this answer



            share|improve this answer










            answered Mar 24 at 11:40









            rmwrmw

            39817




            39817












            • $begingroup$
              Looking at your solution, I realized that I updated $t_{n+1} = t_n +h$ too early. I've made the correction in my answer. Thank you very much!
              $endgroup$
              – mjw
              Mar 24 at 13:40


















            • $begingroup$
              Looking at your solution, I realized that I updated $t_{n+1} = t_n +h$ too early. I've made the correction in my answer. Thank you very much!
              $endgroup$
              – mjw
              Mar 24 at 13:40
















            $begingroup$
            Looking at your solution, I realized that I updated $t_{n+1} = t_n +h$ too early. I've made the correction in my answer. Thank you very much!
            $endgroup$
            – mjw
            Mar 24 at 13:40




            $begingroup$
            Looking at your solution, I realized that I updated $t_{n+1} = t_n +h$ too early. I've made the correction in my answer. Thank you very much!
            $endgroup$
            – mjw
            Mar 24 at 13:40


















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