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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 $y(t)$.

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).

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

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."}]

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]