Mathematica Improved Euler's Method

Question

Population growth in Mathematica with NDSolve:

f[r_, a_, T_] := NDSolve[{x' t] == r*x[t]*(1 - x[t]), x[0] == a}, x, {t, 0, T}]

s1 = f[0.1, 0.5, 30];
s2 = f[0.1, 2, 30];

Plot[{x[t] /. s1, x[t] /. s2, 1}, {t, 0, 30}, PlotRange -> {0, 2}]
Plot[Evaluate[Table[x[t] /. f[0.3, 0.2 n, 10], {n, 1, 10}]], PlotRange -> {0, 2}]

How can we solve it with Improved Euler's method?

P[q_, h_, N_] := (
  u[0] = 1; 
  Do[u[n + 1] = u[n] + h*f[n*h + (h*q/2), u[n] + (h*q/2)*f[n*h, u[n]]], {n, 0, N}]
 )

f[x_, t_] := r*x[t]*(1 - x[t])

Answer 1

generate Improved Euler steps:

makeTableRk2Sub[h_, from_, to_, y0_] := 
 Module[{nSteps = Round[(to - from)/h], data, t, y, k1, k2, predictor,
    tbl},
  Array[y, nSteps, 0];
  Array[t, nSteps, 0];
  y[0] = y0; t[0] = from;

  Do[(*Improved Euler loop*)
   k1 = f[t[n], y[n]];
   predictor = y[n] + h k1;
   t[n + 1] = t[n] + h;
   k2 = f[t[n + 1], predictor];
   y[n + 1] = y[n] + h (1/2*(k1 + k2)),
   {n, 0, nSteps}
   ];
  tbl = Table[{t[n], y[n]}, {n, 0, nSteps}]
  ]

To use (red is NDSolve and blue is improved Euler with large step size)

f[t_, y_] := t^2 + y^2; 
h = 0.3; from = 0; to = 5*h; y0 = 0;
tbl = makeTableRk2Sub[h, from, to, y0];
sol = NDSolve[{y'[t] == t^2 + y[t]^2, y[0] == 0}, y, {t, from, to}];
p1 = ListLinePlot[tbl, Mesh -> All];
p2 = Plot[Evaluate[y[t] /. sol], {t, from, to}, PlotStyle -> Red, 
   Frame -> True, FrameLabel -> {{"y(t)", None}, {"t", 
      "Comparing Improved Euler to NDSolve"}}, GridLines -> Automatic,
    GridLinesStyle -> LightGray, BaseStyle -> 14];

Show[Legended[p2, Style["NDsolve", Red]],Legended[p1, Style["Improved Euler", Blue]]]