The built-in "ExplicitRungeKutta"
method of order 4 is an "embedded" method (i.e., with an embedded error estimation method).
Here is the classical method (with code cribbed from this answer by J.M.):
ClassicalRungeKuttaCoefficients[4, prec_] :=
With[{amat = {{1/2}, {0, 1/2}, {0, 0, 1}},
bvec = {1/6, 1/3, 1/3, 1/6}, cvec = {1/2, 1/2, 1}},
N[{amat, bvec, cvec}, prec]];
Here is a comparison with the built-in method:
NDSolve`EmbeddedExplicitRungeKuttaCoefficients[4, MachinePrecision]
ClassicalRungeKuttaCoefficients[4, MachinePrecision]
(*
{{{0.4`},
{-0.15`, 0.75`},
{0.4318181818181818`, -0.3409090909090909`, 0.9090909090909091`},
{0.1527777777777778`, 0.3472222222222222`, 0.3472222222222222`, 0.1527777777777778`}},
{0.1527777777777778`, 0.3472222222222222`, 0.3472222222222222`, 0.1527777777777778`, 0.`},
{0.4`, 0.6`, 1.`, 1.`},
{0.013269665336144196`, -0.06634832668072098`, 0.06634832668072098`,
0.14596631869758617`, -0.15923598403373035`}}
{{{0.5`},
{0.`, 0.5`},
{0.`, 0.`, 1.`}},
{0.16666666666666666`, 0.3333333333333333`, 0.3333333333333333`, 0.16666666666666666`},
{0.5`, 0.5`, 1.`}}
*)
Here's a way to get the solution steps, based on some of the links I put in a comment above:
vf = v /.
First@NDSolve[{v'[t] + (v'[t])^3 - (t - 1) == 0, v[0] == 1}, {v}, {t, 0, 1},
Method -> {"FixedStep",
Method -> {"ExplicitRungeKutta", "DifferenceOrder" -> 4,
"Coefficients" -> ClassicalRungeKuttaCoefficients}},
StartingStepSize -> 1/10];
Transpose@Flatten[vf[{"Coordinates", {"ValuesOnGrid"}}], 1]
TableForm[%, TableHeadings -> {Range[0, 10], {t, v}}]
(*
{{0., 1.}, {0.1, 0.933905}, {0.2, 0.872309}, {0.3, 0.815591}, {0.4, 0.764202},
{0.5, 0.718679}, {0.6, 0.679664}, {0.7, 0.647912}, {0.8, 0.624276},
{0.9, 0.609622}, {1., 0.604647}}
*)

MaxStepFraction
is1/10
. Pass the optionMaxStepFraction -> 1
and you should be able to get whatever step size you like $\endgroup$ – Michael E2 Apr 6 at 12:41