NOTE
Concerning the transformation to an ODE as
$$ e^{\beta t} \left(\beta ^2 \left(c_0-\epsilon (t)^3\right) \epsilon '(t)+\beta \left(\epsilon (t)^3-c_0\right) \epsilon ''(t)-\beta \left(4 \epsilon (t)^2-2 \epsilon (t)+1\right) \epsilon '(t)^2+2 \epsilon '(t)^3\right)=0 $$
and solving asAnd now follows with the initial conditions extracted from the previousa layman solution $\epsilon_2(t)]$(second order splines) to this problem.
ϵ0fl[a_, =b_, 0.2761;
dϵ0c_, =delta_, 0.3942`;
solt_, n_] := NDSolve[{(E^Sum[(tUnitStep[t β)- (β^2k (c0delta] - ϵ[t]^3)ϵ'[t]UnitStep[t - β (1 - 2 ϵ[t]k + 41) ϵ[t]^2delta]) ϵ'[t]^2(a[k] + 2b[k] ϵ'[t]^3(t - k delta) + βc[k] (-c0t +- ϵ[t]^3)k ϵ''[t]delta)/.parms^2), =={k, 0, ϵ[0]n}]
n === ϵ0,40;
tmax ϵ'[0]= ==10;
delta dϵ0= tmax/4}n;
fl0 = fl[a, ϵb, {tc, 0delta, 5}][[1]];t, n];
ϵtFor[k = Evaluate[ϵ[t]n /.- sol];1, k >= 0, k--,
ϵs fl0 = ϵtfl0 /. {ta[k + 1] -> s};
gr2a[k] =+ Plot[ϵt,delta {t,b[k] 0,+ 5}delta^2 c[k], PlotStyleb[k + 1] -> b[k] + 2 delta c[k]}]
vars = Join[{Thicka[0], Blueb[0]}, PlotRangeTable[c[k], ->{k, All]0, n}]];
Note that the initial condition for $\epsilon'(0)$ was reduced to be solved.
This solution has an error regarding the integral equation shown as follows
dif2
Clear[ϵ]
ϵ[t_] := fl0;
dif = (β*Integrate[ϵsβ*Integrate[ϵ[s]/E^((t - s)*β), {s, 0, t}]) - β*Integrate[ϵs^2β*Integrate[ϵ[s]^2/E^((t - s)*β), {s, 0, t}] - β*ϵt*Integrate[β*ϵ[t]*Integrate[(ϵs^2ϵ[s]^2)/E^((t- s)*β), {s, 0, t}] - ϵ[t] + ϵt^2ϵ[t]^2 + c0;
parms = {c0 -> ϵt0.20, β -> 0.2};
dif0 = dif /. parms;
Plot[dif2points = Table[dif0, {t, 0, 5tmax, delta/2}];
npts = Length[points];
diag = Table[(npts - k + 1), PlotStyle{k, 1, npts}];
obj = points.DiagonalMatrix[diag].points;
sol = NMinimize[obj, vars, Method -> "DifferentialEvolution"];
sol[[1]]
et = fl0 /. sol[[2]];
Plot[et, {Thickt, Blue0, tmax}, PlotRange -> All]All, PlotStyle -> {Thick, Blue}]
NOTE
