This problem can be solved with the Euler wavelets collocation method even in v.8 as follows
J[w_] := 1/(2 + w^2)
eqn = y'[
t] == -Integrate[
y[t1] J[w] Exp[I (0.1 - w) (t1 - t)], {w,
0, \[Infinity]}, {t1, 0, t}]
First step. Let define kernel
by integration on w
Integrate[
Exp[I (-1/10 + w) dt]/(2 + w^2), {w, 0, \[Infinity]},
Assumptions -> {dt > 0}]
(*Out[]= (E^(-((I dt)/
10)) (E^(-Sqrt[2] dt) \[Pi] +
I Sqrt[\[Pi]]
MeijerG[{{1/2}, {}}, {{1/2, 1/2}, {0}}, dt^2/2]))/(2 Sqrt[2])*)
Therefore
kernel[dt_] := (
E^(-((I dt)/
10)) (E^(-Sqrt[2] dt) \[Pi] +
I Sqrt[\[Pi]] MeijerG[{{1/2}, {}}, {{1/2, 1/2}, {0}}, dt^2/2]))/(
2 Sqrt[2])
Second step. Using kernel
and substitution t1 = t s
we transform equation into Fredholm type
eqn = {y'[t] + t NIntegrate[kernel[t - t s] y[t s], {s, 0, 1}] ==
0, y[0] == 1};
Third step. Let define y[t],D[y[t],t]
in the Euler wavelets base as
UE[m_, t_] := EulerE[m, t];
psi[k_, n_, m_, t_] :=
Piecewise[{{2^(k/2) UE[m, 2^k t - 2 n + 1], (n - 1)/2^(k - 1) <=
t < n/2^(k - 1)}, {0, True}}];
PsiE[k_, M_, t_] :=
Flatten[Table[psi[k, n, m, t], {n, 1, 2^(k - 1)}, {m, 0, M - 1}]]
k0 = 3; M0 = 4; With[{k = k0, M = M0},
nn = Length[Flatten[Table[1, {n, 1, 2^(k - 1)}, {m, 0, M - 1}]]]];
dx = 1/(nn); xl = Table[l*dx, {l, 0, nn}]; xcol =
Table[(xl[[l - 1]] + xl[[l]])/2, {l, 2, nn + 1}]; Psijk =
With[{k = k0, M = M0}, PsiE[k, M, t1]]; Int1 =
With[{k = k0, M = M0}, Integrate[PsiE[k, M, t1], t1]];
Psi[y_] := Psijk /. t1 -> y;
int1[y_] := Int1 /. t1 -> y; vary = Array[yy, {nn}];
y[t_] := vary . int1[t] + b0 ; dy[t_] := vary . Psi[t];
Final step. We define system of linear equations and solve it using LinearSolve
int2 = Table[
t NIntegrate[kernel[t - t s] int1[t s], {s, 0, 1}], {t, xcol}];
int0 = Table[t NIntegrate[kernel[t - t s], {s, 0, 1}], {t, xcol}];
eqs = Join[
Table[vary . Psi[xcol[[i]]] + vary . int2[[i]] + b0 int0[[i]] ==
0, {i, nn}], {y[0] == 1}];
var = Join[vary, {b0}]; {vec, mat} = CoefficientArrays[eqs, var];
sol = LinearSolve[mat, -vec];
Visualization
rule = Table[var[[i]] -> sol[[i]], {i, Length[var]}]; Plot[
Evaluate[ReIm[y[t] /. rule]], {t, 0, 1}]
Solve[%, LaplaceTransform[y[t], t, s]]
before doingInverseLaplaceTransform[%, s, t]
. That's not the real problem, though. Since you have two integrals, the solution would involve two Laplace transforms. Note that thew
integral is itself a Laplace transform integral. $\endgroup$t
variables, but I'm out of time for today. $\endgroup$