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4

(* Define function *) Meq = 3/2; g[t_?NumericQ, μ_?NumericQ, σ_?NumericQ] := Meq NIntegrate[(1 - Exp[-t/τ]) E^(-((-μ + Log[τ])^2/(2 σ^2)))/(Sqrt[2 π] σ τ), {τ, 0, ∞}] data = {{0.995, 0.142}, {3.003, 0.2}, {5.908, 0.25}, {10.525, 0.36}, {13.617, 0.498}, {24.321, 0.616}, {33.917, 0.599}, {47.843, 0.7}, {64.172, 0.835}, {91.353, 1.102}, {126.745, 1.083},...


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You're almost there, I think the main trouble you're facing is that you're not yet familiar enough with list manipulation of Mathematica. Check carefully about how I modify the code: int[expr_, {t_, L_, R_, step_}] := step Total[Table[expr, {t, L + step, R, step}]] step = 1/100; bL = 0; bR = 2; grid = Table[i, {i, bL, bR, step}]; eq = {y1''[t] + t^2 y1[t] ...


0

There is an exact solution that coincides with the numerical solution @UlrichNeumann at the first iterations: y[0] = 4 - 3 t; $Assumptions = a > 0 && b > 0; Table[ y[n] = With[{y = y[n - 1]}, 4 - 3 x + a*Integrate[ t (1 - x) (D[y /. x -> t, {t, 2}] - 3/2 (y /. x -> t)^2), {t, 0, x}] + b* Integrate[ x (1 ...


0

Here an approach to solve the problem using NestList sol = NestList[ Function[y,FunctionInterpolation[4 - 3 x + 0.59489439*Integrate[t (1 - x) (D[y[ t], {t, 2}] - 3/2 y[ t]^2), {t, 0, x}] + 0.594894*Integrate[x (1 - t) (D[y[ t], {t, 2}] - 3/2 y[ t]^2), {t, x, 1}], {x, 0,1 }] ] (* Ende Function *) , 4 - 3 # & (* Startwert *) , 10] The iterated ...


0

Is this what you are looking for? y[0,x_] := 1; y[n_,x_] := 1 + Integrate[y[n - 1,x]^2 + t^2, {t, 0, x}] Table[{n,x,y[n,x]},{n,1,3},{x,.1,1,.1}] which gives you {{{1, 0.1, 1.10033}, {1, 0.2, 1.20267}, {1, 0.3, 1.309}, {1, 0.4, 1.42133}, {1, 0.5, 1.5416}, {1, 0.6, 1.672}, {1, 0.7, 1.81433}, {1, 0.8, 1.97067}, {1, 0.9, 2.143}, {1, 1., 2.33333}}, {{2, 0....


0

x = Table[j/100, {j, 1, 100}]; Since x is a List, Map the integration onto the List. Since you are using inexact numbers, NIntegrate is as good as Integrate and is much faster. {#, Integrate[z*(1 - z)*(0.288554*((1 - #/z)^(0.435 - 1)/ Sqrt[#/z])), {z, #, 1}]} &[1/2] // AbsoluteTiming (* {0.524199, {1/2, 0.0838901}} *) {#, NIntegrate[z*...


2

I'm not sure I've understood your NB, but probably this is what you want: ddQ2[x_] = \!\(\*SubsuperscriptBox[\(\[Integral]\), \(x\), \(1\)]\(z \((1 - z)\)*\((0.288554*\*SuperscriptBox[\((1 - \*FractionBox[\(x\), \(z\)])\), \(0.435 - 1\)]/\*SqrtBox[FractionBox[\(x\), \(z\)]])\) \[DifferentialD]z\)\) Then make ListPlot: ListPlot[ddQ2[#] & /@ Table[j/...


3

For Int1, FullSimplify with the assumption that L is real. Int1 = Integrate[ ((1 - Cos[x])/(L^2 (1 - Cos[x]) + 1)^2) Sin[x], {x, 0, Pi}] // FullSimplify[#, Element[L, Reals]] & (* (-1 + 1/(1 + 2 L^2) + Log[1 + 2 L^2])/L^4 *) Also FullSimplify Int2 Int2 = Integrate[((1 - Cos[x])/(L^2 (1 - Cos[x]) + 1)^2) Sin[x], x] // FullSimplify (* (1/(1 + ...


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