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Any help how in Mathematica to solve

y’’[t]+1/2 y’[t]^2 - 1/2 Exp[y[t]]+2 ==0

To get y(t), with no knowledge by any initial conditions for y(t) or y’(t) ?

Edit: I’m making these solutions to reach for initial constraints on y[t] , like y[0], or y’[0] at arbitrary integration constants values.

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    $\begingroup$ What have you tried so far? Please at least write the equation in Mathematica syntax. $\endgroup$
    – Szabolcs
    Apr 20, 2019 at 19:50
  • $\begingroup$ Related: (195654) $\endgroup$
    – Michael E2
    Apr 20, 2019 at 19:54
  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – Kuba
    Apr 21, 2019 at 6:08
  • $\begingroup$ @Kuba, please I want to report replies from one of the members here ( Michael E2) on this thread: mathematica.stackexchange.com/questions/207479/… . His replies are offensive, not polit, and not helpful. $\endgroup$
    – S.S.
    Oct 8, 2019 at 3:23

1 Answer 1

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You can use the parametric function

eq = y''[t] + 1/2 y'[t]^2 - 1/2 Exp[y[t]] + 2 == 0;
ic = {y[0] == a, y'[0] == b};
sol = ParametricNDSolveValue[{eq, ic}, y, {t, 0, 1}, {a, b}]

Table[Plot[Evaluate[Table[sol[a, b][t], {a, -1, 1, 0.5}]], {t, 0, 1}, 
  AxesLabel -> {"t", "y"}, PlotLabel -> Row[{"b = ", b}], 
  PlotLegends -> Automatic], {b, -1, 1, 1}]

fig1

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