The OP seems maybe to want `X`

as a function of `x`

, but the OP gives up an equation that defines `X`

as a function of `t`

. No good clue about how to incorporate `x`

into the solution `X(t)`

as defined by the given equation.

Here’s one way to go about finding `X(t)`

numerically. The equation has a 3-parameter family of solutions, so we seek an order-3 differential equation for the family. This turns out to be easy. Next we need to find initial conditions for `X(t0)`

, `X'(t0)`

, `X''(t0)`

in terms of the parameters `a`

, `b`

, `B`

. This is easy if `t0 = 0`

. We get two solution components `psolM`

/`psolP`

because the initial condition for `X(0)`

is a quadratic equation (`M`

for minus √, `P`

for plus √).

```
eqn = X(t) (X(t) - a) + b Exp(-2 X(t) t) == B;
sys = NestList(D(#, t) &, eqn, 3);
ode = Eliminate(sys, {a, b, B});
ics = Solve(Most@sys /. t -> 0,
NestList(D(#, t) &, X(t), 2) /. t -> 0);
Length@ics (* number of solutions = 2 *)
(* 2 *)
{psolM, psolP} =
ParametricNDSolveValue({ode, #}, X, {t, -2, 2}, {a, b, B},
Method -> "StiffnessSwitching") & /@ (ics /. Rule -> Equal);
nosol = Verbatim(ParametricFunction)(___)(___) ->
Interpolation({{-2., Indeterminate}, {2., Indeterminate}},
InterpolationOrder -> 1);
Manipulate(
Quiet@ListLinePlot({psolM(a, b, B), psolP(a, b, B)} /. nosol,
InterpolationOrder -> 3, PlotRange -> {{-2, 2}, {-5, 5}},
PlotLabel ->
Pane(Style($MessageList, "Label"), {320, 60},
Alignment -> Center)),
{{a, 1}, -4, 4, Appearance -> "Labeled"},
{b, 1, 5, Appearance -> "Labeled"},
{{B, 4}, 1, 5, Appearance -> "Labeled"},
AutorunSequencing -> {{1, 3}, {2, 3}, {3, 3}}
)
```