# Non-linear equation

I have to solve this equation but the problem is X is function of x it means X[x]

$$X(X-a)+ b e^{-2 X t}=B,$$

a,b,B are constants.

How we can I get some result for this equation? Plot as well X(x) but x is also function of t.

• Couldn't find a small x in your equation? Dec 30 '20 at 18:49
• X is X[x] I believe...but I don't understand what the OP wants for a solution....or what they intend to plot with unknown constants. Dec 30 '20 at 18:52
• From the context of the question I think the desired quantity is X(x(t)) and then plot it vs t... This is just a speculation though Dec 30 '20 at 18:58
• People here generally like users to post code as Mathematica code instead of just images or TeX, so they can copy-paste it. It makes it convenient for them and more likely you will get someone to help you. You may find this meta Q&A helpful Dec 30 '20 at 19:52
• You need to formulate better. There is X, x and t. It is not clear what is a function what, and what needs to be found. Dec 31 '20 at 12:21

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}}
]


Mathematica cannot solve your equation for X . Instead try to solve it for t

sol=Solve[X (X - a) + b Exp[-2 X t] == B, t ][[1]] /. C[1] -> 0(*forces real solution*)
(*{t -> Log[b/(B + a X - X^2)]/(2 X)}*)


Now you know t as a function of X. For examplary parameters you can plot the result

ParametricPlot[{t, X} /. sol  /. {a -> 1/5, b -> 1, B -> 3/2}, {X, -2,2}, AxesLabel -> {t, X}]