-1
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$$ \frac{\mathrm du}{\mathrm dt} = 1 - u \mathrm e^{\epsilon(8q-1)} $$ $$ \frac{\mathrm dq}{\mathrm dt} = u \mathrm e^{\epsilon (q-1)} - q $$

$ 0 \leq \epsilon \leq 0.1 $

$u(0) = 0$ and $ q(0)=0$

I try whit this but I can't

https://reference.wolfram.com/language/tutorial/DSolveSystemsOfNonlinearODEs.html

system = {u'[t] + u[t] E^(\[Epsilon] (q - 1)) == 1, u[0] == 0, 
   q'[t] - u[t] E^(\[Epsilon] (q - 1)) + q[t] == 0, q[0] == 0};


sol = DSolve[system, {u, q}, t]

enter image description here

enter image description here

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  • $\begingroup$ Can you provide some of the code you've tried to solve this with so far? $\endgroup$ – user6014 Oct 22 '18 at 22:28
  • $\begingroup$ Sure system = {u'[t] + u[t] E^([Epsilon] (q - 1)) == 1, u[0] == 0, q'[t] - u[t] E^([Epsilon] (q - 1)) + q[t] == 0, q[0] == 0}; sol = DSolve[system, {u, q}, t] $\endgroup$ – Conan Oct 22 '18 at 22:36
  • 2
    $\begingroup$ Use q[t] instead of q in your equations $\endgroup$ – Carl Woll Oct 22 '18 at 22:36
  • $\begingroup$ look at the images, it did not work. $\endgroup$ – Conan Oct 22 '18 at 22:55
  • 1
    $\begingroup$ Clear[q], and use q[0] == 0 instead of q[0] = 0. $\endgroup$ – AccidentalFourierTransform Oct 22 '18 at 23:17
2
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Try this:

Clear[Evaluate[Context[] <> "*"]];
Clear[sol, u, q];
Manipulate[
 sol = NDSolve[{
    u'[t] + u[t]*Exp[\[Epsilon] (q[t] - 1)] == 1,
    q'[t] - u[t]*Exp[\[Epsilon] (q[t] - 1)] + q[t] == 0,
    q[0] == 0,
    u[0] == 0
    },
   {u, q},
   {t, 0, 60}];
 Plot[{u[t],q[t]} /.sol, {t, 0, 60}],
 {{\[Epsilon], 0}, 0, 1}
 ]

enter image description here

EDIT 1

Can you try this? It takes a long time to solve it in my computer. Note that in Mathematica e is represented by Exp[].

DSolve[{
  u'[t] + u[t]*Exp[\[Epsilon] (q[t] - 1)] == 1,
  q'[t] - u[t]*Exp[\[Epsilon] (q[t] - 1)] + q[t] == 0,
  q[0] == 0,
  u[0] == 0
  },
 {u, q}, t]
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  • $\begingroup$ thanks soo much, i try whir Ruge Kutta Method $$E7 = NDSolve[{u'[t] + u[t] E^(\[Epsilon] (q[t] - 1)) - 1 == 0, u[0] == 0, q'[t] - u[t] E^(\[Epsilon] (q[t] - 1)) + q[t] == 0, q[0] == 0}, {u, q}, {t, 0, 6}, Method -> {"ExplicitRungeKutta", "DifferenceOrder" -> 4}];$$ $\endgroup$ – Conan Oct 25 '18 at 22:49

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