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I have the differential equations

{[x]'[t] == -ι - (Δ1 - 1/(Δ2 - ι/2) - ι/2) ι x[t] + (ι y[t])/(Δ2 - ι/2),
 [y]'[t] == -2 ι + (ι x[t])/(Δ2 - ι/2) - (Δ1 -1/(Δ2 - ι/2) - ι/2) ι y[t]}

I solve them with

sol = DSolve[system, {x, y}, t]

How can I plot the solution obtained in the last line?

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  • $\begingroup$ There are things to do after your question is answered. It's a good idea to stay vigilant for some time, better approaches may come later improving over previous replies. Experienced users may point alternatives, caveats or limitations. New users should test answers before voting and wait 24 hours before accepting the best one. Participation is essential for the site, please come back to do your part tomorrow $\endgroup$
    – rhermans
    Commented May 28, 2017 at 9:49

1 Answer 1

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First correct the equations to x'[t]==... and y'[t]==.... Then I propose the following code:

    sol = DSolve[{x'[
                     t] == -ι - (Δ1 - 
    1/(Δ2 - ι/2) - ι/2) ι x[
   t] + (ι y[t])/(Δ2 - ι/2),
  y'[t] == -2 ι + (ι x[t])/(Δ2 - ι/
     2) - (Δ1 - 
    1/(Δ2 - ι/2) - ι/2) ι y[
   t]}, {x, y}, t]

    {xs[t_, Δ1_, Δ2_, ι_, a_, b_], 
  ys[t_, Δ1_, Δ2_, ι_, a_, 
     b_]} = {x[t], y[t]} /. First@sol /. {C[1] -> a, C[2] -> b}

   Plot[Evaluate[{xs[u], ys[u]} /. 
    u -> Sequence[t, 1, 1, 1, 1, 1]], {t, -1, 1}, 
   PlotStyle -> {Blue, Red}]
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  • $\begingroup$ Dear sir, thanks a lot. $\endgroup$
    – Spin
    Commented Nov 14, 2016 at 7:29
  • $\begingroup$ Dear Sir, not working for [Iota]=[Sqrt]{-1}. You have taken [Iota]=1. Please check it. $\endgroup$
    – Spin
    Commented Nov 14, 2016 at 11:45
  • $\begingroup$ @Aril - use Sqrt[-1] vice [Sqrt]{-1} $\endgroup$
    – Bob Hanlon
    Commented Nov 14, 2016 at 16:51
  • $\begingroup$ Dear Sir, Is it possible to plot without using absolute(Abs)? $\endgroup$
    – Spin
    Commented Nov 15, 2016 at 4:21

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