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I have the following problem to solve it is a system of two ode's. I'm given that

q=2/Sqrt[1 + 1600 y]

and then the ode's are as follows.

[x][t] == -0.1 x + 1/(1 + (5 y - (10 y)/Sqrt[1 + 1600 y])^2).
[y][t] == 0.05 x - 0.1 y - (0.03 y + (20 y)/Sqrt[1 + 1600 y])/(0.05 + y).

Since I didn't know how to make mathematica plug in q into the ode's I did it by by copying and pasting i don't know if this is the right method.

equation1 = {x'[t] == (1/(1 + (5*y[t] - 5*(2/(1 + Sqrt[1 + 1600*y[t]]))*y[t])^2)) - 0.1*x[t], y'[t] == 0.05*x[t] - ((10*y[t]*(2/(1 + Sqrt[1 + 1600*y[t]])) + 0.03*y[t])/(0.05 + y[t])) - 0.1*y[t], x[0] == 0, y[0] == 0};
{sx, sy}== NDSolveValue[equation1, {x, y}, {t, 0, 50}]
Plot[{sx[t], sy[t]}, {t, 0, 50}]

and i do get a plot but it is not the correct plot the plot i get is also attached and i have also attached the plot which i am supposed to get.

This is what i need to get:

This is what i need to get

this is what my code gives me:

this is what my code gives me.

I hope this is easy to read and I'm still new to posting questions here and new to mathematica i'm sorry if these seems like a dumb question but i have been stuck on this for a while now. Also if you could give me hints on how i would start a pot of the nullclines and phase portrait anyway thank you very much for any help.

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If I change one of your coefficients, I get the example plot:

equation1 = {x'[
     t] == (1/(1 + (5*y[t] - 
            5*(2/(1 + Sqrt[1 + 1600*y[t]]))*y[t])^2)) - 0.1*x[t], 
   y'[t] ==(*0.05*)0.5*x[t] -                                     (*  <-- N.B.  *)
       ((10*y[t]*(2/(1 + Sqrt[1 + 1600*y[t]])) + 
         0.03*y[t])/(0.05 + y[t])) - 0.1*y[t], x[0] == 0, y[0] == 0};
{sx, sy} = NDSolveValue[equation1, {x, y}, {t, 0, 50}];
Plot[{sx[t], sy[t]}, {t, 0, 50}]

Mathematica graphics

Maybe the 0.05 is a typo?

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  • 1
    $\begingroup$ wow thank you so much i feel like an idiot $\endgroup$ – pkid58 May 5 '15 at 2:57
  • $\begingroup$ @pkid58 You're welcome. $\endgroup$ – Michael E2 May 5 '15 at 2:58

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