Re: Graphing diff. eqs. on the 86


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Re: Graphing diff. eqs. on the 86



According to the manual, tPlot is ignored when t is an axis.
The thing is, tMin is the first t to be evaluated, so QI1 (initial
condition) is defined at tMin.  If you set tMax less than tMin with a
negative tStep, you can see the graph in the other dierction.  But you can't
(in the same equation) go in both directions from your initial condition.
It's called the INITIAL condition, after all.
Now, what you can do, set tMin=0 tMax=-10 tStep=-.2 QI1=0 and trace to t=-10.
Save that Q1 value at t=-10 Q1->A.  Then set tMin=-10 tMax=10 tStep=.2
QI1=A (the old Q1 when t=-10).  This way, t=10 is close to Q1=0.  Just
mentally subtract 10 from t and you have it.

>Hi fellows I hope you can help me,
>
>                                dM
>given the diff.eq.    ----- = 1 - 0.29M    and knowing that M(0)=0
>                                 dt
>
>I translate it to Q'1=1 - 0.29Q1 and enters the initcondition tmin=0, Q|1=0.
>Now the problem is that it only plots for t >= 0. How do I plot the entire
>area -10 < t < 10?
>I tried changing tplot to -10, no help! I also tried changing tmin to -10,
>but it gives a graph through (-10 , 0).
>                                                                1 -
>0.29exp(-0.29 t)
>And yes, the resulting function M(t)=   --------------------------- is
>defined for all real numbers.
>0.29
>Any help is greatly appreciated