Re: Visual verification of solution in DifEq mode in TI-85


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Re: Visual verification of solution in DifEq mode in TI-85



On Sun, 17 May 1998, Roland Bengtsson wrote:

> Is there a smart way to verify a solution for a difequation in TI-85?
> Usual equations is no problem, just to exchange x with the answer but
> difeqs seems to be harder...
>
> How about this one:
>
> y'' + 2y' + 5y = cos x
>
> The answer should be:
>
>       -x
> y = (e  )(A cos (2x) + B sin (2x)) - 0.1 cos x + 0.2 sin x
>
> Now I set y1=<the line above>
>
> der2(y1,x) + 2 der(y1,x) + 5 y1
>
> should be equal to cos x.
> x happens to be 0.123 but this should not affect this.
>
> It doesn't matter if I set A and B to 1, the values still differ.
> Is there another way, send me a mail about it!
>
>
> Thanks Roland
>
> http://www.algonet.se/~roland-b
>
I don't know if its a "smart way" but....
Method I
    graph your solution  y1 = exp(-x) * ( Acos(2x) + Bsin(2x) ) +
0.2cos(x) + 0.1sin(x)  (*note my change from your suggested soln above*)

and the graph y2 = der2(y1,x,x) + 2der1(y1,x,x) + 5y1 - cos(x) + 1
The added 1 is to shift y2 up 1 unit so it can be seen distinctly from the
x-axis.

On the home screen, set  A=B=1 by    1->A  and 1->B
use range:    0 < x < 5
            -.5 < y < 1.5
finally, graphing shows that y2 is identically 1.  This should be visual
confirmation of your soln.


Method II
  stpic of your graph of y1, the candidate soln.
Then, change mode to DifEq.
Enter the problem as:
    Q'1=Q2
    Q'2=cos(t)-2Q2-5Q1
with range:
     0 < t < 5
     0 < x < 5
   -.5 < y < 1.5  as above in method I
Use INITC :
     QI1=1.2
     QI2=1.1 found at x = 0 from the soln being "visualized"
Set AXES:
  x=t
  y=Q1
and graph.
The solution can be traced. But when you RCPIC the graph you saved, you
have the neat visual evidence that your soln is correct.
Have fun!!  Love that DifEq mode on the '85 and '86
Charlie


References: