Can we numerically solve system of First order nonlinear coupled ODE and plot?

Math123 shared this question 3 years ago
Answered

Is it possible to solve numerical first order nonlinear coupled ODE and plot ?

If not pls tell how many types of ODEs can be solved by Geogebra also any other numerical solution possible?

Best Answer
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I can shout eureka?

Files: foro.ggb

Comments (41)

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Please post an example of what you want to solve

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Pls see attachment . I want to solve numerically and plot this also as i change the condition solution change respectively.

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I think this might work :)


https://wiki.geogebra.org/e...

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Thanks,@Michael Borcherds for your reply. I tried but it am not able to use NSolveODE properly. Is it possible for me you can solve me this for acceleration ,velocity and position .

Thanks you very much in advance.

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Pls see attachement. How can i solve also check does this correct otherwise pls correct it and if possible solve this for me. Thanks you very much.

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I need all parameters. Position (x(t),y(t)) , Velocity (V_x(t),V_y(t)) = (x'(t),y'(t)) , acceleration (a_x(t),a_y(t) =(V_x'(t),V_y'(t)) = (x''(t),y''(t))

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Pls see attachment. I tried but there is some problem Pls correct.

1. 't' is not working as a derivative

2. how we get x(t) , y(t) as time 't' passes.

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did you mean something like this?

Files: foro.ggb
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Thanks @mathmagic . . for your reply.. But 't' slider does not work here it is the time derivative of position, velocity and acceleration. I want position x(t) , y(t) ,; Velocity V_x(t) , V_y(t) only by knowing acceleration a_x(t) , a_y(t).. (acceleration is attached) as a function of time. and Point for (x(t) , y(t)) .

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redefine mm as function of t and redefine the limits of slidres

Files: foro.ggb
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Dear @mathmagic i appreciate your support. how can i define x(t) and y(t) from x'(t) ,y'(t). Secondly it is increasing in y(t) as 't' increases . This function should be expected somewhat like parabola . It could be increase in y(t) to some extent with x(t) after certain height then it should decrease y(t) as further 't' increases. I doubt there should be something wrong with attachment.

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It is expected there should be 6 numerical solution (Only by knowing V_x'(t) & V_y'(t).

1. 2 Solutions for : a_x(t) = V_x'(t) = x''(t) & a_y(t) = V_y'(t) = y''(t)

2. 2 Solutions for : V_x(t) = x'(t) & V_y(t) = y'(t)

3. 2 Solutions for : x(t) & y(t)

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Hayat what are initial conditions for the two equations and the values of C and m ?

I suppose that this ode system is very hardy for Geogebra and other CAS (is not a numerical system ).

Best wishes.

Luigi Marino

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I am not sure but it seems that nsolve solves only equations of n-order

the example of the help is:


    f'(t, f, g, h) = g
    g'(t, f, g, h) = h
    h'(t, f, g, h) = -t h + 3t g + 2f + t
equivalent to


    f'''(t, f, f', f'') = -t f'' + 3t f' + 2f + t

I do not know if nsolve can solve your problem. I can not get it


the number of expected solutions is 4 because the solution of ie y'=2y is exp(2x) not two functions

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I can shout eureka?

Files: foro.ggb
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Dear @mathmagic thanks for your time and support. I am analyzing the function. How can i extract V_x'(t) , V_y'(t) as Point and also a_x'(t) , a_y(t) as Point.

Once again i am thankful to you and Geogebra Team .

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numericalint is a locus (path defined with a collection of points)

you can see them with l1 = First(numericalIntegral3, Length(numericalIntegral3)) for V_{x}

Point(numericalIntegral3, sl) with 0<sl<1 run along the locus


y(Point(numericalIntegral3, mm)) is the value of V_{x} with mm=(t-a)/(b-a)

Files: foro.ggb
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mathmagic

the solutions:

1. 2 Solutions for : a_x(t) = V_x'(t) = x''(t) & a_y(t) = V_y'(t) = y''(t)

2. 2 Solutions for : V_x(t) = x'(t) & V_y(t) = y'(t)

3. 2 Solutions for : x(t) & y(t)

have none numbers.

The question of Hayat is not express rigth


.

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Dear @luigi marino i asked bacause a_x(t) = V_x'(t) must give the position , velocity and acceleration along x-axis as 't' passes. similarly a_y(t) for y-axis. So at any moment of 't' i should have one number for each. means

1. a_x(t) = ? , a_y(t) = ?

2. V_x(t) = ? , V_y(t) = ?

3. x(t) = ? , y(t) = ?

My question is that. How can i extract above data from attachment by @mathmagic.

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a,V are in my last attached


the names are ac and v

the animation shows the point the velocity and the aceleration

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I want to know the position components x(t) and y(t), what is point B it represent the position , velocity or acceleration.

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I said you x(t)=y(Point(numericalIntegral1, (t-a)/(b-a)))

y(t)=y(Point(numericalIntegral2, (t-a)/(b-a)))

V_{x}(t)=y(Point(numericalIntegral3, (t-a)/(b-a)))

V_{y}(t)=y(Point(numericalIntegral4, (t-a)/(b-a)))

for aceleration see definition of vector ac in my last attached

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Thanks for your efforts. How can find locus of Points or Curve of (x(t),y(t)) from t [a,b]

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How to find trajectory from of (x(t),y(t)) Line at t [a,b]

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locus(B,t)

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Thanks a lot...

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It's been a great discussion with you @mathmagic . Thanks a lot for your support and Geogebra Team as always.

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How can we find the Point where V_y(t) = 0 , also value of 't' where V_y(t)=0

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Hayat, can you send me the original mechanic problem ?

I try solve this with CAS and RK method.

Best wishes.

Luigi Marino

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Hi @luigi marino thanks for your offer. I almost completed the project and wanted to share my work after completion. Can you pls share your work as i sent a_{x}(t) = V_{x}'(t) , and a_{y}(t) = V_{y}'(t). All the things are related to these two quantities.

I really like to see your work . so i can learn something new that how many ways software can solve the problem.

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luigi marino , I like to see the solution with RK and CAS method. Pls share your work , because all the things come from a_x(t) & a_y(t). you can use these two ODEs to extract all data.

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How to find Point from locus of points like " Locus(B,t) where B(x,y) is Point and 't' is slider "

1. Point where y=0 also at 't' {Like root}

2. extremum from Locus of Points also 't' {Like dy/dx=0}

in same attachment by @mathmagic

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2. Try (change names if neccesary)

C = Element(Sort(First(numericalIntegral4, Length(numericalIntegral4)), abs(y(First(numericalIntegral4, Length(numericalIntegral4))))), 1)

D = ClosestPoint(numericalIntegral4, C)

e = PathParameter(D) (b - a) + a

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Pls see attached . It looks points are not working .

Files: foro.ggb
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Thanks , I need one more point where y(t) = 0 and 't' where y=0

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How can we find y(t) = 0 and 't' where y=0

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same way Y'==0

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Hayat

your problem on motion of projectle with quadratic air resistance is very hard.

The numeric solutions is qualitative for some initial conditions, Geogebra can only graph a model. The difficult is integrate vx and vy from numeric out.

Are you a student ? Look at phisics lecture.

Best wishes.

Luigi

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But i tested this numerical result with LAT, HAT, SAT with closed form solution, it worked very well also i want to add some more forces i tired it looks work well, how can i add them properly. Pls help me to figure it out.

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How can i find Point on x(t) where y(t) = 0. as in attached by mathmagic.

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