Solve differential equation from Matlab

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Sabella Huang
Sabella Huang 2022 年 5 月 27 日
コメント済み: Sabella Huang 2022 年 5 月 29 日
Hello guys,
I want to ask about, how to solve the differential equation in MATLAB for the equation:
dx/dt = (ka*C(xmax-x)-kd*x)/(1+ka*x)
  3 件のコメント
John D'Errico
John D'Errico 2022 年 5 月 27 日
Are ka and kd constants, or is k separate from a and d, so there are three unknown constants. Is xmax known? Is it a value that x may not exceed? Is C a function? Or is C just a known constant?
When you are too vague, you make it impossible for people to help you.



John D'Errico
John D'Errico 2022 年 5 月 27 日
編集済み: John D'Errico 2022 年 5 月 27 日
Making various assumptions, we see the solution as:
syms ka kd C xmax
syms x(t)
dsolve(diff(x) == (ka*C*(xmax-x)-kd*x)/(1+ka*x))
ans = 
Where W0 is the Lambert W function. Actually, the zero'th branch thereof, so lambertw(0,u).
help lambertw
LAMBERTW Lambert's W function. W = LAMBERTW(Z) solves W*exp(W) = Z. W = LAMBERTW(K,Z) is the K-th branch of this multi-valued function. References: [1] Robert M. Corless, G. H. Gonnet, D. E. G. Hare, D. J. Jeffrey, and D. E. Knuth, "On the Lambert W Function", Advances in Computational Mathematics, volume 5, 1996, pp. 329-359. [2] Corless, Jeffrey, Knuth, "A Sequence of Series for The Lambert W Function", ISSAC '97 Documentation for lambertw doc lambertw Other functions named lambertw sym/lambertw
  2 件のコメント
Sabella Huang
Sabella Huang 2022 年 5 月 29 日
ok this work... thx


その他の回答 (1 件)

Bjorn Gustavsson
Bjorn Gustavsson 2022 年 5 月 27 日
For a numerical integration of this ODE look at the help and documentation for ode45 and its siblings, also look at the examples and demos related to these ODE-solvers, you find these in odeexamples. That is perhaps the best way to get started (maybe it will not make you an expert in DE in general, but it will get you going.) You can also read the code of the different demos and adapt them to suit your problem.
For an analytical solution you might have success with dsolve from the symbolic toolbox if you have access to that.


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