dr/dt* ln(a*r*dr/dt)=b/r^7 how to solve this equation

1 Ansicht (letzte 30 Tage)
vishal  vyas
vishal vyas am 20 Feb. 2018
Kommentiert: Torsten am 21 Feb. 2018
kindly help me to solve this equation a = 0.5, b=2, r(0)=1.2
  2 Kommentare
John D'Errico
John D'Errico am 20 Feb. 2018
I'll suggest you probably won't get much of an answer here, because this is not a question about MATLAB. You are asking how to solve that nonlinear differential equation. So it is a question purely about mathematics, on a problem with no clear solution and I will guess no direct analytical solution. You might catch someone here with an idea, but far more likely to get a result is by asking on a site where the question is on-topic.
Walter Roberson
Walter Roberson am 20 Feb. 2018
There is no easy solution for that. The rule is:
r(t) = RootOf(int(P^7*LambertW(1/P^6), P = Z .. 6/5)+2*t)
which is to say that at each point, t, r(t) is the lower bound of the integral P^7*LambertW(1/P^6) such that integrating over P from lower bound to 6/5, plus 2*t, gives 0. (P is an arbitrary variable name here.)

Melden Sie sich an, um zu kommentieren.

Antworten (1)

Roger Stafford
Roger Stafford am 20 Feb. 2018
Here is how I would approach your problem. First we write
a*r*dr/dt*log(a*r*dr/dt) = a*b/r^6
Now define w:
w = log(a*r*dr/dt)
and therefore
a*r*dr/dt = exp(w)
Thus
exp(w)*w = a*b/r^6
Hence
w = lambertw(a*b/r^6)
a*r*dr/dt = exp(lambertw(a*b/r^6))
dr/dt = 1/(a*r)*exp(lambertw(a*b/r^6))
Now finally you have a differential equation in the form that Matlab's ode functions can evaluate numerically, provided you have the lambertw function available.
  1 Kommentar
Torsten
Torsten am 21 Feb. 2018
Alternatively, by setting
y1 = r
y2 = dr/dt,
you can use ODE15S to solve the differential-algebraic system
y1' = y2
y2*log(a*y1*y2)-b/y1^7 = 0
Best wishes
Torsten.

Melden Sie sich an, um zu kommentieren.

Kategorien

Mehr zu Programming finden Sie in Help Center und File Exchange

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by