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

vishal vyas (view profile)

on 20 Feb 2018
Latest activity Commented on by Torsten

Torsten (view profile)

on 21 Feb 2018
kindly help me to solve this equation a = 0.5, b=2, r(0)=1.2

John D'Errico

John D'Errico (view profile)

on 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 (view profile)

on 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.)

Roger Stafford (view profile)

on 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.

Torsten

Torsten (view profile)

on 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.