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I am trying to solve the following differential equation using scipy odeint without much success: import numpy as np from scipy.misc import derivative from scipy.integrate import odeint Imag = 16000. w = 2*np.pi*60 tau = .05 theta = 1.52 phi = theta - np.radians(90) t = np.linspace(0,.1,10000) def Ip(t): return np.sqrt(2)*Imag*(np.sin(w*t+phi-theta)-np.exp(-t/tau)*np.sin(phi-theta)) B = lambda Ip: Ip/(53.05+0.55*abs(Ip)) def L(B): return derivative(B,Ip(t))*377.2 def dI(t): return derivative(Ip,t) def f(y,t): Rb = 8. N = 240. Is = y[0] f0 = (1/(L(B)+0.002))*((dI(t)*L(B)/N)-Rb*y[0]) return [f0]

I have a small problem. I have 2 equation of motion 'ph' and 'ph2' I don´t know how to set ODE to stop calculating 'ph' when x(1)> 0.111 and then starts to calculated 'ph2' again only to 0.111, after that plot 'ph' + 'ph2' to one graph depend on time 'w' i think I have to set some time limitations but dont know how to. I use HELP but no benefit for me. [t,y] = ode45(@ph,[0,w_max],[0,0]); function dx = ph(tt,x) global F1 c m_c Ff p w s ln f_t sig dstr Ren pn Fex Fzmax xz xn l Fz mn Fpp = F1 + c*x(1); if pn<0 pn=abs(pn); end if x(1)<ln pn=spline(w,p,tt)-((2*sig)/dstr*Ren); Fex=3.1416.*f_t.*pn.*

I'm simulating equations of motion for a (somewhat odd) system with mass-springs and double pendulum, for which I have a mass matrix and function f(x), and call ode45 to solve M*x' = f(x,t); I have 5 state variables, q= [ QDot, phi, phiDot, r, rDot]'; (removed Q because nothing depends on it, QDot is current.) Now, to calculate some forces, I would like to also save the calculated values of rDotDot, which ode45 calculates for each integration step, however ode45 doesn't give this back. I've searched around a bit, but the only two solutions I've found are to a) turn this into a 3rd order

Can anybody give me some advice how to solve an ODE in Python that has a time-delay implemented in it? I can't seem to figure out how to do it using scipy.integrate.odeint. What I am looking for should look like: # the constants in the equation b = 1/50 d = 1/75 a = 0.8 G = 10 ** (-2) tau = 0.5 u = [b, d, tau, a, G] # enter initial conditions N0 = 0.1 No0 = 10 w = [N0, No0] def logistic(w, t, u): N, No = w b, d, tau, a, G = u dNdt = b * (No(t) - N(t) ) * (N(t) / No(t) ) - d * N(t - tau) dNodt = G * (a * No(t) - N(t) ) * (N(t) / No(t) ) return [dNdt, dNodt] # create timescale # create timescale

For solving simple ODEs using SciPy, I used to use the odeint function, with form: scipy.integrate.odeint(func, y0, t, args=(), Dfun=None, col_deriv=0, full_output=0, ml=None, mu=None, rtol=None, atol=None, tcrit=None, h0=0.0, hmax=0.0, hmin=0.0, ixpr=0, mxstep=0, mxhnil=0, mxordn=12, mxords=5, printmessg=0)[source] where a simple function to be integrated could include additional arguments of the form: def dy_dt(t, y, arg1, arg2): # processing code here In SciPy 1.0, it seems the ode and odeint funcs have been replaced by a newer solve_ivp method. scipy.integrate.solve_ivp(fun, t_span, y0,

I want to stop solving a differential equation in Matlab if it takes more than a designated amount of time. I tried the following,but it doesn't work... options = odeset('AbsTol',1e-8,'RelTol',1e-5); RUNTIME=5; timerID = tic; while (toc(timerID) < RUNTIME) [t_pae,x_th_pae] = ode15s(@prosomoiwsh,[0 t_end],[80*pi/180;0;130*pi/180;0;th_initial(1);th_initial(2);th_initial(3);th_initial(4)],options); end How can I solve this? UPDATE : I tried what horchler suggested and now my code looks like this : interupt_time = 20; outputFun= @(t,y,flag)interuptFun(t,y,flag,interupt_time); options = odeset(