non linear coupled diff equation

dx/dt=-(r+r1)/L1*x - dy/dt*(L12/L1)

dy/dt= -y*r2 - dx/dt*(L12/L2

r=300 r1= 8.1 L1=13.8 L2 =0.012 L12=0.0013

Kindly guide in Matal core for above equation using ode23 as r2 is function of time

thanks in advance

Sarat Kumar Dash

[Jan]

What is "Matal core"?

[Sarat]

dear Sir

thanks for your resoinse, but I want both x and y and the code in matlab

let me simplyfy the equation as following

dx/dt= -A*x - B*dy/dt

dy/dt= -y*C - dx/dt*D C alone function of time A B D constant

please gice a code in matlab ode23

regards

[Jan]

As written in my answer:

dx/dt=   -A*x - B*dy/dt
dy/dt= -y*C - dx/dt*D

is equivalent to:

dx/dt= -A*x - B * (-y*C + A*x*D) / (1 -  B*D)
dy/dt = (-y*C + A*x*D) / (1 -  B*D)

This should be easy to be implemented, such that an ODE integrator can handle it. Please try it and ask again, if you have a specfic problem.

[Sarat]

thanks for your reply

Answer 1

[Jan]

What about simplifying the equations?

1. dx/dt = -(r+r1)/L1*x - dy/dt * (L12/L1)
2. dy/dt = -y*r2 - dx/dt * (L12/L2)

1. into 2.:
dy/dt = -y*r2 - (-(r+r1) / L1 * x - dy/dt * (L12/L1)) * (L12/L2)
      = -y*r2 + ((r+r1) / L1 * x + dy/dt * (L12/L1)) * (L12/L2)
      = -y*r2 + (r+r1) / L1 * x * (L12 / L2) + dy/dt * L12^2 / (L1*L2)
dy/dt - dy/dt * L12^2 / (L1*L2) = -y*r2 + (r+r1) / L1 * x * (L12 / L2)
dy/dt (1 - L12^2 / (L1*L2))     = -y*r2 + (r+r1) / L1 * x * (L12 / L2)
dy/dt = (-y*r2 + (r+r1) / L1 * x * (L12 / L2)) / (1 - L12^2 / (L1*L2));

Insert this in 1 again.

Do I oversee a point?