non linear coupled diff equation

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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
  4 个评论
Jan
Jan 2021-7-20
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.

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采纳的回答

Jan
Jan 2021-7-20
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?

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