How to Solve an ODE Problem when one of parameters is dependent to derivative?

mahdi_zabchek

Hello Guys!
I have an ODE problem that I'm solving it by MATLAB ODE solvers!
in fact I have a system of non-linear differential equations in one of these equations I have a parameter that it's value is dependent to derivative! the general form of equation is like this (big letter parameters are known!):

dy/dt = A + B + f(C,D,dy/dt)

how can I solve this problem by ode45 or other MATLAB ODE solvers?

hunt_mat

Is the function f known?

mahdi_zabchek

yes! it is.
but it's not reversible

JJacquelin

The ODE : dy/dt = A + B + f(C,D,dy/dt) contains no y and no t. As a consequence dy/dt = constant.
Let X= dy/dt . X is solution of the equation X = A + B + f(C, D, X) wich is not an ODE.
It doesn't matter if the function is not revertsible. We don't need to know the analytical expression of the solution(s) X. We know that dy/dt = constant (or = several different constants if there are several solutions). Each one can be numerically computed, not using an ODE solver, but using an usual numerical equation solver.
The solution(s) is (are) : y(t) = X*t +c
c is a constant to be determined by the boundary condition.

mahdi_zabchek

No! No! it has y and t!
A and B and C and D are NOT constant parameters!
I did't write them because they were not necessary!
in fact You don't need to know what's the equation exactly to answer my question!

My question is simple:

MATLAB ODE solvers solve equations in form of dy/dt=f(t,y) but I want to solve an equation in form of dy/dt=f(t,y,dy/dt) ... How I can do that by MATLAB?

JJacquelin

OK. Sorry for the missunderstanding.
May be, you could use an algorithm of this kind:
Start with given initial values y and t.
Recursive process :
Compute A(y,t), B(y,t), C(y,t) and D(y,t)
Solve X=A+B+f(C,D,X) with a numerical equation solver, introduced as sub-program.
With the computed value X=dy/dt the incrementation of y is done, as well as the incrementation of t.
Then continue the recursive process.

mahdi_zabchek

I'll try it ... thank you so much