Solving Coupled Differential Equation

19 次查看（过去 30 天）
Harshit Agarwal
Harshit Agarwal 2017-10-23
评论： David Goodmanson 2017-10-24
Hello,
I want to solve following differential equation:
(x^2+x+1) / (x^2+x) dx/dt + dy/dt = 1 with constraint x+x^2 = y+y^2
It involves derivatives of both x and y. How can I solve this in Matlab.
Thanks guys in advance!! Cheers

请先登录，再回答此问题。

采纳的回答

Birdman
Birdman 2017-10-23
syms y(t) x(t)
a=(x^2+x+1)/(x^2+x);
%%because of the constraint, x+x^2=y+y^2 ----> x+y=-1. Take the derivative wrt t and you will
%%find x_dot=-y_dot;
eqns=a*diff(x,t)-diff(x,t)==1;
X=dsolve(eqns,t)
Try this.
  6 个评论
显示 4更早的评论
Torsten
Torsten 2017-10-24
编辑：Torsten 2017-10-24
Differentiate the algebraic equation with respect to t.
The differential equation and the differentiated algebraic equation then give you a linear system of equations in the unknowns dx/dt and dy/dt. Solve it explicitly for dx/dt and dy/dt and then use one of the standard ODE integrators.
Or write your system as
M*[dx/dt ; dy/dt] = f(t,x,y)
with
M = [(x^2+x+1)/(x^2+x) 1 ; 0 0]
f = [1 ; x^2+y^2+x*y-10]
and use ODE15S with the state-dependent mass matrix option.
Best wishes
Torsten.
David Goodmanson
David Goodmanson 2017-10-24
Why should x+x^2 = y+y^2 imply x+y = -1 only? x = y also works, in which case
eqns=a*diff(x,t)-diff(x,t)==1;
becomes
eqns=a*diff(x,t)+diff(x,t)==1;
in which case
Warning: Unable to find explicit solution. Returning implicit solution instead.
X = solve(2*x - 2*atanh(2*x + 1) == C2 + t, x)
Not as convenient as the first solution since t is given as a function of x rather than vice versa, but still a solution.

请先登录，再进行评论。

更多回答（0 个）

类别

在 Help Center 和 File Exchange 中查找有关 Numeric Solvers 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by