Solving Coupled Differential Equation

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

Answer 1

[Birdman]

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.

[Harshit Agarwal]

Thanks for the quick answer!! really appreciate that.

How do you arrived at x+x^2=y+y^2 ----> x+y=-1 ? Shouldn't it be dx/dt*(1+2x)=dy/dt*(1+2y)?

[Birdman]

x-y=(y-x)*(y+x),

Division of x-y to y-x brings -1 and therefore;

x+y=-1

[Steven Lord]

x = 42;
y = 42;
L = x + x.^2;
R = y + y.^2;
isequal(L, R) % true
isequal(x+y, -1) % false

Be careful about dividing by 0.

[Harshit Agarwal]

Well, my bad. Let's say if we cannot express dx explicitly in terms of dy, then how to solve this type of equations. For. e.g.,

(x^2+x+1) / (x^2+x) dx/dt + dy/dt = 1 with constraint x^2+y^2+xy=10 (or any constant)

[Torsten]

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]

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.