# How to numerically solve system of equations and differential equations simultaneously?

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Hi all,

I have 3 variables A, B, C, where A and B can be solved by system of equations. For example,

A + B = C + 6

A - B = 2*C

While C should be solved by differential equation,

diff(C,t) == 6*A

This is just a simple example, actually my equations are very complicated. I have tried:

- Using "solve". I think the equations are so complicated that the empty solutions appear. So I tend to find the numerical solution.
- Using "fsolve". The problem is that there is an undefined variable C. Because of the error "FSOLVE requires all values returned by functions to be of data type double.", I can't use symbolic.
- Using "vpasolve". There are similar problems with "fsolve". The error is "Symbolic parameters not supported in nonpolynomial equations.".

Are there other methods I should try? Thank you for any advice.

##### 0 Comments

### Accepted Answer

Askic V
on 19 Apr 2023

Edited: Askic V
on 19 Apr 2023

Does this make sense to you (in your simple example)?

syms A B C(t)

eqn1 = diff(C,t) == 6*A;

C_sol = dsolve(eqn1)

eqn2 = A + B == C_sol + 6;

eqn3 = A - B == 2*C_sol;

[A,B] = equationsToMatrix([eqn2, eqn3], [A, B]);

X = linsolve(A,B)

##### 3 Comments

Askic V
on 19 Apr 2023

### More Answers (2)

Torsten
on 19 Apr 2023

Edited: Torsten
on 19 Apr 2023

MATLAB's ode solvers allow a mixture of differential and algebraic equations as in your case.

These systems are called differential-algebraic equations. For an example, see

Solve Robertson Problem as Semi-Explicit Differential Algebraic Equations (DAEs)

under

Or as a solution for your simple example:

M = [0 0 0;0 0 0; 0 0 1];

tspan = [0 1];

fun = @(t,y) [y(1)+y(2)-y(3)-6;y(1)-y(2)-2*y(3);6*y(1)];

options = odeset('Mass',M,'MStateDependence','none','MassSingular','yes','RelTol',1e-7,'AbsTol',1e-7);

% Starting values for A and B are taken arbitrary ;

% They will be adjusted according to the algebraic equations 1 and 2 by

% ode15s to y0(1) = 4.5 and y0(2) = 2.5 (see below)

y0=[0 0 1];

[T,Y] = ode15s(fun,tspan,y0,options);

plot(T,Y)

Y(1,1)

Y(1,2)

Y(1,3)

grid on

##### 4 Comments

Torsten
on 21 Apr 2023

Edited: Torsten
on 21 Apr 2023

I stick to your program structure of first solving the algebraic equations and inserting the result into the differential equations.

This won't usually work if the algebraic equations are difficult.

You should try my method from above for more complicated systems (i.e. solving algebraic and differential equations all together using ode15s).

syms x1 x2 x3

a = 1;

b = 2;

c = 3;

d = 4;

Eq1 = (a+1i*b)*x1 + 1i/2*x2*c - 1i/2*conj(x2)*d == 0;

Eq2 = (-a+1i*c)*x1 + 1i/2*d*x2 + 1i/2*b*x3 + a*conj(x3) == 0;

Solution = solve([Eq1, Eq2], [x1 x2]);

x1 = matlabFunction(Solution.x1);

x2 = matlabFunction(Solution.x2);

options = odeset('RelTol',1e-10,'AbsTol',1e-10);

fun = @(x,t) 1i*c*x1(x(1));

tspan = [0:0.05: 1];

y0 = 1;

[T,X3] = ode15s(fun,tspan,y0,options);

figure(1)

plot(T,[real(X3),imag(X3),abs(X3)])

figure(2)

plot(T,[real(x1(X3)),imag(x1(X3)),abs(x1(X3))])

figure(3)

plot(T,[real(x2(X3)),imag(x2(X3)),abs(x2(X3))])

Sam Chak
on 19 Apr 2023

Hi @I CHUN LIN

It seems that simple Substitution-and-Elimination method produces the solution

.

Thus, the linear ODE becomes

which indicates that it is possible to find an explicit solution of the differential equation analytically.

syms C(t)

eqn = diff(C, t) == 9*(C + 2);

S = dsolve(eqn)

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