How to solve Coupled Differential Equations

Hi

how to solve that set of equations

y'-x'=2-x

2x'-y'=3+2y

Answer 1

[Star Strider]

Try this —

syms x(t) y(t) x0 y0 
Dx = diff(x);
Dy = diff(y);
ode1 = Dy-Dx == 2 - x

ode1(t) = 

ode2 = 2-Dy - Dy == 3 + 2*y

ode2(t) = 

S = dsolve(ode1, ode2, x(0)==x0, y(0)==y0)

S = struct with fields:
    y: [1×1 sym]
    x: [1×1 sym]

x(t) = simplify(S.x, 500)

x(t) = 

y(t) = simplify(S.y, 500)

y(t) = 

.

[Lorenzo]

There is a way to plot the solutions x(t) and y(t) ?

Thanks

[Walter Roberson]

@Lorenzo

No, it is not possible to plot y(t) . y(t) involves three independent variables: t, x0, and y0. There are no built-in functions for plotting 3 independent variables plus one resultant variable.

[Lorenzo]

Thank you

[Walter Roberson]

Of course, for given x0 and y0, it is easy enough to plot.

syms x(t) y(t) x0 y0 
Dx = diff(x);
Dy = diff(y);
ode1 = Dy-Dx == 2 - x

ode1(t) = 

ode2 = 2-Dy - Dy == 3 + 2*y

ode2(t) = 

S = dsolve(ode1, ode2, x(0)==x0, y(0)==y0)

S = struct with fields:
    y: exp(-t)*(y0 - exp(t)/2 + 1/2)
    x: exp(t)*(x0 - y0/2 + (9*exp(-t))/4 - 9/4) + (exp(-t)*(y0 - exp(t)/2 + 1/2))/2

x(t) = simplify(S.x, 500)

x(t) = 

y(t) = simplify(S.y, 500)

y(t) = 

X0 = -2;
Y0 = 1.5;
sX = subs(x, [x0, y0], [X0, Y0])

sX(t) = 

sY = subs(y, [x0, y0], [X0, Y0])

sY(t) = 

fplot(sX, [0 5])

fplot(sY, [0 5])

Answer 2

[Walter Roberson]

yprime - xprime == 2-x, 2*xprime - yprime == 3 + 2*y
2*yprime - 2*xprime == 4-2*x, 2*xprime - yprime == 3 + 2*y
2*yprime - 2*xprime + 2*xprime - yprime == 4-2*x + 3 + 2*y
yprime == 7 - 2*x + 2*y
7 - 2*x + 2*y - xprime == 2*x
7 - 2*x + 2*y - 2*x == xprime
xprime = 7 - 4*x + 2*y

So...

function dxy = odefun(t, xy)
   dxy = [7 - 4*xy(1) + 2*xy(2); 7 - 2*xy(1) + 2*xy(2)];
end

[Tomasz]

this:

2*yprime - 2*xprime == 4-2*x, 2*xprime - yprime == 3 + 2*y

2*yprime - 2*xprime + 2*xprime - yprime == 4-2*x + 3 + 2*y

is it OK??

[Walter Roberson]

2*yprime - 2*xprime == 4-2*x, 2*xprime - yprime == 3 + 2*y

add the two equations

(2*yprime - 2*xprime) + (2*xprime - yprime) == (4-2*x) + (3 + 2*y)

[Walter Roberson]

syms xprime yprime x y
eqn = [yprime - xprime == 2-x, 2*xprime - yprime == 3 + 2*y]

eqn = 

sol = solve(eqn, [xprime, yprime])

sol = struct with fields:
    xprime: [1×1 sym]
    yprime: [1×1 sym]

sol.xprime

ans = 

sol.yprime

ans = 

Hmmm, where did I go wrong?

[Walter Roberson]

I did make a mistake, but it was the step after you indicated. I copied as 2*x instead of 2-x . The corrected version is

yprime - xprime == 2-x, 2*xprime - yprime == 3 + 2*y
2*yprime - 2*xprime == 4-2*x, 2*xprime - yprime == 3 + 2*y
2*yprime - 2*xprime + 2*xprime - yprime == 4-2*x + 3 + 2*y
yprime == 7 - 2*x + 2*y
7 - 2*x + 2*y - xprime == 2 - x   %corrected
7 - 2*x + 2*y - (2 - x) == xprime
xprime = 5 - x + 2*y

So...

function dxy = odefun(t, xy)
   dxy = [5 - xy(1) + 2*xy(2); 7 - 2*xy(1) + 2*xy(2)];
end

Answer 3

[Sam Chak]

Hi @Lorenzo

Primarily, coupled ordinary differential equations (ODEs) in additive form can be decoupled using the substitution and elimination method (Grade 10 algebra). In this simple system, we can decouple the ODEs simultaneously using the matrix method.

The coupled ODEs

can be expressed in matrix form

and simplified to

.

Because of the square matrix property , we can manipulate the matrix equation to become

so that we can solve the following decoupled ODEs

.

Method 1: Numerical approach

% Describe the ODEs inside the function
function dx = ode(t, x)
    dx(1)   = 5 - 1*x(1) + 2*x(2);  % ODE 1
    dx(2)   = 7 - 2*x(1) + 2*x(2);  % ODE 2
    dx      = [dx(1)
               dx(2)];              % arranged in column vector
end

% Call ode45 to numerically solve the system
tspan   = [0 10];                   % time interval from 0 to 10
xinit   = [1; 0];                   % initial condition at time 0
[t, x]  = ode45(@ode, tspan, xinit);
plot(t, x), grid on, xlabel('Time'), legend('x(t)', 'y(t)')
title('Numerical Solutions')

●

Method 2: Symbolical approach to obtain the analytical solutions (using @Star Strider's code):

%% Requires Symbolic Math Toolbox
syms x(t) y(t) x0 y0
Dx = diff(x);
Dy = diff(y);
ode1 =   Dy - Dx == 2 - x

ode1(t) = 

ode2 = 2*Dx - Dy == 3 + 2*y

ode2(t) = 

S = dsolve(ode1, ode2, x(0)==1, y(0)==0)    % specify the initial condition here

S = struct with fields:
    y: - exp(t/2)*sin((7^(1/2)*t)/2)*((exp(-t/2)*(21*sin((7^(1/2)*t)/2) + 25*7^(1/2)*cos((7^(1/2)*t)/2)))/14 - (17*7^(1/2))/14) - exp(t/2)*cos((7^(1/2)*t)/2)*((exp(-t/2)*(21*co...
    x: - ((3*exp(t/2)*cos((7^(1/2)*t)/2))/4 + (7^(1/2)*exp(t/2)*sin((7^(1/2)*t)/2))/4)*((exp(-t/2)*(21*cos((7^(1/2)*t)/2) - 25*7^(1/2)*sin((7^(1/2)*t)/2)))/14 - 3/2) - ((3*exp(...

x(t) = simplify(S.x, 500)

x(t) = 

y(t) = simplify(S.y, 500)

y(t) = 

fplot(x(t), [0 10]), hold on,  grid on, ylim([-300, 500])
fplot(y(t), [0 10]), hold off, xlabel('Time'), legend('x(t)', 'y(t)')
title('Analytical Solutions')