Linear Algebra : Solution to a system of ODE using matrix method

clc;

clear;

close;

%Read the coefficient matrix

A=input("Enter the coefficent matrix:");

n=length(A);

%Find eigenvalue and eigenvector

% in terms of ModalMatrix=P and SpectralMatrix=D;

[P,D]= eig(A)

PInv=inv(P);

%initial condition for Y'=AY

yx=input("Enter the initial x values:");

yy=input("Enter the initial y values:");

%Y'=AX=P*D*PInv*Y -(1)

%Take X=PInv*Y => X'=PInv*Y'

%Multiply PInv to (1)

%PInv*Y'=D*PInv*Y

%X'=D*X -New System of ODE

%initial condition for X'=DX

xx=yx

xy=PInv*yy'; %Here ' means transpose, to make same dimention to multiply

syms x(t) %For differential Equations X'=DX, take x(t) as dependent variable

for i=1:n

for j=1:n

if i==j

ode=diff(x,t)==D(i,j)*x

cond = x(xx(j))==xy(j);

X(j)=dsolve(ode,cond);

end

end

end

%X- Solution to X'=DX

%We need solution to Y i.e., Y=P*X

Y=vpa(P*X');

Sol=Y';

for i=1:n

fprintf("x%d(t)=%s\n",i,Sol(i))

end

Example

Solve the following system of first-order linear differential equations using matrix method.

$$y'_1=y_1+y_2; y'_2=4y_1-2y_2; y_1 (0)=,y_2 (0)=6$$

Output

Enter the coefficient matrix:[1 1;4 -2]

P =

    0.7071   -0.2425

    0.7071    0.9701

D =

     2     0

     0    -3

Enter the initial x values:[0 0]

Enter the initial y values:[1 6]

ode(t) =

diff(y(t), t) == 2*y(t)

ode(t) =

diff(y(t), t) == -3*y(t)

x1(t)=2.0*exp(2.0*t) - 1.0*exp(-3.0*t)

x2(t)=2.0*exp(2.0*t) + 4.0*exp(-3.0*t)