solve differential equation by method of undetermined coefficient

so given three differential equations to solve undetermined coefficient given initial solutions. not sure how to add initial condition to code . here what i have so far .

initial conditions are x1(0) = 2 ; x2(0) = 6 x3(0) = 1

A = sym([.9375, 0 , 0; .9375 -.625 0; 0 .625 -75/135])

eig(A)

[V,D]=eig(A)

Id3 = sym([1,0,0;0,1,0;0,0,1])

syms lambda

B = lambda*Id3 - A

p = charpoly(A,lambda)

evs = solve(p)

s = null(evs(1)*Id3-A)

ps = vpa(fsolve(@f,[0;0;0]))

m file is

function [ z] = f( x )

z(1) = (15/16)*x(1)+20;

z(2) = (15/16) *x(1)-(5/8)*x(2);

z(3) = (5/8)*x(2)-(75/135)*x(3);

end

Answer 1

[madhan ravi]

tspan=[0 10];
ic=[2;6;1];     %initical conditions
[t,x]=ode45(@myod,tspan,ic);
plot(t,x)
legend('equation 1','equation 2','equation 3')

 function z = myod(t,x)
 z=zeros(3,1);
 z(1) = (15/16)*x(1)+20;
 z(2) = (15/16) *x(1)-(5/8)*x(2);
 z(3) = (5/8)*x(2)-(75/135)*x(3);
 end
 

[madhan ravi]

Brenda Galabe's answer moved here for consistency:

z=zeros(3,1); .... what does that mean

[madhan ravi]

It's just store the variable values called preallocation(search google).

[madhan ravi]

Brenda Galabe's answer moved here for consistency:

ok thank you

[madhan ravi]

Anytime :) , please when you make a comment respond to the one who posts an answer instead of posting a separate answer , thank you for understanding.

[madhan ravi]

or if you want to symbolic approach then:

syms x1(t) x2(t) x3(t)
dx1dt=diff(x1);
dx2dt=diff(x2);
dx3dt=diff(x3);
eqn1 = dx1dt == (15/16)*x1+20;
eqn2 = dx2dt == (15/16) *x1-(5/8)*x2;
eqn3 = dx3dt == (5/8)*x2-(75/135)*x3;
conds=[dx1dt(0)==2;dx2dt(0)==6;dx3dt(0)==1];
[x1(t),x2(t),x3(t)]=dsolve(eqn1,eqn2,eqn3,conds);
ezplot(x1(t))
hold on
ezplot(x2(t))
ezplot(x3(t))
legend('equation 1','equation 2','equation 3')

[madhan ravi]

if you use fplot then replace all the ezplots with:

fplot({x1(t),x2(t),x3(t)})