Runge Kutta, linear differential system

I need to make a matlab program to find the solutions for 0<=t<=1 for

x1'(x) = 2x1 -3x2 +4x3 +0x4 -2x5 -2, x1(0)=1
x2'(x) = x1 - 2x2 - 2x3 +x4 +2x5 -1, x2(0)=-1
x3'(x) = 2x1 + 4x2 +3x3 +x4 -x5 -2, x3(0)=1
x4'(x)=-2x1 +0x2 +x3 + 4x4 -3x5 +1, x4(0)=2
x5'(x)=0x1 -x2 +2x3 -3x4 -2x5 +1, x5(0)=3

and I made a program, and I got an output:
w(1)= 94.70989959 w(2)= -5.06974461 w(3)= 81.70881908 w(4)= 7.03658123 w(5)= 12.84918264

but I have no idea how to tell whether or not I'm right. The other examples I've done have been higher order initial value problems that depended on t... so I used the same kind of scheme... but this one doesn't seem to depend on t... so I'm not sure if I'm suppose to do something different or not?

Is there a way to get an exact solution for this for me to check to see if I'm right or?

Also, for a different example with a higher order equation, my function looked like:
function a=nonf(t,u)
a=zeros(1,length(u));
a(1)=u(2);
a(2)=u(3);
a(3)=2.0*cos(t)+u(1)+u(1)*u(2)+2.0*sin(t)*u(3);

Right now my function for this question looks like:

function a=nonff(t,u, coef)
a=zeros(1,length(u));
for i=1:length(u)
sum=0;
for k=1:length(u)
sum=sum+coef(i,k)*u(k);
end
a(i)= sum+coef(i,length(u)+1);
end

Would that be right, or would I need something like
a(1)=u(2);
a(2)=u(3);
a(3)=u(4)
a(4)=u(5)

Quote:

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Originally Posted by gummy_ratz

I need to make a matlab program to find the solutions for 0<=t<=1 for

x1'(x) = 2x1 -3x2 +4x3 +0x4 -2x5 -2, x1(0)=1
x2'(x) = x1 - 2x2 - 2x3 +x4 +2x5 -1, x2(0)=-1
x3'(x) = 2x1 + 4x2 +3x3 +x4 -x5 -2, x3(0)=1
x4'(x)=-2x1 +0x2 +x3 + 4x4 -3x5 +1, x4(0)=2
x5'(x)=0x1 -x2 +2x3 -3x4 -2x5 +1, x5(0)=3

and I made a program, and I got an output:
w(1)= 94.70989959 w(2)= -5.06974461 w(3)= 81.70881908 w(4)= 7.03658123 w(5)= 12.84918264

but I have no idea how to tell whether or not I'm right. The other examples I've done have been higher order initial value problems that depended on t... so I used the same kind of scheme... but this one doesn't seem to depend on t... so I'm not sure if I'm suppose to do something different or not?

Is there a way to get an exact solution for this for me to check to see if I'm right or?

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A first observation would be: You are asked for a solution on [0,1] not a single point.

A second observation might be: This is a linear 5-th order system, it has a known solution (either in matrix form or by expanding to a single 5-th order linear constant coefficient ODE)

CB

Yeah, my program goes from 0<t<1, that was just my final answer for t=1 when I reached convergence within 8 decimal points. I'm taking numerical analysis now, but I haven't taken differential equations yet, so I'm really not sure, what method could I use to solve this system to see if my runge kutta answer matches up?

Quote:

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Originally Posted by gummy_ratz

Yeah, my program goes from 0<t<1, that was just my final answer for t=1 when I reached convergence within 8 decimal points. I'm taking numerical analysis now, but I haven't taken differential equations yet, so I'm really not sure, what method could I use to solve this system to see if my runge kutta answer matches up?

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Sorry but if you have not done differential equations you are not ready for numerical solution of them. It is not our place to provide a course in DEs.

You could try comparing the RK solution with a solution computed using a simpler method (say Euler) with a smaller step size.

CB

Yeah, I know that there's an easy way to get Maple to find the actual solution to higher order differential equations, and I searched to try to find a way to do it for the linear systems, but I couldn't find one, so I just thought I'd check to see if you guys knew. Thanks, anyways.