Question

This is the problem directly from the homework. I don't need help finding the matrix, I already have that.

Write the system of differential equations
dx/dt=-x+4y-6z
dy/dt=x-y+3z
dz/dt=x-2y+4z
in a matrix form. Find it's general solution, and a solution which at t=1 has the following values of x, y, and z:
x(1)=0 y(1)=0 z(0)=e
Here e is the number exp(1)=2.718...

Answer 1

Which part of the problem are you struggling with? Eigenvalues? Eigenvectors?

Answer 2

i'm struggling with finding the general solution. I have the eigenvalues and eigenvectors, but i don't know how to take them and transform them into the general solution

Answer 3

What did you get for the evalues & evectors?

Answer 4

evalues are 0,1,1 and evectors are (-2,1,1) ; (2,1,0) ; (-3,0,1)

Answer 5

yeah, evalues are right, i just found those using my calculator. so, you know the form of the second solution for the repeated evalues...X2= Kte^(lamda1)t +Pe^(lamda1)t...

Answer 6

So, solution should look like...X = C1X1 + C2X2....X2 is above and X1 is just your first solution...

Answer 7

Sorry for the triple post. I also meant to add that if you're having trouble representing the solution (i.e. you have the evalues and evectors, but just can't figure out how to put it together), you could also solve it using the annihilator method and compare...It becomes a less workable alternative if you have a larger matrix (larger than 3x3) that is more complex though...

Answer 8

why is X2=kte^(lamda1)t+Pe^(lamda1)t...?

Answer 9

should be because you have a repeated evalue (1)...so lets say you solve a homogenous DE by finding the characteristic equation, factoring it, and you get roots m1=m2=3...the solution would be
y(t)=C1e^3t + C2te^3t...

Answer 10

X1=Ke^(lamda1)t ,X2=Kte^(lamda1)t+Pe^(lamda1)t...due to the repeated evalues...The general solution is a linear combination of the solution vectors...Make a little more sense?

Answer 11

yes it does thank you.