Solving a system of ODEs

[jimmycricket]

Given the matrix $b=\begin{pmatrix}-1&0&-1\\-4&3&-1\\0&0&-2\end{pmatrix}$ decide if the system of ODEs, $\frac{dx}{dt}=Bx$ is decoupled. If yes find the general solution x=x_h(t)

Homework Equations

The Attempt at a Solution

I would say the matrix is decoupled since the second equation involving $2x$₂(t) can be solved without the other two equations. Then the third equation can be solved without knowing $x$₁(t). We have:
$x'_1 = -x_1 - x_3 \\ x'_2 = -4x_1 + 3x_2 - x_3 \\ x'_3 = -2x_3$
Im not sure where to go from here.

 

[Filip Larsen]

You may want to look into matrix diagonalization [1]. If B can written as A D A^-1, where D is a diagonal matrix, can you then use this to rewrite you ODE system to a new uncoupled variable basis?

[1] http://en.wikipedia.org/wiki/Diagonalizable_matrix

 

[jimmycricket]

I have found the diagonal matrix,
$$D= \begin{pmatrix} -1 & 0 & 0\\ 0 & 3 & 0\\ 0 & 0 & -2 \end{pmatrix}$$
I thought the matrix B was already uncoupled though. Is this not the case?

 

[Mark44]

Given the matrix $b=\begin{pmatrix}-1&0&-1\\-4&3&-1\\0&0&-2\end{pmatrix}$ decide if the system of ODEs, $\frac{dx}{dt}=Bx$ is decoupled. If yes find the general solution x=x_h(t)

Homework Equations

The Attempt at a Solution

I would say the matrix is decoupled since the second equation involving $2x$₂(t) can be solved without the other two equations.

?
Do you mean the third equation? It involves only x₃' and x₃.

Then the third equation can be solved without knowing $x$₁(t). We have:
$x'_1 = -x_1 - x_3 \\ x'_2 = -4x_1 + 3x_2 - x_3 \\ x'_3 = -2x_3$
Im not sure where to go from here.

I have found the diagonal matrix,
$$D= \begin{pmatrix} -1 & 0 & 0\\ 0 & 3 & 0\\ 0 & 0 & -2 \end{pmatrix}$$
I thought the matrix B was already uncoupled though. Is this not the case?

The system of equations was not uncoupled. The purpose of finding a diagonal matrix that is similar to B gives you a system that is uncoupled. In an uncoupled system, each equation involves only a single variable and its derivative.

 

[jimmycricket]

That was an error on my part, what I meant was the matrix is decoupled since the second equation involving $-2x_3(t)$ can be solved without the other two equations and then we can solve for $x_1(t)$ without knowing $x_2(t)$