Solving Decoupled System of ODEs with Matrix b

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SUMMARY

The discussion centers on determining whether the system of ordinary differential equations (ODEs) represented by the matrix b = \begin{pmatrix}-1&0&-1\\-4&3&-1\\0&0&-2\end{pmatrix} is decoupled. It is concluded that the system is indeed decoupled, as the second equation can be solved independently of the others, and the third equation can be solved without knowledge of x_1(t). The equations derived from the matrix are \dot x_1 = -x_1 - x_3, \dot x_2 = -4x_1 + 3x_2 - x_3, and \dot x_3 = -2x_3.

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jimmycricket
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Homework Statement



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

Homework Equations


The Attempt at a Solution


I would say the matrix is decoupled since the second equation involving [itex]2x[/itex]2(t) can be solved without the other two equations. Then the third equation can be solved without knowing [itex]x[/itex]1(t). We have:

x1(t)-x2(t)-3x3(t)=x'1
2x2(t)=x'2
x2(t)+4x3(t)=x'3

Im not sure where to go from here.
 
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jimmycricket said:

Homework Statement



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

Homework Equations





The Attempt at a Solution


I would say the matrix is decoupled since the second equation involving [itex]2x[/itex]2(t) can be solved without the other two equations. Then the third equation can be solved without knowing [itex]x[/itex]1(t). We have:

x1(t)-x2(t)-3x3(t)=x'1
2x2(t)=x'2
x2(t)+4x3(t)=x'3

Im not sure where to go from here.

How did you get that? You have
[tex] \begin{pmatrix} \dot x_1 \\ \dot x_2 \\ \dot x_3 \end{pmatrix}<br /> = \begin{pmatrix}-1&0&-1\\-4&3&-1\\0&0&-2\end{pmatrix}<br /> \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix}[/tex]
Hence
[tex] \dot x_1 = -x_1 - x_3 \\<br /> \dot x_2 = -4x_1 + 3x_2 - x_3 \\<br /> \dot x_3 = -2x_3[/tex]
 
yes that's what I wrote down on paper. There was a bit of a mistranslation when trying to write it in latex.
 

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