Changing parameters in a 3x3 Matrix trying to find the general solution.
Solution 1
If you look at that matrix right, its determinant is obviously $(b\lambda)(\lambda^2a^2)$, which should make it a breeze to find eigenvalues and eigenvectors.
Solution 2
$X^{'}= \begin{pmatrix} 0&0&a \\ 0&b&0\\ a&0&0 \end{pmatrix}X$ , $X=\begin{pmatrix} x\\y\\z\end{pmatrix}$$$\begin{pmatrix} x'\\y'\\z'\end{pmatrix} =\begin{pmatrix} 0&0&a \\ 0&b&0\\ a&0&0 \end{pmatrix}\begin{pmatrix} x\\y\\z\end{pmatrix}$$then we have $$\begin{cases} x'=az & \text{} \\ y'=by& \text{} \\ z'=ax & \text{} \\ \end{cases}$$ w e have $y=e^{bt+d_1}$$$x'=az\to x''=az'\to x''=a^2x$$ easily find x then by substitution find z
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Faust
Updated on March 05, 2020Comments

Faust over 3 years
Consider the system $$X^{'}= \begin{pmatrix} 0&0&a \\ 0&b&0\\ a&0&0 \end{pmatrix}X$$
depending on the two parameters a and b.
1) find the general solution of this system. \begin{pmatrix} \lambda&0&a \\ 0&(b\lambda)&0\\ a&0&\lambda \end{pmatrix} $\lambda(b\lambda) (\lambda)+ a(b\lambda) (\lambda) = \lambda^{3} +(b+a) \lambda^{2}ab\lambda $
$(\lambda^{2} (b+a) \lambda+ab) (\lambda)$
$(\lambda [(b+a)/2])^{2} = ab + [(a+b)^{2}/4]$
$\lambda_{1} = (ab + [(a+b)/4])^{1/2}$ $\lambda_{2} = (ab + [(a+b)/4])^{1/2}$ $\lambda_{3} = 0$
im not sure how to force the eigenvalues out of this and subbing this mess into that equation for each eigenvalue is there an easier way to compute this?
EDIT as pointed out below this is a mess by expanding down the second column of the matrix we have $\lambda_{1} = b$ $\lambda_{2} = a$ $\lambda_{3} = a$
Where $V_{1}=<0,1,0>$ $V_{2}=<1,0,(1)>$ $V_{3}=<1,0,(1)>$
Why isnt this the general solution?
$C_{1} e^{bt} V_{1}$ + $C_{2} e^{at} V_{2}$ + $C_{3} e^{at} V_{3}$
2) sketch the region in the ab plane where this system has different types of phase portraits.

Faust over 10 yearssorry i don't really understand how u got all those things. or in general what that even says, perhaps its time for me to go to sleep...

Faust over 10 yearsim not sure how you got that but, that would mean $\lambda$ = b,a,a Edit nvm im a moron

Gerry Myerson over 10 yearsEvaluate the determinant by expansion along the second row.

Gerry Myerson over 10 years@julien, $z'=ax$.

Julien over 10 years@GerryMyerson Yes, sure, I added a typo to a typo...Thanks.

Julien over 10 yearsNice observation, Maisam. So doing this, we find necessary conditions on $x,y,z$. And we need to check that they work afterwards. +1.

Faust over 10 yearsi follow the train of logic to get to $x^{''}=a^{2}x$

M.H over 10 years@Faust7:use characteristics polynomial to solve it $(D^2a^2)x=0$

Faust over 10 yearsSorry i have never seen anything like this before i am not sure why your trying to find x?

Faust over 10 yearsare you talking about the vectors?

M.H over 10 years@Faust7:i offer look on this book differential equation and dynamical system an introduction to chaos of Morris sorry i dont have time to explain i must go university

Faust over 10 yearshaha, you have been more then helpful sir thank you very much for all your help!