General Solution of Linear System

2017-12-03 03:05:32

We need to find the general solution of the differential system.

I have

${dx\over dt}=3x-y$

${dy\over dt} = 9x-3y$

The problem is that I only found one eigenvector. Just wondering how I can have a solution with just one eigenvector.

Thanks for any help.

from the first equation we get

$$x''=3x'-y'$$ and $$y=3x-x'$$ then we have $$y'=9x-3(3x-x')$$ and finally

$$x''=3x'-(9x-3(3x-x'))$$

you can also write

$$y'=3(3x-y)=3x'$$

Observe that the coefficient matrix is nilpotent: $$A^2=\begin{bmatrix}3&-1\\9&-3\end{bmatrix}^2=\begin{bmatrix}0&0\\0&0\end{bmatrix}.$$ You might also have noticed this when you found that the only eigenvalue is $0$ with an algebraic multiplicity of two but a geometric multiplicity of one. This means that $$e^{tA} = I+tA+{t^2\over2!}A^2+\cdots = I+tA.$$ Can you complete the solution from here?

More generally, for a $2\times2$ system like this, when you get a repeated eigenvalue $\lambda$, but the matrix isn’t a multiple of the ident

  • from the first equation we get

    $$x''=3x'-y'$$ and $$y=3x-x'$$ then we have $$y'=9x-3(3x-x')$$ and finally

    $$x''=3x'-(9x-3(3x-x'))$$

    you can also write

    $$y'=3(3x-y)=3x'$$

    2017-12-03 04:27:07
  • Observe that the coefficient matrix is nilpotent: $$A^2=\begin{bmatrix}3&-1\\9&-3\end{bmatrix}^2=\begin{bmatrix}0&0\\0&0\end{bmatrix}.$$ You might also have noticed this when you found that the only eigenvalue is $0$ with an algebraic multiplicity of two but a geometric multiplicity of one. This means that $$e^{tA} = I+tA+{t^2\over2!}A^2+\cdots = I+tA.$$ Can you complete the solution from here?

    More generally, for a $2\times2$ system like this, when you get a repeated eigenvalue $\lambda$, but the matrix isn’t a multiple of the identity, you can decompose it into the sum $\lambda I+N$ of a diagonal and nilpotent matrix (i.e., $N=A-\lambda I$) for which $N^2=0$ by Cayley-Hamilton. Since multiples of the identity matrix commute with everything, you then have $$e^{tA}=e^{t(\lambda I+N)}=e^{\lambda tI}e^{tN} = e^{\lambda t}(I+tN).$$

    2017-12-03 04:43:32