- #1

usn7564

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## Homework Statement

Right so I had a standard systems of diff eqs involving repeated eigenvalues. When I find the last vector I have the equation

[tex]\begin{pmatrix}

1 & 1\\

-1 & -1

\end{pmatrix}

\begin{pmatrix}

u_1\\

u_1

\end{pmatrix}

= \begin{pmatrix}

1\\

-1

\end{pmatrix}[/tex]

## The Attempt at a Solution

[tex]u_1 + u_2 = 1[/tex]

Setting [tex]u_2 = 1 - u_1[/tex]

[tex]\bar{u} = \begin{pmatrix}

u_1\\

1-u_1

\end{pmatrix}[/tex]

[tex]\bar{u} = \begin{pmatrix}

u_1\\

1-u_1

\end{pmatrix}

=

\begin{pmatrix}

0\\1

\end{pmatrix}

+

\begin{pmatrix}

u_1\\-u_1

\end{pmatrix}[/tex]

Where the latter vector corresponds to the first answer I got from solving x' = Ax with only one eigenvalue, so ignoring that.

Now my question, the vector (0,1) might as well have been (1,0) which would have solved the system just as well. But this rather radically affects how my general solution looks.

Would I be correct in saying it doesn't matter if I chose (0,1) or (1,0) because

[tex]\bar{x} = c_1e^{rt}\begin{pmatrix}

1\\-1

\end{pmatrix}

+ c_2 ( te^{rt}

\begin{pmatrix}

1\\-1

\end{pmatrix}

+

e^{rt}\begin{pmatrix}

0\\1

\end{pmatrix}

)[/tex]

Will still span the same "e^(rt) space" depending on c_1 and c_2? Just seems odd that the general solution could be so different. ie the above being equivelant to

[tex]\bar{x} = c_1e^{rt}\begin{pmatrix}

1\\-1

\end{pmatrix}

+ c_2 ( te^{rt}

\begin{pmatrix}

1\\-1

\end{pmatrix}

+

e^{rt}\begin{pmatrix}

1\\0

\end{pmatrix}

)[/tex]

?

Last edited: