Differential equation, linear sys. of eq. problem

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  • #1
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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:

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
41,833
963

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]
You mean
[tex]\begin{pmatrix}u_1\\ u_2\end{pmatrix}[/tex]
right?

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 so ignoring that.
I have no idea what you are doing here! Yes, it is true that [tex]u_2= 1- u_1[/tex].
But that because [tex]u_1= 1[/tex] and [tex]u_2= 0[/tex]!

Now my question, the vector (0,1) might as well have been (1,0) which would have solved the system just as well.
No, it doesn't. If your vector is (0, 1), that is if [tex]u_1= 0[/tex] and [tex]u_2= 1[/tex], the first equation, [tex]u_1+ u_2= 1[/tex] (which is the only one you have looked at so far) but your second equation,n [tex]-u_1+ u_2= -1[/tex] would NOT be satisfied: -0+ 1= 1, not -1.

But this rather radically affects how my general solution looks.
What "general solution"? The equation
[tex]\begin{pmatrix}1 & 1 \\ -1 & 1 \end{pmatrix}\begin{pmatrix}u_1 \\ u_2\end{pmatrix}= \begin{pmatrix}1 \\ -1\end{pmatrix}[/tex]
has the unique solution [itex](u_1, u_2)= (1, 0)[/tex]. There is NO "general solution".

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]
?
 
  • #3
63
0
Christ, no matter how many times I look it over longer latex statements get mistakes. Sorry, corrected the matrix now. I'm referring to the general solution of the differential equation I started with which I didn't detail as everything but the u vector was correct in my solution.
 

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