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I am confused on how to do this problem, it states:

Find a fundamental set of solutions and put it in general form for the given system.

$x' = \begin{pmatrix} -1/2 & 1\\ -1 & -1/2 \end{pmatrix}x$

I got the eigen values to be $\lambda_1$ = [ (-1/2) + i ] , $\lambda_2$ = [ (-1/2) - i ] so the corresponding eigen vectors are

For $\lambda_1 = v_1 = \begin{pmatrix} 1 \\ i \end{pmatrix}$

For $\lambda_2 = v_2 = \begin{pmatrix} 1\\ -i \end{pmatrix}$

But here is where I get confused on how to write the general solution using Euler's formula for $e^{it} = cost + isint$, thus we have that

$x_1(t) = e^{-t/2}(cost + isint)\begin{pmatrix} 1\\ i \end{pmatrix}$

The answer is below, but how did they get that?

$x_1(t) = \begin{pmatrix} e^{-t/2} cost\\ -e^{-t/2}sint \end{pmatrix} + i\begin{pmatrix} e^{-t/2}sint\\ e^{-t/2}cost \end{pmatrix} = u(t) + iw(t)$

$u(t) = \begin{pmatrix} e^{-t/2} cost\\ -e^{-t/2}sint \end{pmatrix} , w(t) = \begin{pmatrix} e^{-t/2}sint\\ e^{-t/2}cost \end{pmatrix}$

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The fundamental set of solutions should have two elements. You have only written one. Where's the other one? –  Gerry Myerson Oct 4 '12 at 3:39
    
I edited it above –  diimension Oct 4 '12 at 3:47

1 Answer 1

up vote 1 down vote accepted

Based on your derivations, the general solution has the form

$$ x(t) = c_1 \begin{pmatrix} 1 \\ i \end{pmatrix} {\rm e}^{(-\frac{1}{2}+i)t} + c_2 \begin{pmatrix} 1\\ -i \end{pmatrix} {\rm e}^{(-\frac{1}{2}-i)t}\,, $$

where $c_1,c_2$ are arbitrary constants. Simplifying further

$$ x(t) = c_1 {\rm e}^{-\frac{1}{2}t} \begin{pmatrix} 1 \\ i \end{pmatrix} {\rm e}^{it} + c_2 {\rm e}^{-\frac{1}{2}t} \begin{pmatrix} 1\\ -i \end{pmatrix} {\rm e}^{-it} $$ $$= c_1{\rm e}^{-\frac{1}{2}t} \begin{pmatrix} 1 \\ i \end{pmatrix}(\cos(t)+i\sin(t)) + c_2 {\rm e}^{-\frac{1}{2}t} \begin{pmatrix} 1\\ -i \end{pmatrix} (\cos(t)-i\sin(t))\,. $$

Now, make the above as a linear combination of $\cos(t){\rm e}^{-\frac{1}{2}t}$ and $\sin(t){\rm e}^{-\frac{1}{2}t}$.

$$ A\cos(t){\rm e}^{-\frac{1}{2}t} + B \sin(t){\rm e}^{-\frac{1}{2}t}\,, $$

where $A$ and $B$ are two constant vectors given by

$$ A = c_1\begin{pmatrix} 1 \\ i \end{pmatrix}+c_2 \begin{pmatrix} 1\\ -i \end{pmatrix} \,,\quad B = ic_1\begin{pmatrix} 1 \\ i \end{pmatrix}+ic_2 \begin{pmatrix} 1\\ -i \end{pmatrix} \,. $$

Or, you can write them in terms of new constants,

$$ A= \begin{pmatrix} a_1 \\ a_2 \end{pmatrix}\,, \quad B = \begin{pmatrix} b_1 \\ b_2 \end{pmatrix} \,.$$

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Yea, but how can I put it as a real valued vector functions using Euler's formula indicated in my post? –  diimension Oct 4 '12 at 3:48
    
Thank you very much but when I calculated your answer it didnt match to the answer the book gave me? –  diimension Oct 4 '12 at 4:08
    
@diimension:I added more material. –  Mhenni Benghorbal Oct 4 '12 at 4:14
    
Thank you but im still confused. I got stuck on where A and B are two constants. How can I formally distribute those to my new form? –  diimension Oct 4 '12 at 4:31
1  
@diimension:Are you given initial condition? If yes, then you nedd only to plug in the first equation in my answer and solve the system for $c_1$ and $c_2$, then plug them back in first equation. –  Mhenni Benghorbal Oct 4 '12 at 4:57

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