So I have this question:
Solve the initial value problem:
$\begin{align*} \frac{dx_1}{dt}&=3x_3-2x_4,\\ \frac{dx_2}{dt}&=-2x_3+3x_4,\\ \frac{dx_3}{dt}&=3x_1-2x_2,\\ \frac{dx_4}{dt}&=-2x_1+3x_2,\\ \end{align*} $ with $ \begin{align*} x_1(0) &= 1,\\ x_2(0) &= 2,\\ x_3(0) &= -1,\\ x_4(0) &= 3. \end{align*} $
I know the first step is to form the coefficient matrix and then I have to find the eigenvalues....but I really just know the idea of what I'm supposed to do...I don't actually know how to start. Where do I even begin with this problem?
Edit: Okay so I know
$A=\pmatrix{0&0&3&-2\cr0&0&-2&3\cr3&-2&0&0\cr-2&3&0&0\cr}$ and to find the eigenvalues I find the determinant of this matrix:
$A=\pmatrix{0-\lambda&0&3&-2\cr0&0-\lambda&-2&3\cr3&-2&0-\lambda&0\cr-2&3&0&0-\lambda\cr}$
which gives me the eigenvalues: $\lambda=-5,5,-1,1$
Which after pluggin in and solving gives me the eigenvectors:
$v_1=(1,-1,-1,1)\\ v_2=(-1,1,-1,1)\\ v_3=(-1,-1,1,1)\\ v_4=(1,1,1,1) $
But I just don't know WHAT to do with those vectors...Or am i on the wrong track?
Edit2: Think I figured it out, all thanks to Gerry Myerson
We have $X=\left[ \begin{array}{ccc} c_1e^{a_1t} & -c_2e^{a_2t} & -c_3e^{a_3t} & c_4e^{a_4t} \\ -c_1e^{a_1t} & c_2e^{a_2t} & -c_3e^{a_3t} & c_4e^{a_4t} \\ -c_1e^{a_1t} & -c_2e^{a_2t} & c_3e^{a_3t} & c_4e^{a_4t} \\ c_1e^{a_1t} & c_2e^{a_2t} & c_3e^{a_3t} & c_4e^{a_4t} \end{array} \right]$
Then I just use the inital values:
$X(0)=\left[ \begin{array}{ccc} c_1 & -c_2 & -c_3 & c_4 \\ -c_1 & c_2 & -c_3 & c_4\\ -c_1 & -c_2 & c_3 & c_4 \\ c_1 & c_2 & c_3 & c_4 \end{array} \right]= \left[ \begin{array}{ccc} 1 \\ 2\\ -1 \\ 3 \end{array} \right]$
Solving that gives these values to the constants: $c_1=\frac{3}{4}, c_2=\frac{5}{4}, c_3=-\frac{1}{4}, c_4=\frac{5}{4}$
And finally putting it all together, my solution is: $X=\left[ \begin{array}{ccc} \frac{3}{4}e^{a_1t} -\frac{5}{4}e^{a_2t}+ \frac{1}{4}e^{a_3t}+\frac{5}{4}e^{a_4t} \\ -\frac{3}{4}e^{a_1t}+\frac{5}{4}e^{a_2t}+\frac{1}{4}e^{a_3t}+\frac{5}{4}e^{a_4t} \\ -\frac{3}{4}e^{a_1t}-\frac{5}{4}e^{a_2t}-\frac{1}{4}e^{a_3t}+\frac{5}{4}e^{a_4t} \\ \frac{3}{4}e^{a_1t}+\frac{5}{4}e^{a_2t}-\frac{1}{4}e^{a_3t}+\frac{5}{4}e^{a_4t} \end{array} \right]$