The question is:
Solve the initial value problem:
$\begin{align*} \frac{dx_1}{dt}&=40x_1-6x_2+18x_3,\\ \frac{dx_2}{dt}&=-6x_1+45x_2+12x_3,\\ \frac{dx_3}{dt}&=18x_1+12x_2+13x_3,\\ \end{align*} $ with $ \begin{align*} x_1(0) &= 3,\\ x_2(0) &= -4,\\ x_3(0) &= 12\\ \end{align*} $
So my A matrix is:
$A=\pmatrix{40&-6&18\cr-6&45&12\cr18&12&13\cr}$ and to find the eigenvalues I find the determinant of this matrix:
$A=\pmatrix{40-\lambda&-6&18\cr-6&45-\lambda&12\cr18&12&13-\lambda\cr}$
which gives me these eigenvalues and vectors:
$\lambda _1=0 --- v_1=(-3,-2,6)$
$\lambda_{2,3}=49 --- v_2=(2,0,1)$ and $v_3=(-2,3,0)$
The place I'm having trouble is, from this question I know that the general solution for real eigenvalues is:
$x=c_1e^{a_1t}v_1+c_2e^{a_2t}v_2+c_3e^{a_3t}v_3$
but what would it be for repeated roots? Is it:
$x=c_1e^{a_1t}v_1+c_2e^{a_2t}v_2+c_3te^{a_3t}v_3$