Determine flow curve for vector field I have the following problem:

We have the vector field $V(x,y,z)=\bigg(3x+3z,3y,-3x+3z \bigg)$. Let $r(t)$ denote the flow curve for $V$. At $t=0$ the flow curve passes through the point $(1,2,3)$.
Determine the coordinates for $r(t)$ to the times $t=1$ and $t=3$.

My try
Okay, so let's say our flow curve is in the form: $r(t)=\bigg(x(t),y(t),z(t) \bigg)$. By definition, the flow curve fullfils following.
$$V(r(t))=r'(t) $$
which with our vector field can be written as:
$$\bigg(x'(t),y'(t),z'(t) \bigg)=V(x,y,z) =\bigg(3x(t)+3z(t),3y(t),-3x(t)+3z(t) \bigg)$$
We now have 3 differential equations:
$$x'(t)=3x(t)+3z(t) $$
$$y'(t)=3y(t)$$
$$z'(t)=-3x(t)+3z(t)$$
Including the starting conditions and solving this with Maple gives me: $x(t)=e^{3t}(3\sin(3t)+ \cos(3t)$,
$y(t)=2e^{3t}$,
$z(t)=-3e^{3t}(\sin(3t)-3 \cos(3t))$.
Inserting $t=1$ and $t=3\:$ yields the coordinates:
$$r(1)=\begin{bmatrix}e^3(3\sin(3)+\cos(3))\\2e^3 \\ -e^3(\sin(3)-3\cos(3))= \end{bmatrix}$$
$$r(3)=\begin{bmatrix}e^9(3\sin(9)+\cos(9))\\2e^9 \\ -e^9(\sin(9)-3\cos(9))= \end{bmatrix}$$
However, these aren't very nice results, and I'm starting to doubt my answers. Can anyone confirm my answer, or maybe tell me where I messed up?
 A: This vector field is of the form $A(x)$ where $A$ is a linear matrix, it flow is $r(t)=exp(tA).x$, $r'(t)=A(exp(tA).x)=V(r(t))$. You need to compute $exp(A)$, Try to diagonalize the matrix $A$
A: The equation for $y(t)$ is $y'(t) = 3y(t)$.  The general solution is $y(t) = Ce^{3t}$.  The initial condition $y(0) = 2$ require that $C=2$.  So that takes care of the $y$ coordinate of the integral curve.
As for the other two coordinates, we decouple them by differentiating and re-substituting.
\begin{align*}
     x'' &= 3x' + 3z' \\
         &= 3x' + 3(-3x+3z) \\
         &= 3x' + 3(-3x + (x'-3x)) \\
         &= 6x'-18x
\end{align*}
If $x(t) = e^{rt}$ were a solution, then $r$ would satisfy the characteristic equation
$$
    r^2 - 6r + 18 = 0
$$
This has solutions $r = 3 \pm 3i$.  So the general solution to the $x$ equation is either
$$
    x(t) = C_1 e^{(3+3i)t} + C_2 e^{(3-3i)t}
$$
or, using Euler's identity $e^{i\theta} = \cos \theta + i \sin \theta$,
$$
    x(t) = A e^{3t} \cos 3t + B e^{3t} \sin 3t
$$
The initial condition $x(0) = 1$ requires $A = 1$.  We don't know what $B$ is yet.  But
$$
    z = \frac{1}{3}(x-3x') = B e^{3t}\cos 3t + (2-B) e^{3t} \sin 3t
$$
Since $z(0) = 3$, we know $B=3$.  Putting this together, we have
$$
   r(t) = \begin{bmatrix} x(t) \\ y(t) \\ z(t) \end{bmatrix}
        = \begin{bmatrix} e^{3t} \cos 3t + 3 e^{3t} \sin 3t \\
                         2e^{3t} \\
                         3e^{3t} \cos 3t - e^{3t} \sin 3t
          \end{bmatrix}
$$
