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?

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 $A$
