Help me solve a system of three linear ODEs. I seem to have grasped the $2\times 2$ systems but $3\times 3$ ones give me problems. I solved one $3\times 3$ using the method of elimination. I'm postponing the matrix method for a while.
Here's the system: $$ x' = 3x + y - z $$ $$ y' = x + 2y - z $$ $$ z' = 3x + 3y -z $$
What I do is I turn it into operator notation: $$ (D-3)x - y + z = 0 $$ $$ -x + (D-2)y + z = 0 $$ $$ -3x - 3y + (D+1)z = 0 $$ Then into $2\times 2$ system by eliminating $z$ - first two equations to get $$ (D-2)x + (-D+1)y = 0 $$ and then last two to get: $$ -(D-2)x + (D^2-D+1)y = 0 $$ Now, if I eliminate $x$ from the $2\times 2$ system easily by adding them together I get $$ (D^2-2D+2)y = 0 $$ which gives a complex solution $$ r_1 = 1 + i $$ and $$ r_2 = 1 - i $$ which gives as the solution for $y$: $$ y = C_1 \mathrm{e}^t \cos(t) + C_2 \mathrm{e}^t \sin(t) $$ So, only two constants - $C_1$ and $C_2$. But, it should have three constants since this is $3 \times 3$ system. At least, that's how it was in another solution of $3\times 3$ system that I solved. How do I solve this using elimination method?