Find the GS of the System of DE's $\begin{cases} x' = x-3y\\ y'=3x+7y \end{cases}$ Find the GS of the following system of DE's where the independent variable is $t$ and $x$ and $y$ are the dependent variables
\begin{cases}
x' = x-3y\\
y'=3x+7y
\end{cases}
I know using eigenvalues and eigenvectors or operators is one way to do this. But I wish to double check my answer using a substitution method.
So my work:
The second DE $y'=3x+7y$ can be rewritten as $x = \cfrac{y'}{3}-\cfrac 73y$
then $x' = \cfrac{y''}{3}-\cfrac73y'$
When we plug these values of $x$ and $x'$ into the first DE ($x' = x -3y)$, we get with some rearranging
$\cfrac{y''}{3}-\cfrac83y'+\cfrac{16}{3}y = 0$
Which has a characteristic equation of
$\cfrac{r^2}{3}-\cfrac83r+\cfrac{16}{3} = 0$
with roots $r_1=4$ and $r_2 = 4$
Then the solution for $y$ is $y$ = $C_1e^{4t}+C_2te^{4t}$
Then we back sub to solve for $x$ using $x = \cfrac{y'}{3}$$-\cfrac73y$ with the solution of y we just found.
We get $x =-C_1e^{4t}-C_2te^{4t} + \cfrac{C_2}{3}e^{4t} =-C_1e^{4t}-C_2te^{4t} + C_3e^{4t}$
so the GS to the homo system is
\begin{cases}
x = -C_1e^{4t}-C_2te^{4t}+C_3e^{4t}\\
y = C_1e^{4t}+C_2te^{4t}
\end{cases}
If this solution is right, then I'm confident that I understand how substitution method works for solving DE systems. (Also it would boost my confidence in using the operator method to solve this as I got the same answer as this using the operator method). I'm a little thrown off on the roots being the same but I still think my methodology is still sound. I would appreciate it if someone could tell me if I've got this right cause then I know I completely understand how to solve a system of DE's.
If more work is necessary to show please let me know.

$$\begin{cases}
x = -C_1e^{4t}-C_2te^{4t}+C_3e^{4t}\\
y = C_1e^{4t}+C_2te^{4t}
\end{cases}$$
You should only end with two constants not three.
$$y=C_1e^{4t}+C_2te^{4t}$$
$$y'=4C_1e^{4t}+C_2e^{4t}+4tC_2e^{4t}$$
$$y'=4C_1e^{4t}+C_2e^{4t}(1+4t)$$
And since:
$$x = \cfrac{y'}{3}-\dfrac 73 y$$
You should only have $C_1$ and $C_2$ constant. You applied the method of substitution correctly.
