The Volterra-Lotka model (predator prey equations) -- Linearization ODE 
Q: Find all fixed points of the equation, linearize the equation, substitute the origin point $(0, 0)$ into it and solve the linear version of Volterra-Lotka model.
The system looks like this (where $a,b,c,g,y,x_{0}$ are constants):
\begin{align}
\frac{dx}{dt}&=ax-gx^2-by\left(x-x_0\right)\\
\frac{dy}{dt}&=-cy+dy\left(x-x_0\right)
\end{align}

My take:
Critical point: $$
\begin{pmatrix}0\\ 0\end{pmatrix}
$$
Jacobian Matrix:
$$
\begin{pmatrix}a&bx_0\\ 0&-c-dx_0\end{pmatrix}
$$
$$
P(\lambda) = -(a-\lambda)(c+dx_0 + \lambda) =0  \implies \begin{cases}
\lambda_1 = a \\
\lambda_2 = -c-dx_0
\end{cases}
$$
I get two matrices:
For V1:
$$
\begin{pmatrix}0&bx_0\\ 0&-c-dx_0-a\end{pmatrix}
$$
For V2:
$$
\begin{pmatrix}a+c+dx_{0}&bx_{0}\\ \:\:0&0\end{pmatrix}
$$
Eigenvector 1 :
$$
\begin{pmatrix}1\\ 0\end{pmatrix}
$$
Eigenvector 2 :
$$
\begin{pmatrix}-bx_0\\ a+c+dx_0\end{pmatrix}
$$
So the solution is :
$$
C_1\cdot e^{at}\begin{pmatrix}0\\ 1\end{pmatrix}\:+\:C_2\cdot e^{-t\left(c+dx_0\right)}\cdot \begin{pmatrix}-bx_0\\ a+c+dx_0\end{pmatrix}
$$
I don't think that that's the solution - maybe I have an error with Eigenvector 2 ?
We didn't really went over this material so I'm having a hard time.
 A: I made a few edits to your question for things that looked like typos, because they didn't show up in your final answer. In particular, you had
$$\begin{pmatrix}-\color{red}{3}x_0\\ a+c+dx_0\end{pmatrix}$$
instead of
$$\begin{pmatrix}-\color{red}{b}x_0\\ a+c+dx_0\end{pmatrix}$$ and you had $$\begin{pmatrix}a+c+d\color{red}{g}&b\:\color{red}{g}\\ \:\:0&0\end{pmatrix}$$ instead of $$\begin{pmatrix}a+c+d\color{red}{x_{0}}&b\color{red}{x_{0}}\\ \:\:0&0\end{pmatrix}.$$
There's also a typo in your final answer, in particular you flipped your first eigenvector, so it should be
$$\begin{pmatrix}x\\y\end{pmatrix} = C_1e^{at}\color{red}{\begin{pmatrix}1\\ 0\end{pmatrix}}+C_2 e^{-t\left(c+dx_0\right)}\begin{pmatrix}-bx_0\\ a+c+dx_0\end{pmatrix}.$$
This is indeed the correct answer, and you can verify that it satisfies your linearized system of ODEs:
\begin{align}
\frac{dx}{dt} &= ax + bx_{0}y\\
\frac{dy}{dt} &= -(c+dx_{0})y. 
\end{align}
Differentiating your solution
\begin{align}
\frac{dx}{dt} &=C_{1}ae^{at} + C_{2}bx_{0}(c+dx_{0})e^{-(c+dx_{0})t}\\ 
&= C_{1}ae^{at} + C_{2}bx_{0}(c+dx_{0})e^{-(c+dx_{0})t} + C_{2}bx_{0}ae^{-(c+dx_{0})t} - C_{2}bx_{0}ae^{-(c+dx_{0})t}\\
&=a\left(C_{1}e^{at} - C_{2}bx_{0}e^{-(c+dx_{0})t}\right) + bx_{0}C_{2}(a + c + dx_{0})e^{-(c+dx_{0})t}\\
&=ax + bx_{0}y\\
\frac{dy}{dt} &= -(c+dx_{0})C_{2}(a+c+dx_{0})e^{-(c+dx_{0})t}\\
&=-(c+dx_{0})y.
\end{align}
