Since more than 48 hours has gone by with no additional feedback from the OP, I will finish the problem for the benefit of future readers.
Now, we see that the second equation has the characteristic root:
$$r=3$$
and so the general solution for that equation is:
$$y(t)=c_1e^{3t}$$
Now, substituting this into the first equation, there results:
$$\d{x}{t}=x+2c_1e^{3t}+1$$
Writing this ODE in standard linear form, we obtain:
$$\d{x}{t}-x=2c_1e^{3t}+1$$
We see by inspection that the integrating factor is:
$$\mu(t)=e^{-t}$$
And hence, then ODE becomes:
$$e^{-t}\d{x}{t}-xe^{-t}=2c_1e^{2t}+e^{-t}$$
We may now rewrite the left side:
$$\frac{d}{dt}\left(e^{-t}x\right)=2c_1e^{2t}+e^{-t}$$
Integrating both sides with respect to $t$, we obtain:
$$e^{-t}x=c_1e^{2t}-e^{-t}+c_2$$
And so we obtain:
$$x(t)=c_1e^{3t}+c_2e^{t}-1$$
The general solution is thus:
$$Y(t)=\left(x(t),y(t)\right)=\left(c_1e^{3t}+c_2e^{t}-1,c_1e^{3t}\right)$$
To find the equilibrium points, we equate both derivatives to zero:
$$x+2y+1=0$$
$$3y=0$$
The second equation implies $y=0$ and so the first equation implies $x=-1$, and so the sole equilibrium point is:
$$(x,y)=(-1,0)$$
Using the conventional notation:
$$x_0=x(0)$$ and $$y_0=y(0)$$
we may determine the particular solution satisfying the given initial conditions as follows:
$$x(0)=c_1+c_2-1=-1\implies c_1+c_2=0$$
$$y(0)=c_1=3\implies c_2=-3$$
Hence, we find the particular solution:
$$Y_p(t)=\left(x(t),y(t)\right)=\left(3e^{3t}-3e^{t}-1,3e^{3t}\right)$$
