System of differential equation stability, with initial condition Consider the following system of differential equations:
$$ \dot x(t) = 3x(t) - 2y(t) + 3$$
$$\dot y(t) = 2x(t) - 2y(t) - 1$$
(a) Find the steady state of the system and determine its stability.
(b) Find the particular solution with initial conditions $$x(0) = 0.5$$$$
  y(0) = 0$$
I find the general solution$$ x(t) = 2C_1 e^{2t}+C_2 e^{-t}-4 $$
$$y(t) = C_1 e^{2t}+2C_2 e^{-t}  -4.5$$ 
I can see that $C_1 = 0$ otherwise the terms explode, so:
$$ x(t) = C_2 e^{-t}-4 $$
$$y(t) = 2C_2 e^{-t}  -4.5$$
But now, the initial conditions lead to different particular solutions, I have done the problem several times, and still cannot seem to find out why. Where have I misstepped? 
 A: The steady state of the system is the point where both derivatives are zero simultaneously, i.e. the intersection of the lines $$ 3x-2y+3=0 \text{ and } 2x-2y-1=0 $$
which occurs at the point $(x,y) = (-4,-9/2)$. We make a change of variables so that the equilibrium is at the origin, which should make a homogeneous linear system. Let $\xi = x+4$ and $\eta = y+9/2$. This gives us the system $$ \dot\xi = 3\xi-2\eta \\ \dot\eta =2\xi - 2\eta$$
Stability of this system is determined by the matrix $A = \pmatrix{3&-2\\2&-2}$, which has $\text{tr}A=1 > 0$ and $\text{det}A= -2 < 0$. This means that the equilibrium at $(\xi,\eta) = (0,0)$ is a saddle, which is unstable. Thus, after requiring that the solutions blow up, there will be no solutions with initial conditions off of the stable manifold.
Shifting back into the original coordinates, we draw the same conclusions with $(-4,-9/2)$.
We can see that, after requiring $C_1 = 0$, if we want to satisfy general initial condition $(x_0,y_0)$, we need $$C_2 = x_0 + 4 \\ C_2 = \frac{1}{2}y_0 + \frac{9}{4}$$
which means that only initial conditions on the line $y = 2x+\frac{7}{2}$ will lead to stable solutions, i.e. this line is the stable manifold.
We can clearly see that $0\neq 2(0.5)+\frac{7}{2}$, so the given initial conditions will not lead to a stable solution.
A: Hint :
It is $x(0) = 1/2$, thus :
$$x(0) = 0 \Rightarrow 2C_1 + C_2 -4 = 1/2$$
Also, $y(0) = 0$, thus :
$$y(0) = 0 \Rightarrow C_1 + 2C_2- 9/2 = 0$$
