Which steady state will be attained by these system of equations? I have these two coupled ODEs...
$\dfrac{dc_1}{dt} =- k_1 c_1 c_2  + k_2c_2$ 
$\dfrac{dc_2}{dt} = Y k_1 c_1 c_2 - k_2c_2$ 
$c_1$ has two steady states, $\dfrac{k_2}{k_1}$ and $\dfrac{k_2}{Yk_1}$. 
I solved this problem numerically and it seems that the steady-state solution of $c_1$ is $\dfrac{k_2}{k_1}$. The question I have is that which steady state will be attained by  $c_1$ and why?
 A: The steady state solution of the above continuous system is given by $\dfrac{dc_1}{dt}=0,\dfrac{dc_2}{dt}=0$ simultaneously. Then you have to find out the Jacobian matrix at the corresponding equilibrium points (Steady state solution).By finding Jacobian matrix you can find the local behavior around the equilibrium points.  Then check the eigen value of the Jacobian matrix is either zero, positive or negative? If negative then it is locally stable.
Now for long term behavior of $c_1$ you can find it using Matlab by plotting $c_1$  versus time. I think the system shows different dynamics  for different set of parameters.
A: Not an answer.. But it could help solve the system.
$$\frac {dc_1}{dc_2}=\frac {dc_1}{dt}\frac {dt}{dc_2}=\frac {-k_1c_1+k_2}{Yk_1c_1-k_2}$$
$$\frac {dc_1}{dc_2}=\frac {-k_1c_1+k_2}{Yk_1c_1-k_2}$$
Whish can be solve since it's seperable
$$\int \frac {Yk_1c_1-k_2} {-k_1c_1+k_2}dc_1=\int dc_2=c_2+C$$
$$-Y\int \frac {c_1-k_2/Yk_1} {c_1-k_2/k_1}dc_1=c_2+C$$
Substitute for convenience $\alpha=k_2/k_1$
$$-Y\int \frac {c_1-\alpha/Y} {c_1-\alpha}dc_1=c_2+C$$
Substitute $v=c_1-\alpha$
$$-Y\int \frac {v+\alpha-\alpha/Y} {v}dv=c_2+C$$
Which is easy to integrate...
$$v+\alpha(1- \frac 1Y)\ln|v| =-\frac {c_2}Y+C$$
$$\boxed{c_1+\frac {k_2}{k_1}(1- \frac 1Y)\ln|c_1-\frac{k_2}{k_1}| =-\frac {c_2}Y+K}$$
