Limit in a system I have the following system of differential equations:
I want to find the $\lim_{t\to\infty}a_t$,$\lim_{t\to\infty}b_t$, $\lim_{t\to\infty}c_t$ and $\lim_{t\to\infty}d_t$.
 A: We see that $(ac)'=ca'+ac' = -abc + ac(b-1) = -ac$; thus $ac = C^2e^{-t}$. I choose the integrating constant anticipating the derivations below.
If we take the derivative of $c$ in terms of $a$, we get $c' = -C^2e^{-t}a^{-2}(a'+a)$.
We aim to find an equation only in terms of $a$. Take the derivative of the first equation to get $a''=-ab'-a'b$. Multiply by $a$ to get $aa''=-a^2b'-a'ab=-a^2b'+(a')^2$. Recognize that $a'+b'+c'=-c$, and thus after substituting expressions for $c$ and $c'$ in terms of $a$, $b'=-c-a'-c'=-\left(C^2a^{-1}e^{-t}+a'-C^2e^{-t}a^{-2}(a'+a)\right)$. Substituting this expression into the $a''$ equation and simplifying, we arrive at the following second order nonlinear equation.
$$aa''=-[C^2e^{-t}-a^2]a'+(a')^2$$
I do not think this equation has a simple closed form solution. However, consider a particular solution of form $a=k e^{-\delta t}$. Then $aa'' = k^2 \delta^2 e^{-2\delta t} = (a')^2.$ Substituting this back into the nonlinear differential, we get $C^2e^{-t} = k^2e^{-2\delta t}$, where equating constants gives us a solution $a = Ce^{-t/2}$, and consequently $a=c$. In this scenario, $b=c'/c+1=1/2$, and $d=\int c dt = \gamma_0-2Ce^{-t/2}$. Then we fit back into the initial conditions $a(0)+b(0)+c(0)+d(0)=2C+1/2+\gamma_0-2C=1$ to get $\gamma_0=1/2$
