Analytical solution to first-order rate laws
This is the solution I get:

$$\begin{align}\rho _T(t) &amp;= -\frac{K_2 \rho _{w0} \left(e^{\left(-K_1-K_2-K_3\right) t}-e^{t (-K4-K5)}\right)}{K_1+K_2+K_3-K4-K5}\\\rho _W(t) &amp;= e^{\left(-K_1-K_2-K_3\right) t} \rho _{w0}\\\rho_C(t) &amp;=\tiny{-\frac{\rho _{w0} e^{-t \left(K_1+K_2+K_3+K4+K5\right)} \left(K_2^2 K5 e^{\left(K_1+K_2+K_3\right) t} \left(e^{t (K4+K5)}-1\right)+K_2 \left(K5 (K4+K5) \left(e^{t (K4+K5)}-e^{t \left(K_1+K_2+K_3+K4+K5\right)}\right)+K_1 K5 e^{\left(K_1+K_2+K_3\right) t} \left(e^{t (K4+K5)}-1\right)-K_3 \left(-(K4+2 K5) e^{t \left(K_1+K_2+K_3+K4+K5\right)}+K5 e^{\left(K_1+K_2+K_3\right) t}+(K4+K5) e^{t (K4+K5)}\right)\right)+K_3 (K4+K5) \left(-K_1-K_3+K4+K5\right) \left(e^{t (K4+K5)}-e^{t \left(K_1+K_2+K_3+K4+K5\right)}\right)\right)}{\left(K_1+K_2+K_3\right) (K4+K5) \left(-K_1-K_2-K_3+K4+K5\right)}}\end{align}$$

Solving this by hand isn't that hard, as the coupling is unidrectional. So, we can solve for $\rho_W$ first, and substitute it in the equation for $\frac{d\rho_T}{dt}$.
Now, we have an equation of the form $\frac{dy}{dt}=Ay+Be^{kt}$, which has a solution of the form $y=Ce^{At}+De^{kt}$, where C,D can be found by back-substitution. Now, we subtitute the values for $\rho_T, \rho_W$ in the last equation, and we get an equation of the form $\frac{d\rho_C}{dt}=f(t)$ ($f(t)$ here is some combination of exponentials), which can easily be solved by integrating $f(t)$