Consider this control problem in continuous time, known as Representative Agent Model in macroeconomics:
$$ \max_{c_t,t\ge 0}\int_{0}^{\infty}e^{-\rho t}\ln(c_t)\, \mathrm{d}t,~~~\rho\in (0,1) $$ such that: $$y_t=f(k_t) \\ \dot{k}_t=y_t-c_t\\ k_t \ge 0 \\ k_0 >0 ~~~~\text{given}$$
In this model $k_t$ is the state and $c_t$ is the control variable. We construct the Hamiltonian:
$$H_t = e^{-\rho t}\ln(c_t) + \lambda_t (f(k_t)-c_t)$$
To solve the model we should have $\frac{\partial H}{\partial c_t} = 0 $, $\frac{\partial H}{\partial k_t} = -\dot{\lambda}_t$ and also the transversality condition which is $lim_{t \rightarrow \infty} \lambda_t k_t =0$. If we use these equalities we will have
$$\dot{\lambda}_t = -\lambda_t (\rho + \frac{\dot{c}_t}{c_t})$$
and as a result the growth rate of $c_t$:
$$\frac{\dot{c}_t}{c_t} = f'(k_t)-\rho$$
Now I like to solve this model:
$$ \max_{c_t,t\ge 0}\int_{0}^{\infty}e^{-\rho t}\ln(c_t)\, \mathrm{d}t,~~~\rho\in (0,1) $$ such that: $$y_t=f(k_{1t})+g(k_{2t}) \\ \dot{k}_{1t}=\theta(y_t-c_t), ~~~ \theta\in (0,1)\\ \dot{k}_{2t}=(1-\theta)(y_t-c_t)\\ k_{1t},k_{2t} \ge 0 \\ k_{10},k_{20} >0 ~~~\text{given}$$
This system of differential equations is the result of the first order conditions:
$$ \left[ \matrix{ \theta f' & (1-\theta) f' \\ \theta g' & (1-\theta) g' \\ } \right] \left[ \matrix{ \lambda_{1t}\\ \lambda_{2t}\\ } \right] = - \left[ \matrix{ \dot{\lambda}_{1t}\\ \dot{\lambda}_{2t}\\ } \right] $$
How should I proceed to get the growth rate of $c_t$, like the first model?
Update:
The hamiltonian is:
$$H_t = e^{-\rho t}\ln(c_t) + \theta \lambda_{1t} (f(k_{1t}) + g(k_{2t})-c_t) + (1-\theta)\lambda_{2t} (f(k_{1t}) + g(k_{2t})-c_t)$$
and the FOCs:
$$ \frac{e^{-\rho t}}{c_t}=\theta \lambda_{1t} + (1-\theta)\lambda_{2t}\\ f'(\theta \lambda_{1t} + (1-\theta)\lambda_{2t}) = - \dot{\lambda}_{1t}\\ g'(\theta \lambda_{1t} + (1-\theta)\lambda_{2t}) = - \dot{\lambda}_{2t} $$
OK, I solved it. The answer is:
$$\frac{\dot{c}_t}{c_t} = \theta f'(k_t) + (1-\theta) g'(k_t)-\rho$$