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Well, you can actually solve $k$ analytically. This is just standard first-order linear ODE technique. The solution of the differential equation $ \dot{c}(t) / c(t) = z(t) - \rho $ is $$ c(t) = c_0 \exp\left( \int_0^t z(s) ds - \rho t \right) $$ Plugging this into $ \dot{k}(t) - z(t)k(t) = - c(t) $, multiplying both sides by $ \exp\left( -\int_0^t z(s) ds \...


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