4
We have that:
$$
\dot K = s K^\alpha L_0^b e^{nbt}
$$
Rewriting the differential equation gives:
$$
K^{-\alpha} \frac{dK}{dt} = s L_0^b e^{nbt}
$$
Integrate both sides with respect to $t$ from $0$ to $T$ gives:
$$
\frac{1}{b} [K^b]^T_0 = s L_0^b \frac{1}{nb}[e^{nbt}]^T_0
$$
So:
$$
K^b_T = K^b_0 - \frac{s}{n} L_0^b + \frac{s}{n} L_0^b e^{nbT}
$$
Equivalently:...
1
The direction of $\frac{\partial c^{*}}{\partial n}$ is not ambiguous.
An easy way to show this is taking derivative of $c^*=(1-s)f(k^*)$ so that $\frac{\partial c^{*}}{\partial n}=(1-s)f'\frac{\partial k^{*}}{\partial n}$ and because $f'>0$ and we can prove $\frac{\partial k^{*}}{\partial n}<0$ we thus have $\frac{\partial c^{*}}{\partial n}<0$.
...
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