Population ratio of Birth control to no birth control A country currently has a population of $N_0$ and growth rate of
$a_0$. The country introduces, at $t = 0$, a birth control scheme which
hopes to gradually reduced the growth rate to $a_1 < a_0$ over a period
of time $T$.
Using the formula for birth control which I have already worked out, derive the ratio of the population size with the birth control policy to that without the policy at time T.
$N_{\text{birth control}}(t) = N_0\exp\left[a_0 t - (a_0 - a_1)\frac{t^2}{2T}\right]$
Attempted solution:
Since there are no restrictions or boundaries to the original growth rate, then  I assumed, the formula for no birth control at time T, would just be:
$N_{\text{no birth control}}(t) = N_0\text{exp}[a_0t]$
Then inputting $T$ into the birth control equation, I would get:
$N_{\text{birth control}}(T) = N_0\exp\left[(a_0 + a_1)\frac{T}{2}\right]$
So then I would simply have to find the ratio between the two, resulting in:
$$\frac{N_{\text{birth control}}(T)}{N_{\text{no birth control}}(T)} = \frac{\text{exp}[(a_0 + a_1)T/2]}{\text{exp}[a_0T]}$$
However upon finding the solution, to be:
$$\frac{N_{\text{birth control}}(T)}{N_{\text{no birth control}}(T)} =\exp\left[-\frac{T}{2}(a_0 - a_1)\right]$$
I realise that my formula for no birth control is probably wrong, so could you please explain to me where I went wrong and why it is so?
Thanks in advance
 A: You are right. Note that for the exponential funcition, we have
$$ \exp(x) \cdot \exp(y) = \exp(x+y), \quad \frac 1{\exp(x)} = \exp(-x) $$
Hence, your solution simplifies
\begin{align*}
  \frac{\exp\left(\frac T2 (a_0 + a_1)\right)}{\exp(a_0 T)}
 &= \exp \left(\frac T2(a_0 + a_1)\right) \cdot \exp(-a_0 T)\\
 &= \exp\left(\frac T2(a_0 + a_1)-a_0 T\right)\\
 &= \exp\left(-\frac T2(a_0 - a_1)\right)
\end{align*}
as in the given solution.
A: This seems fine. A tool you could use for checking your formulas is to compute the rate of growth of the population $N$: namely, this rate $r$ is
$$ r = \frac{dN/dt}{N} = \frac{d (\log N)}{dt}.$$
For a population with constant rate of growth $a_0$, this gives $d(\log N)/dt = a_0$, which we integrate as $\log N(t) - \log N(0) = a_0 t$, or $$N(t) = N(0) \exp (a_0 t).$$
For the population with birth control, the rate of growth (during the period $[0, T]$ is an affine function starting at $a_0$ and ending at $a_1$, so it is indeed $a(t) = a_0 + (t/T) (a_1 - a_0)$. We may again integrate the formula $d(\log N)/dt = a_0 + t (a_1-a_0)/T$ as
$$ \log N(t) - \log N(0) = a_0 t + \frac{t^2}{2} \frac{a_1-a_0}{T},$$
which gives your formula for the population with birth control.
However, this formula is only true for $t \in [0,T]$. Once $t > T$, the growth rate is stable at $a_1$, which gives a new formula for $N(t)$. I leave the computation of this formula to you as an exercise! (You should integrate $\log N(t) - \log N(T)$ using the same methods as above).
