Average life expectancy..exponential function Let
$$N_0 = \text{initial number of AIDS patients}$$
$$N= \text{number of patients left}$$
The equation is given by: $$N=N_0\exp(-kt)$$
What is the average life expectancy of one person?
(The answer is $t= \frac1k$)
How did we get to this answer without using expected value and probably/statistics analysis? (differential equations problem)
Thanks in advance.
Edit: 
I know how to come up with the answer using expected value, but the problem is presented as a differential equations one. 

The answer is not very rigorous since, as you know, the DE itself is derived using expectation and average out everything but I think you will get the overall idea.

Let $N(\tau)=N_0-n$, $N(\tau+\delta)=N_0-(n+1)$.
Hence, one life has lapsed in time $\delta$. 
$$\tau=\frac{1}{k}\ln\frac{N_0}{N_0-n}$$
$$\tau+\delta=\frac{1}{k}\ln\frac{N_0}{N_0-(n+1)}$$
$$\delta=\frac{1}{k}\left(\ln\frac{N_0}{N_0-(n+1)}-\ln\frac{N_0}{N_0-n}\right)=\frac{1}{k}\left(\ln\frac{N_0-n}{N_0-(n+1)}\right)$$
Since this must hold true for all $n>1$,
$$\text{life expectancy}=\lim_{n\to \infty}\delta=\lim_{n\to \infty}\frac{1}{k}\left(\ln\frac{N_0-n}{N_0-(n+1)}\right)=\frac1k$$
