00:01
For this problem, we define n to be the number of people present at time t and that n sub 0 is the initial number of people.
00:08
We're given that the rate at which the number of people present at time t is changing over time is proportional to the number of people and then we're also given that after five years the initial number of people doubled.
00:28
So since you want to determine the time it will take for the number of people to triple and quadruple, first thing you have to do is to find n of t and since we're given a separable differential equation, then from there we know that dn over n equals k dt.
00:55
Now if we take the antiderivative both sides, we have natural log of the absolute value of n equals k times t plus let's say a constant c sub 1.
01:13
Solving for n, we would rewrite this in exponential form.
01:20
We have e raised to the natural log of the absolute value of n equal to e raised to k times t plus c sub 1.
01:32
Now that's the same as absolute value of n equal to e raised to k times t times e raised to c sub 1 and then since this is the same as n equal to plus or minus e raised to k times t times e raised to c sub 1, if we replace or represent plus or minus e raised to c sub 1 as a constant c, then n is equal to c times e raised to k times t.
02:14
Now we can find the value of c using the initial value n of 0 equals n sub 0.
02:22
So if n of 0 equals n sub 0, that means n sub 0 equals c times e raised to 0.
02:31
That means c equals n sub 0 which makes n equal to n sub 0 of e raised to k times t.
02:46
And then we can find a value for e sub k using n of 5 equals 2 times n sub 0 since that's given.
02:58
So if n of 5 equals twice n sub 0, that means twice n sub 0 equals n sub 0 times e raised to 5k.
03:13
That's the same as 2 equal to e raised to k raised to the fifth power or that e raised to k is just 2 raised to 1 over 5...