In a city of 5 million people, the number of customers of a new company increases at a rate proportional to the number of people who are not its customers. The company determines that in order for its business to be profitable, it must have at least 1 million customers at any point in time. The company has 1 million customers at the end of the third year of operation, starting from a customer base of 0. Assume that the population remains unchanged at any point in time.

1. Formulate the differential equation based on the text.

The rate of change of the number of customers, $y$, is proportional to the number of people who are not its customers, which is $(5 - y)$. Thus, the differential equation is:
$\frac{dy}{dt} = k(5 - y)$
where $k$ is a positive constant of proportionality.

2. Separate variables and integrate both sides.

$\int \frac{1}{5 - y} \, dy = \int k \, dt$
$-\ln|5 - y| = kt + C$

3. Apply the initial condition to find the constant of integration $C$.

Starting from a customer base of $0$ means $y = 0$ when $t = 0$. Since $5 - y > 0$, we can drop the absolute value:
$-\ln(5 - 0) = k(0) + C$
$C = -\ln 5$

4. Make $y$ the subject to show the required equation.

Substitute $C$ back into the equation:
$-\ln(5 - y) = kt - \ln 5$
$\ln(5 - y) = -kt + \ln 5$
$5 - y = e^{-kt + \ln 5}$
$5 - y = e^{\ln 5} e^{-kt}$
$5 - y = 5e^{-kt}$
$y = 5 - 5e^{-kt} = 5(1 - e^{-kt})$
(Shown)

5. Determine the value of the constant $k$.

The company has 1 million customers at the end of the third year, meaning $y = 1$ when $t = 3$. Substitute these values into the equation:
$1 = 5(1 - e^{-3k})$
$1 - e^{-3k} = 0.2$
$e^{-3k} = 0.8$
$-3k = \ln 0.8$
$k = -\frac{1}{3}\ln 0.8 \approx 0.0744 \text{ (to 3 s.f.)}$

1. Determine the starting point and domain of the function.

For $t = 0$, $y = 5(1 - e^0) = 0$. The context of the problem dictates that time $t \ge 0$, so the graph must start from the origin $(0,0)$.

2. Find the long-term behaviour to determine any asymptotes.

As $t \to \infty$, the term $e^{-kt} \to 0$ (since $k$ is a positive constant). Therefore, $y \to 5(1 - 0) = 5$. The graph has a horizontal asymptote at $y = 5$. The curve approaches this line from below with a decreasing gradient.

1. Find the general solution for $y$ by integrating twice.

$\frac{d^2 y}{dt^2} = -0.1$
$\frac{dy}{dt} = \int -0.1 \,dt = -0.1t + c_1$
$y = \int (-0.1t + c_1) \,dt = -0.05t^2 + c_1t + c_2$

2. Set up simultaneous equations using the values of $y$ at $t=6$ and $t=7$.

At the end of the 6th year, $t = 6$ and $y = 1.8$:
$1.8 = -0.05(6)^2 + 6c_1 + c_2$
$6c_1 + c_2 = 3.6 \quad \text{--- (1)}$
At the end of the 7th year, $t = 7$ and $y = 1.65$:
$1.65 = -0.05(7)^2 + 7c_1 + c_2$
$7c_1 + c_2 = 4.1 \quad \text{--- (2)}$

3. Solve the simultaneous equations to find the specific equation for $y$.

Subtracting (1) from (2):
$c_1 = 4.1 - 3.6 = 0.5$
Substitute $c_1 = 0.5$ into (1):
$6(0.5) + c_2 = 3.6 \implies c_2 = 0.6$
Therefore, the model is:
$y = -0.05t^2 + 0.5t + 0.6$

4. Set $y < 1$ (the profitability threshold) to find the critical value of $t$.

The business is unprofitable when $y < 1$:
$-0.05t^2 + 0.5t + 0.6 < 1$
$-0.05t^2 + 0.5t - 0.4 < 0$
Divide by $-0.05$ (reversing the inequality):
$t^2 - 10t + 8 > 0$
Solving the corresponding quadratic equation $t^2 - 10t + 8 = 0$:
$t = \frac{10 \pm \sqrt{100 - 32}}{2} = 5 \pm \sqrt{17}$
$t \approx 9.123 \quad \text{or} \quad t \approx 0.877$

5. Conclude the correct year based on the condition $t \ge 6$.

Since the model is valid from $t = 6$, we reject $t \approx 0.877$ and require $t > 9.123$.
The number of customers drops below 1 million when $t > 9.123$.
Because $9 < 9.123 \le 10$, this occurs during the 10th year of operation.