3. Population Growth The population of a suburb grows at a rate proportional to the population. Suppose the population doubles in size from 4000 to 8000 in an 18 -month period and continues at the current rate of growth. (a) Write a differential equation that models the population $$ P $$ at time $$ t $$ in months. (b) Find the general solution to the differential equation. (c) Find the particular solution to the differential equation with the initial condition $$ P(0)=4000 $$. (d) What will the population be in 4 years $$ [t=48] $$ ?

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**(a) Differential Equation**

The problem states that the rate of change of the population is proportional to the population. This can be modeled by the differential equation  
$$
\frac{dP}{dt}= kP,
$$
where $$ k $$ is the constant of proportionality.

---

**(b) General Solution**

To solve  
$$
\frac{dP}{dt}= kP,
$$
we separate variables:  
$$
\frac{dP}{P} = k\, dt.
$$
Integrate both sides:  
$$
\int \frac{1}{P}\, dP = \int k\, dt.
$$
This gives  
$$
\ln |P| = kt + C,
$$
where $$ C $$ is the constant of integration. Exponentiating both sides, we obtain:  
$$
P = e^{kt+C} = e^{C}e^{kt}.
$$
Let $$ P_0 = e^C $$. Then the general solution is  
$$
P(t)= P_0 e^{kt}.
$$

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**(c) Particular Solution with $$ P(0)=4000 $$**

Given the initial condition $$ P(0)=4000 $$, substitute $$ t=0 $$ into the general solution:  
$$
4000 = P_0 e^{k\cdot 0} = P_0,
$$
so  
$$
P_0=4000.
$$
Next, we are told that the population doubles from 4000 to 8000 in 18 months. So we have  
$$
8000 = 4000\, e^{18k}.
$$
Divide both sides by 4000:  
$$
2 = e^{18k}.
$$
Taking the natural logarithm of both sides,  
$$
\ln 2 = 18k \quad \Longrightarrow \quad k = \frac{\ln 2}{18}.
$$
Thus, the particular solution is  
$$
P(t)= 4000\, e^{\frac{\ln 2}{18} \, t}.
$$

---

**(d) Population in 4 Years ( $$ t=48 $$ Months)**

To find the population after 48 months, substitute $$ t=48 $$ into the particular solution:  
$$
P(48)= 4000\, e^{\frac{\ln 2}{18}\cdot 48}.
$$
Simplify the exponent:  
$$
\frac{48}{18} = \frac{8}{3},
$$
so the expression becomes  
$$
P(48)= 4000\, e^{\frac{8}{3}\ln 2} = 4000\, \left(e^{\ln 2}\right)^{\frac{8}{3}} = 4000\cdot 2^{\frac{8}{3}}.
$$
Thus, the population in 4 years is  
$$
P(48)= 4000\cdot 2^{\frac{8}{3}}.
$$

To provide an approximate numerical value, note that  
$$
2^{\frac{8}{3}} \approx 6.3496,
$$
so  
$$
P(48) \approx 4000 \times 6.3496 \approx 25,\!398.
$$
**(a) Differential Equation** The population grows at a rate proportional to itself, modeled by $$ \frac{dP}{dt} = kP, $$ where $$ k $$ is the growth rate. **(b) General Solution** The general solution to the differential equation is $$ P(t) = P_0 e^{kt}, $$ where $$ P_0 $$ is the initial population. **(c) Particular Solution** Given $$ P(0) = 4000 $$, the particular solution is $$ P(t) = 4000 e^{\frac{\ln 2}{18} t}. $$ **(d) Population in 4 Years** After 48 months, the population is $$ P(48) = 4000 \times 2^{\frac{8}{3}} \approx 25,\!398. $$

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