Question

The population of the town grows at a rate of 10% per year. Using the differential equation, find how long it will take for the population to grow 4 times.

Answer 1

Hint: An equation involving derivative (derivatives) of the dependent variable with respect to independent variable (variables) is called a differential equation. Example given:
$3\dfrac{{dy}}{{dx}} - 4y = 2x$.
If a quantity y varies with another quantity x, then $\dfrac{{dy}}{{dx}}$ represents the rate of change of y with respect to x. In the given question, the population varies with time.
The variable separable method is used to solve such an equation in which variables can be separated completely i.e. terms containing y should remain with $dy$ and terms containing x should remain with $dx$

Complete step-by-step answer:
Step 1: Make assumptions and state given information.
Let p is the population of the town in a particular year at any time t.
Given that the rate of change of population per year = 10%
Step 2: Form differential equation
$\dfrac{{dp}}{{dt}}$ is the differential rate of change of population with time.
$ \Rightarrow \dfrac{{dp}}{{dt}} = 10\% $ per year,
$\dfrac{{dp}}{{dt}} = \dfrac{{10}}{{100}}p$
Separating the variables:
$10\dfrac{{dp}}{p} = dt$
On integrating both sides
$\int {10\dfrac{{dp}}{p}} = \int {dt} $
Step 3: Find the limits of integration.
Let at t = 0, initial population = $\mathop p\nolimits_0 $
And for the time t, the population grows 4 times = 4$\mathop p\nolimits_0 $
Thus, the integral in step 2:
 \[
  \int\limits_{\mathop p\nolimits_0 }^{4\mathop p\nolimits_0 } {10\dfrac{{dp}}{p}} = \int\limits_0^t {dt} \\
   \Rightarrow \mathop {\left. {10\ln \left( p \right)} \right|}\nolimits_{\mathop p\nolimits_0 }^{4\mathop p\nolimits_0 } = \mathop {\left. t \right|}\nolimits_0^t \\
   \Rightarrow \mathop {\left. {10\ln \left( p \right)} \right|}\nolimits_{\mathop p\nolimits_0 }^{4\mathop p\nolimits_0 } = t \\
 \]
Step 4: Solving the limits
 $
  t = 10\left[ {\ln \left( {4\mathop p\nolimits_0 } \right) - \ln \left( {\mathop p\nolimits_0 } \right)} \right] \\
  {\text{ }} = 10\ln \left( {\dfrac{{4\mathop p\nolimits_0 }}{{\mathop p\nolimits_0 }}} \right) \\
  {\text{ }} = 10\ln 4 \\
  \;{\text{ }} = 10 \times 1.38 \\
 $
$\because t = 13.8$ years.
Final answer: In 13. 8 years (or 13 years and almost 10 months) the certain population of the town will be 4 times its initial population growing at a rate of 10%.

Note: Alternate steps to solve the differential equation from:
$
  \int {10\dfrac{{dp}}{p}} = \int {dt} \\
   \Rightarrow \int {\dfrac{{dp}}{p}} = 0.1\int {dt} \\
 $
Integrating
$\ln p = 0.1t + c$ , where c is a constant
Exponentiation on both sides with $e$
$p = {e^{0.1t + c}} = A{e^{0.1t}}$, where $A = {e^c}$
Let the initial population $\mathop p\nolimits_0 $, then
$\mathop p\nolimits_0 = A$ …… (1)
If the population is increased 4 times in time $t,$then
$4\mathop p\nolimits_0 = A{e^{0.1t}}$
From (1), $\mathop p\nolimits_0 = A$
$4\mathop p\nolimits_0 = \mathop p\nolimits_0 {\text{ }}{e^{0.1t}}$
Taking log on both sides
$\ln 4 = 0.1t$
$
   \Rightarrow t = 10 \times \ln 4 \\
  \because t = 13.8{\text{ years}} \\
 $