Example 9: Rs 1000 growing continuously at 5% a year doubles in $20\log_e 2$ years

Basics you need

Solution

1. Write the rate as a differential equation

   Let $P$ be the principal (in rupees) at time $t$ (in years). Growing continuously at 5% per year means

   $$\frac{dP}{dt} = \frac{5}{100}\times P = \frac{P}{20}.$$
   "Continuously" means the 5% rule holds at every instant, not once a year: the rate of change at time $t$ is 5% of the amount present at time $t$⁠. The book writes $\frac{dp}{dt}$ with a small $p$ for the same quantity $P$⁠.
   A rate of change is a derivative. Basics: e^{kt} and its rate

2. Separate the variables

   The principal is always positive, so we may divide by $P$⁠:

   $$\frac{dP}{P} = \frac{dt}{20}.$$
   Here $g(t) = \frac{1}{20}$ and $h(P) = P$⁠; dividing by $h(P)$ is allowed because $P > 0$⁠. The left side has only $P$⁠, the right side only $t$⁠.
   Section 9.4.1, with $t$ in place of $x$ and $P$ in place of $y$⁠. Lesson: separate, then integrate

3. Integrate both sides, with one constant

   $$\begin{aligned} \int\frac{dP}{P} &= \int\frac{dt}{20} \\ \log P &= \frac{t}{20} + C_1. \end{aligned}$$
   $\int\frac{dP}{P} = \log|P|$⁠, and $|P| = P$ because $P > 0$⁠. $\int\frac{1}{20}\,dt = \frac{t}{20}$⁠. The two constants combine into one, $C_1$⁠.
   Section 7.2, formulas (i) and (xii). Basics: every method ends with an integral

4. Solve for $P$ and rename the constant

   $$\begin{aligned} P &= e^{\frac{t}{20} + C_1} = e^{\frac{t}{20}}\cdot e^{C_1} \\ &= Ce^{\frac{t}{20}},\quad\text{where } C = e^{C_1}.\quad\ldots(1) \end{aligned}$$
   $\log P = k$ means $P = e^k$⁠. Then $e^{a + b} = e^ae^b$ splits off the constant factor $e^{C_1}$⁠. Since $C_1$ can be any real number, $e^{C_1}$ can be any positive number; calling it $C$ gives the general solution in its usual form, still with exactly one arbitrary constant.
   $e^{a + b} = e^ae^b$⁠; renaming $e^{C_1}$ as $C$⁠. Basics: one constant, one condition

5. Use the starting amount

   At $t = 0$⁠, $P = 1000$⁠. In (1): $1000 = Ce^0 = C$⁠. So

   $$P = 1000e^{\frac{t}{20}}.$$
   The condition "Rs 1000 at the start" picks the particular solution; $e^0 = 1$ makes $C$ equal to the starting amount.
   Section 9.3: a particular solution from a condition. Lesson: general and particular solutions

6. Find when the amount has doubled

   Doubled means $P = 2000$⁠:

   $$\begin{aligned} 2000 = 1000e^{\frac{t}{20}} &\;\Rightarrow\; e^{\frac{t}{20}} = 2 \\ &\;\Rightarrow\; \frac{t}{20} = \log_e 2 \\ &\;\Rightarrow\; t = 20\log_e 2. \end{aligned}$$

   Numerically, $t \approx 20 \times 0.6931472 \approx 13.86$ years.

   Taking $\log_e$ undoes $e$⁠: $e^k = 2 \iff k = \log_e 2$⁠. The 1000 cancels, so any amount doubles in the same time at this rate.
   $e^k = 2 \iff k = \log_e 2$ (Chapter 5, Section 5.4). Basics: e^{kt} and its rate

7. Check, and compare with yearly compounding

   For $P = 1000e^{t/20}$⁠: $\frac{dP}{dt} = 1000e^{t/20}\cdot\frac{1}{20} = \frac{P}{20}$⁠, and $P(0) = 1000$⁠. At $t = 20\log_e 2$⁠, $P = 1000e^{\log_e 2} = 2000$⁠.

   If instead 5% were added once a year, after $n$ years the amount would be $1000(1.05)^n$⁠, which doubles when $n = \frac{\log 2}{\log 1.05} \approx 14.21$ years. Continuous growth doubles about a third of a year sooner.

   Substituting back confirms the solution and the condition. The comparison shows why the word "continuously" matters: growth on the growth, every instant, beats adding 5% once a year.
   $\frac{d}{dt}e^{kt} = ke^{kt}$⁠. Basics: e^{kt} is k times itself

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