Exercise 9.3 · Q21: Rs 1000 at 5% continuous growth is $1000e^{0.5}$ after 10 years

Basics you need

Solution

1. Name the variables and write the model

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

   $$\frac{dP}{dt} = \frac{5}{100}\times P = \frac{P}{20}.$$
   At every instant the rate of increase is 5% of the amount present then. The equation has order 1 and degree 1.
   A rate of change is a derivative. Basics: e^{kt} and its rate

2. Separate the variables

   $$\frac{dP}{P} = \frac{dt}{20}.$$
   $h(P) = P$⁠, $g(t) = \frac{1}{20}$⁠. Dividing by $P$ is allowed because the principal is positive.
   Section 9.4.1, with $t$ and $P$ in place of $x$ and $y$⁠. Lesson: separate, then integrate

3. Integrate, and solve for $P$

   $$\begin{aligned} \log P &= \frac{t}{20} + C_1 \\ P &= e^{\frac{t}{20}}\cdot e^{C_1} = Ce^{\frac{t}{20}},\quad C = e^{C_1}. \end{aligned}$$
   $\int\frac{dP}{P} = \log P$ (as $P > 0$⁠) and $\int\frac{dt}{20} = \frac{t}{20}$⁠; one constant $C_1$⁠. Then $\log P = k$ means $P = e^k$⁠, and $e^{a + b} = e^ae^b$ splits off the constant factor, renamed $C$⁠.
   Section 7.2, formula (xii); $e^{a + b} = e^ae^b$⁠. Basics: every method ends with an integral

4. Use the deposit

   At $t = 0$⁠, $P = 1000$⁠: $1000 = Ce^0 = C$⁠. So

   $$P = 1000e^{\frac{t}{20}}.$$
   $e^0 = 1$⁠, so $C$ is the amount deposited. One condition fixes the one constant.
   Section 9.3: a particular solution from a condition. Basics: one constant, one condition

5. The amount after 10 years

   $$\begin{aligned} P(10) &= 1000e^{\frac{10}{20}} = 1000e^{0.5} \\ &= 1000 \times 1.648 = 1648. \end{aligned}$$
   $\frac{10}{20} = 0.5$⁠, and the question gives $e^{0.5} = 1.648$⁠. So the deposit is worth Rs 1648 after 10 years, with the value the question supplies.
   Evaluating the particular solution. Lesson: interest every instant

6. Check: the exact value, and yearly compounding

   $P = 1000e^{t/20}$ gives $\frac{dP}{dt} = \frac{1000}{20}e^{t/20} = \frac{P}{20}$⁠, and $P(0) = 1000$⁠.

   More precisely, $e^{0.5} = 1.6487212\ldots$⁠, so the exact amount is $1000e^{0.5} \approx$ Rs 1648.72. The given value $1.648$ is $e^{0.5}$ cut to three decimal places, which gives Rs 1648.

   With 5% added once a year instead, the amount would be $1000 \times 1.05^{10} \approx$ Rs 1628.89.

   Substituting back confirms the solution and the deposit. The question's $e^{0.5} = 1.648$ explains the textbook's Rs 1648; the 72 paise difference comes only from cutting $e^{0.5}$ short at 1.648 (rounded to three places it would be 1.649). Continuous growth earns about Rs 20 more than yearly compounding over 10 years.
   $\frac{d}{dt}e^{kt} = ke^{kt}$⁠. Basics: e^{kt} is k times itself

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