1.1 Exponential Growth and Decay

Many natural processes follow the pattern: the rate of change of a quantity is proportional to the amount present.

The Model:

$\frac{dN}{dt} = kN$

where $N$ is the quantity at time $t$, and $k$ is the constant of proportionality (positive for growth, negative for decay).

Solution:

Separating variables and integrating:

$\frac{dN}{N} = k \, dt \implies \ln|N| = kt + C \implies N = N_0 e^{kt}$

where $N_0$ is the initial amount at $t = 0$.

Example: Radioactive Decay

A radioactive substance has an initial mass of 200 g. After 8 days, only half remains. Find the mass after 30 days.

Solution:

Given $N_0 = 200$ g, and at $t = 8$ days, $N = 100$ g.

$100 = 200e^{8k} \implies e^{8k} = \frac{1}{2} \implies k = -\frac{1}{8}\ln 2$

At $t = 30$ days:

$N = 200e^{-\frac{30}{8}\ln 2} = 200 \cdot 2^{-\frac{30}{8}} \approx 14.87 \text{ g}$

1.2 Newton's Law of Cooling

The rate of cooling of an object is proportional to the difference between its temperature and the surrounding temperature.

The Model:

$\frac{dT}{dt} = -k(T - T_s)$

where $T$ is the object's temperature, $T_s$ is the surrounding temperature, and $k > 0$.

Solution:

$\frac{dT}{T - T_s} = -k \, dt \implies \ln|T - T_s| = -kt + C$

$\implies T(t) = T_s + (T_0 - T_s)e^{-kt}$

Example:

Oil was heated to 70°C and cooled to 50°C after 6 minutes in surroundings of 25°C. Find the time to cool from 50°C to 40°C.

Solution:

At $t = 6$: $50 = 25 + (70 - 25)e^{-6k}$ $25 = 45e^{-6k} \implies k = \frac{1}{6}\ln\left(\frac{9}{5}\right)$

For $T = 40°C$, $T_0 = 50°C$: $40 = 25 + 25e^{-kt} \implies t = \frac{\ln(25/15)}{k} \approx 5.2 \text{ minutes}$

2.1 Falling Body Problems

Consider a mass $m$ falling under gravity with air resistance proportional to velocity.

The Model (Newton's Second Law):

$m\frac{dv}{dt} = mg - kv$

where $g$ is acceleration due to gravity, and $k$ is the drag coefficient.

Solution:

$\frac{dv}{dt} = g - \frac{k}{m}v$

This is a first-order linear equation. The solution gives velocity approaching terminal velocity $v_t = \frac{mg}{k}$ as $t \to \infty$.

Example:

An object weighing 2.45 N falls from rest. Air resistance is $2v$. Find velocity at time $t$. (Take $g = 9.8$ m/s²)

Solution:

Mass $m = \frac{2.45}{9.8} = 0.25$ kg

$0.25\frac{dv}{dt} = 2.45 - 2v$

$\frac{dv}{dt} = 9.8 - 8v$

Separating variables:

$\int \frac{dv}{9.8 - 8v} = \int dt \implies -\frac{1}{8}\ln|9.8 - 8v| = t + C$

Using $v(0) = 0$: $C = -\frac{1}{8}\ln(9.8)$

$v(t) = 1.225\left(1 - e^{-8t}\right) \text{ m/s}$

2.2 Vibrating Springs (Simple Harmonic Motion)

A mass $m$ attached to a spring with spring constant $k$ oscillates.

The Model:

$m\frac{d^2x}{dt^2} + kx = 0$

or equivalently:

$\frac{d^2x}{dt^2} + \omega^2 x = 0, \quad \text{where } \omega = \sqrt{\frac{k}{m}}$

Solution:

The characteristic equation is $\lambda^2 + \omega^2 = 0$, giving $\lambda = \pm i\omega$.

$x(t) = A\cos(\omega t) + B\sin(\omega t)$

This represents simple harmonic motion with period $T = \frac{2\pi}{\omega}$.

Example:

A 4 kg mass stretches a spring by 0.5 m (requiring 2 N force). Find the position at any time $t$ with $x(0) = 2$ m and $x'(0) = 3$ m/s.

Solution:

$k = \frac{2}{0.5} = 4 \implies \omega = \sqrt{\frac{4}{4}} = 1$

$x(t) = A\cos t + B\sin t$

Using $x(0) = 2$: $A = 2$

Differentiating: $x' = -A\sin t + B\cos t$

Using $x'(0) = 3$: $B = 3$

$\boxed{x(t) = 2\cos t + 3\sin t}$

2.3 Damped Vibrating Springs

Real oscillations experience damping due to friction or resistance.

The Model:

$m\frac{d^2x}{dt^2} + c\frac{dx}{dt} + kx = 0$

where $c$ is the damping constant.

The solution depends on whether the system is underdamped, critically damped, or overdamped.

2.4 Electrical Circuits (RLC)

An RLC series circuit is described by a second-order differential equation.

The Model (for charge $Q$ on capacitor):

$L\frac{d^2Q}{dt^2} + R\frac{dQ}{dt} + \frac{1}{C}Q = E(t)$

For current $I = \frac{dQ}{dt}$:

$L\frac{d^2I}{dt^2} + R\frac{dI}{dt} + \frac{1}{C}I = E'(t)$

Example:

For an RLC circuit with $L = 0.25$ H, $R = 250$ Ω, $C = 29.76 \times 10^{-6}$ F, find current given $I(0) = 0$ and $I'(0) = 34$.

Solution:

The characteristic equation:

$0.25\lambda^2 + 250\lambda + \frac{1}{29.76 \times 10^{-6}} = 0$

Solving gives $\lambda_1 = -160$, $\lambda_2 = -840$

$I(t) = Ae^{-160t} + Be^{-840t}$

Using conditions: $I(0) = 0 \implies A + B = 0$

$I'(t) = -160Ae^{-160t} - 840Be^{-840t}$

$I'(0) = 34 \implies -160A - 840B = 34$

Solving: $A = \frac{1}{20}$, $B = -\frac{1}{20}$

$\boxed{I(t) = \frac{1}{20}\left(e^{-160t} - e^{-840t}\right)}$

1. Identify the physical situation and determine which quantities are changing.
2. State the governing principle (e.g., Newton's second law, Kirchhoff's law, Newton's cooling law).
3. Formulate the differential equation using the appropriate model.
4. Solve the differential equation using separation of variables, integrating factor, or characteristic equation methods.
5. Apply initial or boundary conditions to find particular solutions.
6. Interpret the solution in terms of the original physical problem.

In Tanzania, differential equations are used in everyday contexts such as predicting the growth of savings in mobile money accounts (like M-Pesa) where interest earned is proportional to the current balance, modelling the decay of pesticide effectiveness in agricultural farms, or analyzing the cooling of freshly harvested tea leaves before processing. For example, a small-scale maize farmer in Morogoro could use exponential decay equations to determine how long harvested maize will remain safe in storage based on temperature measurements, applying Newton's law of cooling with the ambient village temperature.