Engineering Mathematics III in Quantity Surveying equips professionals with essential skills to model and solve problems involving rates of change and accumulations, which frequently arise in cost estimation, resource allocation, and project scheduling. Ordinary differential equations (ODEs) form the backbone of these analyses, enabling the prediction of variables such as cost growth over time or material decay. This chapter focuses on first-order differential equations, foundational for modeling simple but critical phenomena in quantity surveying contexts in Kenya.

4.1 First-Order Differential Equations

First-order differential equations describe relationships involving a function and its first derivative, representing rates of change. In quantity surveying, these equations can model cost depreciation, inflation effects on budgets, or resource consumption rates. Mastery of solution methods such as separation of variables and integrating factors allows for practical problem-solving in real project scenarios.

4.1.1 Methods of Solving First-Order Differential Equations

First-order differential equations have the general form

$$ \frac{dy}{dx} = f(x, y) $$

where \(y\) is the dependent variable and \(x\) is the independent variable. Common methods to solve these equations include:

• Separation of variables: When the equation can be written as \(g(y) dy = h(x) dx\).
• Integrating factor method: Used for linear equations of the form \(\frac{dy}{dx} + P(x) y = Q(x)\).
• Exact equations: When the differential equation satisfies the exactness condition.
• Substitution methods: For equations reducible to separable or linear forms.

This section focuses on the first two methods most applicable in quantity surveying calculations.

Worked Examples

Example 1: Solve \(\frac{dy}{dx} = 3y\).

Given: \(\frac{dy}{dx} = 3y\)

Method: Separation of variables.

$$ \frac{dy}{y} = 3 dx $$

Integrate both sides:

$$ \int \frac{1}{y} dy = \int 3 dx $$

$$ \ln |y| = 3x + C $$

Solve for \(y\):

$$ y = Ce^{3x} $$

Answer: \( y = Ce^{3x} \)

Example 2: Solve \(\frac{dy}{dx} + 2y = 4\).

Given: \(\frac{dy}{dx} + 2y = 4\)

Method: Integrating factor.

Calculate integrating factor:

$$ \mu(x) = e^{\int 2 dx} = e^{2x} $$

Multiply through:

$$ e^{2x} \frac{dy}{dx} + 2 e^{2x} y = 4 e^{2x} $$

Left side is derivative:

$$ \frac{d}{dx}(e^{2x} y) = 4 e^{2x} $$

Integrate both sides:

$$ e^{2x} y = \int 4 e^{2x} dx = 2 e^{2x} + C $$

Solve for \(y\):

$$ y = 2 + C e^{-2x} $$

Answer: \( y = 2 + C e^{-2x} \)

Example 3: Solve \(\frac{dy}{dx} = \frac{x}{y}\).

Given: \(\frac{dy}{dx} = \frac{x}{y}\)

Method: Separation of variables.

Rewrite:

$$ y dy = x dx $$

Integrate both sides:

$$ \int y dy = \int x dx $$

$$ \frac{y^2}{2} = \frac{x^2}{2} + C $$

Multiply both sides by 2:

$$ y^2 = x^2 + 2C $$

Answer: \( y^2 = x^2 + 2C \)

4.1.2 Separation of Variables

Separation of variables applies when the differential equation can be expressed as a product of functions solely in \(y\) and \(x\):

$$ \frac{dy}{dx} = g(x) h(y) $$

Rearranged as:

$$ \frac{1}{h(y)} dy = g(x) dx $$

Integrating both sides yields the implicit or explicit solution.

Worked Examples

Example 1: A construction material decays at a rate proportional to its current amount \(y\). If \(\frac{dy}{dt} = -0.05 y\), find \(y(t)\).

Given: \(\frac{dy}{dt} = -0.05 y\), \(y(0) = y_0\)

Separate variables:

$$ \frac{dy}{y} = -0.05 dt $$

Integrate:

$$ \int \frac{1}{y} dy = \int -0.05 dt $$

$$ \ln |y| = -0.05 t + C $$

Exponentiate:

$$ y = Ce^{-0.05 t} $$

Apply initial condition \(y(0) = y_0\):

$$ y_0 = Ce^0 = C $$

So,

$$ y = y_0 e^{-0.05 t} $$

Answer: \( y = y_0 e^{-0.05 t} \)

Example 2: Solve \(\frac{dy}{dx} = \frac{x^2}{y^3}\).

Given: \(\frac{dy}{dx} = \frac{x^2}{y^3}\)

Separate variables:

$$ y^3 dy = x^2 dx $$

Integrate:

$$ \int y^3 dy = \int x^2 dx $$

$$ \frac{y^4}{4} = \frac{x^3}{3} + C $$

Multiply both sides by 4:

$$ y^4 = \frac{4x^3}{3} + 4C $$

Answer: \( y^4 = \frac{4x^3}{3} + 4C \)

Example 3: A quantity surveyor models cost increase \(C\) over time \(t\) with \(\frac{dC}{dt} = k (1000 - C)\), where \(k=0.1\). Find \(C(t)\) if initial cost \(C(0)=200\).

Given: \(\frac{dC}{dt} = 0.1 (1000 - C)\), \(C(0)=200\)

Rewrite:

$$ \frac{dC}{dt} = 0.1 (1000 - C) $$

Separate variables:

$$ \frac{dC}{1000 - C} = 0.1 dt $$

Integrate:

$$ \int \frac{1}{1000 - C} dC = \int 0.1 dt $$

Note: \(\int \frac{1}{a, x} dx = -\ln |a, x|\)

So,

$$ -\ln |1000 - C| = 0.1 t + C_1 $$

Multiply both sides by -1:

$$ \ln |1000 - C| = -0.1 t + C_2 $$

Exponentiate:

$$ 1000 - C = Ce^{-0.1 t} $$

Let constant be \(K = e^{C_2}\):

$$ 1000 - C = K e^{-0.1 t} $$

Solve for \(C\):

$$ C = 1000 - K e^{-0.1 t} $$

Apply initial condition \(C(0) = 200\):

$$ 200 = 1000 - K e^{0} = 1000 - K $$

$$ K = 800 $$

$$ C = 1000 - 800 e^{-0.1 t} $$

Answer: \( C = 1000 - 800 e^{-0.1 t} \)

4.1.3 Integrating Factor

The integrating factor method solves linear first-order differential equations of the form:

$$ \frac{dy}{dx} + P(x) y = Q(x) $$

The integrating factor \(\mu(x)\) is:

$$ \mu(x) = e^{\int P(x) dx} $$

Multiply the entire equation by \(\mu(x)\) to make the left side an exact derivative:

$$ \frac{d}{dx} [\mu(x) y] = \mu(x) Q(x) $$

Then integrate both sides to find \(y\).

Worked Examples

Example 1: Solve \(\frac{dy}{dx} + 3y = 6\).

Given: \(\frac{dy}{dx} + 3y = 6\)

Calculate integrating factor:

$$ \mu(x) = e^{\int 3 dx} = e^{3x} $$

Multiply both sides:

$$ e^{3x} \frac{dy}{dx} + 3 e^{3x} y = 6 e^{3x} $$

Left side is derivative:

$$ \frac{d}{dx} (e^{3x} y) = 6 e^{3x} $$

Integrate both sides:

$$ e^{3x} y = \int 6 e^{3x} dx = 2 e^{3x} + C $$

Solve for \(y\):

$$ y = 2 + C e^{-3x} $$

Answer: \( y = 2 + C e^{-3x} \)

Example 2: Solve \(\frac{dy}{dx} - \frac{2}{x} y = x^2\), \(x > 0\).

Given: \(\frac{dy}{dx} - \frac{2}{x} y = x^2\)

Calculate integrating factor:

$$ \mu(x) = e^{\int -\frac{2}{x} dx} = e^{-2 \ln x} = x^{-2} $$

Multiply both sides:

$$ x^{-2} \frac{dy}{dx} - \frac{2}{x} x^{-2} y = x^{-2} x^2 = 1 $$

Simplify left side:

$$ \frac{d}{dx} (x^{-2} y) = 1 $$

Integrate both sides:

$$ x^{-2} y = \int 1 dx = x + C $$

Solve for \(y\):

$$ y = x^{2} (x + C) = x^{3} + C x^{2} $$

Answer: \( y = x^{3} + C x^{2} \)

Example 3: A project’s cost rate \(C(t)\) satisfies \(\frac{dC}{dt} + 0.2 C = 50\), with \(C(0) = 100\). Find \(C(t)\).

Given: \(\frac{dC}{dt} + 0.2 C = 50\), \(C(0) = 100\)

Calculate integrating factor:

$$ \mu(t) = e^{\int 0.2 dt} = e^{0.2 t} $$

Multiply through:

$$ e^{0.2 t} \frac{dC}{dt} + 0.2 e^{0.2 t} C = 50 e^{0.2 t} $$

Left side is derivative:

$$ \frac{d}{dt} (e^{0.2 t} C) = 50 e^{0.2 t} $$

Integrate both sides:

$$ e^{0.2 t} C = \int 50 e^{0.2 t} dt = 50 \times \frac{e^{0.2 t}}{0.2} + K = 250 e^{0.2 t} + K $$

Solve for \(C\):

$$ C = 250 + K e^{-0.2 t} $$

Apply initial condition \(C(0) = 100\):

$$ 100 = 250 + K e^{0} = 250 + K $$

$$ K = -150 $$

$$ C = 250 - 150 e^{-0.2 t} $$

Answer: \( C = 250 - 150 e^{-0.2 t} \)

Chapter Summary

This chapter focused on solving ordinary differential equations, beginning with first-order differential equations and exploring methods such as separation of variables and the integrating factor technique. It then advanced to second-order differential equations, distinguishing between homogeneous and non-homogeneous types and demonstrating how to approach their solutions. The importance of boundary conditions was emphasized, showing how initial or boundary values are applied to determine specific solutions to differential equations. Finally, the chapter highlighted practical applications of ordinary differential equations in real-life scenarios, including physics and engineering contexts such as particle motion and electrical circuits. Through these topics, students gained comprehensive skills in formulating, solving, and applying ordinary differential equations to engineering problems.