Water engineering professionals frequently encounter dynamic systems where flow rates, pressures, and contaminant concentrations change over time or space. Understanding how to model and solve these changes accurately is essential for designing and managing water supply, treatment, and distribution systems. This chapter focuses on solving ordinary differential equations (ODEs), which describe such dynamic processes in water engineering, using methods tailored to first and second order equations. Mastery of these techniques enables engineers to predict system behavior and optimize performance in real-world Kenyan water projects.

3.1 First order differential equations

First order differential equations arise in water engineering when modeling processes such as the rate of change of contaminant concentration in a reservoir, flow velocity changes, or water volume in tanks over time. These equations involve the first derivative of the unknown function and are fundamental for transient analysis. The method of undetermined coefficients provides a systematic approach to solving linear first order ODEs with constant coefficients and specific forcing functions common in engineering contexts.

3.1.1 Method of undetermined coefficient

The method of undetermined coefficients solves linear first order differential equations with constant coefficients and forcing functions that are polynomials, exponentials, sines, cosines, or combinations thereof. The general form is:

$$ \frac{dy}{dx} + py = g(x) $$

where \(p\) is a constant and \(g(x)\) is a known function. The solution is the sum of the complementary function (CF) and a particular integral (PI). The CF solves the homogeneous equation (\(dy/dx + py = 0\)), and the PI is guessed based on the form of \(g(x)\) with unknown coefficients determined by substitution.

Worked Examples

Example 1: Flow rate change in a water tank.

Given: The flow rate \(Q(t)\) satisfies

$$ \frac{dQ}{dt} + 3Q = 6 $$

Find \(Q(t)\).

Step 1: Complementary function solves

$$ \frac{dQ}{dt} + 3Q = 0 $$

Step 2: Solve homogeneous equation:

$$ \frac{dQ}{dt} = -3Q $$

Separate variables:

$$ \frac{dQ}{Q} = -3 dt $$

Integrate:

$$ \ln|Q| = -3t + C $$

Exponentiate:

$$ Q_c = Ce^{-3t} $$

Step 3: Guess particular integral \(Q_p = A\) (constant), substitute into original:

$$ 0 + 3A = 6 \implies A = 2 $$

Step 4: General solution:

$$ Q(t) = Ce^{-3t} + 2 $$

Answer: \(Q(t) = Ce^{-3t} + 2\) (units as per \(Q\), e.g., m³/s)

Example 2: Contaminant concentration with exponential input.

Given:

$$ \frac{dy}{dx} + 2y = e^{3x} $$

Find \(y(x)\).

Step 1: CF solves

$$ \frac{dy}{dx} + 2y = 0 $$

Step 2: Solve homogeneous:

$$ \frac{dy}{dx} = -2y $$

Integrate:

$$ \ln|y| = -2x + C $$

$$ y_c = Ce^{-2x} $$

Step 3: Guess PI \(y_p = Ae^{3x}\), substitute:

$$ \frac{d}{dx}(Ae^{3x}) + 2Ae^{3x} = e^{3x} $$

$$ 3Ae^{3x} + 2Ae^{3x} = e^{3x} $$

$$ 5Ae^{3x} = e^{3x} \implies A = \frac{1}{5} $$

Step 4: General solution:

$$ y = Ce^{-2x} + \frac{1}{5} e^{3x} $$

Answer: \(y = Ce^{-2x} + \frac{1}{5} e^{3x}\)

Example 3: Flow velocity with sinusoidal forcing

Given:

$$ \frac{dv}{dt} + 4v = \sin 2t $$

Find \(v(t)\).

Step 1: CF solves

$$ \frac{dv}{dt} + 4v = 0 $$

Step 2: Solve homogeneous:

$$ v_c = Ce^{-4t} $$

Step 3: Guess PI:

Since RHS is \(\sin 2t\), try

$$ v_p = A \cos 2t + B \sin 2t $$

Step 4: Differentiate:

$$ \frac{dv_p}{dt} = -2A \sin 2t + 2B \cos 2t $$

Step 5: Substitute into original:

$$ (-2A \sin 2t + 2B \cos 2t) + 4 (A \cos 2t + B \sin 2t) = \sin 2t $$

Group terms:

$$ (2B + 4A) \cos 2t + (-2A + 4B) \sin 2t = \sin 2t $$

Step 6: Equate coefficients:

For \(\cos 2t\):

$$ 2B + 4A = 0 $$

For \(\sin 2t\):

$$ -2A + 4B = 1 $$

Step 7: Solve system:

From first:

$$ 2B = -4A \implies B = -2A $$

Substitute into second:

$$ -2A + 4(-2A) = 1 $$

$$ -2A - 8A = 1 \implies -10A = 1 \implies A = -\frac{1}{10} $$

Then:

$$ B = -2 \times (-\frac{1}{10}) = \frac{1}{5} $$

Step 8: PI:

$$ v_p = -\frac{1}{10} \cos 2t + \frac{1}{5} \sin 2t $$

Step 9: General solution:

$$ v = Ce^{-4t} - \frac{1}{10} \cos 2t + \frac{1}{5} \sin 2t $$

Answer: \(v = Ce^{-4t} - \frac{1}{10} \cos 2t + \frac{1}{5} \sin 2t\)

Example 4: Pump discharge with polynomial forcing

Given:

$$ \frac{dQ}{dt} + 5Q = t^2 $$

Find \(Q(t)\).

Step 1: CF solves

$$ \frac{dQ}{dt} + 5Q = 0 $$

Step 2: CF:

$$ Q_c = Ce^{-5t} $$

Step 3: Guess PI:

Since RHS is polynomial \(t^2\), try

$$ Q_p = At^2 + Bt + C $$

Step 4: Differentiate:

$$ \frac{dQ_p}{dt} = 2At + B $$

Step 5: Substitute:

$$ (2At + B) + 5(At^2 + Bt + C) = t^2 $$

Expand:

$$ 2At + B + 5At^2 + 5Bt + 5C = t^2 $$

Group terms:

$$ 5A t^2 + (2A + 5B) t + (B + 5C) = t^2 + 0 t + 0 $$

Step 6: Equate coefficients:

$$ 5A = 1 \implies A = \frac{1}{5} $$

$$ 2A + 5B = 0 \implies 2 \times \frac{1}{5} + 5B = 0 \implies \frac{2}{5} + 5B = 0 \implies B = -\frac{2}{25} $$

$$ B + 5C = 0 \implies -\frac{2}{25} + 5C = 0 \implies C = \frac{2}{125} $$

Step 7: PI:

$$ Q_p = \frac{1}{5} t^2 - \frac{2}{25} t + \frac{2}{125} $$

Step 8: General solution:

$$ Q = Ce^{-5t} + \frac{1}{5} t^2 - \frac{2}{25} t + \frac{2}{125} $$

Answer: \(Q = Ce^{-5t} + \frac{1}{5} t^2 - \frac{2}{25} t + \frac{2}{125}\)

Example 5: Reservoir water level with exponential decay input

Given:

$$ \frac{dh}{dt} + 0.1 h = 5 e^{-0.2 t} $$

Find \(h(t)\).

Step 1: CF:

$$ h_c = Ce^{-0.1 t} $$

Step 2: Guess PI:

Try

$$ h_p = Ae^{-0.2 t} $$

Step 3: Differentiate:

$$ \frac{dh_p}{dt} = -0.2 A e^{-0.2 t} $$

Step 4: Substitute:

$$ -0.2 A e^{-0.2 t} + 0.1 A e^{-0.2 t} = 5 e^{-0.2 t} $$

Simplify:

$$ (-0.2 A + 0.1 A) e^{-0.2 t} = 5 e^{-0.2 t} $$

$$ -0.1 A = 5 \implies A = -50 $$

Step 5: General solution:

$$ h = Ce^{-0.1 t} - 50 e^{-0.2 t} $$

Answer: \(h = Ce^{-0.1 t} - 50 e^{-0.2 t}\)

Chapter Summary

This chapter covered solving ordinary differential equations, beginning with first order differential equations and the method of undetermined coefficients to find particular solutions. It then progressed to second order differential equations, emphasizing the method of undetermined coefficients for solving non-homogeneous cases. Further exploration of first order differential equations included additional solution techniques and applications relevant to engineering contexts. The study of second order differential equations continued, focusing on more complex forms and solution strategies. Throughout the chapter, emphasis was placed on understanding the structure of differential equations and applying systematic methods to obtain explicit solutions. The methods presented are foundational for modelling and solving real-world problems in civil engineering and other technical fields. Mastery of these techniques enables accurate analysis of dynamic systems and processes described by ordinary differential equations.