Solve ordinary differential equations is a fundamental skill in architectural mathematics, crucial for modelling physical phenomena such as heat transfer in building materials, structural load responses, and environmental system dynamics. In Kenya’s built environment sector, architects frequently encounter first-order differential equations when analysing time-dependent processes in building design and construction management. Mastery of these equations enables precise predictions and optimized architectural solutions in real-world projects. This chapter focuses on solving first-order ordinary differential equations (ODEs) using methods relevant to architectural applications.
First-order differential equations describe relationships involving a function and its first derivative. In architecture, these equations model processes like cooling rates of building materials, pollutant dispersion in ventilation systems, or water flow rates in drainage design. Understanding how to solve these equations provides architects with analytical tools to predict system behaviours and improve design efficiency.
First-order ODEs can be solved by various methods depending on their form. The most common methods include separation of variables, integrating factors, and exact equations. Each method applies to specific equation types, and selecting the right method is essential for efficient solution.
The general form of a first-order ODE is:
$$\frac{dy}{dx} + P(x)y = Q(x)$$
where \(y\) is the dependent variable, and \(x\) is the independent variable.
Example 1: An architect models the cooling of a concrete slab where the temperature \(T\) changes with time \(t\) according to \(\frac{dT}{dt} = -0.1T\). Solve for \(T(t)\).
Given: \(\frac{dT}{dt} = -0.1T\)
This is separable.
$$\frac{dT}{dt} = -0.1T$$
Example 2: The ventilation system pollutant concentration \(C\) satisfies \(\frac{dC}{dt} + 2C = 5\). Solve for \(C(t)\).
Given: \(\frac{dC}{dt} + 2C = 5\)
This is linear and requires integrating factor.
Example 3: Water flow velocity \(v\) changes with distance \(x\) as \(\frac{dv}{dx} = \frac{v}{x} + x^2\). Solve for \(v(x)\).
Given: \(\frac{dv}{dx} - \frac{1}{x}v = x^2\)
This is linear, requiring integrating factor.
Separation of variables applies when the differential equation can be expressed as the product of a function of \(y\) and a function of \(x\):
$$\frac{dy}{dx} = g(x)h(y)$$
The equation is rearranged to isolate \(y\) terms on one side and \(x\) terms on the other:
$$\frac{1}{h(y)} dy = g(x) dx$$
Integration on both sides yields the solution.
Example 1: Cooling of a concrete slab with temperature \(T\) satisfies \(\frac{dT}{dt} = -0.1T\).
Given: \(\frac{dT}{dt} = -0.1T\)
Step 1: Separate variables
$$\frac{1}{T} dT = -0.1 dt$$
Step 2: Integrate both sides
$$\int \frac{1}{T} dT = \int -0.1 dt$$
Step 3: Integration
$$\ln |T| = -0.1 t + C$$
Step 4: Solve for \(T\)
$$T = Ce^{-0.1 t}$$
Answer: \(T = Ce^{-0.1 t}\), where \(C\) is a constant determined by initial conditions.
Example 2: The rate of change of pollutant concentration \(C\) in a room satisfies \(\frac{dC}{dt} = -k C\), where \(k=0.05\) min\(^{-1}\).
Given: \(\frac{dC}{dt} = -0.05 C\)
Step 1: Separate variables
$$\frac{1}{C} dC = -0.05 dt$$
Step 2: Integrate
$$\int \frac{1}{C} dC = \int -0.05 dt$$
Step 3: Integration
$$\ln |C| = -0.05 t + C_1$$
Step 4: Solve for \(C\)
$$C = C_2 e^{-0.05 t}$$
Answer: \(C = C_2 e^{-0.05 t}\), where \(C_2\) depends on initial concentration.
Example 3: The humidity level \(H\) in a building satisfies \(\frac{dH}{dt} = 3(H_{max} - H)\), where \(H_{max} = 80\%\).
Given: \(\frac{dH}{dt} = 3(80 - H)\)
Step 1: Rewrite
$$\frac{dH}{dt} = 240 - 3H$$
Step 2: Separate variables
$$\frac{dH}{240 - 3H} = dt$$
Step 3: Integrate left side
Let \(u = 240 - 3H\), then \(du = -3 dH\), so \(dH = -\frac{1}{3} du\)
$$\int \frac{dH}{240 - 3H} = \int \frac{-\frac{1}{3} du}{u} = -\frac{1}{3} \int \frac{1}{u} du = -\frac{1}{3} \ln|u| + C$$
Step 4: Integrate right side
$$\int dt = t + C_1$$
Step 5: Combine and solve
$$-\frac{1}{3} \ln|240 - 3H| = t + C_2$$
Step 6: Multiply both sides by -3
$$\ln|240 - 3H| = -3t + C_3$$
Step 7: Exponentiate both sides
$$240 - 3H = C_4 e^{-3t}$$
Step 8: Solve for \(H\)
$$H = \frac{240 - C_4 e^{-3t}}{3}$$
Answer: \(H = 80 - \frac{C_4}{3} e^{-3t}\), with \(C_4\) from initial conditions.
Example 4: The rate of heat loss \(Q\) from a wall satisfies \(\frac{dQ}{dt} = -k Q^2\), where \(k=0.01\).
Given: \(\frac{dQ}{dt} = -0.01 Q^2\)
Step 1: Separate variables
$$\frac{dQ}{Q^2} = -0.01 dt$$
Step 2: Integrate
$$\int Q^{-2} dQ = \int -0.01 dt$$
Step 3: Integration
$$-Q^{-1} = -0.01 t + C$$
Step 4: Solve for \(Q\)
$$\frac{1}{Q} = 0.01 t + C_1$$
Answer: \(Q = \frac{1}{0.01 t + C_1}\)
Example 5: The moisture absorption \(M\) in bricks satisfies \(\frac{dM}{dt} = k (M_s - M)\), where \(M_s = 15\%\), \(k=0.2\).
Given: \(\frac{dM}{dt} = 0.2 (15 - M)\)
Step 1: Rewrite
$$\frac{dM}{dt} = 3 - 0.2 M$$
Step 2: Separate variables
$$\frac{dM}{3 - 0.2 M} = dt$$
Step 3: Integrate left side
Let \(u = 3 - 0.2 M\), \(du = -0.2 dM\), so \(dM = -\frac{1}{0.2} du = -5 du\)
$$\int \frac{dM}{3 - 0.2 M} = \int \frac{-5 du}{u} = -5 \int \frac{1}{u} du = -5 \ln|u| + C$$
Step 4: Integrate right side
$$\int dt = t + C_1$$
Step 5: Combine
$$-5 \ln|3 - 0.2 M| = t + C_2$$
Step 6: Solve for \(M\)
$$\ln|3 - 0.2 M| = -\frac{t}{5} + C_3$$
$$3 - 0.2 M = C_4 e^{-\frac{t}{5}}$$
$$M = \frac{3 - C_4 e^{-\frac{t}{5}}}{0.2}$$
Answer: \(M = 15 - 5 C_4 e^{-\frac{t}{5}}\)
When a first-order ODE is linear but not separable, the integrating factor method is used. The general form is:
$$\frac{dy}{dx} + P(x) y = Q(x)$$
The integrating factor \( \mu(x) \) is:
$$\mu(x) = e^{\int P(x) dx}$$
Multiplying the entire equation by \(\mu(x)\) transforms the left side into the derivative of \(\mu(x) y\), allowing direct integration.
Example 1: The pollutant concentration \(C\) satisfies:
$$\frac{dC}{dt} + 2C = 5$$
Given: \(P(t) = 2\), \(Q(t) = 5\)
Step 1: Find integrating factor
$$\mu(t) = e^{\int 2 dt} = e^{2t}$$
Step 2: Multiply both sides
$$e^{2t} \frac{dC}{dt} + 2 e^{2t} C = 5 e^{2t}$$
Step 3: Recognize left side as derivative
$$\frac{d}{dt} (e^{2t} C) = 5 e^{2t}$$
Step 4: Integrate both sides
$$\int \frac{d}{dt} (e^{2t} C) dt = \int 5 e^{2t} dt$$
$$e^{2t} C = \frac{5}{2} e^{2t} + K$$
Step 5: Solve for \(C\)
$$C = \frac{5}{2} + K e^{-2t}$$
Answer: \(C = \frac{5}{2} + K e^{-2t}\)
Example 2: The water flow velocity \(v\) satisfies:
$$\frac{dv}{dx} - \frac{1}{x} v = x^2$$
Given: \(P(x) = -\frac{1}{x}\), \(Q(x) = x^2\)
Step 1: Find integrating factor
$$\mu(x) = e^{\int -\frac{1}{x} dx} = e^{-\ln|x|} = \frac{1}{x}$$
Step 2: Multiply both sides
$$\frac{1}{x} \frac{dv}{dx} - \frac{1}{x^2} v = x^2 \cdot \frac{1}{x} = x$$
Step 3: Left side as derivative
$$\frac{d}{dx} ( \frac{v}{x} ) = x$$
Step 4: Integrate both sides
$$\int \frac{d}{dx} ( \frac{v}{x} ) dx = \int x dx$$
$$\frac{v}{x} = \frac{x^2}{2} + C$$
Step 5: Solve for \(v\)
$$v = x ( \frac{x^2}{2} + C ) = \frac{x^3}{2} + C x$$
Answer: \(v = \frac{x^3}{2} + C x\)
Example 3: Heat loss \(Q\) satisfies:
$$\frac{dQ}{dt} + 0.1 Q = 20$$
Given: \(P(t) = 0.1\), \(Q(t) = 20\)
Step 1: Find integrating factor
$$\mu(t) = e^{\int 0.1 dt} = e^{0.1 t}$$
Step 2: Multiply both sides
$$e^{0.1 t} \frac{dQ}{dt} + 0.1 e^{0.1 t} Q = 20 e^{0.1 t}$$
Step 3: Left side as derivative
$$\frac{d}{dt} (e^{0.1 t} Q) = 20 e^{0.1 t}$$
Step 4: Integrate both sides
$$\int \frac{d}{dt} (e^{0.1 t} Q) dt = \int 20 e^{0.1 t} dt$$
$$e^{0.1 t} Q = \frac{20}{0.1} e^{0.1 t} + K = 200 e^{0.1 t} + K$$
Step 5: Solve for \(Q\)
$$Q = 200 + K e^{-0.1 t}$$
Answer: \(Q = 200 + K e^{-0.1 t}\)
Example 4: Moisture content \(M\) satisfies:
$$\frac{dM}{dt} + 0.2 M = 3$$
Given: \(P(t) = 0.2\), \(Q(t) = 3\)
Step 1: Find integrating factor
$$\mu(t) = e^{\int 0.2 dt} = e^{0.2 t}$$
Step 2: Multiply both sides
$$e^{0.2 t} \frac{dM}{dt} + 0.2 e^{0.2 t} M = 3 e^{0.2 t}$$
Step 3: Left side as derivative
$$\frac{d}{dt} (e^{0.2 t} M) = 3 e^{0.2 t}$$
Step 4: Integrate both sides
$$\int \frac{d}{dt} (e^{0.2 t} M) dt = \int 3 e^{0.2 t} dt$$
$$e^{0.2 t} M = \frac{3}{0.2} e^{0.2 t} + C = 15 e^{0.2 t} + C$$
Step 5: Solve for \(M\)
$$M = 15 + C e^{-0.2 t}$$
Answer: \(M = 15 + C e^{-0.2 t}\)
Example 5: Temperature \(T\) satisfies:
$$\frac{dT}{dt} + 0.3 T = 10$$
Given: \(P(t) = 0.3\), \(Q(t) = 10\)
Step 1: Find integrating factor
$$\mu(t) = e^{\int 0.3 dt} = e^{0.3 t}$$
Step 2: Multiply both sides
$$e^{0.3 t} \frac{dT}{dt} + 0.3 e^{0.3 t} T = 10 e^{0.3 t}$$
Step 3: Left side as derivative
$$\frac{d}{dt} (e^{0.3 t} T) = 10 e^{0.3 t}$$
Step 4: Integrate both sides
$$\int \frac{d}{dt} (e^{0.3 t} T) dt = \int 10 e^{0.3 t} dt$$
$$e^{0.3 t} T = \frac{10}{0.3} e^{0.3 t} + K = \frac{100}{3} e^{0.3 t} + K$$
Step 5: Solve for \(T\)
$$T = \frac{100}{3} + K e^{-0.3 t}$$
Answer: \(T = \frac{100}{3} + K e^{-0.3 t}\)
Solve the first-order differential equation \(\frac{dy}{dx} = 4y\) given \(y(0) = 3\). (5 marks)
The pollutant concentration \(C\) in a room satisfies \(\frac{dC}{dt} + 3C = 12\). Find \(C(t)\) if \(C(0) = 0\). (6 marks)
A wall loses heat such that \(\frac{dQ}{dt} = -0.05 Q^2\). Find \(Q(t)\) given \(Q(0) = 20\). (7 marks)
Water velocity \(v\) in a pipe satisfies \(\frac{dv}{dx} - \frac{2}{x} v = x^3\). Solve for \(v(x)\). (7 marks)
The moisture content \(M\) in bricks satisfies \(\frac{dM}{dt} + 0.1 M = 5\). Find \(M(t)\) given \(M(0) = 0\). (6 marks)
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Create a free accountThis chapter focused on solving ordinary differential equations, beginning with first-order differential equations and the main methods used to solve them, including separation of variables and integrating factors. It then explored second-order differential equations, distinguishing between homogeneous and non-homogeneous types and how their solutions differ. The importance of boundary conditions was emphasized, showing how initial or boundary values are applied to obtain specific solutions for differential equations. Finally, the chapter highlighted practical applications of ordinary differential equations in real-life scenarios such as particle motion and electrical circuits, demonstrating their relevance in physics and engineering contexts. Through these topics, students gained essential skills for formulating and solving differential equations that model various architectural and engineering problems.
An architect models the rate of change of temperature \(T\) in a building wall as \(\frac{dT}{dt} = -0.5 T\). Given that the initial temperature is \(T(0) = 40^\circ C\), find the temperature after 3 hours. (2 marks)
Solve the differential equation \(\frac{dy}{dx} = 3y\) with the initial condition \(y(0) = 2\). (2 marks)