Differential Equations - Study Notes
Differential equations are fundamental to civil engineering practice, appearing in structural dynamics, fluid mechanics, heat transfer, and geotechnical analysis. This chapter provides systematic coverage of ordinary differential equations (ODEs), solution methods, and engineering applications aligned with professional licensure requirements. Mastery of differential equations is essential for solving real-world problems in design and analysis—from modeling settlement in foundations to analyzing seismic response of buildings. These notes emphasize problem-solving techniques, worked examples using SI units, and practical applications relevant to Philippine construction standards (NSCP 2015) and engineering practice.
1. Classification and Fundamentals of Differential Equations
A differential equation is a mathematical statement that relates an unknown function to its derivatives. In engineering, we typically work with ordinary differential equations (ODEs) where the unknown is a function of a single independent variable (usually time or position).
Order and Degree:
- The order of a differential equation is determined by the highest derivative present. A first-order equation contains dy/dx; a second-order equation contains d²y/dx².
- The degree is the power (exponent) of the highest-order derivative, after the equation has been made polynomial in all its derivatives (clearing radicals and fractions involving derivatives).
Example classifications:
- dy/dx + 5y = 0 is first-order, first-degree
- d²y/dx² − 3dy/dx + 2y = sin(x) is second-order, first-degree
- (d²y/dx²)³ + (dy/dx)² = x is second-order, third-degree
Linear vs. Nonlinear:
A differential equation is linear if the dependent variable y and all its derivatives appear to the first power only and do not multiply each other.
- Linear: y'' + 2y' + 3y = x
- Nonlinear: y² + (y')² = 1 or y·y' = x
Ordinary vs. Partial:
Ordinary differential equations (ODEs) involve derivatives with respect to one independent variable. Partial differential equations (PDEs) involve partial derivatives with respect to multiple independent variables. This chapter focuses on ODEs.
Solution and Initial Conditions:
The solution to a differential equation is a function (or family of functions) that satisfies the equation. A general solution contains arbitrary constants equal in number to the order of the equation. A particular solution results from applying initial conditions (ICs) or boundary conditions (BCs) to determine the constants.
For example, if y'' − y = 0, the general solution is y = C₁eˣ + C₂e⁻ˣ. If we know y(0) = 1 and y'(0) = 0, we can find C₁ and C₂ to get the particular solution: y = (eˣ + e⁻ˣ)/2 = cosh(x).
Examples
Classify the following equation: (d²y/dx²) + 3(dy/dx)² + xy = 0
Solution
Step 1: Identify the highest derivative. We have d²y/dx², so the order is 2.
Step 2: Examine the power of the highest-order derivative. The term d²y/dx² appears to the first power.
Step 3: Check if the equation is polynomial in derivatives. Yes, all derivatives appear polynomially.
Step 4: The degree is the power of d²y/dx², which is 1.
Conclusion: This is a second-order, first-degree differential equation. It is also nonlinear because of the (dy/dx)² term.
The general solution to dy/dx = 2x is y = x² + C. Find the particular solution if y(1) = 3.
Solution
Step 1: Write the general solution with the arbitrary constant.
y = x² + C
Step 2: Apply the initial condition y(1) = 3.
3 = (1)² + C
3 = 1 + C
Step 3: Solve for C.
C = 2
Step 4: Write the particular solution.
y = x² + 2
Verification: dy/dx = 2x ✓ and y(1) = 1 + 2 = 3 ✓
Key points
- Order: highest derivative present; degree: highest power of that derivative
- Linear equations have dependent variable and derivatives to first power only
- General solution has n arbitrary constants for an nth-order equation
- Initial conditions (n conditions) give a unique particular solution
- Ordinary DEs have one independent variable; partial DEs have multiple
- Verification: substitute solution back into original equation
2. First-Order Differential Equations: Separable Type
The separable differential equation is the most straightforward first-order type to solve. It has the general form:
dy/dx = g(x)·h(y)
The key principle is that we can separate variables—collect all terms involving y on one side and all terms involving x on the other side—then integrate both sides independently.
Solution Method for Separable Equations:
- Write the equation in the form dy/dx = g(x)·h(y)
- Separate variables: (1/h(y))·dy = g(x)·dx
- Integrate both sides: ∫(1/h(y))dy = ∫g(x)dx + C
- Solve for y (if possible) to get the general solution
- Apply initial conditions to find the arbitrary constant and obtain the particular solution
This technique appears frequently in civil engineering applications such as:
- Radioactive decay: dN/dt = −λN (where λ is the decay constant)
- Population growth: dP/dt = kP
- Water drainage from a tank: dh/dt = −a√(2gh)
Common Engineering Application—Drainage of Tanks:
When liquid drains from a cylindrical tank through an orifice, the rate of outflow is proportional to the square root of the depth: dh/dt = −C√h, where h is depth and C is a constant depending on orifice size and gravitational acceleration.
Separating: dh/√h = −C dt
Integrating: 2√h = −Ct + K
This allows prediction of drain time for design purposes.
Examples
Solve dy/dx = 3xy with initial condition y(0) = 2
Solution
Step 1: Identify as separable. We have dy/dx = 3xy = (3x)(y)
Step 2: Separate variables. Move all y terms to left, all x terms to right.
(1/y)dy = 3x dx
Step 3: Integrate both sides.
∫(1/y)dy = ∫3x dx
ln|y| = (3x²)/2 + C
Step 4: Solve for y (exponentiate both sides).
|y| = e^((3x²)/2 + C) = e^C · e^(3x²/2)
Let A = ±e^C (arbitrary constant)
y = A·e^(3x²/2)
Step 5: Apply initial condition y(0) = 2.
2 = A·e^0 = A
Step 6: Write the particular solution.
y = 2e^(3x²/2)
A cylindrical tank with cross-sectional area A = 2 m² drains through an orifice. The drainage rate follows dh/dt = −0.5√h, where h is depth in meters and t is time in seconds. Starting with h(0) = 4 m, find the time to drain to h = 0.25 m.
Solution
Step 1: Recognize as separable equation.
dh/dt = −0.5√h
Step 2: Separate variables.
dh/√h = −0.5 dt
Step 3: Integrate both sides.
∫h^(−1/2) dh = ∫−0.5 dt
2h^(1/2) = −0.5t + C
2√h = −0.5t + C
Step 4: Apply initial condition h(0) = 4.
2√4 = −0.5(0) + C
2(2) = C
C = 4
Step 5: Write the general solution with known constant.
2√h = −0.5t + 4
Step 6: Find time when h = 0.25 m.
2√0.25 = −0.5t + 4
2(0.5) = −0.5t + 4
1 = −0.5t + 4
0.5t = 3
t = 6 seconds
Conclusion: The tank drains from 4 m to 0.25 m in 6 seconds.
A radioactive material has 500 g initially. After 10 years, 400 g remains. How much will remain after 30 years? (Use the decay model dN/dt = −λN)
Solution
Step 1: Recognize the separable decay equation.
dN/dt = −λN
Step 2: Separate variables.
dN/N = −λ dt
Step 3: Integrate both sides.
∫(1/N)dN = ∫−λ dt
ln|N| = −λt + C₁
Step 4: Exponentiate to solve for N.
N = e^(−λt + C₁) = e^C₁ · e^(−λt)
N = N₀e^(−λt), where N₀ = e^C₁
Step 5: Use initial condition N(0) = 500.
500 = N₀e^0 → N₀ = 500
Step 6: Use second condition N(10) = 400 to find λ.
400 = 500e^(−λ·10)
0.8 = e^(−10λ)
ln(0.8) = −10λ
−0.2231 = −10λ
λ = 0.02231 year⁻¹
Step 7: Find N(30).
N(30) = 500e^(−0.02231·30)
N(30) = 500e^(−0.6693)
N(30) = 500(0.512)
N(30) = 256 g
Conclusion: After 30 years, 256 g of the radioactive material remains.
Key points
- Separable equations: dy/dx = g(x)·h(y); separate and integrate independently
- Separate: collect all y terms on left, all x terms on right
- Integrate: ∫(1/h(y))dy = ∫g(x)dx + C
- Always solve for C using initial conditions to get particular solution
- Common in growth/decay, fluid flow, and thermal problems
- Check solution by substituting back into original equation
3. First-Order Linear Differential Equations
A first-order linear differential equation has the standard form:
dy/dx + P(x)y = Q(x)
where P(x) and Q(x) are functions of x alone (not y). This type is fundamental in engineering and appears in:
- Transient electrical circuit analysis
- Thermal problems involving convective cooling
- Fluid mechanics with linear resistance
- Dynamics with viscous damping
The Integrating Factor Method:
The solution method uses an integrating factor μ(x), which is cleverly designed so that the left side becomes the derivative of a product:
μ(x) = e^(∫P(x)dx)
The complete solution procedure:
- Write the equation in standard form: dy/dx + P(x)y = Q(x)
- Calculate the integrating factor: μ(x) = e^(∫P(x)dx)
- Multiply the entire equation by μ(x): μ(dy/dx) + μP(x)y = μQ(x)
- Recognize that the left side is d/dx[μy], so: d/dx[μy] = μQ(x)
- Integrate both sides: μy = ∫μQ(x)dx + C
- Solve for y: y = (1/μ)[∫μQ(x)dx + C]
- Apply initial conditions to find C
Why This Works—Mathematical Insight:
The integrating factor μ is constructed such that μ · (dy/dx + Py) becomes d/dx[μy]. By the product rule:
d/dx[μy] = μ(dy/dx) + (dμ/dx)y
For this to equal μ(dy/dx) + μPy, we need (dμ/dx)y = μPy, or dμ/dx = μP.
Solving: dμ/μ = P dx, so ln(μ) = ∫P dx, giving μ = e^(∫P dx).
Civil Engineering Example—Cooling of Concrete:
Newton's law of cooling: dT/dt + hT = hT_ambient, where h is a heat transfer coefficient. Using the integrating factor method allows us to predict concrete temperature during curing, essential for design of curing schedules in NSCP 2015 specifications.
Examples
Solve dy/dx + 2y = 6 with y(0) = 1
Solution
Step 1: Identify as first-order linear in standard form.
dy/dx + P(x)y = Q(x), where P(x) = 2, Q(x) = 6
Step 2: Calculate the integrating factor.
μ(x) = e^(∫2 dx) = e^(2x)
Step 3: Multiply the entire equation by μ(x).
e^(2x)(dy/dx) + e^(2x)(2y) = e^(2x)(6)
Step 4: Recognize the left side as a derivative product.
d/dx[e^(2x)·y] = 6e^(2x)
Step 5: Integrate both sides.
∫d/dx[e^(2x)·y] dx = ∫6e^(2x) dx
e^(2x)·y = 6·(e^(2x)/2) + C
e^(2x)·y = 3e^(2x) + C
Step 6: Solve for y.
y = 3 + Ce^(−2x)
Step 7: Apply initial condition y(0) = 1.
1 = 3 + Ce^0
1 = 3 + C
C = −2
Step 8: Write the particular solution.
y = 3 − 2e^(−2x)
Verification: dy/dx = 4e^(−2x); 4e^(−2x) + 2(3 − 2e^(−2x)) = 4e^(−2x) + 6 − 4e^(−2x) = 6 ✓
An RL circuit has L = 0.5 H, R = 10 Ω, and applied voltage E(t) = 20 V. Starting with i(0) = 0, find the current i(t). The governing equation is: L(di/dt) + Ri = E
Solution
Step 1: Write the equation and convert to standard form.
0.5(di/dt) + 10i = 20
Divide by 0.5: di/dt + 20i = 40
Step 2: Identify P(t) = 20, Q(t) = 40
Step 3: Calculate the integrating factor.
μ(t) = e^(∫20 dt) = e^(20t)
Step 4: Multiply the entire equation by μ.
e^(20t)(di/dt) + e^(20t)(20i) = e^(20t)(40)
Step 5: Recognize the derivative product.
d/dt[e^(20t)·i] = 40e^(20t)
Step 6: Integrate both sides.
e^(20t)·i = 40·(e^(20t)/20) + C
e^(20t)·i = 2e^(20t) + C
Step 7: Solve for i.
i(t) = 2 + Ce^(−20t)
Step 8: Apply initial condition i(0) = 0.
0 = 2 + Ce^0
C = −2
Step 9: Write the particular solution.
i(t) = 2(1 − e^(−20t)) A
Physical interpretation: Current approaches the steady-state value of E/R = 20/10 = 2 A asymptotically as t → ∞. The time constant is τ = L/R = 0.5/10 = 0.05 s.
Solve dy/dx + (y/x) = sin(x) with y(π/2) = 0
Solution
Step 1: Identify as first-order linear.
dy/dx + P(x)y = Q(x), where P(x) = 1/x, Q(x) = sin(x)
Step 2: Calculate the integrating factor.
μ(x) = e^(∫(1/x) dx) = e^(ln|x|) = x (taking x > 0)
Step 3: Multiply the entire equation by μ(x) = x.
x(dy/dx) + x(y/x) = x·sin(x)
x(dy/dx) + y = x·sin(x)
Step 4: Recognize as d/dx[xy].
d/dx[xy] = x·sin(x)
Step 5: Integrate both sides.
xy = ∫x·sin(x) dx
Use integration by parts: u = x, dv = sin(x)dx
du = dx, v = −cos(x)
xy = −x·cos(x) + ∫cos(x) dx
xy = −x·cos(x) + sin(x) + C
Step 6: Solve for y.
y = (−x·cos(x) + sin(x) + C)/x
y = −cos(x) + sin(x)/x + C/x
Step 7: Apply initial condition y(π/2) = 0.
0 = −cos(π/2) + sin(π/2)/(π/2) + C/(π/2)
0 = 0 + 2/π + 2C/π
−2/π = 2C/π
C = −1
Step 8: Write the particular solution.
y = −cos(x) + sin(x)/x − 1/x
Key points
- Standard form: dy/dx + P(x)y = Q(x); P and Q are functions of x only
- Integrating factor: μ(x) = e^(∫P(x)dx)
- Multiply entire equation by μ to get: d/dx[μy] = μQ
- Integrate: μy = ∫μQ dx + C
- Solve for y: y = (1/μ)[∫μQ dx + C]
- Apply initial conditions to determine the arbitrary constant C
- Check: substitute solution back into original equation
4. Exact Differential Equations
An exact differential equation has the form:
M(x, y)dx + N(x, y)dy = 0
It is called exact when it represents the total (exact) differential of some function F(x, y):
dF = (∂F/∂x)dx + (∂F/∂y)dy
Comparing with M dx + N dy = 0, exactness requires:
∂M/∂y = ∂N/∂x
If this condition is satisfied, a solution exists in implicit form:
F(x, y) = C
Solution Method for Exact Equations:
- Write the equation in the form M dx + N dy = 0
- Test for exactness by verifying ∂M/∂y = ∂N/∂x
- If exact, find F by: F(x, y) = ∫M(x, y)dx (treating y as constant) + g(y)
- Determine g(y) by: ∂F/∂y = N, so g'(y) = N − ∂/∂y[∫M dx]
- Integrate g'(y) to find g(y)
- The solution is F(x, y) = C
- Apply initial conditions to find C
Why Exactness Matters in Engineering:
Exact equations arise naturally in thermodynamics, mechanics, and fluid flow where conservation laws (energy, mass, momentum) lead to exact differentials. In NSCP 2015 soil mechanics applications, exact equations appear in consolidation theory and seepage analysis.
Alternative Approach—Finding the Potential Function:
For an exact equation, there exists a "potential function" F such that M = ∂F/∂x and N = ∂F/∂y. Finding F directly (rather than verifying exactness) can sometimes be quicker if the equation has a recognizable form.
Examples
Solve (2x + y)dx + (x + 3y)dy = 0
Solution
Step 1: Write in standard form M dx + N dy = 0.
M(x, y) = 2x + y
N(x, y) = x + 3y
Step 2: Test for exactness.
∂M/∂y = ∂(2x + y)/∂y = 1
∂N/∂x = ∂(x + 3y)/∂x = 1
Since ∂M/∂y = ∂N/∂x = 1, the equation is EXACT.
Step 3: Find F(x, y) by integrating M with respect to x (treat y as constant).
F(x, y) = ∫(2x + y)dx = x² + xy + g(y)
where g(y) is an unknown function of y only.
Step 4: Differentiate F with respect to y and set equal to N.
∂F/∂y = x + g'(y) = N = x + 3y
Step 5: Solve for g'(y).
g'(y) = 3y
Step 6: Integrate to find g(y).
g(y) = ∫3y dy = (3y²)/2
Step 7: Write the complete F.
F(x, y) = x² + xy + (3y²)/2
Step 8: The implicit solution is F(x, y) = C.
x² + xy + (3y²)/2 = C
Or: 2x² + 2xy + 3y² = 2C (multiply by 2 to clear fractions)
Verification: ∂F/∂x = 2x + y = M ✓; ∂F/∂y = x + 3y = N ✓
Solve (6xy + 2y)dx + (3x² + 2x)dy = 0 with y(1) = 2
Solution
Step 1: Identify M and N.
M(x, y) = 6xy + 2y = y(6x + 2)
N(x, y) = 3x² + 2x = x(3x + 2)
Step 2: Test for exactness.
∂M/∂y = 6x + 2
∂N/∂x = 6x + 2
Since ∂M/∂y = ∂N/∂x, the equation is EXACT.
Step 3: Find F by integrating M with respect to x.
F(x, y) = ∫(6xy + 2y)dx = 6y·(x²/2) + 2y·x + g(y)
F(x, y) = 3x²y + 2xy + g(y)
Step 4: Differentiate F with respect to y.
∂F/∂y = 3x² + 2x + g'(y) = N = 3x² + 2x
Step 5: Solve for g'(y).
g'(y) = 0
Step 6: Integrate.
g(y) = 0 (constant can be absorbed into the C)
Step 7: Write F.
F(x, y) = 3x²y + 2xy
Step 8: Implicit solution.
3x²y + 2xy = C
Step 9: Apply initial condition y(1) = 2.
3(1)²(2) + 2(1)(2) = C
6 + 4 = C
C = 10
Step 10: Particular solution.
3x²y + 2xy = 10
Or, factoring: xy(3x + 2) = 10
Key points
- Exact form: M dx + N dy = 0 with ∂M/∂y = ∂N/∂x
- Test: Check if ∂M/∂y = ∂N/∂x; if yes, equation is exact
- Solution approach: Find F(x, y) such that ∂F/∂x = M and ∂F/∂y = N
- Implicit solution: F(x, y) = C
- Exactness fails if ∂M/∂y ≠ ∂N/∂x (may need integrating factor to convert)
- Appears in thermodynamic and conservative force problems
5. Higher-Order Linear Differential Equations with Constant Coefficients
Higher-order linear differential equations with constant coefficients appear extensively in structural dynamics, vibration analysis, and control systems. The general form is:
a_n(d^n y/dx^n) + a_(n-1)(d^(n-1) y/dx^(n-1)) + ... + a_1(dy/dx) + a_0·y = f(x)
where a_0, a_1, ..., a_n are constants and f(x) is the forcing function.
Homogeneous Case (f(x) = 0):
The most important case for initial applications is the homogeneous equation:
ay'' + by' + cy = 0
Solution approach uses the characteristic equation:
am² + bm + c = 0
The roots m_1, m_2 determine the form of the general solution:
Case 1: Distinct Real Roots (m_1 ≠ m_2)
y = C_1·e^(m_1·x) + C_2·e^(m_2·x)
Case 2: Repeated Real Root (m_1 = m_2 = m)
y = (C_1 + C_2·x)·e^(m·x)
Case 3: Complex Conjugate Roots (m = α ± βi)
y = e^(α·x)·(C_1·cos(β·x) + C_2·sin(β·x))
Alternatively: y = e^(α·x)·(A·cos(β·x + φ))
Why This Works—Theory:
We assume a solution of the form y = e^(mx). Substituting into ay'' + by' + cy = 0:
a·m²·e^(mx) + b·m·e^(mx) + c·e^(mx) = 0
e^(mx)(am² + bm + c) = 0
Since e^(mx) ≠ 0, we must have am² + bm + c = 0, the characteristic equation.
Civil Engineering Applications:
- Structural Dynamics (Undamped): For a building's fundamental vibration mode:
m·ẍ + k·x = 0
Characteristic: m·λ² + k = 0 → λ = ±i√(k/m) = ±iω₀
Solution: x(t) = A·cos(ω₀·t + φ) — harmonic oscillation at natural frequency ω₀ = √(k/m) rad/s
- Damped Vibrations: m·ẍ + c·ẋ + k·x = 0
Characteristic: m·λ² + c·λ + k = 0
Discriminant determines underdamped (complex roots), critically damped (repeated root), or overdamped (distinct real roots)
Critical damping (c_crit = 2√(mk)) is important in seismic design per NSCP 2015.
- Beam Deflection (Euler-Bernoulli): EI·d⁴y/dx⁴ = w(x)
For statically indeterminate beams, the solution involves solving a 4th-order homogeneous equation first.
Examples
Solve y'' − 5y' + 6y = 0 with y(0) = 1, y'(0) = 2
Solution
Step 1: Write the characteristic equation.
m² − 5m + 6 = 0
Step 2: Solve for roots (factor).
(m − 2)(m − 3) = 0
m₁ = 2, m₂ = 3
Step 3: Write the general solution (distinct real roots).
y = C₁e^(2x) + C₂e^(3x)
Step 4: Apply first initial condition y(0) = 1.
1 = C₁e^0 + C₂e^0
1 = C₁ + C₂ ... (Equation 1)
Step 5: Find y' and apply second initial condition.
y' = 2C₁e^(2x) + 3C₂e^(3x)
At x = 0: y'(0) = 2 = 2C₁ + 3C₂ ... (Equation 2)
Step 6: Solve the system of equations.
From Eq. 1: C₁ = 1 − C₂
Substitute into Eq. 2: 2(1 − C₂) + 3C₂ = 2
2 − 2C₂ + 3C₂ = 2
C₂ = 0
Therefore: C₁ = 1
Step 7: Write the particular solution.
y = e^(2x)
Verification: y'' − 5y' + 6y = 4e^(2x) − 5(2e^(2x)) + 6e^(2x) = 4e^(2x) − 10e^(2x) + 6e^(2x) = 0 ✓
Solve y'' − 4y' + 4y = 0 with y(0) = 2, y'(0) = 1
Solution
Step 1: Write the characteristic equation.
m² − 4m + 4 = 0
Step 2: Solve for roots (perfect square).
(m − 2)² = 0
m = 2 (repeated root)
Step 3: Write the general solution (repeated root).
y = (C₁ + C₂x)e^(2x)
Step 4: Apply first initial condition y(0) = 2.
2 = (C₁ + C₂·0)e^0
2 = C₁ ... (Equation 1)
Step 5: Find y' and apply second initial condition.
y' = C₂e^(2x) + (C₁ + C₂x)·2e^(2x)
y' = [C₂ + 2(C₁ + C₂x)]e^(2x)
y' = [C₂ + 2C₁ + 2C₂x]e^(2x)
At x = 0: y'(0) = 1 = (C₂ + 2C₁)e^0
1 = C₂ + 2C₁ ... (Equation 2)
Step 6: Substitute C₁ = 2 into Eq. 2.
1 = C₂ + 2(2)
1 = C₂ + 4
C₂ = −3
Step 7: Write the particular solution.
y = (2 − 3x)e^(2x)
Physical interpretation: In structural damping, repeated roots indicate critical damping—the system returns to equilibrium without oscillation.
A simply-supported beam section undergoes free vibration: d²x/dt² + 16x = 0 with x(0) = 0.05 m and dx/dt(0) = 0. Find x(t). (This models a building sway mode, where 4 rad/s is the natural frequency.)
Solution
Step 1: Write the characteristic equation.
m² + 16 = 0
m² = −16
m = ±4i
Step 2: Identify as complex roots α ± βi with α = 0, β = 4.
Step 3: Write the general solution (complex roots).
x(t) = e^(0·t)(C₁cos(4t) + C₂sin(4t))
x(t) = C₁cos(4t) + C₂sin(4t)
Step 4: Apply first initial condition x(0) = 0.05.
0.05 = C₁cos(0) + C₂sin(0)
0.05 = C₁
Step 5: Find dx/dt and apply second initial condition.
dx/dt = −4C₁sin(4t) + 4C₂cos(4t)
At t = 0: dx/dt(0) = 0 = −4C₁sin(0) + 4C₂cos(0)
0 = 4C₂
C₂ = 0
Step 6: Write the particular solution.
x(t) = 0.05cos(4t) m
Step 7: Interpret the result.
Amplitude: 0.05 m (5 cm)
Angular frequency: ω₀ = 4 rad/s
Period: T = 2π/4 = π/2 ≈ 1.57 s
The building sways with amplitude 5 cm at a period of 1.57 seconds—a typical low-rise building fundamental period (NSCP 2015 Table 6-2).
A structure undergoes damped vibration: d²x/dt² + 2(dx/dt) + 10x = 0 with x(0) = 1 m, dx/dt(0) = 0. Find x(t) and describe the motion.
Solution
Step 1: Write the characteristic equation.
m² + 2m + 10 = 0
Step 2: Use the quadratic formula.
m = (−2 ± √(4 − 40))/2 = (−2 ± √(−36))/2
m = (−2 ± 6i)/2 = −1 ± 3i
Step 3: Identify α = −1, β = 3.
Step 4: Write the general solution.
x(t) = e^(−t)(C₁cos(3t) + C₂sin(3t))
Step 5: Apply first initial condition x(0) = 1.
1 = e^0(C₁cos(0) + C₂sin(0))
1 = C₁
Step 6: Find dx/dt.
dx/dt = −e^(−t)(C₁cos(3t) + C₂sin(3t)) + e^(−t)(−3C₁sin(3t) + 3C₂cos(3t))
dx/dt = e^(−t)[−C₁cos(3t) − C₂sin(3t) − 3C₁sin(3t) + 3C₂cos(3t)]
dx/dt = e^(−t)[(−C₁ + 3C₂)cos(3t) + (−C₂ − 3C₁)sin(3t)]
Step 7: Apply second initial condition dx/dt(0) = 0.
0 = e^0[(−C₁ + 3C₂)cos(0) + (−C₂ − 3C₁)sin(0)]
0 = −C₁ + 3C₂
With C₁ = 1: 0 = −1 + 3C₂ → C₂ = 1/3
Step 8: Write the particular solution.
x(t) = e^(−t)(cos(3t) + (1/3)sin(3t)) m
Step 9: Describe the motion.
- The exponential factor e^(−t) causes the amplitude to decay at rate 1/s
- The oscillation occurs at frequency 3 rad/s
- As t → ∞, x(t) → 0
- This is underdamped motion: oscillates while decaying (damping ratio ζ = α/√(α² + β²) = 1/√(1+9) ≈ 0.316 < 1)
- Critical damping would be c_crit = 2√(km) = 2√10 ≈ 6.32; actual c = 2 < c_crit
Key points
- Homogeneous: ay'' + by' + cy = 0; assumes y = e^(mx)
- Characteristic equation: am² + bm + c = 0; solve for roots m
- Distinct real roots m₁, m₂: y = C₁e^(m₁x) + C₂e^(m₂x)
- Repeated root m: y = (C₁ + C₂x)e^(mx)
- Complex roots α ± βi: y = e^(αx)(C₁cos(βx) + C₂sin(βx))
- General solution has constants equal to equation order
- Apply n initial conditions to find n constants (nth-order equation)
- Appears in structural vibrations, electrical circuits, thermal transients
6. Non-Homogeneous Equations and Particular Solutions
When a differential equation includes a non-zero forcing function (right-hand side term), we have a non-homogeneous (or inhomogeneous) equation:
ay'' + by' + cy = f(x)
The general solution consists of two parts:
y_general = y_homogeneous + y_particular
where:
- y_homogeneous (y_h) is the general solution to the homogeneous equation (ay'' + by' + cy = 0)
- y_particular (y_p) is any single solution to the full non-homogeneous equation
The homogeneous part is solved using the characteristic equation method from Section 5. The challenge is finding y_p.
Method of Undetermined Coefficients (for constant-coefficient equations with simple f(x)):
This method works when f(x) is a polynomial, exponential, sine, cosine, or combination thereof.
The procedure:
- Identify the form of f(x)
- Make an initial guess for y_p based on the form of f(x) (see table below)
- If any term in your guess appears in y_h, multiply the entire guess by x (or x² if necessary)
- Substitute y_p, y'_p, and y''_p into the non-homogeneous equation
- Equate coefficients on both sides to solve for unknown constants in y_p
- Combine: y = y_h + y_p
- Apply initial conditions to find arbitrary constants in y_h
Guess Table for y_p (Method of Undetermined Coefficients):
f(x) Form → Initial Guess for y_p (before checking resonance):
- f(x) = P_n(x) (polynomial of degree n) → y_p = A_n·x^n + A_(n-1)·x^(n-1) + ... + A_0
- f(x) = e^(αx) → y_p = A·e^(αx)
- f(x) = sin(βx) or cos(βx) → y_p = A·sin(βx) + B·cos(βx)
- f(x) = P_n(x)·e^(αx) → y_p = (A_n·x^n + ... + A_0)·e^(αx)
- f(x) = P_n(x)·sin(βx) or similar → y_p = (A_n·x^n + ... + A_0)·sin(βx) + (B_n·x^n + ... + B_0)·cos(βx)
- Sums: If f(x) = f₁(x) + f₂(x), then y_p = y_(p1) + y_(p2)
Resonance Adjustment:
If any term in the initial guess for y_p is already part of y_h (a solution to the homogeneous equation), multiply the entire guess by x (or x² if the repetition is worse). This occurs when the forcing frequency matches a natural frequency—critical in vibration analysis.
Physical Significance in Civil Engineering:
- Forced Vibrations (NSCP 2015 Seismic Design): Buildings subjected to ground motion f(t) follow:
m·ẍ + c·ẋ + k·x = F_ground(t)
- Steady-State Response: The particular solution represents the steady-state response to continuous forcing (wind, traffic, machinery). The homogeneous part (transient response) decays if damping is present.
- Resonance: If the forcing frequency matches the natural frequency ω₀ = √(k/m), the system resonates—amplitude grows without bound (in undamped case) or reaches a peak in the damped case. NSCP 2015 mandates avoiding resonance through proper design of T (period).
Alternative Method—Variation of Parameters:
For non-homogeneous equations where f(x) doesn't fit the undetermined coefficients method, the general variation of parameters technique works:
If y₁, y₂ are solutions to the homogeneous equation, then:
y_p = −y₁∫(y₂·f/W)dx + y₂∫(y₁·f/W)dx
where W = y₁y'₂ − y'₁y₂ is the Wronskian (a measure of linear independence). This is more labor-intensive but universally applicable.
Examples
Solve y'' − 3y' + 2y = x with y(0) = 0, y'(0) = 0
Solution
Step 1: Find y_h by solving the characteristic equation.
m² − 3m + 2 = 0
(m − 1)(m − 2) = 0
m = 1, 2
y_h = C₁e^x + C₂e^(2x)
Step 2: Find y_p using undetermined coefficients.
f(x) = x (degree 1 polynomial)
Initial guess: y_p = Ax + B
y'_p = A
y''_p = 0
Step 3: Check for resonance.
The terms in y_p (Ax, B) do not appear in y_h, so no adjustment needed.
Step 4: Substitute into the non-homogeneous equation.
0 − 3(A) + 2(Ax + B) = x
−3A + 2Ax + 2B = x
2Ax + (−3A + 2B) = x + 0
Step 5: Equate coefficients.
Coefficient of x: 2A = 1 → A = 1/2
Constant term: −3A + 2B = 0 → −3(1/2) + 2B = 0 → B = 3/4
Step 6: Write y_p.
y_p = (1/2)x + 3/4
Step 7: General solution.
y = C₁e^x + C₂e^(2x) + (1/2)x + 3/4
Step 8: Apply initial condition y(0) = 0.
0 = C₁e^0 + C₂e^0 + 0 + 3/4
0 = C₁ + C₂ + 3/4
C₁ + C₂ = −3/4 ... (Eq. 1)
Step 9: Find y' and apply y'(0) = 0.
y' = C₁e^x + 2C₂e^(2x) + 1/2
At x = 0: 0 = C₁ + 2C₂ + 1/2
C₁ + 2C₂ = −1/2 ... (Eq. 2)
Step 10: Solve the system.
From Eq. 1: C₁ = −3/4 − C₂
Substitute into Eq. 2: (−3/4 − C₂) + 2C₂ = −1/2
−3/4 + C₂ = −1/2
C₂ = −1/2 + 3/4 = 1/4
C₁ = −3/4 − 1/4 = −1
Step 11: Particular solution.
y = −e^x + (1/4)e^(2x) + (1/2)x + 3/4
Solve y'' − 4y = 3e^(2x) with y(0) = 0, y'(0) = 2
Solution
Step 1: Find y_h.
m² − 4 = 0 → m = ±2
y_h = C₁e^(2x) + C₂e^(−2x)
Step 2: Find y_p using undetermined coefficients.
f(x) = 3e^(2x)
Initial guess: y_p = Ae^(2x)
Step 3: Check for resonance.
The term e^(2x) DOES appear in y_h (as C₁e^(2x)), so we multiply the guess by x:
y_p = Axe^(2x)
Step 4: Calculate derivatives.
y'_p = Ae^(2x) + Ax·2e^(2x) = Ae^(2x)(1 + 2x)
y''_p = A·2e^(2x)(1 + 2x) + Ae^(2x)·2 = Ae^(2x)[2(1 + 2x) + 2]
y''_p = Ae^(2x)(2 + 4x + 2) = Ae^(2x)(4 + 4x)
Step 5: Substitute into y'' − 4y = 3e^(2x).
Ae^(2x)(4 + 4x) − 4Axe^(2x) = 3e^(2x)
Ae^(2x)(4 + 4x − 4x) = 3e^(2x)
4Ae^(2x) = 3e^(2x)
4A = 3
A = 3/4
Step 6: Write y_p.
y_p = (3/4)xe^(2x)
Step 7: General solution.
y = C₁e^(2x) + C₂e^(−2x) + (3/4)xe^(2x)
Step 8: Apply y(0) = 0.
0 = C₁ + C₂ + 0
C₁ + C₂ = 0 ... (Eq. 1)
Step 9: Find y' and apply y'(0) = 2.
y' = 2C₁e^(2x) − 2C₂e^(−2x) + (3/4)e^(2x) + (3/4)x·2e^(2x)
y' = 2C₁e^(2x) − 2C₂e^(−2x) + (3/4)e^(2x)(1 + 2x)
At x = 0: 2 = 2C₁ − 2C₂ + 3/4
2C₁ − 2C₂ = 2 − 3/4 = 5/4
C₁ − C₂ = 5/8 ... (Eq. 2)
Step 10: Solve.
From Eq. 1: C₁ = −C₂
Substitute into Eq. 2: −C₂ − C₂ = 5/8 → −2C₂ = 5/8 → C₂ = −5/16
C₁ = 5/16
Step 11: Particular solution.
y = (5/16)e^(2x) − (5/16)e^(−2x) + (3/4)xe^(2x)
A one-story building is modeled as: m·ẍ + c·ẋ + k·x = F₀·cos(ωt), where m = 10,000 kg, c = 50,000 N·s/m, k = 1,000,000 N/m, F₀ = 10,000 N, ω = 4 rad/s. Find the steady-state displacement (particular solution only). [This models wind or earthquake forcing.]
Solution
Step 1: Divide by m to standard form.
ẍ + 5ẋ + 100x = 1·cos(4t)
(where 1 = F₀/m = 10,000/10,000)
Step 2: Set up the particular solution.
For f(t) = 1·cos(4t), guess:
x_p = A·cos(4t) + B·sin(4t)
Step 3: Calculate derivatives.
ẋ_p = −4A·sin(4t) + 4B·cos(4t)
ẍ_p = −16A·cos(4t) − 16B·sin(4t)
Step 4: Substitute into ẍ + 5ẋ + 100x = cos(4t).
[−16A·cos(4t) − 16B·sin(4t)] + 5[−4A·sin(4t) + 4B·cos(4t)] + 100[A·cos(4t) + B·sin(4t)] = cos(4t)
Step 5: Collect cos(4t) and sin(4t) terms.
cos(4t): −16A + 20B + 100A = 1 → 84A + 20B = 1
sin(4t): −16B − 20A + 100B = 0 → −20A + 84B = 0
Step 6: Solve the system.
From sin equation: 84B = 20A → B = (20/84)A = (5/21)A
Substitute into cos equation: 84A + 20(5/21)A = 1
84A + (100/21)A = 1
A[84 + 100/21] = 1
A[(1764 + 100)/21] = 1
A(1864/21) = 1
A = 21/1864 ≈ 0.01127 m
B = (5/21)·(21/1864) = 5/1864 ≈ 0.002682 m
Step 7: Amplitude of steady-state response.
Amplitude = √(A² + B²) = √[(21/1864)² + (5/1864)²]
Amplitude = √[(441 + 25)/1864²] = √(466)/1864 ≈ 0.01147 m ≈ 11.47 mm
Step 8: Phase angle.
tan(φ) = B/A = (5/21) → φ = arctan(5/21) ≈ 13.4°
Step 9: Steady-state solution.
x_p(t) ≈ 0.01147·cos(4t − 13.4°) m
Or: x_p(t) = 0.01127·cos(4t) + 0.002682·sin(4t) m
Step 10: Engineering interpretation.
- The building oscillates with the applied frequency 4 rad/s (T = 2π/4 ≈ 1.57 s)
- Maximum displacement ≈ 11.47 mm
- Phase lag of 13.4° means the building's response lags the force
- This is safe; frequency 4 rad/s is far from natural frequency ω₀ = √(k/m) = √100 = 10 rad/s (resonance would cause much larger amplitude)
- NSCP 2015 requires the building period away from 0.5–2.0 s critical range to minimize seismic response
Key points
- Non-homogeneous: ay'' + by' + cy = f(x) with f(x) ≠ 0
- General solution: y = y_h (homogeneous) + y_p (particular)
- y_h from characteristic equation; y_p by undetermined coefficients or variation of parameters
- Undetermined coefficients guess based on form of f(x); adjust if resonance occurs
- Resonance: multiply guess by x if a forcing term matches a natural frequency
- Steady-state response given by y_p; transient response by y_h
- Apply initial conditions to the complete solution y = y_h + y_p
7. Applications of Differential Equations in Civil Engineering
Differential equations are the language of engineering analysis. This section covers key applications directly relevant to civil engineering practice and the PRC licensure exam.
7.1 Growth and Decay—Radioactive Contamination and Material Degradation
The differential equation dy/dt = ky models exponential growth (k > 0) or decay (k < 0). General solution: y(t) = y₀e^(kt)
Applications:
- Radioactive decay: Half-life problems (carbon dating, nuclear waste disposal)
- Material degradation: Concrete strength loss, reinforcement corrosion (relevant to NSCP 2015 durability requirements)
- Pollutant concentration: Decrease of contaminants in soil or water
Example: A concrete beam loses strength at rate proportional to current strength. If initial strength is f_c = 30 MPa and strength loss rate is 2% per year, find the remaining strength after 20 years.
Solution: df_c/dt = −0.02·f_c → f_c(t) = 30e^(−0.02t)
After 20 years: f_c(20) = 30e^(−0.4) ≈ 20.1 MPa
7.2 Newton's Law of Cooling—Concrete Curing and Temperature Control
The governing equation: dT/dt = −k(T − T_ambient)
General solution: T(t) = T_ambient + (T₀ − T_ambient)e^(−kt)
where T₀ is initial temperature and k is the cooling rate constant.
Applications:
- Concrete curing: Predicting internal temperature during hydration (NSCP 2015 Section 6.4 on curing)
- Thermal load analysis: Daily temperature variation in structures
- Steel fabrication: Cooling rate effects on mechanical properties
Example: Fresh concrete is poured at T₀ = 60°C when ambient is 25°C. The cooling constant k = 0.1 hr⁻¹. Find temperature after 4 hours.
Solution: T(t) = 25 + (60 − 25)e^(−0.1·4) = 25 + 35e^(−0.4) ≈ 25 + 35(0.6703) ≈ 48.5°C
7.3 Mixing Problems—Water Quality and Material Composition
For a tank where liquid enters and exits, the rate equation is:
dQ/dt = (rate in) − (rate out)
where Q is the quantity of solute (salt, contaminant, etc.).
Typical setup:
- Tank volume: V (constant)
- Inflow rate: r_in L/s at concentration c_in
- Outflow rate: r_out L/s
- Tank concentration at time t: C(t) = Q(t)/V
Differential equation: dQ/dt = c_in·r_in − (Q/V)·r_out
Applications:
- Water treatment plant design: Pollutant removal or additive mixing
- Wastewater treatment: Contaminant reduction in tanks
- Deicing salt application: Salt concentration in groundwater (environmental concern in cold climates)
Example: A 1000 L tank initially contains pure water. Saltwater (0.1 kg/L) enters at 10 L/min; mixture exits at 10 L/min. Find salt content Q(t) and time to reach 50 kg salt.
Solution:
dQ/dt = 0.1(10) − (Q/1000)(10) = 1 − Q/100
This is first-order linear: dQ/dt + Q/100 = 1
Integrating factor: μ = e^(t/100)
Q·e^(t/100) = ∫e^(t/100) dt = 100e^(t/100) + C
With Q(0) = 0: C = −100
Q(t) = 100(1 − e^(−t/100))
For Q = 50: 50 = 100(1 − e^(−t/100)) → e^(−t/100) = 0.5 → t = 100·ln(2) ≈ 69.3 min
7.4 Structural Dynamics and Vibration Analysis
Free vibration (no external force): m·ẍ + c·ẋ + k·x = 0
Classification:
- Underdamped (ζ < 1): Oscillates with decaying amplitude; most common in structures
- Critically damped (ζ = 1): Returns to rest without oscillating; ideal for seismic dampers
- Overdamped (ζ > 1): Returns slowly without oscillating; energy dissipation slow
where damping ratio ζ = c/(2√(mk)) and natural frequency ω₀ = √(k/m)
NSCP 2015 Applications:
- Fundamental period: T = 2π/ω₀; NSCP 2015 Section 6.2 limits T to avoid resonance
- Seismic response: Ground acceleration a_g excites the building; m·ẍ + c·ẋ + k·x = −m·a_g(t)
- Wind sway: Dynamic response to wind gusts modeled as periodic forcing
- Base isolation: Reduced k or increased ζ lowers response amplitude
Example: A 3-story building has m = 300 metric tons, k = 10 MN/m, c = 0.5 MN·s/m. Find the natural frequency, damping ratio, and period.
Solution:
- m = 300,000 kg
- k = 10,000,000 N/m
- c = 500,000 N·s/m
ω₀ = √(k/m) = √(10,000,000/300,000) = √(33.33) ≈ 5.77 rad/s
ζ = c/(2√(mk)) = 500,000/(2√(300,000 × 10,000,000)) = 500,000/(2 × 1,732,050.8) ≈ 0.144
T = 2π/ω₀ = 2π/5.77 ≈ 1.09 s
Interpretation: The building has a period of 1.09 s in underdamped mode (ζ = 0.144 < 1). From NSCP 2015 seismic maps, if the building is in a region with high response at 1.0 s, this period could lead to excessive sway—design mitigation (increase damping or stiffness) would be needed.
7.5 Fluid Flow and Drainage Design
Orifice flow: dh/dt = −C√(2gh), where h is water depth, C is a discharge coefficient function of orifice geometry, and g is gravitational acceleration.
Applications:
- Tank and basin drainage: Design of catch basins, reservoirs, and water retention systems
- Slope drainage: Surface and subsurface water removal to prevent slope failure
- Foundation design: Drainage systems to control hydrostatic pressure (NSCP 2015 Section 4 on geotechnical design)
Example: A cylindrical basin (A = 5 m²) with h₀ = 2 m drains through an orifice with C = 0.15 m^(5/2)/(s√m). Find drain time.
Solution: From dh/dt = −0.15√(2·9.81·h) = −0.15√(19.62h)
Separate: dh/√h = −0.15√19.62 dt
2√h = −0.15√19.62·t + K
With h(0) = 2: K = 2√2 = 2.828
2√h = −0.667t + 2.828
Drain time (h → 0): 0 = −0.667t + 2.828 → t ≈ 4.24 s (approximately; need numerical integration for exact value)
7.6 Seepage and Groundwater Flow (Geotechnical Applications)
Darcy's law and the diffusivity equation for saturated flow in soil:
∂h/∂t = D(∂²h/∂x²)
where h is hydraulic head and D is hydraulic diffusivity (often treated as constant).
For 1D consolidation (Terzaghi's theory), similar diffusivity equation governs excess pore pressure:
∂u/∂t = c_v(∂²u/∂z²)
where u is excess pore pressure and c_v is coefficient of consolidation.
Applications:
- Settlement predictions: Time for consolidation to complete under a new building load
- Piping analysis: Seepage path and gradient calculation to prevent internal erosion
- Dewatering: Drawdown prediction around pumping wells
Time factor (NSCP 2015 geotechnical design): T_v = c_v·t/H², where H is drainage path length, gives degree of consolidation U. Typical values: U = 50% at T_v = 0.197, U = 90% at T_v = 0.848.
Examples
A reinforced concrete structure in a tropical marine environment experiences strength loss due to corrosion. Initial 28-day compressive strength f_c = 35 MPa. Environmental testing shows the strength degradation follows df_c/dt = −0.035·f_c per year. Estimate the effective design strength after 20 years of service. What is the half-life of the concrete's strength (time to reach 50% of original)?
Solution
Part 1: Find strength after 20 years
Step 1: Recognize as exponential decay.
df_c/dt = −0.035·f_c
Solution: f_c(t) = f_{c0}·e^(−0.035t)
with f_{c0} = 35 MPa
Step 2: Calculate f_c at t = 20 years.
f_c(20) = 35·e^(−0.035·20) = 35·e^(−0.7)
f_c(20) = 35·(0.4966) ≈ 17.38 MPa
Step 3: Percentage retention.
(17.38/35) × 100% ≈ 49.7% ≈ 50% (half strength after ~20 years)
Part 2: Find half-life
Step 1: Set f_c(t_half) = 0.5·f_{c0}
0.5·f_{c0} = f_{c0}·e^(−0.035·t_half)
0.5 = e^(−0.035·t_half)
ln(0.5) = −0.035·t_half
−0.6931 = −0.035·t_half
t_half = 0.6931/0.035 ≈ 19.8 years
Conclusion: The half-life is approximately 20 years. NSCP 2015 durability design (Section 4.3) recommends material selection and protective measures to limit this degradation in marine environments. After 20 years, the strength drops to 17.38 MPa—a 50% loss that may compromise the original design assumptions.
A building has mass m = 200 metric tons and lateral stiffness k = 8 MN/m. Design a damping system to achieve critical damping (ζ = 1) for earthquake resilience. Find: (a) required damping coefficient c_crit, (b) natural frequency ω₀, (c) the period T, and (d) describe the motion after an impulse.
Solution
Part (a): Required damping coefficient for critical damping
Step 1: Recall critical damping condition.
ζ = 1 means c = c_crit = 2√(km)
Step 2: Calculate with given values.
m = 200,000 kg
k = 8,000,000 N/m
c_crit = 2√(200,000 × 8,000,000)
c_crit = 2√(1.6 × 10¹²)
c_crit = 2 × 1.265 × 10⁶ = 2.53 × 10⁶ N·s/m ≈ 2.53 MN·s/m
Part (b): Natural frequency
ω₀ = √(k/m) = √(8,000,000/200,000) = √40 ≈ 6.32 rad/s
Part (c): Period
T = 2π/ω₀ = 2π/6.32 ≈ 0.994 s ≈ 1.0 s
Part (d): Motion after an impulse (e.g., earthquake ground acceleration)
For critically damped motion x(t) = (C₁ + C₂·t)e^(ω₀·t) = (C₁ + C₂·t)e^(6.32t) (if we consider free response after initial disturbance).
More precisely, after ground shaking stops:
- The building displacement returns to zero WITHOUT oscillation
- Return is as fast as possible (fastest no-overshoot response)
- Residual motion stops quickly—important for tall buildings where oscillation causes secondary damage
Civil Engineering Note: Critical damping is ideal for seismic isolation systems and dampers (e.g., tuned mass dampers, magnetorheological dampers) per NSCP 2015 Chapter 17 (Seismic Design). In practice, structures are often slightly underdamped (ζ ≈ 0.05–0.1) due to material damping; additional dampers increase ζ toward critical for improved resilience.
Step 5: Practical implementation
A damping coefficient c_crit = 2.53 MN·s/m could be achieved via:
- Viscous fluid dampers: sized to dissipate energy proportional to velocity
- Friction dampers: slide at controlled force
- Magnetorheological dampers: variable damping via electromagnetic field (modern buildings)
Design ensures 1-second period matches NSCP 2015 design spectrum less resonant frequency region.
A clay soil layer (H = 3 m double drainage, i.e., H = 1.5 m single drainage) has coefficient of consolidation c_v = 0.04 m²/year. A new building foundation applies a load causing excess pore pressure u₀. Calculate: (a) time to 50% consolidation (U = 50%), (b) time to 90% consolidation, and (c) predict when settlement reaches 80% of ultimate settlement. [From NSCP 2015 geotechnical design guidelines]
Solution
Background: Terzaghi's 1D consolidation equation gives degree of consolidation U as a function of time factor:
T_v = c_v·t/H² (where H is drainage path length)
For U = 50%: T_v ≈ 0.197
For U = 90%: T_v ≈ 0.848
For U = 80%: T_v ≈ 0.568 (from consolidation tables)
Part (a): Time for 50% consolidation
Step 1: Use T_v = c_v·t/H² with T_v = 0.197
0.197 = (0.04 m²/year)·t/(1.5 m)²
0.197 = 0.04·t/2.25
t = (0.197 × 2.25)/0.04 = 0.443/0.04 = 11.08 years
Part (b): Time for 90% consolidation
With T_v = 0.848:
0.848 = 0.04·t/(1.5)²
t = (0.848 × 2.25)/0.04 = 1.908/0.04 = 47.7 years
Part (c): Time for 80% settlement
With T_v = 0.568:
0.568 = 0.04·t/(1.5)²
t = (0.568 × 2.25)/0.04 = 1.278/0.04 = 31.95 years ≈ 32 years
Conclusion and NSCP 2015 Compliance:
- 50% of primary consolidation occurs in ~11 years (relatively fast)
- 90% of primary consolidation occurs in ~48 years (slow tail of settlement)
- At 32 years, 80% of total settlement has occurred—building may settle at 80% after 3 decades
- NSCP 2015 Foundation Code (Chapter 4) requires settlement calculations and often limits differential settlement to ≤50 mm for buildings
This consolidation timeline affects:
- Design of utility connections (allow slack to accommodate future settlement)
- Monitoring programs (establish settlement benchmarks)
- Long-term serviceability (crack control in structures sensitive to differential movement)
- If settlement is unacceptable, mitigation includes improved drainage (increase c_v), preloading, or deep pile foundations to bypass compressible layer
Key points
- Growth/decay: y = y₀e^(kt); find k from known data point, then extrapolate
- Cooling: T(t) = T_amb + (T₀ − T_amb)e^(−kt); predicts transient thermal response
- Mixing: dQ/dt = (rate in) − (rate out); use first-order linear method
- Vibration: m·ẍ + c·ẋ + k·x = 0; characteristic equation gives oscillation behavior
- NSCP 2015: Period T = 2π/√(k/m); avoid resonance bands for seismic safety
- Drainage: dh/dt ∝ √h (orifice); separable equation → time to drain
- Consolidation: ∂u/∂t = c_v·∂²u/∂z²; predicts settlement time via time factor T_v
- Verify solutions against physical constraints (e.g., decay should approach zero)
Diagrams
Differential Equation Classification and Solution Strategy
This flowchart guides the problem-solver through differential equation identification and solution method selection. Starting from a given DE, students classify by type (ODE vs. PDE, order, first-order category, or higher-order with/without forcing), then apply the appropriate solution technique. Branches lead to specific characteristic equations, integrating factors, or particular solution guessing strategies. The final verification step ensures correctness before claiming a solution.
First-Order Differential Equations: Methods and Solution Paths
A comprehensive mind map organizing first-order differential equation types by form, solution method, and civil engineering applications. The map shows how to identify separable, linear, and exact equations, and branches into practical applications (growth/decay in population and materials, cooling of concrete, mixing in treatment tanks, and drainage design). This visual reinforces the decision-making process: recognize form → select method → apply solution steps → interpret physically.
Solution Workflow for Second-Order Linear Homogeneous DE with Initial Conditions
This sequence diagram traces the step-by-step workflow for solving a second-order linear homogeneous differential equation with constant coefficients using initial conditions. It shows the logical flow from writing the characteristic equation, analyzing its roots (three cases: distinct real, repeated, complex), selecting the corresponding general solution form, applying two initial conditions to create a system of equations, solving for the constants C₁ and C₂, and finally obtaining the unique particular solution. The diagram reinforces the importance of root analysis in determining solution form—a critical decision point in the problem-solving process.
Damping States in Structural Vibration: Free Response Classification
This state diagram illustrates the three fundamental damping regimes in structural dynamics determined by the damping ratio ζ. Each state represents a different response behavior to disturbances (earthquakes, wind, impulses). The diagram shows transitions as damping is increased, and identifies the characteristic solution form for each state. The underdamped state (ζ < 1) is most common in civil structures but can lead to resonance problems; critically damped (ζ = 1) is the design ideal for seismic dampers per NSCP 2015, providing fastest return without overshoot; overdamped (ζ > 1) provides slower but very stable response. This visual clarifies why solution form changes discontinuously from oscillatory (complex roots) to non-oscillatory (real repeated or distinct roots).
In summary
Differential equations are essential mathematical tools for civil engineering analysis. This chapter covered classification (order, degree, linear/nonlinear), first-order solution methods (separable, linear, exact), higher-order linear equations with constant coefficients (via characteristic equation), non-homogeneous equations (undetermined coefficients), and practical applications including growth/decay, Newton's cooling, mixing problems, structural vibrations (critical to NSCP 2015 seismic design), fluid drainage, and geotechnical consolidation. Key competencies for the PRC Civil Engineer Licensure Examination include: (1) identifying equation type and selecting appropriate solution method; (2) solving the characteristic equation and interpreting root types (distinct real, repeated, complex); (3) applying integrating factors for first-order linear equations; (4) handling resonance in forced vibrations; (5) connecting mathematical solutions to physical behavior (damping, oscillation, decay); and (6) interpreting results in the context of NSCP 2015 design standards. Mastery requires extensive practice with worked problems, careful attention to initial conditions, and verification of solutions. The transition from abstract differential equations to concrete engineering problems (predicting settlement time, seismic response, drainage duration) is where deep understanding emerges—practice these applications repeatedly for licensure exam readiness.