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CELE Engineering Mathematics — Differential EquationsDetailed Explanation

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Want to really understand Differential Equations before tackling CELE Engineering Mathematics questions? This detailed explanation breaks down every key concept, shows you why it matters for the CELE 2026, and walks through the reasoning Professional Regulation Commission (PRC) — Board of Civil Engineering expects on high-difficulty questions.

Exam context

Professional Regulation Commission (PRC) — Board of Civil Engineering runs the Civil Engineer Licensure Examination on May and November 2026. Its Engineering Mathematics section sits under a "Core" weighting, and Differential Equations is the 7th chapter in the 10-chapter CELE Engineering Mathematics rotation. The CELE passing mark is 70% weighted average, no sub-test below 50%, and the most recent 2026 paper drew about a meaningful share of questions from Engineering Mathematics.

Differential Equations - Detailed Explanation

Differential equations (DEs) are among the most frequently tested topics in the PRC Civil Engineer Licensure Examination under Engineering Mathematics. A differential equation is any equation that relates an unknown function to one or more of its derivatives. In civil engineering practice, DEs model real-world phenomena such as structural deflection, seepage flow, heat transfer through building envelopes, consolidation settlement, population growth in water demand forecasting, and radioactive decay of tracers used in hydraulic studies. Mastery of DEs is therefore both an examination necessity and a professional asset. This chapter covers: (1) classification of DEs, (2) solution methods for first-order DEs – separable, linear, and exact types, (3) higher-order linear DEs with constant coefficients, (4) engineering applications, and (5) an introduction to Laplace transforms. Every concept is illustrated with board-style worked examples in SI units at PRC CE examination difficulty level.

Concepts

Classification of Differential Equations

Before selecting a solution method, you must correctly classify the DE. The three key classification axes are: 1. ORDER – The order of a DE equals the highest derivative that appears. For example, dy/dx = 3x is first-order; d²y/dx² + 5y = 0 is second-order; d³y/dx³ – y = x is third-order. 2. DEGREE – The degree is the power (exponent) of the highest-order derivative, provided the equation is polynomial in its derivatives (i.e., radicals and fractions involving derivatives have been cleared). Example: (y'')³ + 2y' = 0 is second-order, third-degree. Caution: sin(y') = x has no defined degree because it is not polynomial in y'. 3. LINEARITY – A DE is linear if the dependent variable y and all its derivatives appear to the first power only, with no products among them. The general form of a linear nth-order ODE is: a_n(x)y^(n) + a_(n-1)(x)y^(n-1) + … + a_1(x)y' + a_0(x)y = g(x). Non-linear examples: y·y' = x (product of y and y'), (y')² = 1 (derivative squared), sin(y) = x (transcendental function of y). Additionally: Ordinary DEs (ODEs) have one independent variable; Partial DEs (PDEs) have two or more (e.g., the 2-D heat equation). The PRC CE board exam focuses almost exclusively on ODEs.

Examples

Always check degree on the HIGHEST-order derivative term after clearing fractions/radicals in the derivative terms. Here d²y/dx² is cubed, so degree = 3. Nonlinearity is triggered by the cube of the derivative.

Scenario

Classify: (d²y/dx²)³ + 4(dy/dx) – 2y = cos x

Solution

Order: 2 (highest derivative is d²y/dx²) Degree: 3 (the second derivative is raised to the 3rd power) Linearity: NON-LINEAR (because (d²y/dx²)³ — the derivative is raised to a power greater than 1) Type: ODE (one independent variable x)

This is the standard form for a first-order linear ODE: dy/dx + P(x)y = Q(x), where P(x) = 3 and Q(x) = e^x. Recognizing this form immediately tells you to use the integrating factor method.

Scenario

Classify: dy/dx + 3y = e^x

Solution

Order: 1 Degree: 1 Linearity: LINEAR (y and dy/dx appear to the first power; no products between them) Type: ODE — specifically a first-order linear ODE

This is the exact equation governing large-amplitude pendulum oscillations. Nonlinearity arises from sin(y), not from the derivative. For small angles, sin(y) ≈ y, linearising the equation — a common engineering approximation tested on boards.

Scenario

Classify: d²y/dx² + sin(y) = 0 (simple pendulum equation)

Solution

Order: 2 Degree: 1 (d²y/dx² appears to the first power) Linearity: NON-LINEAR (sin(y) is a transcendental, nonlinear function of y) Type: ODE

Applications

  • Structural beam deflection: EI(d⁴y/dx⁴) = w(x) — fourth-order linear ODE.
  • Groundwater flow (Darcy's law): first-order linear ODE for hydraulic head.
  • Consolidation theory (Terzaghi): parabolic PDE (treated as ODE in 1-D review problems).
  • Vibration of structures: second-order linear ODE with constant coefficients.
  • Water quality mixing in tanks: first-order linear ODE for concentration.

Misconceptions

  • Degree is NOT the same as order. Order = which derivative; degree = what power that derivative is raised to.
  • A DE with non-constant coefficients a_n(x) can still be LINEAR — linearity refers to y and its derivatives, not to the coefficient functions.
  • sin(y) makes a DE nonlinear in y, but sin(x) (a function of the independent variable only) does NOT affect linearity.
  • Clearing fractions/radicals in derivatives before determining degree is mandatory — many students skip this step and get the wrong degree.

Related Concepts

  • General vs particular solution
  • Initial value problems (IVP) and boundary value problems (BVP)
  • Separable ODEs
  • Linear first-order ODEs
  • Characteristic equation of higher-order ODEs

Common Exam Questions

Example

What is the order and degree of: (y''')² + 4y'' – y = 0? Answer: Order 3, Degree 2, Nonlinear (degree > 1 means nonlinear in the highest derivative).

Approach

Identify highest derivative → order. Raise the highest-order term to the power shown → degree. Check for products/transcendentals of y or its derivatives → linearity.

Question Type

Classification – multiple choice

Example

Is y·(dy/dx) = x linear? No — y is multiplied by dy/dx.

Approach

Check if dependent variable and all its derivatives appear only to the first power and are never multiplied together.

Question Type

Identify linear vs non-linear

Key Points To Remember

  • Order = highest derivative present; Degree = exponent of that highest derivative (only defined when polynomial in derivatives).
  • Linear means y and all derivatives appear to the first power, never multiplied together.
  • Clear all radicals or fractions containing derivatives before determining degree.
  • ODE = one independent variable; PDE = two or more independent variables.
  • Board exam classification questions often present a DE and ask for order, degree, or linearity — answer all three to avoid partial-credit traps.

First-Order Separable Differential Equations

A separable DE has the form dy/dx = g(x)·h(y), meaning the right-hand side factors into a function of x alone times a function of y alone. The solution strategy is: STEP 1 – Separate variables: Rearrange so all y-terms (including dy) are on the left and all x-terms (including dx) are on the right: dy / h(y) = g(x) dx STEP 2 – Integrate both sides: ∫ dy/h(y) = ∫ g(x) dx STEP 3 – Add one constant of integration (C) on one side only. STEP 4 – Apply initial condition (if given) to solve for C. STEP 5 – Solve for y explicitly if required (implicit solutions are sometimes acceptable). IMPORTANT: Division by h(y) is only valid if h(y) ≠ 0. Always check if h(y) = 0 gives additional (singular) solutions.

Examples

Key technique: after ln|y| = …, exponentiate to remove the natural log. The ±e^C₁ is combined into a single arbitrary constant C. Always substitute the IC into the GENERAL solution (with C still present) to find C.

Scenario

BOARD-STYLE: Solve dy/dx = xy, given y(0) = 2.

Solution

STEP 1 – Separate: dy/y = x dx STEP 2 – Integrate both sides: ∫ dy/y = ∫ x dx ln|y| = x²/2 + C₁ STEP 3 – Exponentiate: |y| = e^(x²/2 + C₁) = e^C₁ · e^(x²/2) y = C·e^(x²/2) where C = ±e^C₁ (absorb sign into C) STEP 4 – Apply IC: y(0) = 2 2 = C·e^0 = C → C = 2 FINAL ANSWER: y = 2e^(x²/2)

Two initial conditions determine both C and k. Always identify the two knowns: the initial amount (IC at t = 0, gives C) and a second data point (gives k). The decay constant k is negative — confirm this as a check.

Scenario

BOARD-STYLE: A radioactive substance decays at a rate proportional to the amount present. Initially 500 g are present; after 10 years, 400 g remain. Find the amount after 25 years.

Solution

MODEL: dA/dt = kA (k < 0 for decay) SEPARATE: dA/A = k dt INTEGRATE: ln|A| = kt + C₁ → A = Ce^(kt) IC 1: A(0) = 500 → C = 500 → A = 500e^(kt) IC 2: A(10) = 400 → 400 = 500e^(10k) e^(10k) = 400/500 = 0.8 10k = ln(0.8) = –0.2231 k = –0.02231 per year FIND A(25): A(25) = 500e^(–0.02231×25) = 500e^(–0.5579) A(25) = 500 × 0.5724 ≈ 286.2 g

This is an implicit general solution. The IC allows us to select the correct sign for the square root (positive branch, since initial value is positive). Implicit solutions are acceptable on board exams unless the problem explicitly asks you to solve for y.

Scenario

Solve: dy/dx = x/y with y(1) = 3.

Solution

STEP 1 – Separate: y dy = x dx STEP 2 – Integrate: y²/2 = x²/2 + C₁ → y² = x² + C (where C = 2C₁) STEP 3 – Apply IC: (3)² = (1)² + C → 9 = 1 + C → C = 8 PARTICULAR SOLUTION: y² = x² + 8 or y = √(x² + 8) (taking positive root since y(1) = 3 > 0)

Applications

  • Radioactive decay and carbon-14 dating in geotechnical investigations.
  • Population growth models for water supply demand forecasting.
  • Velocity of a falling body with air resistance: m(dv/dt) = mg – kv (separable when k is constant).
  • Seepage quantity through earth dams using Darcy's law.
  • Rate of chemical reactions in water treatment plants.

Misconceptions

  • Adding TWO constants of integration (one per side) — you must combine them into a single constant C.
  • Applying the initial condition BEFORE integrating — always integrate first.
  • Forgetting the ± when taking square roots or the absolute value when integrating 1/y → ln|y|.
  • Writing k as positive for decay — k must be NEGATIVE for a decreasing quantity.
  • Not checking if h(y) = 0 gives a valid singular solution (e.g., y = 0 for dy/dx = xy).

Related Concepts

  • Exponential growth and decay
  • Newton's Law of Cooling (a separable DE)
  • Linear first-order ODEs
  • Integration techniques (partial fractions, substitution)
  • Natural logarithm and exponential functions

Common Exam Questions

Example

Solve dy/dx = (x+1)/(y–1), y(0) = 2. Separate: (y–1)dy = (x+1)dx. Integrate: y²/2 – y = x²/2 + x + C. IC: 4/2 – 2 = 0 + 0 + C → C = 0. Answer: y² – 2y = x² + 2x, or (y–1)² = (x+1)² + const.

Approach

Separate → integrate → apply IC → solve for C → write final answer.

Question Type

Find the particular solution given IC

Example

Radium has a half-life of 1600 years. What fraction remains after 4000 years? k = –ln2/1600. Fraction = e^(k×4000) = 2^(–4000/1600) = 2^(–2.5) ≈ 0.1768 or 17.68%.

Approach

Write A = A₀e^(kt). Use half-life condition: A₀/2 = A₀e^(k·t_half) → k = –ln2/t_half. Then evaluate at the required time.

Question Type

Half-life / decay problems

Example

A population doubles in 10 years (k = ln2/10). When is it 5 times the original? t = ln5/k = ln5/(ln2/10) = 10·ln5/ln2 ≈ 23.22 years.

Approach

Set A(t) = nA₀, solve for t: nA₀ = A₀e^(kt) → t = ln(n)/k.

Question Type

Growth to a multiple

Key Points To Remember

  • Separation requires the RHS to factor as g(x)·h(y) — verify this factorability first.
  • Include only ONE constant of integration C on one side after integrating.
  • Apply the initial condition AFTER integrating, not before.
  • Check for singular (equilibrium) solutions where h(y) = 0.
  • The general solution contains C; a particular solution has C determined by an IC.
  • For decay/growth problems, dy/dt = ky is the prototypical separable DE.

First-Order Linear Differential Equations

The standard form is: dy/dx + P(x)·y = Q(x) If Q(x) = 0, it is homogeneous; if Q(x) ≠ 0, it is non-homogeneous (or forced). SOLUTION METHOD — Integrating Factor: STEP 1 – Write in standard form (coefficient of dy/dx must be 1). STEP 2 – Identify P(x) and Q(x). STEP 3 – Compute the integrating factor: μ(x) = e^(∫P(x)dx) (No constant of integration needed here — absorb into the final C.) STEP 4 – Multiply both sides by μ(x): μ·y' + μ·P·y = μ·Q(x) The left side is the exact derivative: d/dx[μ·y] = μ·Q(x) STEP 5 – Integrate both sides: μ·y = ∫μ·Q(x)dx + C STEP 6 – Solve for y: y = (1/μ)·[∫μ·Q(x)dx + C] STEP 7 – Apply IC to find C. WHY IT WORKS: The integrating factor μ converts the left side into the derivative of a product (product rule in reverse). This is the key insight.

Examples

Note how the integrating factor e^(2x) converts the left side. Integration by parts is often needed on the RHS when Q(x) is a polynomial. The answer has two parts: the particular solution (2x – 1) and the homogeneous solution (2e^(–2x)).

Scenario

BOARD-STYLE: Solve dy/dx + 2y = 4x, y(0) = 1.

Solution

STEP 1 – Standard form: already in standard form. P(x) = 2, Q(x) = 4x. STEP 2 – Integrating factor: μ = e^(∫2 dx) = e^(2x) STEP 3 – Multiply by μ: e^(2x)·dy/dx + 2e^(2x)·y = 4x·e^(2x) d/dx[e^(2x)·y] = 4x·e^(2x) STEP 4 – Integrate RHS (by parts: u = 4x, dv = e^(2x)dx): ∫4x·e^(2x)dx = 4x·(e^(2x)/2) – 4·(e^(2x)/4) = 2x·e^(2x) – e^(2x) + C = e^(2x)(2x – 1) + C STEP 5 – So: e^(2x)·y = e^(2x)(2x–1) + C y = 2x – 1 + Ce^(–2x) STEP 6 – Apply IC: 1 = 2(0) – 1 + C → C = 2 FINAL ANSWER: y = 2x – 1 + 2e^(–2x)

The equilibrium (steady-state) salt content = 1000 g (= 200 L × 5 g/L). As t→∞, A→1000 g. At t = 30 min, the tank has reached about 45% of steady state. Tank problems always follow: dA/dt = Rate_in – Rate_out.

Scenario

BOARD-STYLE (Mixing Tank): A 200-L tank initially contains pure water. Brine with salt concentration 5 g/L enters at 4 L/min; the well-mixed solution leaves at 4 L/min. Find the salt content (g) after 30 min.

Solution

Let A(t) = salt in tank (g) at time t (min). Volume = constant = 200 L (inflow rate = outflow rate) Rate in = 5 g/L × 4 L/min = 20 g/min Rate out = A/200 g/L × 4 L/min = A/50 g/min DE: dA/dt = 20 – A/50 Rewrite: dA/dt + A/50 = 20 → FIRST-ORDER LINEAR. P = 1/50, Q = 20. μ = e^(∫(1/50)dt) = e^(t/50) Multiply: d/dt[e^(t/50)·A] = 20e^(t/50) Integrate: e^(t/50)·A = 20·50·e^(t/50) + C = 1000e^(t/50) + C A = 1000 + Ce^(–t/50) IC: A(0) = 0 → 0 = 1000 + C → C = –1000 A(t) = 1000(1 – e^(–t/50)) A(30) = 1000(1 – e^(–30/50)) = 1000(1 – e^(–0.6)) = 1000(1 – 0.5488) = 451.2 g

The key step is converting to standard form by dividing by the coefficient of dy/dx. Here P(x) = –1/x, giving μ = 1/x (a simple inverse function). Always simplify μ to its lowest form before proceeding.

Scenario

Solve: x·dy/dx – y = x², x > 0, with y(1) = 0.

Solution

STEP 1 – Standard form (divide by x): dy/dx – (1/x)y = x P(x) = –1/x, Q(x) = x STEP 2 – μ = e^(∫–1/x dx) = e^(–ln x) = e^(ln(1/x)) = 1/x STEP 3 – Multiply by 1/x: d/dx[(1/x)y] = x·(1/x) = 1 STEP 4 – Integrate: y/x = x + C y = x² + Cx STEP 5 – IC: 0 = 1 + C → C = –1 FINAL ANSWER: y = x² – x

Applications

  • Mixing problems in water treatment and reservoir management.
  • RL and RC circuit analysis (dI/dt + (R/L)I = V/L).
  • Newton's Law of Cooling: dT/dt + kT = kT_s (linear in T).
  • Beam deflection with non-uniform loading (after twice integrating EI y'' = M).
  • Groundwater contaminant transport: first-order linear DE for concentration.

Misconceptions

  • Not converting to standard form first — if the coefficient of dy/dx is not 1, you MUST divide through before identifying P(x).
  • Computing μ with a constant of integration — μ = e^(∫P dx), no +C inside the exponent.
  • Forgetting to divide by μ at the end to get y (writing μy = … as the final answer).
  • In mixing problems, Rate_out concentration = A(t)/V(t) — if volume changes (inflow ≠ outflow), V is a function of t.
  • Confusing Newton's Law of Cooling sign: dT/dt = –k(T – T_s) means the object approaches T_s, so if T > T_s the derivative is negative (cooling). k > 0 always.

Related Concepts

  • Separable equations (Newton's law of cooling is separable)
  • Product rule of differentiation (basis of integrating factor method)
  • Integration by parts (needed when Q(x) is polynomial × exponential)
  • Mixing/tank problems
  • RL/RC circuit equations

Common Exam Questions

Example

See mixing tank example above — this exact problem type appears on PRC CE boards.

Approach

Set up dA/dt = Rate_in – Rate_out. Identify P and Q. Use integrating factor. Apply IC A(0) = initial amount.

Question Type

Find y(t) for a mixing problem

Example

dy/dx + (3/x)y = x². P = 3/x, μ = x³. d/dx[x³y] = x⁵. x³y = x⁶/6 + C. y = x³/6 + C/x³.

Approach

Rewrite in standard form → compute μ = e^(∫P dx) → multiply → integrate RHS → divide by μ.

Question Type

Find the general solution of a first-order linear ODE

Example

Object cools from 90°C to 70°C in 10 min in a 20°C room. Find temperature after 20 min. T(t) = 20 + 70e^(–kt). 70 = 70e^(–10k) — wait, use T(0) = 90: T = 20 + 70e^(–kt). At t=10, T=70: 70 = 20 + 70e^(–10k) → e^(–10k) = 50/70 → k = ln(7/5)/10. T(20) = 20 + 70e^(–20k) = 20 + 70(50/70)² = 20 + 70(25/49) ≈ 55.7°C.

Approach

dT/dt = –k(T – T_s). Rearrange: dT/dt + kT = kT_s. Solve by IF or recognize as separable: dT/(T – T_s) = –k dt.

Question Type

Newton's Law of Cooling — find temperature at time t

Key Points To Remember

  • Always rewrite the equation in standard form dy/dx + P(x)y = Q(x) before computing μ.
  • The integrating factor is μ = e^(∫P dx) — no constant of integration in the exponent.
  • After multiplying by μ, the left side ALWAYS becomes d/dx[μy] — use this fact, do not expand again.
  • The complete solution is y = (1/μ)∫μQ dx + C/μ — both the particular and homogeneous parts.
  • For mixing/tank problems, set up rate-in minus rate-out to get Q(x).
  • If P(x) is a constant P, then μ = e^(Px) — a simple exponential.

Exact Differential Equations

An exact DE has the form M(x,y)dx + N(x,y)dy = 0 where ∂M/∂y = ∂N/∂x. This condition means M dx + N dy is an exact differential of some function F(x,y), i.e., dF = M dx + N dy. SOLUTION METHOD: STEP 1 – Check exactness: compute ∂M/∂y and ∂N/∂x. If equal, the equation is exact. STEP 2 – Find F such that ∂F/∂x = M. Integrate M with respect to x (treating y as a constant): F(x,y) = ∫M dx + g(y) where g(y) is an unknown function of y only. STEP 3 – Use ∂F/∂y = N to find g'(y): ∂/∂y[∫M dx] + g'(y) = N → g'(y) = N – ∂/∂y[∫M dx] STEP 4 – Integrate g'(y) to find g(y). STEP 5 – Write the solution: F(x,y) = C. NOTE: If not exact, an integrating factor (different from the linear ODE type) may make it exact — this advanced step occasionally appears on the boards. ALTERNATIVE: You may also integrate N with respect to y first, then use ∂F/∂x = M to find h(x). Both approaches give the same F.

Examples

The exactness check ∂M/∂y = ∂N/∂x = 2x confirms we can proceed. The unknown g(y) in the integration step is the key distinguishing feature of exact equations. The final answer is always F(x,y) = C in implicit form.

Scenario

BOARD-STYLE: Solve (2xy + 3)dx + (x² – 4y)dy = 0.

Solution

M = 2xy + 3, N = x² – 4y ∂M/∂y = 2x; ∂N/∂x = 2x ✓ EXACT Find F: ∂F/∂x = M = 2xy + 3 Integrate w.r.t. x: F = x²y + 3x + g(y) Use ∂F/∂y = N: ∂/∂y[x²y + 3x + g(y)] = x² + g'(y) = x² – 4y g'(y) = –4y g(y) = –2y² F(x,y) = x²y + 3x – 2y² GENERAL SOLUTION: x²y + 3x – 2y² = C

When g'(y) = 0, the g(y) term contributes nothing and can be set to zero (its value is absorbed into the arbitrary constant C). This is a common simplification on boards.

Scenario

Check if (y² + 2x)dx + (2xy)dy = 0 is exact and solve.

Solution

M = y² + 2x, N = 2xy ∂M/∂y = 2y; ∂N/∂x = 2y ✓ EXACT Integrate M w.r.t. x: F = xy² + x² + g(y) ∂F/∂y = 2xy + g'(y) = 2xy → g'(y) = 0 → g(y) = 0 (constant, absorbed into C) SOLUTION: xy² + x² = C

Applications

  • Thermodynamic state equations (work and heat as inexact differentials vs exact differentials of state functions).
  • Conservative force fields in structural mechanics.
  • Potential flow in fluid mechanics.
  • Checking if a differential form corresponds to a potential function.

Misconceptions

  • Confusing ∂M/∂y with ∂M/∂x — the partial derivatives must be cross-partials: M differentiated w.r.t. y, and N w.r.t. x.
  • Forgetting that g(y) is a function of y only, not of x — if you get an expression involving x when solving for g'(y), you made an error.
  • Writing the answer as M = 0 or N = 0 instead of F(x,y) = C.
  • Skipping the exactness check — always verify before using this method.

Related Concepts

  • Partial differentiation
  • Total differential and exact differentials
  • Conservative vector fields
  • Potential functions
  • Integrating factors for non-exact equations

Common Exam Questions

Example

(3x²y + 2)dx + (x³ + 4y)dy = 0. ∂M/∂y = 3x², ∂N/∂x = 3x². Exact. F = x³y + 2x + 2y² = C.

Approach

Compute ∂M/∂y and ∂N/∂x. If equal, integrate M w.r.t. x, differentiate w.r.t. y, match with N to find g(y), write F = C.

Question Type

Test exactness and solve

Key Points To Remember

  • Test: ∂M/∂y = ∂N/∂x is NECESSARY AND SUFFICIENT for exactness in a simply-connected domain.
  • The solution is F(x,y) = C where dF = M dx + N dy — an implicit equation.
  • When integrating M w.r.t. x, treat y as a constant; the 'constant' of integration is g(y).
  • Verify by differentiating your F and confirming ∂F/∂x = M and ∂F/∂y = N.
  • If ∂M/∂y ≠ ∂N/∂x, the equation is NOT exact — try separable or linear methods instead, or find an integrating factor.

Higher-Order Linear ODEs with Constant Coefficients

The general form is: aₙy^(n) + aₙ₋₁y^(n-1) + … + a₁y' + a₀y = f(x) The COMPLETE (general) solution = y_h (homogeneous/complementary solution) + y_p (particular solution). For y_h (when f(x) = 0): Write the CHARACTERISTIC EQUATION by replacing y^(k) with m^k: aₙmⁿ + aₙ₋₁mⁿ⁻¹ + … + a₁m + a₀ = 0 The roots of this polynomial equation determine the form of y_h: CASE 1 – DISTINCT REAL ROOTS m₁, m₂, …, mₙ: y_h = C₁e^(m₁x) + C₂e^(m₂x) + … + Cₙe^(mₙx) CASE 2 – REPEATED REAL ROOT m (multiplicity r): y_h = (C₁ + C₂x + C₃x² + … + Cᵣxʳ⁻¹)e^(mx) CASE 3 – COMPLEX CONJUGATE ROOTS α ± βi: y_h = e^(αx)(C₁cos βx + C₂sin βx) If this pair is repeated (double complex), multiply by x: e^(αx)[(C₁ + C₂x)cos βx + (C₃ + C₄x)sin βx] For y_p (particular solution): Use the Method of Undetermined Coefficients (for polynomial, exponential, sine/cosine forcing) or Variation of Parameters (for general f(x)). Note: On PRC CE boards, most questions test the HOMOGENEOUS case (finding y_h and identifying root types). Second-order equations (n = 2) are most common.

Examples

Both roots are distinct and real — the simplest case. Each root mₖ contributes one term C_k·e^(m_k·x) to the solution. Two roots → two arbitrary constants → second-order ODE. ✓

Scenario

BOARD-STYLE: Solve y'' – 5y' + 6y = 0.

Solution

CHARACTERISTIC EQUATION: m² – 5m + 6 = 0 FACTOR: (m – 2)(m – 3) = 0 ROOTS: m₁ = 2, m₂ = 3 (CASE 1: Distinct real roots) GENERAL SOLUTION: y = C₁e^(2x) + C₂e^(3x)

The repeated root m = 3 (twice) means the second solution is x·e^(3x) instead of another e^(3x). A critical board exam pitfall: writing y = C₁e^(3x) + C₂e^(3x) = (C₁+C₂)e^(3x) = Ae^(3x) — this gives only ONE arbitrary constant and is WRONG for a 2nd-order ODE.

Scenario

BOARD-STYLE: Solve y'' – 6y' + 9y = 0.

Solution

CHARACTERISTIC EQUATION: m² – 6m + 9 = 0 FACTOR: (m – 3)² = 0 ROOT: m = 3 (CASE 2: Repeated real root, multiplicity 2) GENERAL SOLUTION: y = (C₁ + C₂x)e^(3x)

This equation models undamped free vibration. The characteristic roots ±βi give purely oscillatory solutions. α = 0 means no growth or decay — constant amplitude oscillation. If α < 0, amplitude decays (damped vibration); if α > 0, amplitude grows (unstable).

Scenario

BOARD-STYLE: Solve y'' + 4y = 0.

Solution

CHARACTERISTIC EQUATION: m² + 4 = 0 m² = –4 → m = ±2i COMPLEX ROOTS: α = 0, β = 2 (CASE 3: Pure imaginary roots) GENERAL SOLUTION: y = e^(0·x)(C₁cos2x + C₂sin2x) = C₁cos2x + C₂sin2x This represents SIMPLE HARMONIC OSCILLATION with angular frequency β = 2.

Discriminant b² – 4ac = 4 – 20 = –16 < 0 confirms complex roots. α = –1 (negative real part) means the amplitude decays exponentially — damped oscillation. This models a damped spring-mass system in structural dynamics. The damped natural frequency is β = 2 rad/s.

Scenario

BOARD-STYLE: Solve y'' + 2y' + 5y = 0 (damped oscillation).

Solution

CHARACTERISTIC EQUATION: m² + 2m + 5 = 0 QUADRATIC FORMULA: m = [–2 ± √(4 – 20)] / 2 = [–2 ± √(–16)] / 2 = –1 ± 2i COMPLEX ROOTS: α = –1, β = 2 GENERAL SOLUTION: y = e^(–x)(C₁cos2x + C₂sin2x)

For a second-order ODE, two ICs are needed to determine both C₁ and C₂. Always differentiate the general solution BEFORE applying the IC on y', not after substituting C values.

Scenario

Solve y'' + y' – 6y = 0 with y(0) = 1, y'(0) = 0.

Solution

CHAR. EQN: m² + m – 6 = 0 → (m+3)(m–2) = 0 → m = –3, 2 y = C₁e^(–3x) + C₂e^(2x) IC 1: y(0) = 1 → C₁ + C₂ = 1 IC 2: y'(x) = –3C₁e^(–3x) + 2C₂e^(2x) y'(0) = 0 → –3C₁ + 2C₂ = 0 → C₂ = 3C₁/2 Substitute: C₁ + 3C₁/2 = 1 → 5C₁/2 = 1 → C₁ = 2/5 C₂ = 1 – 2/5 = 3/5 PARTICULAR SOLUTION: y = (2/5)e^(–3x) + (3/5)e^(2x)

Applications

  • Free vibration of structures: m·ÿ + c·ẏ + k·y = 0 (mass-spring-dashpot).
  • Beam deflection: EI·y⁴ = w(x) — fourth-order ODE with constant coefficients.
  • Electrical circuits: L·dI²/dt² + R·dI/dt + (1/C)·I = dV/dt.
  • Column buckling (Euler): EI·y'' + P·y = 0 — eigenvalue problem.
  • Wave propagation and resonance in structural analysis.

Misconceptions

  • For repeated roots: writing y = C₁e^(mx) + C₂e^(mx) — this reduces to ONE constant, not two. CORRECT: y = (C₁ + C₂x)e^(mx).
  • For complex roots α ± βi: writing y = e^(αx)(C₁e^(βx) + C₂e^(–βx)) — this is valid but non-standard; always convert to trig form using Euler's formula.
  • Confusing α and β: α is the REAL part (in the exponent of e), β is the IMAGINARY part (in cos/sin).
  • Applying only ONE initial condition to a second-order ODE — you need TWO ICs (y and y' at the same point) to get two equations for C₁ and C₂.
  • Errors in the quadratic formula — especially the sign under the radical and the ± correctly applied.

Related Concepts

  • Quadratic formula and polynomial factoring
  • Euler's formula: e^(iβx) = cos(βx) + i·sin(βx)
  • Simple harmonic motion and damped oscillation
  • Particular solutions via undetermined coefficients
  • Superposition principle for linear ODEs

Common Exam Questions

Example

y'' + 6y' + 9y = 0. Char. eqn: m² + 6m + 9 = (m+3)² = 0. Repeated root m = –3. y = (C₁ + C₂x)e^(–3x).

Approach

Write char. eqn. → solve for m using factoring or quadratic formula → identify root type → write y_h.

Question Type

Find y_h given a second-order ODE

Example

Roots: m = 1, –2, 3. Char. eqn: (m–1)(m+2)(m–3) = 0 → m³ – 2m² – 5m + 6 = 0. ODE: y''' – 2y'' – 5y' + 6y = 0. General soln: C₁eˣ + C₂e^(–2x) + C₃e^(3x).

Approach

From roots, write each factor (m – m_i) → multiply factors → expand to get char. eqn → write ODE.

Question Type

Given roots, write the ODE or solution

Example

y'' + 4y' + 4y = 0: disc = 16–16 = 0 → critically damped. y'' + 4y' + 3y = 0: disc = 16–12 = 4 > 0 → overdamped. y'' + 4y' + 8y = 0: disc = 16–32 < 0 → underdamped.

Approach

Compute discriminant b² – 4ac. Positive → overdamped (2 distinct real); zero → critically damped (repeated); negative → underdamped (complex).

Question Type

Identify type of motion from ODE coefficients

Key Points To Remember

  • Characteristic equation: substitute m^k for y^(k). The polynomial in m has the same coefficients as the ODE.
  • Three root cases: distinct real → simple exponentials; repeated real → multiply by x powers; complex → e^(αx) times trig functions.
  • For roots α ± βi: α is the real part (controls growth/decay), β is the imaginary part (controls oscillation frequency).
  • Pure imaginary roots ±βi (α = 0): y = C₁cos(βx) + C₂sin(βx) — simple harmonic oscillation.
  • Total solution y = y_h + y_p satisfies the full (non-homogeneous) ODE.
  • Number of arbitrary constants in y_h = order of the ODE.

Engineering Applications of Differential Equations

The PRC CE board exam regularly features application problems. The key skill is translating a word problem into a DE, then solving it. Here are the four most common application categories: 1. GROWTH AND DECAY: dy/dt = ky. Solution: y = y₀e^(kt). k > 0 for growth; k < 0 for decay. Radioactive decay: k = –ln2/t_half. Half-life: t_half = –ln2/k. 2. NEWTON'S LAW OF COOLING: dT/dt = –k(T – T_s), where T_s = ambient (surrounding) temperature and k > 0. Solution: T(t) = T_s + (T₀ – T_s)e^(–kt). As t→∞, T→T_s. 3. MIXING (TANK) PROBLEMS: dA/dt = Rate_in – Rate_out Rate_in = (conc_in)(flow_rate_in) Rate_out = (A/V)(flow_rate_out) where V = volume at time t. If inflow rate ≠ outflow rate, V(t) changes: V(t) = V₀ + (r_in – r_out)t 4. MOTION/MECHANICS: Newton's second law F = ma = m(dv/dt). With air resistance proportional to velocity: m(dv/dt) = mg – kv (falling body). Terminal velocity: v_terminal = mg/k (when dv/dt = 0). FOR BOARD EXAMS: Always (a) define the variable and its units, (b) write the governing DE, (c) solve, (d) apply ICs, (e) evaluate at the required time or condition.

Examples

The temperature at t = 15 min is 61.3°C — still cooling toward 25°C. Note: e^(–1.4514) ≈ (19/31)³ as a quick check: (19/31)³ = (0.6129)³ ≈ 0.2303. The small discrepancy is from rounding in k. Always use the exact expression for k when computing later times.

Scenario

BOARD-STYLE (Newton's Cooling): A metal plate at 180°C is placed in a room at 25°C. After 5 minutes, its temperature is 120°C. Find the temperature after 15 minutes.

Solution

MODEL: dT/dt = –k(T – 25) SEPARATE: dT/(T–25) = –k dt INTEGRATE: ln|T – 25| = –kt + C₁ T – 25 = Ce^(–kt) T(t) = 25 + Ce^(–kt) IC 1: T(0) = 180 → 180 = 25 + C → C = 155 T(t) = 25 + 155e^(–kt) IC 2: T(5) = 120 → 120 = 25 + 155e^(–5k) 95 = 155e^(–5k) → e^(–5k) = 95/155 = 19/31 –5k = ln(19/31) → k = –ln(19/31)/5 = ln(31/19)/5 = 0.09676 min⁻¹ FIND T(15): T(15) = 25 + 155e^(–0.09676×15) = 25 + 155e^(–1.4514) = 25 + 155 × 0.2341 = 25 + 36.28 = 61.3°C

Terminal velocity = mg/k — you can find this WITHOUT solving the DE by setting dv/dt = 0. The time to reach terminal velocity is approximately t = 5/k = 5/0.2 = 25 s (five time constants). At t = 5 s (one time constant), the person is at about 63% of terminal velocity.

Scenario

BOARD-STYLE (Motion): A 70-kg person jumps from a plane. Air resistance = 14v N (v in m/s). Find: (a) velocity at t = 5 s, (b) terminal velocity.

Solution

MODEL: m(dv/dt) = mg – kv; m = 70 kg, g = 9.81 m/s², k = 14 N·s/m 70(dv/dt) = 70(9.81) – 14v = 686.7 – 14v dv/dt + 0.2v = 9.81 LINEAR DE: P = 0.2, Q = 9.81; μ = e^(0.2t) d/dt[e^(0.2t)v] = 9.81e^(0.2t) e^(0.2t)v = 9.81/0.2 · e^(0.2t) + C = 49.05e^(0.2t) + C v = 49.05 + Ce^(–0.2t) IC: v(0) = 0 → 0 = 49.05 + C → C = –49.05 v(t) = 49.05(1 – e^(–0.2t)) m/s (a) v(5) = 49.05(1 – e^(–1.0)) = 49.05(1 – 0.3679) = 49.05(0.6321) = 31.0 m/s (b) Terminal: as t→∞, e^(–0.2t)→0, v_terminal = 49.05 m/s Check: v_terminal = mg/k = 70(9.81)/14 = 686.7/14 = 49.05 m/s ✓

Applications

  • Carbon-14 dating used to estimate age of ancient timber in heritage structures.
  • Cooling of freshly cast concrete monitored by DPWH for quality control.
  • Water tank mixing problems in MWSS network analysis.
  • Free-fall and impact velocity for structural impact loading assessment.
  • Population growth for urban planning and civil infrastructure demand.

Misconceptions

  • In mixing problems, if the tank starts with a brine (non-zero initial concentration), then A(0) ≠ 0 — include this in the IC.
  • In Newton's Cooling, k is always POSITIVE. The negative sign in –k(T – T_s) accounts for the direction of heat transfer.
  • Terminal velocity is INDEPENDENT of initial conditions — it depends only on m, g, and k.
  • For growth/decay problems, y₀ is the value at t = 0, not at some other reference time. If the reference is at t ≠ 0, shift the time variable.

Related Concepts

  • Separable and linear first-order ODEs
  • Exponential functions and natural logarithms
  • Newton's Second Law of Motion
  • Conservation of mass (mixing problems)
  • Thermodynamics (heat transfer, Fourier's law)

Common Exam Questions

Example

When does A(t) = 1000(1 – e^(–t/50)) reach 600 g? 600 = 1000(1–e^(–t/50)) → e^(–t/50) = 0.4 → t = 50ln(2.5) ≈ 45.8 min.

Approach

Set y(t) = target value in the solved equation. Solve for t using natural log.

Question Type

Find time for quantity to reach a given value

Example

Uranium-238 half-life = 4.47 × 10⁹ yr. Fraction remaining after 1.0 × 10⁹ yr: e^(–ln2/4.47×10⁹ × 10⁹) = 2^(–1/4.47) ≈ 0.857 or 85.7%.

Approach

y/y₀ = e^(kt). For half-life: set y/y₀ = 0.5. For fraction remaining after time T: compute e^(kT) where k = –ln2/t_half.

Question Type

Half-life and fraction remaining

Key Points To Remember

  • Growth/decay: y = y₀e^(kt). Use two data points to find both y₀ and k if neither is at t = 0.
  • Newton's cooling: T approaches T_s exponentially. k > 0 always.
  • Half-life formula: t_half = ln2/|k| = ln2/λ (λ = decay constant).
  • Mixing: rate_out concentration = A(t)/V(t). Volume is constant only when inflow rate = outflow rate.
  • Terminal velocity: set dv/dt = 0 in the equation of motion.
  • For ALL applications: check units at every step to catch setup errors.

Practice Problems

Even though dy/dx and y appear linearly (first power, no products), the presence of (d²y/dx²)⁴ makes the equation nonlinear. A single nonlinear term makes the entire DE nonlinear. This is a quick classification problem frequently used as item 1 in DE sections of board exams.

Problem

PROBLEM 1 (Classification): Classify the following DE — state order, degree, linearity: (d²y/dx²)⁴ + 3x(dy/dx) – 2y = e^x.

Solution

Order: 2 (highest derivative is d²y/dx²) Degree: 4 (the second derivative is raised to the 4th power) Linearity: NONLINEAR (the term (d²y/dx²)⁴ is a power > 1 of the highest derivative) Type: ODE

Both sides have the same integrable form 1/(1 + u²) → arctan(u). The IC gives C = 0 immediately. The result y = x is elegant and verifiable: dy/dx = 1 = (1+x²)/(1+x²) ✓. This type of problem tests recognition of standard integral forms.

Problem

PROBLEM 2 (Separable): Solve dy/dx = (1+y²)/(1+x²) with y(0) = 0.

Solution

Separate: dy/(1+y²) = dx/(1+x²) Integrate both sides: arctan(y) = arctan(x) + C Apply IC: arctan(0) = arctan(0) + C → 0 = 0 + C → C = 0 Solution: arctan(y) = arctan(x) Simplify: y = x (taking tan of both sides — valid since tan(arctan(x)) = x) FINAL ANSWER: y = x

The integrating factor 1/x² uses the property that e^(–2 ln x) = x⁻². After multiplying, the left side becomes d/dx[y/x²] — always verify this by expanding using the product rule. Integration of sin x is –cos x (no by-parts needed here). The general solution has two parts: –x² cos x (particular) and Cx² (homogeneous).

Problem

PROBLEM 3 (Linear ODE): Solve dy/dx – (2/x)y = x² sin x, for x > 0.

Solution

Standard form: dy/dx – (2/x)y = x² sin x P(x) = –2/x, Q(x) = x² sin x Integrating factor: μ = e^(∫–2/x dx) = e^(–2 ln x) = e^(ln x⁻²) = x⁻² = 1/x² Multiply by μ: (1/x²)y' – (2/x³)y = sin x d/dx[y/x²] = sin x Integrate: y/x² = –cos x + C y = x²(–cos x + C) = –x² cos x + Cx² FINAL ANSWER: y = –x² cos x + Cx²

Always present the solution as F(x,y) = C. The answer can be verified by implicit differentiation: differentiating x²y – x³ + y² = C with respect to x gives (2xy + x²dy/dx – 3x² + 2y·dy/dx) = 0, which rearranges to (2xy–3x²)dx + (x²+2y)dy = 0 ✓.

Problem

PROBLEM 4 (Exact): Solve (2xy – 3x²)dx + (x² + 2y)dy = 0.

Solution

M = 2xy – 3x², N = x² + 2y ∂M/∂y = 2x; ∂N/∂x = 2x ✓ EXACT Integrate M w.r.t. x: F = x²y – x³ + g(y) ∂F/∂y = x² + g'(y) = N = x² + 2y g'(y) = 2y → g(y) = y² F = x²y – x³ + y² GENERAL SOLUTION: x²y – x³ + y² = C or equivalently: y² + x²y – x³ = C

For a third-order ODE, the characteristic equation is a cubic polynomial. Factoring by grouping is the key technique. Three distinct real roots → three simple exponential terms. The general solution has THREE arbitrary constants (matching the order of 3). Board exam tip: try rational roots ±1, ±2, ±4 first using the Rational Root Theorem.

Problem

PROBLEM 5 (Higher-Order): Find the general solution of y''' – y'' – 4y' + 4y = 0.

Solution

CHARACTERISTIC EQUATION: m³ – m² – 4m + 4 = 0 Factor by grouping: m²(m – 1) – 4(m – 1) = (m² – 4)(m – 1) = (m–2)(m+2)(m–1) = 0 ROOTS: m₁ = 1, m₂ = 2, m₃ = –2 (all distinct real roots) GENERAL SOLUTION: y = C₁eˣ + C₂e^(2x) + C₃e^(–2x)

This represents damped oscillation with α = –3 (exponential decay) and β = 2 (angular frequency). The amplitude decays as e^(–3x) while the system oscillates at 2 rad/unit. The two ICs y(0) and y'(0) correctly determine both C₁ and C₂. This type of problem models underdamped structural vibration.

Problem

PROBLEM 6 (Higher-Order with Complex Roots): Solve y'' + 6y' + 13y = 0, y(0) = 2, y'(0) = 0.

Solution

CHARACTERISTIC EQUATION: m² + 6m + 13 = 0 QUADRATIC FORMULA: m = [–6 ± √(36–52)]/2 = [–6 ± √(–16)]/2 = –3 ± 2i ROOTS: α = –3, β = 2 (complex conjugate pair) y = e^(–3x)(C₁cos2x + C₂sin2x) IC 1: y(0) = 2 → e⁰(C₁·1 + C₂·0) = 2 → C₁ = 2 y' = –3e^(–3x)(C₁cos2x+C₂sin2x) + e^(–3x)(–2C₁sin2x+2C₂cos2x) y'(0) = –3(C₁) + 2C₂ = 0 → –3(2) + 2C₂ = 0 → C₂ = 3 PARTICULAR SOLUTION: y = e^(–3x)(2cos2x + 3sin2x)

k ≈ 0.0147/yr means roughly 1.47% annual growth rate. For 2030 (t = 20 yr from 2010): P ≈ 16.1 million. Quick check: doubling time = ln2/k = 0.6931/0.01470 ≈ 47 years — so by 2057, population would double to about 24 million. Civil engineers use this to size water supply systems, roads, and sewage treatment plants per design horizons required by DPWH and LWUA.

Problem

PROBLEM 7 (Application – Growth): The population of Metro Manila in 2010 was 12 million. By 2020, it grew to 13.9 million. Assuming exponential growth, estimate the population in 2030.

Solution

MODEL: P(t) = P₀e^(kt), where t = years after 2010 IC 1: P(0) = 12 (million) → P₀ = 12 P(t) = 12e^(kt) IC 2: P(10) = 13.9 → 13.9 = 12e^(10k) e^(10k) = 13.9/12 = 1.1583 10k = ln(1.1583) = 0.14699 k = 0.014699 per year (≈ 1.47% annual growth rate) P(2030) = P(20) = 12e^(0.014699×20) = 12e^(0.29398) = 12 × 1.3415 = 16.1 million

Variable volume mixing requires carrying V(t) = 50 + 2t through the entire problem — the most challenging tank problem type. The integrating factor (50+2t)^(3/2) arises from the integral of 3/(50+2t). Note: if inflow rate = outflow rate (5=5), V = constant = 50 L — a simpler problem. The final salt content 82.32 g in 100 L represents 41.2% of what would be present at equilibrium (2 g/L × 100 L = 200 g) — the short fill time (25 min) explains the lower salt accumulation.

Problem

PROBLEM 8 (Application – Mixing with Variable Volume): A 100-L tank initially contains 50 L of pure water. Brine (2 g/L) enters at 5 L/min while the solution drains at 3 L/min. How much salt is in the tank when the tank is full (100 L)?

Solution

Let A(t) = salt (g) at time t (min). Volume: V(t) = 50 + (5–3)t = 50 + 2t L Tank is full when V = 100: 50 + 2t = 100 → t = 25 min Rate in = 2 g/L × 5 L/min = 10 g/min Rate out = [A/(50+2t)] g/L × 3 L/min = 3A/(50+2t) g/min DE: dA/dt = 10 – 3A/(50+2t) Rewrite: dA/dt + [3/(50+2t)]A = 10 → LINEAR, P = 3/(50+2t) IF: μ = e^(∫3/(50+2t)dt) = e^((3/2)ln|50+2t|) = (50+2t)^(3/2) d/dt[(50+2t)^(3/2) A] = 10(50+2t)^(3/2) Integrate RHS: ∫10(50+2t)^(3/2)dt = 10 · (50+2t)^(5/2)/(5/2·2) = (50+2t)^(5/2) (50+2t)^(3/2) A = (50+2t)^(5/2) + C A = (50+2t) + C/(50+2t)^(3/2) IC: A(0) = 0 → 0 = 50 + C/(50)^(3/2) → C = –50·(50)^(3/2) = –50^(5/2) = –(50)^(5/2) (50)^(5/2) = 50² × √50 = 2500 × 7.071 = 17677.7 A(t) = (50+2t) – 17677.7/(50+2t)^(3/2) At t = 25 min (tank full, V = 100 L): A(25) = 100 – 17677.7/(100)^(3/2) = 100 – 17677.7/1000 = 100 – 17.68 = 82.32 g Salt concentration at full tank: 82.32 g / 100 L = 0.823 g/L

Exam Preparation Tips

  • CLASSIFICATION DRILL: Practice classifying at least 20 DEs for order, degree, and linearity before the exam. This is guaranteed quick points — do not miss them.
  • STANDARD FORMS: Memorize the three standard forms: (1) separable: M(y)dy = N(x)dx; (2) linear: y' + P(x)y = Q(x); (3) exact: ∂M/∂y = ∂N/∂x. Recognizing the form immediately tells you the method.
  • CHARACTERISTIC EQUATION: Practice the characteristic equation approach until it is automatic. Know all three cases (distinct real, repeated, complex) and the corresponding solution forms by heart — this appears on EVERY board exam.
  • GROWTH AND DECAY FORMULA SHEET: Write y = y₀e^(kt), half-life t₁/₂ = ln2/|k|, and doubling time t_d = ln2/k on your scratch paper at the start of the exam for quick reference.
  • CHECK YOUR ANSWER: For separable DEs, differentiate your solution and substitute back into the DE. This catches sign errors and missing constants. Budget 30 seconds per problem for verification.
  • BOARD EXAM TIME STRATEGY: In the PRC CE board exam, DE problems typically appear in 3–5 items per subject exam. Spend no more than 3 minutes per item — practice solving common types under timed conditions.
  • INTEGRATING FACTOR KEY STEP: After computing μ, immediately write d/dx[μy] = μQ(x). Do NOT expand and try to simplify again — this wastes time and introduces errors.
  • INITIAL CONDITIONS: Always apply the IC to the COMPLETE general solution (after all integration is done) to find C. Never substitute IC values before integrating.
  • COMPLEX ROOTS: Use the quadratic formula discriminant b²–4ac to quickly determine root type. Negative discriminant → complex roots → trig form. Practice computing α and β: m = –b/2a ± (√|disc|/2a)i.
  • MIXING PROBLEMS SETUP: For any tank problem, immediately write dA/dt = (conc_in)(rate_in) – (A/V)(rate_out). Determine if V is constant or variable. This setup is 50% of the problem.
  • NEWTON'S COOLING: The solution T(t) = T_s + (T₀ – T_s)e^(–kt) should be memorized. Given T₀ (initial), T_s (ambient), and one other data point, you can always find k and then T at any time.
  • REPEATED ROOT TRAP: If the characteristic equation gives a repeated root m (e.g., (m–3)² = 0), NEVER write C₁e^(3x) + C₂e^(3x). The correct answer is (C₁ + C₂x)e^(3x). This is one of the most common errors on PRC CE board exams.
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In summary

Differential equations form a cornerstone of engineering mathematics and are consistently present in the PRC Civil Engineer Licensure Examination. The chapter revolves around a structured approach: (1) classify the DE to identify its order, degree, and linearity; (2) select the appropriate solution method — separable variables, integrating factor for linear DEs, or the exactness test for exact equations; (3) for higher-order constant-coefficient equations, solve the characteristic equation and apply the correct solution form based on root type (distinct real, repeated, or complex); and (4) translate engineering word problems into DEs and solve them with proper initial or boundary conditions. For the PRC CE board exam, the highest-yield topics are: the characteristic equation and its three root cases (30% of DE items), exponential growth and decay applications (25%), and first-order linear ODE solutions (20%). These three areas alone account for approximately 75% of differential equation questions on the licensure examination. The most common traps are: forgetting the x-multiplier for repeated roots, applying initial conditions before integrating, writing two constants of integration instead of one, and confusing the sign of k in decay problems. Eliminating these errors through deliberate practice will significantly improve your board exam performance. As future licensed civil engineers, your ability to model and solve differential equations underpins work in structural dynamics (NSCP 2015 Section 208 — earthquake response analysis), hydraulics (Bernoulli and continuity in unsteady flow), geotechnical engineering (Terzaghi consolidation), and environmental engineering (water quality modeling under RA 9275 — the Philippine Clean Water Act). Mastery of this topic is both an examination requirement and a professional responsibility under RA 544 (Civil Engineering Act of the Philippines). Study the worked examples thoroughly, practice under timed conditions, and verify every answer by substitution. Kaya ninyo ito — you have what it takes to pass the board exam!

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