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GELE MathematicsDifferential EquationsRevision Notes

Revision notes for GELE Mathematics — Differential Equations. Short, focused, and designed for the week before exam day. Use these when you are already familiar with the chapter and need a quick refresh on the high-yield items Professional Regulation Commission (PRC) — Board of Geodetic Engineering tests.

Exam context

On the GELE 2026, the Mathematics subtest carries a "Core" weight in Professional Regulation Commission (PRC) — Board of Geodetic Engineering's pattern. Differential Equations lands at position 7th out of 10 in the standard review order. Target score is 70% weighted average, no sub-test below 50%, and roughly a meaningful share of items come from Mathematics on a typical GELE paper.

Differential Equations - Revision Notes

Differential equations (DEs) are among the most frequently tested topics in the Engineering Mathematics portion of the PRC Civil Engineer Licensure Examination. A differential equation relates an unknown function to one or more of its derivatives. In civil engineering practice, DEs model consolidation settlement (Terzaghi's theory), groundwater flow (Darcy's Law in transient conditions), structural dynamics, heat transfer in concrete curing, and population-based traffic load projections. Mastery of classification, solution methods, and standard engineering applications is essential for exam success. These revision notes cover all examinable subtopics: classification, first-order methods (separable, linear, exact), higher-order linear constant-coefficient equations, and applied problems (growth/decay, Newton's cooling, mixing tanks).

Sections

Formulas

Example

y'' + 4y' + 3y = 0 is a 2nd-order, 1st-degree, linear ODE with constant coefficients.

Formula

F(x, y, y', y'', ..., y^(n)) = 0

Variables

x = independent variable; y = dependent variable; y', y'', y^(n) = first, second, nth derivatives

Application

General form of an nth-order ODE

Exam Tips

  • PRC multiple-choice items often ask: 'What is the order and degree of the following DE?' — practice identifying these rapidly.
  • If derivatives appear inside a radical or as a denominator, always clear them algebraically before classifying.
  • Linear vs. nonlinear: scan the equation for y², (y')², y·y', sin(y), e^y — any of these signals nonlinearity.

Key Points

  • Order is the highest derivative present in the equation. Degree is the power of the highest-order derivative after the equation is made polynomial in its derivatives (radicals and fractions in derivatives must be cleared first).
  • An ordinary differential equation (ODE) involves derivatives with respect to a single independent variable. A partial differential equation (PDE) involves partial derivatives with respect to two or more independent variables.
  • A DE is linear if the dependent variable and all its derivatives appear to the first power only, with coefficients that are functions of the independent variable alone. Any product of the unknown function with its derivative, or any nonlinear function of the unknown (e.g., sin y, y²), makes it nonlinear.
  • For the PRC exam, focus on ODEs — first-order and second-order linear with constant coefficients.
  • The general solution contains as many arbitrary constants as the order. A particular solution applies initial or boundary conditions to fix those constants.

Definitions

Term

Order

Definition

The order of a DE is the order of the highest derivative present.

Importance

Determines the number of arbitrary constants in the general solution and the number of initial/boundary conditions required.

Term

Degree

Definition

The degree is the exponent of the highest-order derivative after clearing radicals and fractions involving derivatives.

Importance

Degree must be a positive integer for the concept to apply; PDEs and equations with irrational derivatives have undefined degree under this definition.

Term

General Solution

Definition

The solution containing the full set of arbitrary constants (n constants for an nth-order ODE).

Importance

Provides the family of all solutions; initial conditions select one particular member.

Term

Particular Solution

Definition

A solution obtained by assigning specific numerical values to the arbitrary constants using initial or boundary conditions.

Importance

This is the actual engineering answer — it satisfies both the DE and the specified conditions.

Section Title

1. Classification of Differential Equations

Common Mistakes

  • Confusing order and degree: y'' + (y')³ = 0 is 2nd-order (highest derivative is y'') but 1st-degree in y'' (y'' appears to the first power after the equation is written as a polynomial in y'').
  • Forgetting to clear radicals before stating the degree: √(y'') = x is rewritten as y'' = x² (degree 1), not degree 1/2.
  • Calling a 2nd-order equation 'degree 2' simply because it is second-order — order and degree are different properties.
  • Claiming an ODE is nonlinear because the coefficients contain x (e.g., xy' + y = 0 is still linear — nonlinearity refers to y and its derivatives, not x).

Formulas

Example

dy/dx = 2xy → ∫ dy/y = ∫ 2x dx → ln|y| = x² + C → y = Ae^(x²), where A = e^C

Formula

∫ dy/h(y) = ∫ g(x) dx + C

Variables

h(y) = function of y only; g(x) = function of x only; C = constant of integration

Application

Separable first-order ODE: dy/dx = g(x)·h(y)

Example

dy/dx + (2/x)y = x² → P = 2/x → μ = e^(∫2/x dx) = e^(2 ln x) = x² → d(x²y)/dx = x⁴ → x²y = x⁵/5 + C → y = x³/5 + C/x²

Formula

μ(x) = e^(∫P(x) dx)

Variables

P(x) = coefficient of y in standard form; μ = integrating factor

Application

Converts dy/dx + P(x)y = Q(x) to d(μy)/dx = μQ(x)

Example

Given μ = x² and Q = x²: y·x² = ∫x⁴ dx = x⁵/5 + C → y = x³/5 + C·x⁻²

Formula

y·μ = ∫ Q(x)·μ dx + C

Variables

y = unknown function; μ = integrating factor; Q(x) = right-hand side of standard form

Application

Complete solution formula for linear first-order ODE after multiplying through by μ

Example

(2xy) dx + (x² + 1) dy = 0: ∂M/∂y = 2x, ∂N/∂x = 2x → exact → F = x²y + y = C

Formula

∂M/∂y = ∂N/∂x

Variables

M = coefficient of dx; N = coefficient of dy in M dx + N dy = 0

Application

Test for exactness of a first-order ODE

Exam Tips

  • Decision rule: (1) Can you separate all y's from all x's? → Separable. (2) Is it dy/dx + P(x)y = Q(x)? → Integrating factor. (3) Does ∂M/∂y = ∂N/∂x? → Exact.
  • Always present the answer as y = f(x) unless instructed otherwise; board problems nearly always want the explicit form.
  • When initial conditions are given (e.g., y(0) = 1), solve for C immediately after integrating — do not carry C to the final answer.
  • For exam speed: memorize that μ = e^(∫P dx) and d(μy)/dx = μQ; avoid re-deriving this under time pressure.

Key Points

  • Three main solution methods for first-order ODEs on the PRC exam: (1) Separation of Variables, (2) Linear First-Order (Integrating Factor), (3) Exact Equations.
  • Always identify the type first before choosing the solution method — this is the most important skill.
  • Separable: all y-terms (including dy) on one side, all x-terms (including dx) on the other, then integrate both sides.
  • Linear first-order: standard form is dy/dx + P(x)y = Q(x). The integrating factor μ = e^(∫P dx) converts the left side into d(μy)/dx.
  • Exact: M dx + N dy = 0 is exact when ∂M/∂y = ∂N/∂x. Solution found by integrating M with respect to x (or N with respect to y), then adjusting for the remaining variable.
  • Bernoulli equation dy/dx + P(x)y = Q(x)yⁿ is nonlinear but is reduced to a linear DE by the substitution v = y^(1-n).

Definitions

Term

Separable Equation

Definition

A first-order ODE that can be written in the form dy/dx = g(x)·h(y), allowing all y-terms to move to one side and all x-terms to the other.

Importance

Most common type on the PRC exam; appears in growth/decay and cooling problems.

Term

Integrating Factor (μ)

Definition

A function μ(x) = e^(∫P dx) that, when multiplied through a linear first-order ODE, makes the left side an exact derivative d(μy)/dx.

Importance

Transforms any linear first-order ODE into a directly integrable form.

Term

Exact Equation

Definition

M dx + N dy = 0 is exact if ∂M/∂y = ∂N/∂x throughout the region, meaning there exists a function F(x,y) such that dF = M dx + N dy.

Importance

Solution is F(x,y) = C; saves work because no integrating factor is needed.

Section Title

2. First-Order Differential Equations

Common Mistakes

  • Forgetting the constant of integration when integrating both sides of a separable equation — this constant is essential for applying the initial condition.
  • In the integrating factor method, applying μ only to one side: multiply BOTH sides (left side and Q(x)) by μ.
  • Errors in ∫P dx for the integrating factor when P involves absolute values or requires substitution.
  • For exact equations: integrating M w.r.t. x and then forgetting that the 'constant' may actually be a function of y; always differentiate F w.r.t. y and equate to N to find that function.
  • Dividing by h(y) when separating variables without checking if h(y) = 0 — this can lose singular solutions.

Formulas

Example

y'' - 5y' + 6y = 0 → m² - 5m + 6 = 0 → (m-2)(m-3) = 0 → m = 2, 3 (distinct real roots)

Formula

am² + bm + c = 0

Variables

a, b, c = constant coefficients; m = characteristic root

Application

Characteristic equation for ay'' + by' + cy = 0; roots determine the form of the general solution

Example

m = 2, 3 → y = C₁e^(2x) + C₂e^(3x)

Formula

y = C₁e^(m₁x) + C₂e^(m₂x)

Variables

m₁, m₂ = distinct real roots of the characteristic equation; C₁, C₂ = arbitrary constants

Application

General solution when characteristic equation has two distinct real roots

Example

m² - 4m + 4 = 0 → (m-2)² = 0 → m = 2 (repeated) → y = (C₁ + C₂x)e^(2x)

Formula

y = (C₁ + C₂x)e^(mx)

Variables

m = repeated real root; C₁, C₂ = arbitrary constants

Application

General solution when characteristic equation has a repeated (equal) real root

Example

m² + 4 = 0 → m = ±2i (α=0, β=2) → y = C₁cos(2x) + C₂sin(2x)

Formula

y = e^(αx)[C₁cos(βx) + C₂sin(βx)]

Variables

α ± βi = complex conjugate roots; α = real part; β = imaginary part (β > 0)

Application

General solution when characteristic equation yields complex conjugate roots

Example

y'' - y = e^(2x): yc = C₁e^x + C₂e^(-x); guess yp = Ae^(2x) → A=1/3 → y = C₁e^x + C₂e^(-x) + (1/3)e^(2x)

Formula

y = yc + yp

Variables

yc = complementary function (homogeneous solution); yp = particular integral

Application

Complete general solution of a non-homogeneous linear ODE

Exam Tips

  • Memorize the three root cases and their solution forms — this is tested almost every board exam cycle.
  • For 'solve with initial conditions y(0)=a, y'(0)=b': write the general solution first, then differentiate, then apply both conditions simultaneously.
  • Discriminant shortcut: b² - 4ac > 0 → distinct real; = 0 → repeated; < 0 → complex conjugate.
  • The method of undetermined coefficients guess table: f(x) = polynomial of degree n → guess Aₙxⁿ + ... + A₀; f(x) = e^(kx) → guess Ae^(kx); f(x) = sin(kx) or cos(kx) → guess A·sin(kx) + B·cos(kx).

Key Points

  • For a homogeneous linear ODE with constant coefficients (right side = 0), the solution method is: assume y = e^(mx), substitute, and solve the resulting characteristic (auxiliary) equation.
  • The nature of the roots of the characteristic equation completely determines the form of the general solution.
  • Three root cases: (a) distinct real roots, (b) repeated real roots, (c) complex conjugate roots — each gives a different solution structure.
  • For non-homogeneous equations (right side ≠ 0), the general solution = complementary function (yc) + particular integral (yp). The complementary function is the homogeneous solution; the particular integral is found by undetermined coefficients or variation of parameters.
  • The most common second-order case on the PRC exam: ay'' + by' + cy = f(x).
  • Undetermined coefficients: guess yp based on the form of f(x) — polynomial, exponential, sine/cosine, or combinations.

Definitions

Term

Characteristic Equation

Definition

The algebraic equation obtained by substituting y = e^(mx) into a homogeneous linear ODE with constant coefficients; its roots determine the solution form.

Importance

Central tool for solving all constant-coefficient linear ODEs; must be solved accurately to get the correct solution form.

Term

Complementary Function (yc)

Definition

The general solution of the associated homogeneous equation (right side set to zero).

Importance

Always required as part of the complete solution for non-homogeneous equations.

Term

Particular Integral (yp)

Definition

Any single solution of the non-homogeneous equation; found by undetermined coefficients or variation of parameters.

Importance

Accounts for the forcing function f(x) in the non-homogeneous equation.

Term

Undetermined Coefficients

Definition

A method of finding yp by guessing a function of the same form as f(x) (with unknown coefficients A, B, ...) and substituting into the ODE to solve for those coefficients.

Importance

Fast and direct; applicable when f(x) is a polynomial, exponential, sine, cosine, or sum/product of these.

Section Title

3. Higher-Order Linear ODEs with Constant Coefficients

Common Mistakes

  • Using y = C₁e^(m₁x) + C₂e^(m₂x) when roots are repeated — the correct form is (C₁ + C₂x)e^(mx).
  • For complex roots α ± βi, writing y = C₁e^(αx)cos(βx) + C₂e^(αx)sin(βx) is correct, but students often forget the e^(αx) factor when α ≠ 0.
  • When f(x) = Ae^(mx) and m is already a root of the characteristic equation, the usual guess for yp must be multiplied by x (resonance case).
  • Sign errors when computing the discriminant b² - 4ac: a negative discriminant means complex roots with β = √(4ac - b²)/(2a).
  • Forgetting to find C₁ and C₂ using initial conditions when a particular solution is requested.
  • Dropping the particular integral and presenting only the complementary function as the final answer.

Formulas

Example

Radioactive material: 100 g decays to 50 g in 1600 years (half-life of Radium-226). Find the amount left after 4000 years. k = -ln2/1600 = -4.332×10⁻⁴ yr⁻¹. y(4000) = 100·e^(-4.332×10⁻⁴ × 4000) = 100·e^(-1.7329) = 17.68 g

Formula

y(t) = y₀·e^(kt)

Variables

y₀ = initial value at t=0; k = growth constant (k>0) or decay constant (k<0); t = time

Application

Exponential growth (population, bacteria) or radioactive decay

Example

A concrete structural element at 80°C is placed in a 20°C environment. After 30 min, T = 60°C. Find T after 1 hour. Step 1: 60 = 20 + 60·e^(-30k) → e^(-30k) = 40/60 = 2/3 → k = ln(3/2)/30 = 0.01352 min⁻¹. Step 2: T(60) = 20 + 60·e^(-0.01352×60) = 20 + 60·(4/9) = 20 + 26.67 = 46.67°C

Formula

T(t) = Ts + (T₀ - Ts)·e^(-kt)

Variables

T(t) = temperature at time t; Ts = surrounding (ambient) temperature; T₀ = initial temperature; k > 0

Application

Newton's Law of Cooling — temperature of an object approaching ambient

Example

Tank contains 100 L of pure water. Brine of 0.5 kg/L flows in at 4 L/min; well-mixed solution exits at 4 L/min (V constant = 100 L). dQ/dt = 0.5(4) - (Q/100)(4) = 2 - Q/25. This is a linear first-order ODE: dQ/dt + Q/25 = 2. μ = e^(t/25). Q = 50 + Ce^(-t/25). Q(0)=0 → C = -50. Q(t) = 50(1 - e^(-t/25)) kg.

Formula

dQ/dt = (Cᵢₙ × Rᵢₙ) - (Q/V × Rₒᵤₜ)

Variables

Q = amount of substance in tank (kg); Cᵢₙ = inflow concentration (kg/L); Rᵢₙ = inflow rate (L/min); V = tank volume (L); Rₒᵤₜ = outflow rate (L/min)

Application

Mixing tank problem: salt content in a brine tank over time

Example

If k = 4.332×10⁻⁴ yr⁻¹, then T₁/₂ = 0.6931/4.332×10⁻⁴ = 1600 years (confirming Radium-226 data).

Formula

T₁/₂ = ln(2)/|k|

Variables

T₁/₂ = half-life; |k| = magnitude of decay constant

Application

Radioactive decay, half-life problems

Exam Tips

  • For growth/decay board problems: (1) Write y = y₀e^(kt); (2) Use the given data point to find k; (3) Substitute the required time to find the answer.
  • For Newton's cooling: rewrite as dT/dt + k·T = k·Ts (linear form) or solve directly as separable. Both work — use whichever is faster for you.
  • Mixing tank problems almost always result in a linear first-order ODE — apply the integrating factor method directly.
  • Always verify the sign of k against the physical situation: cooling means T decreases → k must be positive in the formula T = Ts + (T₀-Ts)e^(-kt).
  • Doubling time Td = ln(2)/k is frequently asked; memorize this alongside the half-life formula.

Key Points

  • Growth and Decay: dy/dt = ky. k > 0 → growth (population, compound interest); k < 0 → decay (radioactivity, concrete carbonation). Solution: y = y₀e^(kt).
  • Half-life T₁/₂ = ln(2)/|k|. Doubling time Td = ln(2)/k.
  • Newton's Law of Cooling: dT/dt = -k(T - Ts), where Ts is the surrounding temperature and k > 0. Solution: T - Ts = (T₀ - Ts)e^(-kt). Used in forensic engineering and material temperature monitoring.
  • Mixing Problems (Tank Problems): d(Q)/dt = (rate in) - (rate out). Set up the ODE from mass balance; Q is the amount of substance (e.g., salt in kg) in the tank at time t.
  • Simple Harmonic Motion (undamped): d²x/dt² + ω²x = 0 → x = C₁cos(ωt) + C₂sin(ωt), where ω = √(k/m) for a spring-mass system.
  • Laplace Transform: converts a linear ODE with initial conditions into an algebraic equation in the s-domain; the solution is found algebraically and inverted back. Powerful for systems with discontinuous or impulsive forcing.

Definitions

Term

Exponential Growth/Decay Model

Definition

A first-order separable ODE dy/dt = ky whose solution y = y₀e^(kt) models quantities that change at a rate proportional to their current value.

Importance

Fundamental model in engineering: radioactivity, population, consolidation pore pressure dissipation (Terzaghi), concrete strength gain.

Term

Newton's Law of Cooling

Definition

States that the rate of change of temperature of an object is proportional to the difference between its temperature and the ambient temperature: dT/dt = -k(T - Ts).

Importance

Applied in civil engineering to concrete curing temperature monitoring, fire safety analysis, and HVAC system design.

Term

Mixing Problem

Definition

A first-order linear ODE derived from mass balance: rate of accumulation = rate in - rate out, applied to a well-mixed tank or reservoir.

Importance

Models pollutant transport in water treatment, salt concentration in irrigation tanks, and chemical dosing in water supply.

Term

Half-Life

Definition

The time required for a decaying quantity to reduce to exactly one-half of its initial value; T₁/₂ = ln(2)/|k|.

Importance

Key parameter in radioactive decay calculations and first-order degradation processes.

Section Title

4. Engineering Applications of Differential Equations

Common Mistakes

  • Growth/Decay: forgetting that decay requires k < 0; using a positive k and then subtracting — be consistent with the sign convention in the ODE setup.
  • Newton's cooling: writing dT/dt = k(T - Ts) with k > 0 (wrong sign) leading to an unbounded (growing) temperature — physically impossible for cooling. Always use dT/dt = -k(T - Ts) with k > 0.
  • Mixing problems: using the wrong volume V when inflow rate ≠ outflow rate (the volume is not constant in that case — V must be expressed as a function of t).
  • Forgetting to convert units consistently — e.g., mixing rate in L/min vs. L/hr, or half-life in years vs. seconds.
  • In the mixing tank, confusing Q (total amount in kg) with concentration (kg/L); the outflow concentration is Q/V, not Q.

Formulas

Example

L{e^(2t)} = ∫₀^∞ e^(-st)·e^(2t) dt = ∫₀^∞ e^(-(s-2)t) dt = 1/(s-2), valid for s > 2

Formula

L{f(t)} = F(s) = ∫₀^∞ e^(-st)·f(t) dt

Variables

s = complex frequency parameter (Re(s) > 0 for convergence); F(s) = Laplace transform of f(t)

Application

Definition of the Laplace transform; used to convert time-domain ODEs to s-domain algebraic equations

Example

If y(0)=1: L{y'} = sY(s) - 1

Formula

L{y'(t)} = s·Y(s) - y(0)

Variables

Y(s) = L{y(t)}; y(0) = initial condition on y

Application

Transform of first derivative; incorporates initial condition into the s-domain equation

Example

y'' - 3y' + 2y = 0, y(0)=1, y'(0)=0 → [s²Y - s - 0] - 3[sY - 1] + 2Y = 0 → Y(s²-3s+2) = s-3 → Y = (s-3)/[(s-1)(s-2)] → partial fractions → Y = 2/(s-1) - 1/(s-2) → y = 2e^t - e^(2t)

Formula

L{y''(t)} = s²·Y(s) - s·y(0) - y'(0)

Variables

Y(s) = L{y(t)}; y(0) and y'(0) = initial conditions

Application

Transform of second derivative; used for 2nd-order ODE problems with initial conditions

Exam Tips

  • Maintain a compact Laplace transform table on your reference sheet for the board exam.
  • When doing partial fractions, verify by recombining — this is a quick self-check under exam conditions.
  • For PRC exam: Laplace problems usually involve at most 2nd-order ODEs with polynomial/exponential right-hand sides — master the standard five transform pairs and derivative rules.

Key Points

  • The Laplace transform converts a linear ODE with initial conditions directly into an algebraic equation in the complex frequency domain (s-domain), eliminating the need to find the complementary function and particular integral separately.
  • Definition: L{f(t)} = F(s) = ∫₀^∞ e^(-st)f(t) dt.
  • Key transforms to memorize: L{1} = 1/s; L{t^n} = n!/s^(n+1); L{e^(at)} = 1/(s-a); L{sin(bt)} = b/(s²+b²); L{cos(bt)} = s/(s²+b²).
  • Transform of derivatives: L{f'(t)} = sF(s) - f(0); L{f''(t)} = s²F(s) - sf(0) - f'(0). This is how initial conditions are automatically incorporated.
  • After transforming to the s-domain, solve the algebraic equation for F(s), then use inverse Laplace transform (partial fractions) to recover y(t).
  • On the PRC exam, Laplace problems are typically straightforward: transform a simple ODE, solve for F(s), and invert using standard table entries.

Definitions

Term

Laplace Transform

Definition

An integral transform L{f(t)} = F(s) that converts a function of time t into a function of complex frequency s, enabling algebraic manipulation of ODEs.

Importance

Particularly powerful for ODEs with discontinuous or impulsive inputs (step functions, delta functions) and for systems with non-zero initial conditions.

Term

Inverse Laplace Transform

Definition

The operation L⁻¹{F(s)} = f(t) that recovers the time-domain function from its s-domain representation, typically using partial fraction decomposition and a table of standard transforms.

Importance

The final step in the Laplace method; requires proficiency with partial fractions.

Section Title

5. Laplace Transform Method (Introduction)

Common Mistakes

  • Forgetting to include initial condition terms when transforming derivatives — L{y'} = sY - y(0), not just sY.
  • Errors in partial fraction decomposition when inverting F(s) — practice this algebra step thoroughly.
  • Applying Laplace transform to nonlinear ODEs (it only works for linear ODEs).
  • Confusing the transform pairs: L{sin(bt)} = b/(s²+b²) vs L{cos(bt)} = s/(s²+b²) — note the numerator differs.

Connections

  • Differential equations are the mathematical language of Newton's Second Law (F = ma = m d²x/dt²), which underlies structural dynamics, seismic response analysis, and vibration theory in NSCP 2015 provisions for dynamic loading.
  • The exponential decay model y = y₀e^(-kt) directly appears in Terzaghi's consolidation theory, where excess pore water pressure dissipates over time — a critical concept for settlement analysis of soft clays common in Philippine coastal construction projects.
  • The characteristic equation approach for second-order ODEs connects directly to the eigenvalue problems in matrix algebra (another PRC exam topic) and to the determination of natural frequencies in structural dynamics.
  • Integration techniques (separable and linear ODEs) reinforce skills from Integral Calculus — specifically integration by parts, substitution, and partial fractions — all of which are also standalone PRC exam topics.
  • Laplace transforms use complex numbers (another Engineering Math topic) and require partial fraction decomposition (from Algebra), reinforcing the interconnected nature of the Engineering Mathematics board exam subjects.
  • Mixing tank problems model the same governing equations as first-order chemical reactions in environmental engineering (water quality, pollutant transport) and hydraulics (gradually varied flow approximations).
  • Newton's Law of Cooling appears in thermal analysis relevant to concrete mix design — specifically the monitoring of heat of hydration in mass concrete (foundations, dams, bridge piers) per ACI 318 provisions for temperature differential limits.
  • The general solution structure (complementary + particular) parallels the superposition principle in structural analysis (dead load + live load + lateral load responses), reinforcing physical intuition for the mathematical framework.

Exam Strategy

For the PRC Civil Engineer Licensure Examination, Differential Equations problems appear in the Engineering Mathematics subject. The most frequently tested subtopics are: (1) Classification of DEs — expect 1-2 items asking for order, degree, and type; (2) Separable first-order ODEs — usually 2-3 items, often set in growth/decay or cooling contexts; (3) Linear first-order ODEs with integrating factors — 1-2 items; (4) Second-order constant-coefficient ODEs — 2-3 items focused on the characteristic equation and solution form; and (5) Applied word problems — 2-3 items on population growth, radioactive decay, Newton's cooling, or mixing tanks. Strategy: (a) Quickly identify the DE type — this is the most critical skill. (b) For multiple-choice, eliminate wrong answer forms immediately (e.g., if roots are complex, any answer without sine/cosine terms is wrong). (c) For numerical applied problems, always write the governing ODE first, solve it symbolically, then substitute numbers — this reduces arithmetic errors. (d) Manage time by solving pure-mathematics classification and characteristic-equation items first (fastest), then tackle word problems. (e) Always apply the initial condition if given — an answer with an unevaluated constant C will not match any multiple-choice option. (f) Double-check signs in Newton's cooling (negative k in the exponent) and decay problems (negative k value).

Quick Review Questions

What is the order and degree of the differential equation: (y'')² + 3(y')³ = xy?

The highest derivative present is y'' (second derivative), so the order is 2. The equation is already polynomial in its derivatives. The highest-order derivative y'' appears to the power of 2 (from the term (y'')²), so the degree is 2. Note: the term (y')³ does not affect the degree because y' is not the highest-order derivative.

Solve the separable ODE: dy/dx = 3x²y, with initial condition y(0) = 2.

Separate: dy/y = 3x² dx. Integrate both sides: ln|y| = x³ + C. Exponentiate: y = Ae^(x³) where A = e^C. Apply y(0) = 2: 2 = A·e⁰ = A. Therefore y = 2e^(x³).

Find the integrating factor for the linear ODE: dy/dx + (3/x)y = x².

Standard form: dy/dx + P(x)y = Q(x) with P(x) = 3/x. Integrating factor: μ = e^(∫P dx) = e^(∫3/x dx) = e^(3 ln|x|) = e^(ln x³) = x³. (Valid for x > 0.)

Identify the solution form for y'' - 6y' + 9y = 0.

Characteristic equation: m² - 6m + 9 = 0 → (m - 3)² = 0 → m = 3 (repeated root). For a repeated root m, the solution is y = (C₁ + C₂x)e^(mx) = (C₁ + C₂x)e^(3x).

A population of bacteria is 500 at t = 0 and 2000 at t = 4 hours. What is the population at t = 8 hours?

y = y₀e^(kt). At t=4: 2000 = 500e^(4k) → e^(4k) = 4 → 4k = ln4 → k = ln4/4. At t=8: y = 500e^(k×8) = 500e^(2ln4) = 500×(e^(ln4))² = 500×16 = 8000. Alternatively: the population quadruples every 4 hours, so at t=8, it quadruples again: 2000×4 = 8000.

What are the roots of the characteristic equation m² + 6m + 13 = 0, and what is the corresponding solution form?

Using the quadratic formula: m = (-6 ± √(36-52))/2 = (-6 ± √(-16))/2 = (-6 ± 4i)/2 = -3 ± 2i. These are complex conjugate roots with α = -3 and β = 2. The solution is y = e^(αx)(C₁cos βx + C₂sin βx) = e^(-3x)(C₁cos2x + C₂sin2x).

A substance decays according to dy/dt = -0.02y. How long does it take for 80% of the substance to decay (leaving 20% remaining)?

Solution: y = y₀e^(-0.02t). For 20% remaining: 0.20y₀ = y₀e^(-0.02t) → e^(-0.02t) = 0.20 → -0.02t = ln(0.20) = -1.6094 → t = 1.6094/0.02 = 80.47 ≈ 80.5 time units.

Check whether the equation (2x + y) dx + (x + 3y²) dy = 0 is exact.

M = 2x + y, N = x + 3y². Check: ∂M/∂y = 1 and ∂N/∂x = 1. Since ∂M/∂y = ∂N/∂x = 1 throughout the domain, the equation is exact. The solution is F(x,y) = x² + xy + y³ = C.

Using Laplace transforms, find L{3e^(2t) - 2sin(3t)}.

Using linearity and standard transform pairs: L{e^(at)} = 1/(s-a) → L{3e^(2t)} = 3/(s-2). L{sin(bt)} = b/(s²+b²) → L{2sin(3t)} = 2·3/(s²+9) = 6/(s²+9). Therefore F(s) = 3/(s-2) - 6/(s²+9).

A tank initially contains 200 L of pure water. Brine with a salt concentration of 0.1 kg/L flows in at 5 L/min, and the well-mixed solution flows out at 5 L/min. Set up the ODE for Q(t), the amount of salt (in kg) at time t.

Rate in = 0.1 kg/L × 5 L/min = 0.5 kg/min. Rate out = (Q/200 kg/L) × 5 L/min = Q/40 kg/min. Mass balance: dQ/dt = 0.5 - Q/40 → dQ/dt + Q/40 = 0.5. This is a linear first-order ODE. Solution: Q(t) = 20(1 - e^(-t/40)) kg, using Q(0) = 0.

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