Study guide
2.2 Differential Equations & Mathematical Modeling
PRC Mechanical Engineer Licensure Examination MELE
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1 Chapter 1: Licensing Overview, Law & Professional Ethics
1.1 MELE Facts, RA 8495 & Board Organization 1.2 Code of Ethics for Mechanical Engineers & Professional Practice 1.3 Philippine Mechanical Engineering Code (PME Code) Governance & Legal Mandates
2 Chapter 2: Advanced Mathematics & Engineering Statistics
2.1 Differential & Integral Calculus in Engineering Applications 2.2 Differential Equations & Mathematical Modeling 2.3 Applied Engineering Statistics & Quality Control Probability
3 Chapter 3: Engineering Thermodynamics
3.1 Thermodynamic State Postulate, Properties & P-V-T Equations of State 3.2 First Law Analysis for Open & Closed Thermal Systems 3.3 Second Law, Entropy, Exergy & Reversibility Analysis
4 Chapter 4: Fluid Mechanics & Hydraulic Engineering
4.1 Fluid Statics, Manometry, Hydrostatic Force & Buoyancy 4.2 Fluid Kinematics, Dynamics & Bernoulli Equation 4.3 Pipe Flow, Friction Losses & Piping Network Analysis
5 Chapter 5: Heat Transfer & Thermal Systems
5.1 Conduction Heat Transfer & Thermal Insulation Design 5.2 Natural & Forced Convection Heat Transfer Correlations 5.3 Thermal Radiation & Industrial Heat Exchanger Design
6 Chapter 6: Mechanics & Strength of Materials
6.1 Statics & Dynamics of Rigid Bodies & Engineering Linkages 6.2 Normal, Shear & Triaxial Stress-Strain Relationships 6.3 Bending Stresses, Torsion & Column Buckling Analysis
7 Chapter 7: Engineering Economics & Financial Analysis
7.1 Time Value of Money & Cash Flow Equivalence Formulas 7.2 Asset Depreciation, Taxation & Economic Evaluation Methods 7.3 Break-Even Analysis, Payback Period & EOQ Inventory Models
8 Chapter 8: Engineering Materials & Metallurgy
8.1 Ferrous & Non-Ferrous Metals, Alloys & Polymers 8.2 Phase Diagrams, Iron-Carbon System & Heat Treatment 8.3 Mechanical Testing, Fatigue Failure & Fracture Mechanics
9 Chapter 9: Machine Elements & Stress Analysis
9.1 Failure Theories, Static & Fluctuating Stress Design 9.2 Power Transmission Shafts, Keys & Rigid/Flexible Couplings 9.3 Spur, Helical, Bevel & Worm Gear Design Calculations
10 Chapter 10: Power Transmission Components
10.1 Belts, V-Belts, Chains, Wire Ropes & Flywheels 10.2 Rolling Element & Hydrodynamic Journal Bearings 10.3 Mechanical Springs, Clutches & Industrial Brakes
11 Chapter 11: Manufacturing, Fasteners & GD&T
11.1 Threaded Fasteners, Rivets & Welded Joint Analysis 11.2 Manufacturing Processes, Machine Tools & Machining Calculations 11.3 Engineering Drawing, Limits, Fits, Tolerances & GD&T
12 Chapter 12: Power Plants, Boilers & Thermal Power Cycles
12.1 Steam Power Plants & Rankine Cycles 12.2 Steam Generators, Boilers & Combustion Engineering 12.3 Internal Combustion Engines: SI & CI Cycles 12.4 Gas Turbine Power Plants & Combined Cycles
13 Chapter 13: Renewable Energy, Fluid Machinery & Plant Systems
13.1 Hydroelectric, Geothermal & Renewable Energy Power Plants 13.2 Pumps & Hydraulic Machinery 13.3 Air Compressors, Fans & Blowers
14 Chapter 14: Refrigeration, HVAC & Building Mechanical Systems
14.1 Refrigeration Systems, Cycles & Refrigerants 14.2 Psychrometrics & Air Conditioning Systems 14.3 Piping/Ductwork Design, PME Code & Industrial Safety Management
2.2 Differential Equations & Mathematical Modeling
Key Takeaways Explain This Topic AI
First-order ODEs model physical transient phenomena like transient quenching heat transfer and hydraulic tank discharge.
Homogeneous second-order linear ODEs govern free mechanical vibrations classified by damping ratio zeta into overdamped, critically damped, and underdamped regimes.
Underdamped systems oscillate at the damped natural frequency omega_d = omega_n * sqrt(1 - zeta^2).
Forced harmonic vibrations experience maximum amplitude magnification near resonance (excitation frequency ratio r approx 1).
Laplace transforms convert dynamic differential equations into simple algebraic expressions for efficient initial value problem resolution.
Last updated: July 2026
- 2.2 Differential Equations & Mathematical Modeling
- Introduction to Engineering Differential Equations
Differential equations (DEs) form the backbone of mechanical engineering dynamics, fluid mechanics, heat transfer, and control systems. Physical laws—such as Newton's Second Law of Motion ( F = m a F = m a F = ma ), Fourier's Law of Conduction ( q = − k A d T d x q = -k A \frac{dT}{dx} q = − k A d x d T ), and Torricelli's Law of Fluid Flow—are naturally formulated as differential equations. The PRC MELE tests candidates on classifying DEs, solving first-order transient equations, evaluating second-order vibration dynamics, and utilizing Laplace transforms for initial value problems.
1. Classification & First-Order ODEs
An Ordinary Differential Equation (ODE) contains functions of a single independent variable and their derivatives. The order is the highest derivative present; the degree is the exponent of the highest-order derivative after removing fractional exponents.
Common First-Order Solution Techniques:
Separable Variables : d y d x = g ( x ) h ( y ) ⟹ ∫ d y h ( y ) = ∫ g ( x ) d x + C \frac{dy}{dx} = g(x)h(y) \implies \int \frac{dy}{h(y)} = \int g(x) dx + C d x d y = g ( x ) h ( y ) ⟹ ∫ h ( y ) d y = ∫ g ( x ) d x + C .
First-Order Linear ODEs : Standard Form: d y d x + P ( x ) y = Q ( x ) \frac{dy}{dx} + P(x)y = Q(x) d x d y + P ( x ) y = Q ( x ) Integrating Factor: I ( x ) = e ∫ P ( x ) d x I(x) = e^{\int P(x) dx} I ( x ) = e ∫ P ( x ) d x .
General Solution: y ( x ) = 1 I ( x ) ∫ I ( x ) Q ( x ) d x + C I ( x ) y(x) = \frac{1}{I(x)} \int I(x) Q(x) dx + \frac{C}{I(x)} y ( x ) = I ( x ) 1 ∫ I ( x ) Q ( x ) d x + I ( x ) C .
Exact ODEs : M ( x , y ) d x + N ( x , y ) d y = 0 M(x, y)dx + N(x, y)dy = 0 M ( x , y ) d x + N ( x , y ) d y = 0 is exact if ∂ M ∂ y = ∂ N ∂ x \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} ∂ y ∂ M = ∂ x ∂ N .
Bernoulli ODEs : Non-linear form d y d x + P ( x ) y = Q ( x ) y n \frac{dy}{dx} + P(x)y = Q(x)y^n d x d y + P ( x ) y = Q ( x ) y n . Solved by substituting u = y 1 − n u = y^{1-n} u = y 1 − n to transform into a linear ODE in u u u .
Worked Numerical Example: Quenching Heat Transfer (Newton's Law of Cooling)
A solid steel component at an initial temperature of 800 ∘ C 800^\circ\text{C} 80 0 ∘ C is quenched in an oil bath maintained at a constant temperature T m = 30 ∘ C T_m = 30^\circ\text{C} T m = 3 0 ∘ C . Newton's Law of Cooling states that the rate of heat loss is proportional to the temperature difference: d T d t = − k ( T − T m ) \frac{dT}{dt} = -k(T - T_m) d t d T = − k ( T − T m )
If the temperature of the casting drops to 400 ∘ C 400^\circ\text{C} 40 0 ∘ C after t = 5 minutes t = 5 \text{ minutes} t = 5 minutes , determine:
The cooling constant k k k in min − 1 \text{min}^{-1} min − 1 .
The total time required for the casting to cool down to 100 ∘ C 100^\circ\text{C} 10 0 ∘ C .
Step 1: Separate Variables and Integrate ∫ d T T − 30 = − k ∫ d t ⟹ ln ( T − 30 ) = − k t + C ⟹ T ( t ) − 30 = C ′ e − k t \int \frac{dT}{T - 30} = -k \int dt \implies \ln(T - 30) = -kt + C \implies T(t) - 30 = C' e^{-kt} ∫ T − 30 d T = − k ∫ d t ⟹ ln ( T − 30 ) = − k t + C ⟹ T ( t ) − 30 = C ′ e − k t
Step 2: Apply Initial Condition T ( 0 ) = 800 ∘ C T(0) = 800^\circ\text{C} T ( 0 ) = 80 0 ∘ C 800 − 30 = C ′ e 0 ⟹ C ′ = 770 ∘ C 800 - 30 = C' e^0 \implies C' = 770^\circ\text{C} 800 − 30 = C ′ e 0 ⟹ C ′ = 77 0 ∘ C T ( t ) = 30 + 770 e − k t T(t) = 30 + 770 e^{-kt} T ( t ) = 30 + 770 e − k t
Step 3: Calculate Cooling Constant k k k using T ( 5 ) = 400 ∘ C T(5) = 400^\circ\text{C} T ( 5 ) = 40 0 ∘ C 400 = 30 + 770 e − 5 k ⟹ 370 = 770 e − 5 k ⟹ e − 5 k = 370 770 = 0.48052 400 = 30 + 770 e^{-5k} \implies 370 = 770 e^{-5k} \implies e^{-5k} = \frac{370}{770} = 0.48052 400 = 30 + 770 e − 5 k ⟹ 370 = 770 e − 5 k ⟹ e − 5 k = 770 370 = 0.48052 − 5 k = ln ( 0.48052 ) = − 0.7329 ⟹ k = 0.1466 min − 1 -5k = \ln(0.48052) = -0.7329 \implies k = 0.1466 \text{ min}^{-1} − 5 k = ln ( 0.48052 ) = − 0.7329 ⟹ k = 0.1466 min − 1
Step 4: Find Time t t t to reach 100 ∘ C 100^\circ\text{C} 10 0 ∘ C 100 = 30 + 770 e − 0.1466 t ⟹ 70 = 770 e − 0.1466 t ⟹ e − 0.1466 t = 70 770 = 0.09091 100 = 30 + 770 e^{-0.1466 t} \implies 70 = 770 e^{-0.1466 t} \implies e^{-0.1466 t} = \frac{70}{770} = 0.09091 100 = 30 + 770 e − 0.1466 t ⟹ 70 = 770 e − 0.1466 t ⟹ e − 0.1466 t = 770 70 = 0.09091 − 0.1466 t = ln ( 0.09091 ) = − 2.3979 ⟹ t = − 2.3979 − 0.1466 = 16.36 minutes -0.1466 t = \ln(0.09091) = -2.3979 \implies t = \frac{-2.3979}{-0.1466} = 16.36 \text{ minutes} − 0.1466 t = ln ( 0.09091 ) = − 2.3979 ⟹ t = − 0.1466 − 2.3979 = 16.36 minutes
The cooling constant is 0.1466 min − 1 0.1466 \text{ min}^{-1} 0.1466 min − 1 and the casting reaches 100 ∘ C 100^\circ\text{C} 10 0 ∘ C in 16.36 minutes 16.36 \text{ minutes} 16.36 minutes .
2. Second-Order Linear ODEs & Mechanical Vibrations
Translational mechanical systems consisting of mass m m m , viscous damper c c c , and spring constant k k k subjected to external force F ( t ) F(t) F ( t ) are modeled by: m d 2 x d t 2 + c d x d t + k x = F ( t ) or x ¨ + c m x ˙ + k m x = F ( t ) m m \frac{d^2x}{dt^2} + c \frac{dx}{dt} + k x = F(t) \quad \text{or} \quad \ddot{x} + \frac{c}{m}\dot{x} + \frac{k}{m}x = \frac{F(t)}{m} m d t 2 d 2 x + c d t d x + k x = F ( t ) or x ¨ + m c x ˙ + m k x = m F ( t )
Homogeneous Case ( F ( t ) = 0 F(t) = 0 F ( t ) = 0 ): Free Vibrations
Assume solution x ( t ) = e r t x(t) = e^{rt} x ( t ) = e r t , yielding the Characteristic Equation: m r 2 + c r + k = 0 ⟹ r 1 , 2 = − c ± c 2 − 4 m k 2 m m r^2 + c r + k = 0 \implies r_{1,2} = \frac{-c \pm \sqrt{c^2 - 4mk}}{2m} m r 2 + cr + k = 0 ⟹ r 1 , 2 = 2 m − c ± c 2 − 4 mk
Key Mechanical Definitions:
Undamped Natural Frequency: ω n = k m \omega_n = \sqrt{\frac{k}{m}} ω n = m k (rad/s)
Critical Damping Coefficient: c c = 2 k m = 2 m ω n c_c = 2\sqrt{km} = 2m\omega_n c c = 2 km = 2 m ω n (N·s/m)
Damping Ratio: ζ = c c c = c 2 k m \zeta = \frac{c}{c_c} = \frac{c}{2\sqrt{km}} ζ = c c c = 2 km c
The Three Damping Regimes:
Damping Condition Damping Ratio ζ \zeta ζ Characteristic Roots r 1 , 2 r_{1,2} r 1 , 2 Dynamic Response x ( t ) x(t) x ( t )
Overdamped ζ > 1 \zeta > 1 ζ > 1 Real & Distinct Exponential return to zero (no oscillation)
Critically Damped ζ = 1 \zeta = 1 ζ = 1 Real & Repeated x ( t ) = ( C 1 + C 2 t ) e − ω n t x(t) = (C_1 + C_2 t)e^{-\omega_n t} x ( t ) = ( C 1 + C 2 t ) e − ω n t (Fastest non-oscillatory return)
Underdamped ζ < 1 \zeta < 1 ζ < 1 Complex Conjugates ( − α ± i ω d -\alpha \pm i\omega_d − α ± i ω d ) x ( t ) = X e − ζ ω n t cos ( ω d t − ϕ ) x(t) = X e^{-\zeta \omega_n t} \cos(\omega_d t - \phi) x ( t ) = X e − ζ ω n t cos ( ω d t − ϕ ) (Decaying oscillation)
Where the Damped Natural Frequency is: ω d = ω n 1 − ζ 2 \omega_d = \omega_n \sqrt{1 - \zeta^2} ω d = ω n 1 − ζ 2
Worked Numerical Example: Damped Mass-Spring Response
An industrial compressor mount is modeled as a mass m = 10 kg m = 10 \text{ kg} m = 10 kg , spring stiffness k = 1000 N/m k = 1000 \text{ N/m} k = 1000 N/m , and damper coefficient c = 40 N ⋅ s/m c = 40 \text{ N}\cdot\text{s/m} c = 40 N ⋅ s/m . Calculate:
Undamped natural frequency ω n \omega_n ω n .
Damping ratio ζ \zeta ζ and dynamic classification.
Damped natural frequency ω d \omega_d ω d .
Step 1: Compute Undamped Frequency ω n \omega_n ω n ω n = k m = 1000 10 = 100 = 10.0 rad/s \omega_n = \sqrt{\frac{k}{m}} = \sqrt{\frac{1000}{10}} = \sqrt{100} = 10.0 \text{ rad/s} ω n = m k = 10 1000 = 100 = 10.0 rad/s
Step 2: Compute Critical Damping c c c_c c c and Damping Ratio ζ \zeta ζ c c = 2 k m = 2 1000 × 10 = 2 10000 = 200 N ⋅ s/m c_c = 2\sqrt{km} = 2\sqrt{1000 \times 10} = 2\sqrt{10000} = 200 \text{ N}\cdot\text{s/m} c c = 2 km = 2 1000 × 10 = 2 10000 = 200 N ⋅ s/m ζ = c c c = 40 200 = 0.20 \zeta = \frac{c}{c_c} = \frac{40}{200} = 0.20 ζ = c c c = 200 40 = 0.20
Since ζ = 0.20 < 1.0 \zeta = 0.20 < 1.0 ζ = 0.20 < 1.0 , the system is Underdamped .
Step 3: Compute Damped Natural Frequency ω d \omega_d ω d ω d = ω n 1 − ζ 2 = 10.0 1 − ( 0.20 ) 2 = 10.0 1 − 0.04 = 10.0 0.96 = 9.798 ≈ 9.80 rad/s \omega_d = \omega_n \sqrt{1 - \zeta^2} = 10.0 \sqrt{1 - (0.20)^2} = 10.0 \sqrt{1 - 0.04} = 10.0 \sqrt{0.96} = 9.798 \approx 9.80 \text{ rad/s} ω d = ω n 1 − ζ 2 = 10.0 1 − ( 0.20 ) 2 = 10.0 1 − 0.04 = 10.0 0.96 = 9.798 ≈ 9.80 rad/s
In cycles per second (Hz): f d = ω d 2 π = 9.798 2 π = 1.56 Hz f_d = \frac{\omega_d}{2\pi} = \frac{9.798}{2\pi} = 1.56 \text{ Hz} f d = 2 π ω d = 2 π 9.798 = 1.56 Hz .
3. Forced Vibrations & Resonance
When subjected to harmonic excitation F ( t ) = F 0 cos ( ω t ) F(t) = F_0 \cos(\omega t) F ( t ) = F 0 cos ( ω t ) , the steady-state response amplitude X X X is given by: X = F 0 / k ( 1 − r 2 ) 2 + ( 2 ζ r ) 2 X = \frac{F_0/k}{\sqrt{\left(1 - r^2\right)^2 + \left(2\zeta r\right)^2}} X = ( 1 − r 2 ) 2 + ( 2 ζ r ) 2 F 0 / k
where r = ω ω n r = \frac{\omega}{\omega_n} r = ω n ω is the frequency ratio.
Resonance occurs when excitation frequency ω ≈ ω n \omega \approx \omega_n ω ≈ ω n ( r ≈ 1 r \approx 1 r ≈ 1 ), leading to extremely large vibration amplitudes that can cause catastrophic structural failure.
4. Laplace Transforms in Dynamic Systems
The Laplace Transform converts linear differential equations in time domain t t t into algebraic equations in complex frequency domain s s s : L { f ( t ) } = F ( s ) = ∫ 0 ∞ e − s t f ( t ) d t \mathcal{L}\{f(t)\} = F(s) = \int_{0}^{\infty} e^{-st} f(t) dt L { f ( t )} = F ( s ) = ∫ 0 ∞ e − s t f ( t ) d t
Common Operational Properties:
- L { 1 } = 1 s \mathcal{L}\{1\} = \frac{1}{s} L { 1 } = s 1
- L { e a t } = 1 s − a \mathcal{L}\{e^{at}\} = \frac{1}{s - a} L { e a t } = s − a 1
L { sin ( ω t ) } = ω s 2 + ω 2 \mathcal{L}\{\sin(\omega t)\} = \frac{\omega}{s^2 + \omega^2} L { sin ( ω t )} = s 2 + ω 2 ω
L { cos ( ω t ) } = s s 2 + ω 2 \mathcal{L}\{\cos(\omega t)\} = \frac{s}{s^2 + \omega^2} L { cos ( ω t )} = s 2 + ω 2 s
First Derivative : L { y ˙ } = s Y ( s ) − y ( 0 ) \mathcal{L}\{\dot{y}\} = sY(s) - y(0) L { y ˙ } = s Y ( s ) − y ( 0 )
Second Derivative : L { y ¨ } = s 2 Y ( s ) − s y ( 0 ) − y ˙ ( 0 ) \mathcal{L}\{\ddot{y}\} = s^2Y(s) - s y(0) - \dot{y}(0) L { y ¨ } = s 2 Y ( s ) − sy ( 0 ) − y ˙ ( 0 )
MELE Exam Tips & Shortcuts
Fast Damping Ratio Check : Compute c 2 − 4 m k c^2 - 4mk c 2 − 4 mk . If positive ⟹ \implies ⟹ Overdamped; if zero ⟹ \implies ⟹ Critically Damped; if negative ⟹ \implies ⟹ Underdamped.
Logarithmic Decrement : Measure peak decay δ = ln ( x 1 x 2 ) = 2 π ζ 1 − ζ 2 \delta = \ln\left(\frac{x_1}{x_2}\right) = \frac{2\pi \zeta}{\sqrt{1-\zeta^2}} δ = ln ( x 2 x 1 ) = 1 − ζ 2 2 π ζ . For small damping ( ζ < 0.1 \zeta < 0.1 ζ < 0.1 ), δ ≈ 2 π ζ \delta \approx 2\pi \zeta δ ≈ 2 π ζ .
Laplace Initial Values : Use the Initial Value Theorem lim t → 0 f ( t ) = lim s → ∞ s F ( s ) \lim_{t \to 0} f(t) = \lim_{s \to \infty} s F(s) lim t → 0 f ( t ) = lim s → ∞ s F ( s ) and Final Value Theorem lim t → ∞ f ( t ) = lim s → 0 s F ( s ) \lim_{t \to \infty} f(t) = \lim_{s \to 0} s F(s) lim t → ∞ f ( t ) = lim s → 0 s F ( s ) to quickly check transient solutions without inverse transforms.