2.2 Differential Equations & Mathematical Modeling

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 ($F = m a$), Fourier's Law of Conduction ($q = -k A \frac{dT}{dx}$), 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.

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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:

1. Separable Variables: $\frac{dy}{dx} = g(x)h(y) \implies \int \frac{dy}{h(y)} = \int g(x) dx + C$.
2. First-Order Linear ODEs: Standard Form: $\frac{dy}{dx} + P(x)y = Q(x)$ Integrating Factor: $I(x) = e^{\int P(x) dx}$. General Solution: $y(x) = \frac{1}{I(x)} \int I(x) Q(x) dx + \frac{C}{I(x)}$.
3. Exact ODEs: $M(x, y)dx + N(x, y)dy = 0$ is exact if $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$.
4. Bernoulli ODEs: Non-linear form $\frac{dy}{dx} + P(x)y = Q(x)y^n$. Solved by substituting $u = y^{1-n}$ to transform into a linear ODE in $u$.

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Worked Numerical Example: Quenching Heat Transfer (Newton's Law of Cooling)

A solid steel component at an initial temperature of $800^\circ\text{C}$ is quenched in an oil bath maintained at a constant temperature $T_m = 30^\circ\text{C}$. Newton's Law of Cooling states that the rate of heat loss is proportional to the temperature difference: $\frac{dT}{dt} = -k(T - T_m)$ If the temperature of the casting drops to $400^\circ\text{C}$ after $t = 5 \text{ minutes}$, determine:

1. The cooling constant $k$ in $\text{min}^{-1}$.
2. The total time required for the casting to cool down to $100^\circ\text{C}$.

Step 1: Separate Variables and Integrate $\int \frac{dT}{T - 30} = -k \int dt \implies \ln(T - 30) = -kt + C \implies T(t) - 30 = C' e^{-kt}$

Step 2: Apply Initial Condition $T(0) = 800^\circ\text{C}$ $800 - 30 = C' e^0 \implies C' = 770^\circ\text{C}$ $T(t) = 30 + 770 e^{-kt}$

Step 3: Calculate Cooling Constant $k$ using $T(5) = 400^\circ\text{C}$ $400 = 30 + 770 e^{-5k} \implies 370 = 770 e^{-5k} \implies e^{-5k} = \frac{370}{770} = 0.48052$ $-5k = \ln(0.48052) = -0.7329 \implies k = 0.1466 \text{ min}^{-1}$

Step 4: Find Time $t$ to reach $100^\circ\text{C}$ $100 = 30 + 770 e^{-0.1466 t} \implies 70 = 770 e^{-0.1466 t} \implies e^{-0.1466 t} = \frac{70}{770} = 0.09091$ $-0.1466 t = \ln(0.09091) = -2.3979 \implies t = \frac{-2.3979}{-0.1466} = 16.36 \text{ minutes}$

The cooling constant is $0.1466 \text{ min}^{-1}$ and the casting reaches $100^\circ\text{C}$ in $16.36 \text{ minutes}$.

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2. Second-Order Linear ODEs & Mechanical Vibrations

Translational mechanical systems consisting of mass $m$, viscous damper $c$, and spring constant $k$ subjected to external force $F(t)$ are modeled by: $m \frac{d^2x}{dt^2} + c \frac{dx}{dt} + k x = F(t) \quad \text{or} \quad \ddot{x} + \frac{c}{m}\dot{x} + \frac{k}{m}x = \frac{F(t)}{m}$

Homogeneous Case ($F(t) = 0$): Free Vibrations

Assume solution $x(t) = e^{rt}$, yielding the Characteristic Equation: $m r^2 + c r + k = 0 \implies r_{1,2} = \frac{-c \pm \sqrt{c^2 - 4mk}}{2m}$

Key Mechanical Definitions:

• Undamped Natural Frequency: $\omega_n = \sqrt{\frac{k}{m}}$ (rad/s)
• Critical Damping Coefficient: $c_c = 2\sqrt{km} = 2m\omega_n$ (N·s/m)
• Damping Ratio: $\zeta = \frac{c}{c_c} = \frac{c}{2\sqrt{km}}$

The Three Damping Regimes:

Where the Damped Natural Frequency is: $\omega_d = \omega_n \sqrt{1 - \zeta^2}$

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Worked Numerical Example: Damped Mass-Spring Response

An industrial compressor mount is modeled as a mass $m = 10 \text{ kg}$, spring stiffness $k = 1000 \text{ N/m}$, and damper coefficient $c = 40 \text{ N}\cdot\text{s/m}$. Calculate:

1. Undamped natural frequency $\omega_n$.
2. Damping ratio $\zeta$ and dynamic classification.
3. Damped natural frequency $\omega_d$.

Step 1: Compute Undamped Frequency $\omega_n$ $\omega_n = \sqrt{\frac{k}{m}} = \sqrt{\frac{1000}{10}} = \sqrt{100} = 10.0 \text{ rad/s}$

Step 2: Compute Critical Damping $c_c$ and Damping Ratio $\zeta$ $c_c = 2\sqrt{km} = 2\sqrt{1000 \times 10} = 2\sqrt{10000} = 200 \text{ N}\cdot\text{s/m}$ $\zeta = \frac{c}{c_c} = \frac{40}{200} = 0.20$

Since $\zeta = 0.20 < 1.0$, the system is Underdamped.

Step 3: Compute Damped Natural Frequency $\omega_d$ $\omega_d = \omega_n \sqrt{1 - \zeta^2} = 10.0 \sqrt{1 - (0.20)^2} = 10.0 \sqrt{1 - 0.04} = 10.0 \sqrt{0.96} = 9.798 \approx 9.80 \text{ rad/s}$

In cycles per second (Hz): $f_d = \frac{\omega_d}{2\pi} = \frac{9.798}{2\pi} = 1.56 \text{ Hz}$.

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3. Forced Vibrations & Resonance

When subjected to harmonic excitation $F(t) = F_0 \cos(\omega t)$, the steady-state response amplitude $X$ is given by: $X = \frac{F_0/k}{\sqrt{\left(1 - r^2\right)^2 + \left(2\zeta r\right)^2}}$ where $r = \frac{\omega}{\omega_n}$ is the frequency ratio.

Resonance occurs when excitation frequency $\omega \approx \omega_n$ ($r \approx 1$), leading to extremely large vibration amplitudes that can cause catastrophic structural failure.

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4. Laplace Transforms in Dynamic Systems

The Laplace Transform converts linear differential equations in time domain $t$ into algebraic equations in complex frequency domain $s$: $\mathcal{L}\{f(t)\} = F(s) = \int_{0}^{\infty} e^{-st} f(t) dt$

Common Operational Properties:

• $\mathcal{L}{1} = \frac{1}{s}$
• $\mathcal{L}{e^{at}} = \frac{1}{s - a}$
• $\mathcal{L}{\sin(\omega t)} = \frac{\omega}{s^2 + \omega^2}$
• $\mathcal{L}{\cos(\omega t)} = \frac{s}{s^2 + \omega^2}$
• First Derivative: $\mathcal{L}{\dot{y}} = sY(s) - y(0)$
• Second Derivative: $\mathcal{L}{\ddot{y}} = s^2Y(s) - s y(0) - \dot{y}(0)$

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MELE Exam Tips & Shortcuts

• Fast Damping Ratio Check: Compute $c^2 - 4mk$. If positive $\implies$ Overdamped; if zero $\implies$ Critically Damped; if negative $\implies$ Underdamped.
• Logarithmic Decrement: Measure peak decay $\delta = \ln\left(\frac{x_1}{x_2}\right) = \frac{2\pi \zeta}{\sqrt{1-\zeta^2}}$. For small damping ($\zeta < 0.1$), $\delta \approx 2\pi \zeta$.
• Laplace Initial Values: Use the Initial Value Theorem $\lim_{t \to 0} f(t) = \lim_{s \to \infty} s F(s)$ and Final Value Theorem $\lim_{t \to \infty} f(t) = \lim_{s \to 0} s F(s)$ to quickly check transient solutions without inverse transforms.