1.3 Ordinary Differential Equations and Laplace Transforms

1.3 Ordinary Differential Equations and Laplace Transforms

Differential equations are named explicitly in the NCEES Mathematics specification (area 1, sub-topic B). On the FE Other Disciplines exam they rarely ask you to derive a solution from scratch — they ask you to classify the equation, pick the right solution form, and evaluate a constant from an initial condition. Every dynamic model elsewhere in this guide reduces to one of the forms below: a mixing tank, an RC or RL circuit transient, a cooling body, a decaying isotope, and a damped mass-spring system.

Classification: What Kind of Equation Is This?

Separable First-Order Equations

If the equation can be written $\dfrac{dy}{dx} = g(x)h(y)$, separate and integrate:

$\int \frac{dy}{h(y)} = \int g(x)\,dx + C$

The single most valuable case on the FE exam is exponential growth and decay:

$\frac{dy}{dt} = k y \quad\Longrightarrow\quad y(t) = y_0 e^{kt}$

With $k < 0$ this is decay, and the half-life relationship follows directly from setting $y = y_0/2$:

$t_{1/2} = \frac{\ln 2}{|k|} = \frac{0.693}{|k|}$

Cross-topic payoff: this identical formula appears in the Safety, Health, and Environment area for radiation half-life, in Chemistry for first-order reaction kinetics, and in Basic Electrical Engineering for RC discharge with $|k| = 1/(RC)$. Learn it once, score it three times.

First-Order Linear Equations: Integrating Factor

Put the equation in standard form, then multiply through by the integrating factor $\mu(x)$:

$\frac{dy}{dx} + p(x)\,y = q(x), \qquad \mu(x) = e^{\int p(x)\,dx}$

$y(x) = \frac{1}{\mu(x)}\left[\int \mu(x)\,q(x)\,dx + C\right]$

For the very common constant-coefficient case $y' + a y = b$ (a tank filling, a capacitor charging, a body cooling toward ambient), the solution is the sum of a steady-state value and a decaying transient:

$y(t) = \frac{b}{a} + \left(y_0 - \frac{b}{a}\right)e^{-at}$

Read that structure. The first term is where the system ends up; the second is how it gets there, decaying with time constant $\tau = 1/a$. After $t = \tau$ the transient has fallen to 36.8% of its initial size; after $3\tau$, to about 5%.

Worked Example: Newton's Law of Cooling

A steel casting at $500^\circ\text{C}$ is placed in a $25^\circ\text{C}$ room. After 10 minutes it has cooled to $350^\circ\text{C}$. What is its temperature after 30 minutes?

Newton's law of cooling gives $\dfrac{dT}{dt} = -k,(T - T_\infty)$, so with $\theta = T - T_\infty$:

$\theta(t) = \theta_0 e^{-kt}, \qquad \theta_0 = 500 - 25 = 475^\circ\text{C}$

Find $k$ from the 10-minute data point, where $\theta(10) = 350 - 25 = 325^\circ\text{C}$:

$\frac{325}{475} = e^{-10k} \;\Rightarrow\; -10k = \ln(0.6842) = -0.3795 \;\Rightarrow\; k = 0.03795\ \text{min}^{-1}$

At $t = 30$ min:

$\theta(30) = 475\,e^{-(0.03795)(30)} = 475\,e^{-1.1385} = 475(0.3203) = 152.1^\circ\text{C}$

$T(30) = 152.1 + 25 = \boxed{177^\circ\text{C}}$

Trap: the exponential decays toward zero in $\theta$, not in $T$. Candidates who write $T = 500e^{-kt}$ and forget to subtract the ambient temperature get $160^\circ\text{C}$ — which is offered as a distractor.

Second-Order Linear Equations with Constant Coefficients

For $a y'' + b y' + c y = 0$, substitute $y = e^{rx}$ to get the characteristic equation:

$a r^2 + b r + c = 0, \qquad r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

The discriminant decides everything:

The Mass-Spring-Damper Standard Form

Writing $m\ddot{x} + c\dot{x} + kx = 0$ in normalized form exposes the two parameters the exam asks about:

$\ddot{x} + 2\zeta\omega_n \dot{x} + \omega_n^2 x = 0, \qquad \omega_n = \sqrt{\frac{k}{m}}, \qquad \zeta = \frac{c}{2\sqrt{km}}$

• $\zeta = 0$: undamped, oscillates forever at $\omega_n$
• $0 < \zeta < 1$: underdamped, oscillates at the damped frequency $\omega_d = \omega_n\sqrt{1-\zeta^2}$
• $\zeta = 1$: critically damped
• $\zeta > 1$: overdamped

Non-Homogeneous Equations

The complete solution is $y = y_h + y_p$: the homogeneous solution above plus a particular solution matched to the forcing function.

Trap: if the trial form already appears in $y_h$ (resonance), multiply the trial by $x$ before solving for the coefficients.

Laplace Transforms

The Reference Handbook supplies a transform table, so the method is mechanical: transform, solve algebraically for $Y(s)$, invert.

$\mathcal{L}\{f(t)\} = F(s) = \int_0^\infty f(t)e^{-st}\,dt$

The two derivative properties are what make it useful — initial conditions enter automatically:

$\mathcal{L}\{f'(t)\} = sF(s) - f(0), \qquad \mathcal{L}\{f''(t)\} = s^2F(s) - sf(0) - f'(0)$

Two shortcuts worth memorizing: the final value theorem $\displaystyle\lim_{t\to\infty} f(t) = \lim_{s\to 0} sF(s)$, and the initial value theorem $\displaystyle\lim_{t\to 0^+} f(t) = \lim_{s\to\infty} sF(s)$. Both let you answer "what is the steady-state response?" without inverting anything.

Worked Example: Step Response by Laplace Transform

Solve $y' + 3y = 6$ with $y(0) = 0$.

Transform both sides:

$sY(s) - 0 + 3Y(s) = \frac{6}{s} \;\Rightarrow\; Y(s)(s+3) = \frac{6}{s} \;\Rightarrow\; Y(s) = \frac{6}{s(s+3)}$

Partial fractions: $\dfrac{6}{s(s+3)} = \dfrac{2}{s} - \dfrac{2}{s+3}$. Inverting term by term:

$y(t) = 2 - 2e^{-3t}$

Check against the standard form $y = b/a + (y_0 - b/a)e^{-at}$ with $a = 3$, $b = 6$, $y_0 = 0$: $y = 2 + (0-2)e^{-3t}$. Same answer. Confirm the steady state with the final value theorem: $\lim_{s\to0} s\cdot\frac{6}{s(s+3)} = \frac{6}{3} = 2$. ✓

Additional First-Order ODE Practice

Separable First-Order ODEs

An ODE is separable if it can be written as $\frac{dy}{dx} = g(x)h(y)$, solved by separating variables: $\int \frac{1}{h(y)} dy = \int g(x) dx + C$

First-Order Linear ODEs (Integrating Factor Method)

Standard form: $\frac{dy}{dx} + P(x)y = Q(x)$

1. Compute integrating factor $I(x) = e^{\int P(x)dx}$.
2. Multiply standard ODE by $I(x)$ to obtain $\frac{d}{dx}[I(x)y] = I(x)Q(x)$.
3. Integrate both sides: $y(x) = \frac{1}{I(x)} \left[ \int I(x) Q(x) dx + C \right]$

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Additional Second-Order ODE Practice

Standard homogeneous form: $a \frac{d^2 y}{dt^2} + b \frac{dy}{dt} + c y = 0$

Substitute trial solution $y = e^{rt}$ to form the characteristic (auxiliary) equation: $a r^2 + b r + c = 0 \implies r_{1,2} = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Non-Homogeneous Second-Order ODEs

For $a y'' + b y' + c y = f(t)$, the total solution is $y(t) = y_h(t) + y_p(t)$, where $y_h(t)$ is the general solution to the homogeneous equation and $y_p(t)$ is a particular solution found using the Method of Undetermined Coefficients.

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Worked Engineering Problems

Worked Example 1: Transient Fluid Tank Mixing (First-Order ODE)

Problem: A 100-liter storage tank initially contains pure water ($y(0) = 0\text{ kg}$ of salt). Salt brine containing $0.5\text{ kg/L}$ concentration flows into the tank at $4\text{ L/min}$. The well-mixed solution flows out at $4\text{ L/min}$. Set up and solve the differential equation for salt mass $y(t)$ in the tank at any time $t$.

Solution:

Step 1: Set up mass balance rate equation $\frac{dy}{dt} = \text{Rate In} - \text{Rate Out}$. $\text{Rate In} = (0.5\text{ kg/L}) \times (4\text{ L/min}) = 2.0\text{ kg/min}$ $\text{Rate Out} = \left(\frac{y(t)}{100}\text{ kg/L}\right) \times (4\text{ L/min}) = 0.04 y(t)\text{ kg/min}$ $\frac{dy}{dt} = 2 - 0.04y \implies \frac{dy}{dt} + 0.04y = 2$

Step 2: Identify integrating factor $I(t)$. $P(t) = 0.04 \implies I(t) = e^{\int 0.04 dt} = e^{0.04t}$

Step 3: Multiply and integrate. $\frac{d}{dt}\left[ e^{0.04t} y \right] = 2 e^{0.04t}$ $e^{0.04t} y = \int 2 e^{0.04t} dt = \frac{2}{0.04} e^{0.04t} + C = 50 e^{0.04t} + C$ $y(t) = 50 + C e^{-0.04t}$

Step 4: Apply initial condition $y(0) = 0$. $0 = 50 + C e^0 \implies C = -50$ $y(t) = 50(1 - e^{-0.04t})\text{ kg}$

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Worked Example 2: Damped Mass-Spring Mechanical System (Second-Order ODE)

Problem: A $1\text{ kg}$ mass on a shock absorber system satisfies $y'' + 6y' + 13y = 0$, where $y(t)$ is displacement in meters. Given initial displacement $y(0) = 2\text{ m}$ and initial velocity $y'(0) = 0\text{ m/s}$, find $y(t)$.

Solution:

Step 1: Solve characteristic equation $r^2 + 6r + 13 = 0$. $r = \frac{-6 \pm \sqrt{36 - 4(1)(13)}}{2} = \frac{-6 \pm \sqrt{36 - 52}}{2} = \frac{-6 \pm \sqrt{-16}}{2} = -3 \pm 2i$ Here $\alpha = -3$ and $\beta = 2$ (Underdamped response).

Step 2: Form general underdamped solution. $y(t) = e^{-3t}(C_1 \cos 2t + C_2 \sin 2t)$

Step 3: Apply initial condition $y(0) = 2$. $y(0) = e^0 (C_1 \cos 0 + C_2 \sin 0) = C_1 = 2$

Step 4: Differentiate $y(t)$ and apply $y'(0) = 0$. $y'(t) = -3e^{-3t}(C_1 \cos 2t + C_2 \sin 2t) + e^{-3t}(-2C_1 \sin 2t + 2C_2 \cos 2t)$ $y'(0) = -3(1)(C_1) + 1(2C_2) = -3(2) + 2C_2 = -6 + 2C_2 = 0 \implies C_2 = 3$

Final Solution: $y(t) = e^{-3t}(2 \cos 2t + 3 \sin 2t)\text{ m}$