Differential Equations

Differential equations model how engineering quantities change — RC/RL circuit currents, spring-mass vibrations, reactor concentrations, transient heat conduction. The FE expects you to identify the ODE type, pick the matching solution method, and apply initial conditions, leaning on the Laplace-transform and solution-form tables in the Handbook.

First-Order ODEs

Separable

Writable as dy/dx = f(x)·g(y); separate and integrate: ∫dy/g(y) = ∫f(x) dx.

Worked example — separable IVP. Solve dy/dx = 3x²y with y(0) = 2. Separate: dy/y = 3x² dx. Integrate both sides: ln|y| = x³ + C, so y = A·e^(x³). Apply y(0) = 2: A·e⁰ = A = 2. Thus y = 2e^(x³). Note the exponent is x³ (the antiderivative of 3x²), not 3x — confusing these is the usual mistake.

First-Order Linear

Standard form dy/dx + P(x)y = Q(x). Multiply by the integrating factor μ(x) = e^∫P(x) dx, then y = (1/μ)∫μ Q dx.

Worked example — integrating factor. Solve dy/dx + 2y = 6. Here P = 2 so μ = e^(2x). Then y = e^(−2x)∫6e^(2x) dx = e^(−2x)(3e^(2x) + C) = 3 + Ce^(−2x). As x → ∞ the transient Ce^(−2x) decays and y approaches the steady-state value 3.

Exact

M(x,y)dx + N(x,y)dy = 0 is exact when ∂M/∂y = ∂N/∂x; then a potential F(x,y) exists with ∂F/∂x = M and ∂F/∂y = N, and the solution is F(x,y) = C. To build F, integrate M with respect to x, then differentiate the result with respect to y and match it to N to recover the y-only term.

Recognizing the Type Quickly

On the FE the hardest part is classification, so scan in this order: (1) Can you split into f(x)·g(y)? → separable. (2) Is it linear in y of the form y′ + P(x)y = Q(x)? → integrating factor. (3) Otherwise test ∂M/∂y = ∂N/∂x for exactness. A first-order linear ODE with constant coefficient and constant forcing (like the integrating-factor example above) always settles to a steady state plus a decaying transient — a pattern that mirrors RC-circuit charging and first-order sensor response, both of which the FE poses as word problems.

Second-Order Linear ODEs (Constant Coefficients)

General form ay″ + by′ + cy = f(x). For the homogeneous case (f = 0), solve the characteristic equation ar² + br + c = 0. The discriminant b² − 4ac decides the solution shape:

Worked example — distinct real roots. Solve y″ − 5y′ + 6y = 0. Characteristic equation r² − 5r + 6 = (r−2)(r−3) = 0 gives r = 2, 3. Solution: y = C₁e^(2x) + C₂e^(3x).

For the non-homogeneous case, the full solution is y = yₕ + yₚ. Guess yₚ by undetermined coefficients:

If a guessed term already appears in yₕ, multiply the guess by x (or x²) to keep it independent.

For a mechanical analogy, b² − 4ac < 0 corresponds to an underdamped vibration (oscillating decay), = 0 is critically damped, and > 0 is overdamped — useful for recognizing which root case a physical problem implies.

Laplace Transforms

The Laplace transform converts a time function f(t) into F(s) = ∫₀^∞ f(t)e^(−st) dt, turning calculus into algebra. The Handbook prints the transform pairs and properties; your job is to apply them.

Derivative properties: ℒ{f′} = sF(s) − f(0); ℒ{f″} = s²F(s) − s f(0) − f′(0). s-shift: ℒ{e^(at)f(t)} = F(s−a).

Procedure

1. Laplace-transform both sides of the ODE.
2. Substitute the initial conditions.
3. Solve algebraically for Y(s).
4. Inverse-transform (match a table pair) to get y(t).

Worked example — Laplace solution. Solve y″ + 4y = 0 with y(0) = 1, y′(0) = 0.

• Transform: s²Y − s(1) − 0 + 4Y = 0.
• Collect: Y(s²+4) = s, so Y(s) = s/(s²+4).
• Match the pair s/(s²+ω²) with ω² = 4 → ω = 2.
• Inverse: y(t) = cos 2t.

The transform method shines for problems with discontinuous forcing (step or impulse inputs) where undetermined coefficients become awkward.

Quick transform recall. For f(t) = e^(3t), match e^(at) → 1/(s−a) with a = 3, giving F(s) = 1/(s−3), valid for s > 3. The sign trap: it is s minus a, so e^(+3t) → 1/(s−3), not 1/(s+3).

Why Engineers Use Laplace

Beyond the algebra, the Laplace domain exposes a system's poles — the values of s that make the denominator of Y(s) zero. Poles in the left half-plane (negative real part) mean the time response decays and the system is stable; poles on the imaginary axis give sustained oscillation; right-half-plane poles mean the response grows without bound. This connects directly to the second-order root cases above: complex characteristic roots α ± βi become poles at exactly s = α ± βi, so the discriminant test and the pole location carry the same information. The FE may ask you to read stability straight off the denominator of a transfer function rather than to fully invert the transform.