Differential Equations

Key Takeaways

  • First-order ODEs: separable, linear (integrating factor), and exact equations are the main types tested.
  • Second-order linear ODEs with constant coefficients use the characteristic equation ar² + br + c = 0.
  • The characteristic equation roots determine the solution form: real distinct, repeated, or complex conjugate.
  • Laplace transforms convert differential equations to algebraic equations, simplifying the solution process.
  • Initial value problems (IVPs) require specific initial conditions to find the particular solution.
  • The FE Reference Handbook contains Laplace transform pairs — know how to use the table.
Last updated: March 2026

Differential Equations

Differential equations describe how quantities change and are fundamental to modeling engineering systems — from circuit analysis to heat transfer to vibrations.

First-Order Ordinary Differential Equations (ODEs)

Separable Equations

An ODE is separable if it can be written as: dydx=f(x)g(y)\frac{dy}{dx} = f(x) \cdot g(y)

Solution method: Separate variables and integrate both sides: dyg(y)=f(x)dx\int \frac{dy}{g(y)} = \int f(x) \, dx

Example: dy/dx = xy → ∫dy/y = ∫x dx → ln|y| = x²/2 + C → y = Aeˣ²/²

First-Order Linear Equations

Standard form: dy/dx + P(x)y = Q(x)

Solution using integrating factor μ(x): μ(x)=eP(x)dx\mu(x) = e^{\int P(x) \, dx} y=1μ(x)μ(x)Q(x)dxy = \frac{1}{\mu(x)} \int \mu(x) Q(x) \, dx

Example: dy/dx + 2y = 6

  • P(x) = 2, so μ = e²ˣ
  • y = e⁻²ˣ ∫ 6e²ˣ dx = e⁻²ˣ (3e²ˣ + C) = 3 + Ce⁻²ˣ

Exact Equations

An equation M(x,y)dx + N(x,y)dy = 0 is exact if ∂M/∂y = ∂N/∂x.

Solution: Find F(x,y) such that ∂F/∂x = M and ∂F/∂y = N.

Second-Order Linear ODEs with Constant Coefficients

The general form: ay+by+cy=f(x)ay'' + by' + cy = f(x)

Homogeneous Case (f(x) = 0)

Solve the characteristic equation: ar² + br + c = 0

Root TypeConditionGeneral Solution
Real, distinct (r₁ ≠ r₂)b² - 4ac > 0y = C₁eʳ¹ˣ + C₂eʳ²ˣ
Real, repeated (r₁ = r₂ = r)b² - 4ac = 0y = (C₁ + C₂x)eʳˣ
Complex conjugate (α ± βi)b² - 4ac < 0y = eᵅˣ(C₁ cos βx + C₂ sin βx)

Non-Homogeneous Case (f(x) ≠ 0)

General solution = Homogeneous solution (yₕ) + Particular solution (yₚ)

Method of Undetermined Coefficients — guess yₚ based on f(x):

f(x)Guess for yₚ
Constant kA
Polynomial (degree n)Aₙxⁿ + Aₙ₋₁xⁿ⁻¹ + ... + A₀
eᵅˣAeᵅˣ
sin(βx) or cos(βx)A cos(βx) + B sin(βx)
eᵅˣ sin(βx)eᵅˣ(A cos βx + B sin βx)

Important: If your guess duplicates a term in yₕ, multiply by x (or x² if needed) to ensure independence.

Laplace Transforms

The Laplace transform converts a function of time f(t) to a function of complex frequency F(s): L{f(t)}=F(s)=0f(t)estdt\mathcal{L}\{f(t)\} = F(s) = \int_0^{\infty} f(t) e^{-st} \, dt

Common Laplace Transform Pairs

f(t)F(s)
11/s
t1/s²
tⁿn!/sⁿ⁺¹
eᵅᵗ1/(s-a)
sin(ωt)ω/(s²+ω²)
cos(ωt)s/(s²+ω²)
eᵅᵗ sin(ωt)ω/((s-a)²+ω²)
eᵅᵗ cos(ωt)(s-a)/((s-a)²+ω²)

Key Properties

PropertyTime Domains-Domain
Linearityaf(t) + bg(t)aF(s) + bG(s)
First Derivativef'(t)sF(s) - f(0)
Second Derivativef''(t)s²F(s) - sf(0) - f'(0)
Time Shiftf(t-a)u(t-a)e⁻ᵃˢF(s)
s-Shifteᵅᵗf(t)F(s-a)

Solving ODEs with Laplace Transforms

  1. Take the Laplace transform of both sides of the ODE
  2. Substitute initial conditions
  3. Solve algebraically for Y(s)
  4. Take the inverse Laplace transform to find y(t)

Example: Solve y'' + 4y = 0 with y(0) = 1, y'(0) = 0

Step 1: s²Y(s) - s(1) - 0 + 4Y(s) = 0 Step 2: Y(s)(s² + 4) = s Step 3: Y(s) = s/(s² + 4) Step 4: y(t) = cos(2t)

Test Your Knowledge

What is the general solution of y'' - 5y' + 6y = 0?

A
B
C
D
Test Your Knowledge

What is the Laplace transform of f(t) = e³ᵗ?

A
B
C
D
Test Your Knowledge

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

A
B
C
D