Difficulty: Intermediate to Advanced | Prerequisites: First-Order ODEs, Higher-Order ODEs, basic linear algebra (eigenvalues, determinants)
Laplace transforms convert differential equations into algebraic equations, which is especially powerful when initial conditions are given from the start. Systems of ODEs use matrix methods to handle coupled equations. Together, these are the most advanced analytical tools in MAT 2388 and tend to carry the heaviest weight on the final exam.
Laplace transforms turn differential equations into algebra by moving from the time domain t to the frequency domain s. You transform the equation, solve for Y(s) algebraically, then inverse-transform back to y(t). Heaviside and Dirac delta functions extend this to piecewise and impulsive forcing. Systems of ODEs rewrite coupled equations in matrix form and solve via eigenvalues, connecting ODE methods to linear algebra.
Laplace transform
An integral transform that converts f(t) into F(s) via L{f(t)} = ∫₀^∞ e^(–st) f(t) dt. Think of it as translating a differential equation into an algebraic one.
Inverse Laplace transform
The reverse operation: given F(s), recover f(t). Usually done by manipulating F(s) to match entries in a standard transform table.
Laplace transform table
A reference list of known transform pairs (f(t) ↔ F(s)). The core strategy is to reshape your expression until it matches something on the table.
Heaviside step function u(t – c)
A function that is 0 for t < c and 1 for t ≥ c. Used to model sudden switches or piecewise-defined forcing functions.
Dirac delta function δ(t – t₀)
An idealised impulse: zero everywhere except at t = t₀, with total integral equal to 1. Models instantaneous shocks (a hammer strike, a voltage spike).
Eigenvalue
A scalar λ such that A̅v̅ = λv̅ for some non-zero vector v̅. In systems of ODEs, eigenvalues determine the exponential behaviour of solutions.
Eigenvector
The vector v̅ paired with an eigenvalue. It gives the direction of each solution component.
System of ODEs
A set of coupled differential equations written in matrix form as y̅' = Ay̅. Solving means finding eigenvalues and eigenvectors of A.
The Laplace transform of f(t) is defined as L{f(t)} = F(s) = ∫₀^∞ e^(–st) f(t) dt.
Linearity. L{af(t) + bg(t)} = aL{f(t)} + bL{g(t)}. You can split a complex function into simpler parts, transform each one, and add the results.
Table-driven workflow. In practice, you rarely compute the integral from scratch. Instead, manipulate your expression (multiply by 1, add 0, complete the square, do partial fractions) until it matches a known entry on the transform table.
Common pairs to memorise. L{1} = 1/s, L{t^n} = n!/s^(n+1), L{e^(at)} = 1/(s–a), L{sin(bt)} = b/(s²+b²), L{cos(bt)} = s/(s²+b²).
The real power of Laplace transforms is that differentiation in t becomes multiplication by s.
L{f'(t)} = sL{f(t)} – f(0) = sF(s) – f(0)
L{f''(t)} = s²F(s) – sf(0) – f'(0)
The initial conditions f(0) and f'(0) are baked into the algebra from the start.
Solving an IVP step by step:
Let L{y} = Y(s). Write out the Laplace transform of each derivative.
Transform the whole equation. Apply L to both sides. Substitute the derivative formulas and plug in the initial conditions.
Isolate Y(s). Rearrange algebraically so Y(s) is alone on one side.
Partial fractions. Decompose the right side into simpler fractions that match the transform table.
Inverse transform. Convert each fraction back to the time domain using the table. The result is y(t).
Step 4 is where most of the work lives. Comfort with partial fraction decomposition is essential.
Heaviside functions model piecewise behaviour: a forcing term that switches on or off at a given time.
u(t – c): Equals 0 for t < c, equals 1 for t ≥ c. Turns a function "on" at time c.
1 – u(t – c): Equals 1 for t < c, equals 0 for t ≥ c. Turns a function "off" at time c.
u(t – a) – u(t – b): A "window" function, equals 1 for a ≤ t < b, zero outside.
f(t) shifted: To apply f only after time c, write f(t–c) · u(t–c).
Laplace transform: L{f(t–c) u(t–c)} = e^(–cs) F(s). The exponential e^(–cs) is the signature of a time shift in the s-domain.
Strategy: Express any piecewise function in terms of Heaviside functions, then transform each piece using the shift formula.
The Dirac delta models an instantaneous impulse.
Definition. δ(t – t₀) is zero everywhere except at t = t₀, with ∫δ(t – t₀) dt = 1.
Laplace transform. When t₀ > 0: L{δ(t – t₀)} = e^(–st₀). When t₀ = 0: L{δ(t)} = 1.
Solving IVPs with Dirac delta. If the right side of the ODE contains δ(t – t₀), simply take the Laplace transform of the entire equation and solve as normal. The delta's transform is just the exponential e^(–st₀), which is straightforward to handle.
These rules handle cases where the s-domain expression is hard to invert directly.
Derivative rule. L⁻¹{F'(s)} = –t f(t). More generally, L⁻¹{F^(n)(s)} = (−1)^n t^n f(t).
Practical use. If F(s) resists inverse transformation, try differentiating or integrating it in the s-domain to get a simpler expression, then adjust.
Laguerre differential equations use this for variable coefficients of the form (at + b). The key identities:
L{t y'(t)} = –d/ds [sY(s) – y(0)] = –Y(s) – sY'(s)
L{t y''(t)} = –d/ds [s²Y(s) – sy(0) – y'(0)] = –2sY(s) – s²Y'(s) + y(0)
These convert an ODE with variable coefficients into a first-order ODE in Y(s), which may be easier to solve.
Converting a higher-order ODE to a system. For a second-order equation, define u₁ = y and u₂ = y' = u₁'. Then u₂' = y''. Rewrite the original equation in terms of u₁ and u₂, isolate u₂', and express the result as a vector equation u̅' = Au̅.
Solving y̅' = Ay̅ (eigenvalue method):
Find eigenvalues. Compute det(A – λI) = 0 and solve for λ.
Find eigenvectors. For each λ, solve (A – λI)v̅ = 0.
Write the solution. Each eigenvalue-eigenvector pair contributes a term c₁ v̅₁ e^(λ₁t).
Case 1: λ₁ ≠ λ₂ (distinct eigenvalues). The general solution is y̅ = c₁ v̅₁ e^(λ₁t) + c₂ v̅₂ e^(λ₂t).
Case 2: λ₁ = λ₂ (repeated eigenvalue). A generalised eigenvector is needed. The solution involves a term with te^(λt).
MoUC and VOP can also be applied to systems in matrix form (sections 12.1 and 12.2 in the course), extending the scalar methods to vector equations.
L\{f(t)\} = F(s) = \int_0^\infty e^{-st} f(t)\,dtL\{f'(t)\} = sF(s) - f(0)L\{f''(t)\} = s^2 F(s) - sf(0) - f'(0)L\{f(t-c)\,u(t-c)\} = e^{-cs}F(s)L\{\delta(t - t_0)\} = e^{-st_0} \quad (t_0 > 0)L^{-1}\{F^{(n)}(s)\} = (-1)^n t^n f(t)\det(A - \lambda I) = 0 \quad \text{(eigenvalue equation)}Students often forget to include the initial-condition terms when transforming derivatives. L{y''} is s²Y(s) – sy(0) – y'(0), not just s²Y(s).
When using Heaviside functions, a common error is forgetting to shift the function argument. The transform formula requires f(t–c) u(t–c), not f(t) u(t–c).
Students sometimes confuse the Dirac delta with a finite pulse. The delta has zero width and infinite height; its area is 1. It is an idealisation, not a rectangle.
In systems of ODEs, a repeated eigenvalue does not simply double the first solution. You need a generalised eigenvector, and the solution picks up a factor of t.
⚠️ Solving IVPs with Laplace transforms is a near-certain final exam question. The full pipeline (transform, isolate Y(s), partial fractions, inverse transform) must be fluent.
⚠️ Heaviside functions regularly appear as piecewise forcing terms. Expect to convert a described scenario into step functions and then transform.
⚠️ Eigenvalue problems for 2×2 systems are a standard question. Know how to compute det(A – λI) = 0 quickly.
⚠️ Partial fraction decomposition is the bottleneck of most Laplace problems. Algebra errors here cascade through the rest of the solution.
Fill in the blank: L{e^(3t)} = ___. (1/(s–3))
True or false: L{y''} = s²Y(s). (False, it is s²Y(s) – sy(0) – y'(0).)
Fill in the blank: L{δ(t – 5)} = ___. (e^(–5s))
True or false: u(t–3) equals 1 when t = 2. (False, u(t–3) = 0 when t < 3.)
Fill in the blank: to find eigenvalues of A, solve det(___) = 0. (A – λI)
Q: Use Laplace transforms to solve y'' + 4y = 0, y(0) = 1, y'(0) = 0.
A: Transform: s²Y – s(1) – 0 + 4Y = 0, so Y(s²+4) = s, giving Y = s/(s²+4). Inverse transform: y(t) = cos(2t).
Q: Write f(t) = {0 for t < 2, t² for t ≥ 2} using Heaviside functions.
A: f(t) = t² u(t–2). For the Laplace transform, rewrite as (t–2+2)² u(t–2) and expand before applying the shift formula.
Q: Find the eigenvalues of A = [[3, 1], [0, 2]].
A: det(A – λI) = (3–λ)(2–λ) – 0 = 0. Eigenvalues: λ = 3 and λ = 2.
Laplace transforms provide a third route (alongside MoUC and VOP from Doc 2) to solve non-homogeneous equations, and they are the only analytical method that handles discontinuous or impulsive forcing directly. Systems of ODEs connect to every eigenvalue problem in linear algebra and reappear in control theory, circuit analysis, and any setting where multiple quantities evolve together. Numerical methods (Doc 4) approximate these solutions when the transforms or eigenvalues become intractable.
Related Terms / Search Tags
Laplace transform, inverse Laplace, transform table, partial fractions, initial value problem, IVP, Heaviside function, step function, unit step, Dirac delta, impulse function, Laguerre, variable coefficients, system of ODEs, eigenvalue, eigenvector, matrix exponential, coupled equations, MAT 2388, s-domain, frequency domain