Laplace transforms
The idea
The Laplace transform converts a function of time y(t) into a function of a new variable s, defined as the integral of e^(−st) y(t) from t = 0 to infinity. Its power for differential equations is that it converts calculus into algebra: differentiation in t becomes multiplication by s, with the transform of y' equal to sY(s) − y(0). Initial conditions enter automatically at the start instead of being fitted at the end — a major advantage for engineering problems with switches, impulses, and piecewise forcing.
The workflow is a round trip: transform the whole equation, solve algebraically for Y(s), then invert back to y(t) using a table of standard pairs — 1/s pairs with the constant 1, 1/(s + a) with e^(−at), and partial fractions break complicated Y(s) into those recognizable pieces. Most of the labor lives in the partial-fraction step, which is pure algebra.
Two cautions. The transform is linear, but it does not turn products into products — the transform of a product of two time functions is not the product of their transforms. And the s-domain is bookkeeping, not physics: Y(s) has no direct physical reading; only after inversion does the answer mean anything in time.
Worked example
Use the Laplace transform to solve y' + 3y = 6 with y(0) = 1.
- Transform every term using linearity and the derivative rule: y' becomes sY − y(0) = sY − 1, the term 3y becomes 3Y, and the constant 6 becomes 6/s.
- Assemble the algebraic equation: sY − 1 + 3Y = 6/s, so (s + 3)Y = 1 + 6/s = (s + 6)/s.
- Solve for the transform: Y(s) = (s + 6)/(s(s + 3)).
- Split by partial fractions: write (s + 6)/(s(s + 3)) = A/s + B/(s + 3); clearing denominators gives s + 6 = A(s + 3) + Bs, and choosing s = 0 yields A = 2 while s = −3 yields −3B = 3, so B = −1.
- Invert term by term from the table: 2/s returns the constant 2, and −1/(s + 3) returns −e^(−3t), so y(t) = 2 − e^(−3t).
- Verify in the time domain: y(0) = 2 − 1 = 1, and y' + 3y = 3e^(−3t) + 6 − 3e^(−3t) = 6 — the solution rises from 1 toward the steady state 2, as a stable first-order system driven by a constant should.
Answer. y(t) = 2 − e^(−3t), which starts at 1 and settles at the steady-state value 2.
Check your understanding
- Why does turning differentiation into multiplication by s make differential equations easier rather than just different?
- How do initial conditions enter a Laplace solution, and why is that more convenient than fitting constants afterward?
- What role do partial fractions play in the inversion step, and why must the table pairs be matched exactly?
- Which kinds of forcing functions — switches, impulses, ramps — make the Laplace approach especially attractive, and why?
Build the foundations first
Laplace transforms builds on these concepts. If any feel shaky, start there.