The Laplace transform converts f(t) to F(s) = ∫₀^∞ e^(-st)f(t)dt, mapping time-domain differential equations to frequency-domain algebra. Key properties: linearity, the derivative rule L[f'(t)] = sF(s) - f(0), and shifting theorems. These transform initial conditions into the equation automatically, making Laplace transforms powerful for solving IVPs, especially with discontinuous forcing functions.
You've already studied improper integrals and know that ∫₀^∞ e^(−at) dt converges for a > 0. The Laplace transform wraps that convergence trick into a machine for solving differential equations. Given a function f(t) defined for t ≥ 0, the transform is ℒ{f}(s) = F(s) = ∫₀^∞ e^(−st) f(t) dt. The exponential e^(−st) acts as a damping factor: for large enough s, it forces the integral to converge even if f(t) grows (as long as f doesn't grow faster than some exponential). The output F(s) is a new function of the parameter s.
The whole point of the transform is the derivative rule: ℒ{f′(t)} = sF(s) − f(0). Differentiation in t becomes multiplication by s in the s-domain, with the initial condition f(0) appearing algebraically. For a second derivative: ℒ{f″(t)} = s²F(s) − sf(0) − f′(0). Every derivative lowers the problem by one degree. So a second-order ODE like f″ + 3f′ + 2f = g(t) with initial conditions f(0) = a, f′(0) = b transforms into a purely algebraic equation in F(s) and G(s) = ℒ{g}. Solve for F(s), then transform back. The initial conditions are absorbed automatically — no separate step needed to impose them.
Linearity ℒ{αf + βg} = αF + βG follows directly from linearity of integration and lets you handle sums of functions term by term. The s-shifting theorem says ℒ{e^(at)f(t)} = F(s − a): multiplying by an exponential in t shifts the argument in s. This handles forcing functions with exponential growth or decay. The t-shifting theorem ℒ{u_c(t)f(t − c)} = e^(−cs)F(s) handles functions that "switch on" at time t = c, where u_c is the unit step function. This is where the Laplace transform genuinely outperforms variation of parameters — discontinuous forcing functions like step functions and impulses are handled with almost no extra complexity.
The Laplace transform establishes convergence for Re(s) > some abscissa of convergence. For polynomials, ℒ{tⁿ} = n!/s^(n+1) for s > 0; for exponentials, ℒ{e^(at)} = 1/(s − a) for s > a. These basic transforms, combined with the linearity and shifting properties, form a table that covers almost every forcing function you'll encounter. The transform method's procedure is always the same: transform the ODE → solve the algebraic equation for F(s) → apply partial fractions and the table → invert back to f(t).