turning an IVP into algebra
Strip the Laplace method down to its single big idea and this is it: a calculus problem becomes an algebra problem. A differential equation asks you to find an unknown function — a hard, infinite-dimensional question. The transform converts that very question into finding an unknown number-valued expression Y(s) through ordinary algebraic manipulation. That conversion is the whole reason the method is powerful.
Concretely, watch what each ingredient becomes. The unknown function y(t) becomes the unknown transform Y(s). A derivative y' becomes multiplication by s (with an initial value attached); y'' becomes multiplication by s^2. The forcing function g(t) becomes its transform G(s). So an equation like a y'' + b y' + c y = g(t) — a relation among functions and their derivatives — turns into (a s^2 + b s + c) Y(s) = G(s) + (initial-condition terms), a relation among polynomials in s. Every derivative, the thing that made it a differential equation, has been replaced by a power of s, the thing that makes it merely algebraic.
Once you are looking at a polynomial equation, all the leverage of algebra is available: you just divide to get Y(s) = [G(s) + initial terms] / (a s^2 + b s + c). The denominator a s^2 + b s + c is no coincidence — it is the characteristic polynomial of the equation, so the s-domain quietly encodes the same structure (roots, stability, natural frequencies) you would analyze by classical means. This is the deep payoff: the difficulty of solving an IVP has been transferred from calculus, where it is genuinely hard, to algebra, where it is routine.
The IVP y'' - y = 1, y(0) = 0, y'(0) = 0 transforms to (s^2 Y - 0 - 0) - Y = 1/s, i.e. (s^2 - 1) Y = 1/s. A differential equation has become the algebraic equation Y = 1/(s(s^2 - 1)), ready for partial fractions.
Derivatives become powers of s; the equation becomes pure algebra in Y(s).
The trick only flattens LINEAR equations with constant coefficients this cleanly. A nonlinear term like y^2 or y y' has no simple transform — L{y^2} is not Y(s)^2 — so the algebraization breaks down and the method does not directly apply.