The rule that does all the work
Guide one promised it and guide two stocked the table around it; now we meet the single fact that makes the Laplace transform worth learning. It is the transform of a derivative: if Y(s) is the transform of y(t), then L{y'} = s * Y(s) - y(0). Read it as a trade. On the left is a derivative — a calculus operation. On the right there is no derivative at all: just a multiplication by s and the subtraction of one number, the starting value y(0). The hard thing on this bank became an easy thing on that bank.
Two things deserve a second look. First, differentiation has turned into multiplication by s — that is the whole calculus-into-algebra promise, made concrete in one line. Second, and this is the part that sets the method apart, the number y(0) shows up uninvited. You did not bolt it on; the rule itself reached into your problem and grabbed the initial value. That is what people mean when they say the Laplace method has the initial conditions built in — they ride along inside the algebra from the first step rather than waiting at the end.
Where y(0) comes from
The rule is not a spell — it falls out of a single integration by parts, and seeing that once will fix it in memory forever. By definition L{y'} is the integral from 0 to infinity of e^(-s t) * y'(t) dt. Integration by parts trades the derivative off y and onto the weight e^(-s t): you differentiate the e^(-s t) (giving a factor -s) and integrate the y'. Out drops a boundary term [e^(-s t) * y(t)] evaluated from 0 to infinity, plus s times the integral of e^(-s t) * y(t) — and that leftover integral is exactly s * Y(s).
Now evaluate that boundary term, and watch the initial condition appear. At the top end, t going to infinity, the weight e^(-s t) crushes y(t) down to zero — this is exactly the exponential-order assumption earning its keep, since it guarantees y cannot outrun the decaying exponential. At the bottom end, t = 0, the weight is e^(0) = 1, so the term is simply y(0). The boundary term is therefore 0 - y(0) = -y(0). Assemble the pieces: L{y'} = -y(0) + s*Y(s). There is the rule, born from one honest calculation, with y(0) arriving as the lower limit of an integral.
Climbing to higher derivatives
A second-order ODE needs the transform of y'', and you get it by applying the first-derivative rule twice — no new idea, just the same move bootstrapped. Think of y'' as the derivative of the function (y'). The rule says L{(y')'} = s * L{y'} - y'(0). Now substitute what you already know for L{y'}, namely s*Y - y(0), and expand. The result is L{y''} = s^2 * Y - s * y(0) - y'(0). Notice the pattern forming: a power of s for the derivative order, and one initial value peeled off per order — this time both y(0) and y'(0).
L{y'} = s*Y - y(0)
L{y''} = s^2*Y - s*y(0) - y'(0)
L{y'''} = s^3*Y - s^2*y(0) - s*y'(0) - y''(0)
General:
L{y^(n)} = s^n*Y - s^(n-1)*y(0) - s^(n-2)*y'(0) - ... - y^(n-1)(0)
Mnemonic: powers of s count DOWN from n-1 ;
initial values count UP from y(0) to y^(n-1)(0).The general formula L{y^(n)} = s^n * Y - s^(n-1) * y(0) - ... - y^(n-1)(0) is worth reading as bookkeeping rather than memorising. An nth-order initial value problem needs exactly n initial values — y(0) through y^(n-1)(0) — to have one definite answer, and the formula consumes precisely those n numbers, no more, no fewer. That is not a coincidence; it is the same count showing up on both sides. The transform is quietly accounting for every degree of freedom the equation has.
Turning a whole equation into algebra
Here is the rule doing its real job. Take a concrete IVP: y'' + 3 y' + 2 y = 0 with y(0) = 1 and y'(0) = 0. Transform every term, leaning on the linearity you met in guide one so you can go piece by piece. The y'' becomes s^2 Y - s*y(0) - y'(0); the 3 y' becomes 3(sY - y(0)); the 2 y becomes 2Y; and the zero on the right stays zero. Every derivative has vanished, replaced by powers of s and the known numbers y(0) and y'(0).
Substitute the actual values y(0) = 1, y'(0) = 0 and collect. The transformed equation reads (s^2 + 3s + 2) Y - (s + 3) = 0, an ordinary algebraic equation in the single unknown Y. No derivatives, no integrals — just a linear equation you solve by rearranging: Y = (s + 3) / (s^2 + 3s + 2). The differential equation has been transferred entirely into algebra, initial conditions and all. The bracket s^2 + 3s + 2 is, satisfyingly, the very same characteristic polynomial you would have written by guessing y = e^(rt) — the two methods are looking at one structure from two sides.
The mirror rule, and where to be careful
There is a companion worth knowing, because integration shows up as often as differentiation in physics. The transform of an integral is the mirror image of the derivative rule: where differentiating multiplied by s, integrating divides by s. Concretely, the transform of the running integral of y from 0 to t equals Y(s)/s. The symmetry is no accident — integration and differentiation are inverse operations in t, so they become multiplication and division by s in the s-domain. Charge accumulating on a capacitor, distance piling up from velocity: division by s is how the s-domain says 'accumulate'.
Two honest cautions before you trust the derivative rule blindly. First, it is a one-sided transform: the integral starts at 0, so everything is anchored to behaviour from time zero onward, and the y(0) it carries is the value at that left edge — there is no symmetric term from the other end because the upper limit decays away. Second, the rule assumes y is differentiable in the ordinary sense between jumps; if y itself jumps at some interior time, the plain formula misses a contribution there, and you handle that case with the step-function machinery of a later guide rather than this bare rule.
Step back and see what you now hold. One short rule, L{y'} = sY - y(0), simultaneously kills the derivatives and feeds in the initial data, and its higher-order ladder does the same for any order you meet. Apply it across an equation and the whole IVP collapses to algebra in Y(s) — the heart of the master algorithm. You have built steps one and two of that four-step plan. What remains is the return trip: splitting Y(s) and reading y(t) back out, which the next guide hands you in full.