The Operational Rules

Linearity: the rule you already trust

In the previous guide you met the Laplace transform as a definite integral, L{f}(s) = integral from 0 to infinity of e^{-st} f(t) dt — a machine that eats a function of time t and hands back a function of a new variable s. Computing that integral term by term, every single time, would be exhausting. The whole power of the transform is that you almost never go back to the integral. Instead you build a short table of basic transforms once, then move everything around with a small set of operational rules. This guide is that toolbox, and the first tool is the one you can already feel in your hands.

The Laplace transform is linear: L{a f(t) + b g(t)} = a L{f}(s) + b L{g}(s) for any constants a, b. This is not a new miracle — it falls straight out of the definite integral being linear, which you have trusted since Volume I. The integral of a sum is the sum of the integrals, and constants slide out front; the transform inherits both habits whole. So to transform 3 t^2 - 5 sin(t), you do not touch the defining integral at all: look up L{t^2} and L{sin t} in your table, then assemble 3 L{t^2} - 5 L{sin t}. Linearity is what lets a table of a dozen entries cover an infinite zoo of combinations.

Shifting in s: multiply by an exponential in time

The next rule answers a question you will ask constantly: what does multiplying f(t) by an exponential e^{at} do to its transform? The first shifting theorem gives the clean answer — it simply shifts the transform sideways in s. Precisely, L{e^{at} f(t)}(s) = F(s - a), where F(s) = L{f}(s). Damp or amplify a signal by e^{at} in time, and its whole transform slides by a along the s-axis. The proof is one line and worth seeing: the e^{at} fuses with the e^{-st} already in the defining integral, turning e^{-st} e^{at} into e^{-(s-a)t}, which is exactly the original integrand evaluated at s - a instead of s.

This little rule is worth its weight in gold because exponentials-times-something are everywhere in applications. A damped oscillation e^{-t} cos(omega t) is the bread and butter of every circuit and every shock absorber. You already know L{cos(omega t)} = s/(s^2 + omega^2). To get the damped version, you do not redo any integral — you just replace s by s + 1 (here a = -1): L{e^{-t} cos(omega t)} = (s + 1)/((s + 1)^2 + omega^2). Every e^{at} in the time world becomes an honest shift s -> s - a in the transform world, and that single substitution handles an enormous family of decaying and growing signals at no extra cost.

Shifting in time, and scaling

The first theorem shifted s; its mirror image shifts t. Suppose you delay a signal — nothing happens until time a, then your function f starts running, shifted over. The honest way to write a delayed-and-switched-on signal uses the Heaviside step u(t - a), which is 0 before a and 1 after. The second shifting theorem then says L{f(t - a) u(t - a)} = e^{-as} F(s). A pure delay of a in time becomes multiplication by the factor e^{-as} in the transform. Delays are messy in the time domain and trivial in the s-domain — this is exactly why Laplace methods shine for systems that switch on and off.

One more reshaping rule rounds out the geometry. The scaling theorem tells you what stretching the time axis does: L{f(at)}(s) = (1/a) F(s/a) for a positive constant a. Compress time by a factor a and the transform stretches in s by the same factor, with a 1/a out front to keep the bookkeeping honest — a clean reciprocal trade between the two domains. Together, the two shifts and this scaling let you slide, delay, and rescale a signal and immediately know its transform from a single table entry, without ever reopening the improper integral that defines L.

The crown jewel: transforms of derivatives

Now for the rule that justifies the whole enterprise. Everything so far has been convenient; this is the rule that turns calculus into algebra. The transform of a derivative is L{f'(t)}(s) = s F(s) - f(0). Read it slowly: differentiating in time becomes multiplying by s in the transform world — plus a correction term f(0) that remembers where the function started. Differentiation, that hard limiting operation from Volume I, collapses into ordinary multiplication. That is the magic the entire rung is built on.

Where does the f(0) come from? Straight from integration by parts, the Volume I move you reach for whenever a derivative sits inside an integral. Apply parts to integral from 0 to infinity of e^{-st} f'(t) dt: the boundary term [e^{-st} f(t)] from 0 to infinity contributes -f(0) at the lower limit (and vanishes at infinity for well-behaved f), while the leftover integral rebuilds s F(s). So the seemingly magical s F(s) - f(0) is just integration by parts wearing a tuxedo. The lesson is honest and reassuring: nothing new was smuggled in — this is first-year calculus, deployed cleverly.

First derivative:   L{f'}      = s F(s) - f(0)
Second derivative:  L{f''}     = s^2 F(s) - s f(0) - f'(0)
n-th derivative:    L{f^(n)}   = s^n F(s)
                                 - s^(n-1) f(0)
                                 - s^(n-2) f'(0)
                                 - ...
                                 - f^(n-1)(0)

Pattern:  each derivative  d/dt  ->  multiply by s,
          and one initial value is peeled off per order.

Stare at the second-derivative line and you can already see why a whole differential equation dissolves. Take a forced oscillator y'' + omega^2 y = g(t) with starting data y(0) and y'(0). Transform every term using linearity and the derivative rule: the y'' becomes s^2 Y(s) - s y(0) - y'(0), the y becomes Y(s), and g becomes G(s). The differential equation has turned into a plain algebraic equation in Y(s), and — crucially — the initial conditions y(0) and y'(0) walked in through the front door, baked into the algebra from the start rather than fitted at the end. Solve for Y(s) by ordinary algebra, then invert. The next guide makes this its whole method, solving initial-value problems by Laplace transform.

The mirror jewel: transforms of integrals

If differentiation multiplies by s, you can guess its mirror image: integration ought to divide by s, and it does. The transform of an integral says L{ integral from 0 to t of f(tau) dtau }(s) = F(s)/s, with no boundary correction because the integral starts at zero. This is no coincidence — it is the Fundamental Theorem of Calculus reflected into the s-domain. Differentiating and integrating are inverse operations in time; multiplying and dividing by s are inverse operations in the transform. The transform takes the deep symmetry of Volume I calculus and renders it as the cheapest arithmetic there is.

These two rules together are why the Laplace transform is the engineer's wand. A circuit law might say the voltage across a capacitor is proportional to the accumulated charge, which is the integral of the current — and the inductor's voltage is proportional to the rate of change of current, a derivative. A single circuit can therefore mix derivatives and integrals of the same unknown in one equation, an integro-differential equation that looks fearsome in time. Transform it, and every derivative becomes times-s, every integral becomes divided-by-s, and the whole tangle becomes one rational equation in F(s) that a teenager could rearrange. Convert, do algebra, convert back — the entire forest of calculus operations flattens into a single algebraic plain.

Honesty about the fine print is owed here. These derivative and integral rules assume f is well enough behaved for the transform to exist and for the boundary term to vanish at infinity — concretely, that f grows no faster than some exponential, and that the relevant derivative is itself transformable. That is exactly the existence condition from the previous guide doing quiet work here. Most functions in mechanics and circuits satisfy it easily, but the rules are theorems with hypotheses, not unconditional spells. When a forcing term is a step or an impulse, you must also handle the derivative of a discontinuous object carefully — a subtlety the later guide on switches and the Dirac delta takes head-on.

The whole toolbox, and how to wield it

Step back and the operational rules form one coherent dictionary between two worlds. In the time domain you have addition, exponential modulation, delay, time-stretching, differentiation, and integration; each translates into a single tidy move on the s-side. That dictionary is what you actually use to solve a problem, and the procedure never varies:

1. Transform the whole problem into the s-domain. Apply linearity term by term, the derivative rule to every y', y'', and the integral rule to any accumulated quantity — the initial conditions enter here automatically.
2. Solve the resulting algebraic equation for the unknown transform F(s) or Y(s). No calculus here at all — just multiply out, collect terms, and divide.
3. Reshape the answer into table-ready pieces. Split by partial fractions; recognize an s -> s - a pattern as the first shifting theorem; spot an e^{-as} factor as a delay from the second shifting theorem.
4. Invert each piece back to time with the inverse transform, again term by term, and add. The sum is your solution y(t), initial conditions already satisfied.

Notice how much of the real work lives in step 3 — the reshaping. The forward rules are mechanical; the art is recognizing a transform you have on the way back, and the two shifting theorems are your sharpest tools for that. A factor e^{-2s} in F(s) is not a difficulty to be integrated away — it is a flag that says "a delayed copy switching on at t = 2," read off instantly via the second theorem. Master these few rules until they are reflexes, and the Laplace method stops feeling like magic and starts feeling like what it is: a precise, honest translation that lets ordinary algebra do the heavy lifting that calculus used to demand.