Why a table, not the integral
In the previous guide you met the machine itself: the Laplace transform feeds a time function f(t) through the integral F(s) = integral from 0 to infinity of e^(-st) f(t) dt and hands back a function F(s) living in the s-domain. That integral is the definition — but you will almost never compute it again. The whole point of the method is to do the hard integrals once, for a small set of standard functions, write the answers down, and then just look them up. This guide builds that lookup table and, just as important, teaches you to read it in reverse.
Think of it exactly like a multiplication table. Nobody re-derives 7 times 8 from repeated addition each time it comes up; they recall 56. In the same spirit, nobody integrates e^(-st) e^(at) dt from scratch when they can recall the row L{e^(at)} = 1/(s - a). A working transform table — captured by table lookup — has just two columns: a time function f(t) on the left, its transform F(s) on the right. Master a dozen rows and a couple of rules, and you can transform essentially every function an introductory ODE will throw at you.
The rows you simply memorize
Start with the powers of t. The power rule says L{t^n} = n! / s^(n + 1), where n! (n factorial) is n times (n - 1) times ... times 1. Watch the pattern: L{1} = 1/s (take n = 0, since 0! = 1), L{t} = 1/s^2, L{t^2} = 2/s^3, L{t^3} = 6/s^4. Each extra power of t pushes the power of s in the denominator up by one and drags a factorial along. A polynomial in t therefore transforms into a sum of simple powers of 1/s — clean and mechanical.
Next, the exponential — the natural language of growth and decay. The exponential rule is L{e^(at)} = 1 / (s - a). It comes straight from the definition: integral from 0 to infinity of e^(-st) e^(at) dt = integral of e^(-(s - a) t) dt = 1/(s - a), valid for s > a. Notice what the transform did with the constant a: it slid the simple pole of L{1} = 1/s from s = 0 over to s = a. That single shift is a preview of a rule we will meet shortly — the whole table is more connected than it first looks.
Then the wiggle. The sine and cosine rules are L{sin(at)} = a / (s^2 + a^2) and L{cos(at)} = s / (s^2 + a^2). They share the denominator s^2 + a^2 — your eye should learn to read that shape as 'something is oscillating at frequency a' — and differ only in the numerator: a bare a for sine, an s for cosine. Their close relatives, the hyperbolic functions, differ by just one sign: L{cosh(at)} = s/(s^2 - a^2) and L{sinh(at)} = a/(s^2 - a^2), with a minus where the trig pair had a plus. A plus means oscillation; a minus means exponential growth and decay.
f(t) F(s) = L{f(t)} notes
---- -------------- -----
1 1/s power rule, n = 0
t 1/s^2
t^n n! / s^(n+1) n = 0, 1, 2, ...
e^(at) 1 / (s - a) s > a
sin(at) a / (s^2 + a^2) '+' => oscillation
cos(at) s / (s^2 + a^2)
sinh(at) a / (s^2 - a^2) '-' => growth/decay
cosh(at) s / (s^2 - a^2)Two rules that stretch the table
A table of eight rows would be useless if you could only transform exactly those eight functions. What makes it powerful is a pair of rules that let one row cover a whole family. The first is linearity: the transform of a sum is the sum of the transforms, and a constant multiplier passes straight through, so L{3 t^2 - 5 sin(2t)} = 3 (2/s^3) - 5 (2/(s^2 + 4)). You break a messy function into table pieces, transform each piece, and add. This is the same fairness that lets you convert a shopping list item by item or total-first — either way, same answer.
The second rule explains the shifting you already glimpsed. The first shifting theorem (the s-shift) says L{e^(at) f(t)} = F(s - a): multiplying a time function by e^(at) slides its whole transform sideways, replacing every s by s - a. This single rule generates a huge chunk of any real table for free. A damped oscillation e^(-bt) sin(ct), the bread and butter of vibrations and circuits, is just sin(ct) shifted: L{e^(-bt) sin(ct)} = c / ((s + b)^2 + c^2). You did not integrate anything — you took the sine row and slid it. Recognizing this shifted shape is exactly what you will need to read the table backward later.
Reading the table backward
Here is the part beginners underrate. The table is symmetric: every row works in both directions. Reading f(t) -> F(s) is the forward transform; reading F(s) -> f(t) is the inverse transform, and it is where the actual answer to your differential equation comes from. The Laplace method takes your problem into the s-domain, does easy algebra there, and then you must come back — and coming back means recognizing an F(s) as the right column of some row and reading off the left column. The forward trip is a formality; the return trip is where you earn your solution.
So you must train your eye to recognize standard shapes. See 1/(s - 3) and read e^(3t). See 2/(s^2 + 4) and read sin(2t). See s/(s^2 + 9) and read cos(3t). The hardest backward reads are the shifted ones: an expression like (s + 1)/((s + 1)^2 + 4) is not in the bare table, but the s-shift tells you it is cos(2t) damped by e^(-t). The trick is to complete the square in the denominator until you see a clean (s - a)^2 + b^2, then undo the shift. Pattern recognition, not integration, is the real skill of the inverse.
The row that the whole method needs
There is one more row, and it is the reason any of this is worth doing for differential equations. The transform of a derivative is L{f'(t)} = s F(s) - f(0). Read it slowly: the dreaded operation — differentiation — becomes, in the s-domain, almost nothing more than multiplying by s. Calculus turns into algebra. The second derivative builds on it: L{f''(t)} = s^2 F(s) - s f(0) - f'(0). Each derivative you take in time is one more factor of s in the s-domain, minus a correction made of starting values.
Look at what those f(0) and f'(0) terms are doing — this is no accident. They are precisely the initial conditions of your problem, and the derivative rule pulls them into the algebra automatically. That is the quiet superpower of the method: where classical techniques solve a differential equation first and only afterward fit the constants, the Laplace transform carries the starting data along from the very first line. The whole story of why and how — the IVP-to-algebra translation — is the heart of the next guide, but this row is where it begins.