transform of derivatives
This single rule is the whole reason the Laplace transform solves differential equations. It tells you what happens to the transform when you differentiate in time — and the answer is that differentiation almost turns into multiplication by s, which is what flattens a calculus problem into an algebra problem.
The first-derivative rule is L{f'(t)} = s F(s) - f(0). Each time you differentiate, you pick up another factor of s and subtract an initial value: for the second derivative, L{f''(t)} = s^2 F(s) - s f(0) - f'(0). The pattern continues for higher orders, always carrying along the initial values f(0), f'(0), and so on. Crucially, the initial conditions are baked into the transform automatically — you never solve for them separately as constants of integration. That is the great convenience of the method.
When you apply this to a linear constant-coefficient equation like y'' + 3 y' + 2 y = g(t), every derivative is replaced by powers of s times Y(s) minus the relevant initial values, and the differential equation becomes a single algebraic equation for Y(s). You solve for Y(s) by ordinary algebra, then invert. The initial values f(0) and f'(0) are not optional decorations — they are the very quantities that make the answer satisfy the initial-value problem, and dropping them is the most damaging mistake in the whole method.
For y' + 2 y = 0 with y(0) = 5, transforming gives s Y - 5 + 2 Y = 0, so Y = 5/(s + 2), and inverting gives y(t) = 5 e^{-2 t}.
The initial value 5 enters directly through the derivative rule, so the solution already satisfies y(0) = 5 with no extra constant to fix.
The f(0) terms use the value just after t = 0; for problems with jumps or impulses at the origin this distinction between f(0-) and f(0+) matters, and physicists usually take the value the instant before the input acts.