The definition gives a small table. A few rules stretch it a long way, and one of them, the transform of a derivative, is the reason the transform solves differential equations.
Key idea
$$\mathcal{L}\{y'\} = sY - y(0),$$ $$\mathcal{L}\{y''\} = s^{2}Y - sy(0) - y'(0).$$ First shift: $\mathcal{L}\{e^{at}f(t)\} = F(s - a)$. Multiplying by $t$: $\mathcal{L}\{tf(t)\} = -F'(s)$.
Why it works
Integrate by parts with $u = e^{-st}$, $dv = y'\,dt$: $$\int_0^{\infty} e^{-st}y'\,dt = \left[e^{-st}y\right]_0^{\infty} + s\int_0^{\infty} e^{-st}y\,dt = -y(0) + sY,$$ since $e^{-st}y \to 0$ for large $s$. Apply this to $y'$ to get $\mathcal{L}\{y''\}$ $= s(sY - y(0)) - y'(0)$. The shift rule holds because $e^{-st}e^{at} = e^{-(s - a)t}$, and the $t$ rule because $\frac{d}{ds}e^{-st} = -te^{-st}$.
Worked example: A shifted cosine
Find $\mathcal{L}\{e^{-2t}\cos 3t\}$.
- $\mathcal{L}\{\cos 3t\} = \frac{s}{s^{2} + 9}$.
- Multiplying by $e^{-2t}$ replaces $s$ by $s + 2$.
Answer: $\frac{s + 2}{(s + 2)^{2} + 9}$
Worked example: Transform an initial value problem
Find $Y(s)$ for $y'' + 3y' + 2y = 0$, $y(0) = 1$, $y'(0) = 0$.
- $\mathcal{L}\{y''\} = s^{2}Y - s$ and $\mathcal{L}\{y'\} = sY - 1$.
- $(s^{2}Y - s) + 3(sY - 1) + 2Y$ $= 0$.
- $(s^{2} + 3s + 2)Y = s + 3$.
Answer: $Y = \frac{s + 3}{s^{2} + 3s + 2}$
Worked example: Multiplying by $t$
Find $\mathcal{L}\{t\sin 2t\}$.
- $F(s) = \frac{2}{s^{2} + 4}$, so $F'(s) = -\frac{4s}{(s^{2} + 4)^{2}}$.
- $\mathcal{L}\{t\sin 2t\} = -F'(s)$.
Answer: $\frac{4s}{(s^{2} + 4)^{2}}$
Worked example: A table entry from the derivative rule
Use $\mathcal{L}\{\sin bt\} = \frac{b}{s^{2} + b^{2}}$ and the derivative rule to find $\mathcal{L}\{\cos bt\}$.
- $f = \sin bt$ has $f' = b\cos bt$ and $f(0) = 0$.
- $\mathcal{L}\{f'\} = sF - f(0)$: $b\,\mathcal{L}\{\cos bt\} = \frac{bs}{s^{2} + b^{2}}$.
- Divide by $b$.
Answer: $\mathcal{L}\{\cos bt\} = \frac{s}{s^{2} + b^{2}}$, matching the table
Common mistake
Dropping the initial values: writing $\mathcal{L}\{y''\} = s^{2}Y$. The terms $-sy(0) - y'(0)$ are where the initial conditions enter the problem; without them every solution would start at rest. Check by counting: a second-order equation needs two initial values, and both must appear in your transformed equation.
| $f(t)$ | $F(s)$ |
|---|---|
| $e^{at}\sin bt$ | $\frac{b}{(s - a)^{2} + b^{2}}$ |
| $e^{at}\cos bt$ | $\frac{s - a}{(s - a)^{2} + b^{2}}$ |
| $t^{n}e^{at}$ | $\frac{n!}{(s - a)^{n+1}}$ |
| $t\sin bt$ | $\frac{2bs}{(s^{2} + b^{2})^{2}}$ |
The rules agree with each other, which makes a good check. For $te^{2t}$, the shift rule turns $\mathcal{L}\{t\} = \frac{1}{s^{2}}$ into $\frac{1}{(s - 2)^{2}}$, and the $t$ rule gives $-\frac{d}{ds}\frac{1}{s - 2} = \frac{1}{(s - 2)^{2}}$: the same answer. The derivative rule extends to any order, $\mathcal{L}\{y^{(n)}\}$ $= s^{n}Y - s^{n-1}y(0) - \dots$ $- y^{(n-1)}(0)$, and running it backwards gives the integral rule $\mathcal{L}\left\{\int_0^{t} f(\tau)\,d\tau\right\} = \frac{F(s)}{s}$: integrating in $t$ divides by $s$.