Topic · Differential Equations

The Laplace Transform

An integral transform that converts a function of time $f(t)$ into a function of a complex variable $F(s)$. Its single magic trick: it turns differentiation into multiplication, so linear differential equations with initial conditions become algebraic equations — solve, then translate back.

15 min read Prereqs: Second-order linear ODEs, Improper integrals Updated 2026·05·17

What you'll leave with

  • The definition $\mathcal{L}\{f\}(s) = \int_0^\infty e^{-st} f(t)\,dt$, and the half-dozen transforms you actually memorize.
  • Why the transform is useful: $\mathcal{L}\{f'\} = sF(s) - f(0)$ converts derivatives into multiplication, and initial conditions ride along for free.
  • The three core structural rules — linearity, first shift, second shift — and the convolution theorem.
  • The full workflow, executed end-to-end on a discontinuously forced oscillator.

1. The definition

Laplace transform

For a function $f(t)$ defined on $t \geq 0$, the Laplace transform is the function of $s$ given by

$$ \mathcal{L}\{f(t)\}(s) \;=\; F(s) \;=\; \int_0^\infty e^{-st} f(t)\,dt, $$

defined for every $s$ (real or complex) for which the improper integral converges.

Think of $\mathcal{L}$ as a machine. You feed in a function of time; it returns a function of a new variable $s$. The factor $e^{-st}$ does the work: when $s$ is large enough, it dampens $f(t)$ fast enough that the integral converges, no matter how badly behaved $f$ was at infinity. For most functions in this course, $s$ is a positive real number, and we won't fuss about the complex case.

Two examples by hand, just so the formula stops looking abstract:

Constant $f(t) = 1$.

$$ \mathcal{L}\{1\} \;=\; \int_0^\infty e^{-st}\,dt \;=\; \left[ -\tfrac{1}{s} e^{-st} \right]_0^\infty \;=\; \frac{1}{s}, \quad s > 0. $$

Exponential $f(t) = e^{at}$.

$$ \mathcal{L}\{e^{at}\} \;=\; \int_0^\infty e^{-(s-a)t}\,dt \;=\; \frac{1}{s - a}, \quad s > a. $$

Notice how the time-domain variable $t$ has vanished, replaced entirely by $s$. That swap is the whole point.

Notation

Lowercase letters for time-domain ($f(t)$, $y(t)$); uppercase for the transform ($F(s)$, $Y(s)$). The convention is universal and worth honoring — it keeps "which side of the transform am I on?" obvious at a glance.

2. Why it exists: derivatives become multiplication

The Laplace transform earns its keep with a single identity. Apply integration by parts to $\mathcal{L}\{f'(t)\}$, assuming $f$ is well-behaved enough that the boundary term at infinity vanishes:

$$ \mathcal{L}\{f'(t)\} \;=\; \int_0^\infty e^{-st} f'(t)\,dt \;=\; \big[e^{-st} f(t)\big]_0^\infty + s\!\int_0^\infty e^{-st} f(t)\,dt \;=\; sF(s) - f(0). $$

Read what just happened. The derivative of $f$ in time-land has become multiplication by $s$ in transform-land, with the initial value $f(0)$ falling out as a constant term. Calculus has turned into algebra.

Apply the rule again to get the second derivative:

$$ \mathcal{L}\{f''(t)\} \;=\; s^2 F(s) - s\,f(0) - f'(0). $$

And in general,

$$ \mathcal{L}\{f^{(n)}(t)\} \;=\; s^n F(s) - s^{n-1} f(0) - s^{n-2} f'(0) - \cdots - f^{(n-1)}(0). $$
Why this is a big deal

A linear ODE with constant coefficients is, after transforming, a linear algebraic equation in $F(s)$. Initial conditions are no longer a side problem to be patched in at the end — they enter the algebra automatically. That is the entire reason engineers reach for Laplace before any other method.

3. The table you actually memorize

You don't compute these integrals every time. Six or seven entries cover the overwhelming majority of problems; everything else is a derived consequence of these plus the rules in the next section.

$f(t)$$F(s) = \mathcal{L}\{f\}$Region of convergence
$1$$\dfrac{1}{s}$$s > 0$
$t^n$  ($n = 0, 1, 2, \ldots$)$\dfrac{n!}{s^{n+1}}$$s > 0$
$e^{at}$$\dfrac{1}{s - a}$$s > a$
$\sin(\omega t)$$\dfrac{\omega}{s^2 + \omega^2}$$s > 0$
$\cos(\omega t)$$\dfrac{s}{s^2 + \omega^2}$$s > 0$
$u(t - a)$  (unit step)$\dfrac{e^{-as}}{s}$$s > 0$
$\delta(t - a)$  (impulse)$e^{-as}$all $s$

The unit step $u(t - a)$ is zero before $t = a$ and one after — the cleanest way to express "something switches on at time $a$." The Dirac delta $\delta(t - a)$ models an instantaneous, infinitely concentrated impulse at $t = a$. Both transforms are simple in $s$-land because the exponential $e^{-st}$ tames them on contact.

Region of convergence

The integral defining $F(s)$ converges only for $s$ above some threshold (set by how fast $f$ grows). In practice you can ignore this — once $F(s)$ is a rational function of $s$, you treat it as defined everywhere except its poles and use partial fractions without further comment.

4. Linearity, shifts, convolution

Linearity

Integration is linear, so the transform is linear:

$$ \mathcal{L}\{a f(t) + b g(t)\} \;=\; a F(s) + b G(s). $$

This is what lets you decompose a complicated $F(s)$ into partial fractions and invert each piece separately.

First shift theorem (shift in $s$)

Multiplying $f(t)$ by an exponential shifts $F(s)$:

$$ \mathcal{L}\{e^{at} f(t)\} \;=\; F(s - a). $$

So $\mathcal{L}\{e^{at}\cos(\omega t)\} = \dfrac{s - a}{(s - a)^2 + \omega^2}$ — you take the transform of $\cos(\omega t)$ and replace every $s$ with $s - a$.

Second shift theorem (shift in $t$)

Delaying $f$ by $a$ and switching it on with a unit step multiplies its transform by $e^{-as}$:

$$ \mathcal{L}\{u(t - a)\,f(t - a)\} \;=\; e^{-as} F(s). $$

This is the rule that handles discontinuous forcing. When you invert and see an $e^{-as}$ factor on a known $F(s)$, the answer is the inverse of $F$, shifted right by $a$, and switched on at $t = a$.

Convolution theorem

The convolution of $f$ and $g$ is

$$ (f * g)(t) \;=\; \int_0^t f(\tau)\, g(t - \tau)\,d\tau. $$

Convolution in time is multiplication in $s$:

$$ \mathcal{L}\{f * g\} \;=\; F(s)\,G(s). $$

Going the other way: if you can recognize $F(s) G(s)$ in your transform, you can write the inverse as a convolution integral. For linear systems this is the formal statement of "output equals input convolved with impulse response."

Don't confuse the two shifts

First shift moves things in $s$ (multiply by exponential in $t$ ⇒ shift in $s$). Second shift moves things in $t$ (multiply by exponential in $s$ ⇒ shift in $t$, gated by a unit step). They look symmetric but apply in opposite situations.

5. The workflow: transform, solve, invert

Every Laplace solution of an initial-value problem follows the same three-step pattern. The transform takes you out of time-land, where calculus is hard, into $s$-land, where the problem is algebra; you solve there, then invert.

TIME DOMAIN (t) ODE in y(t) ay″ + by′ + cy = g(t) y(0), y′(0) given Solution y(t) s-DOMAIN Algebraic equation P(s)·Y(s) = G(s) + IC Solved for Y(s) Y(s) = G(s)/P(s) + … ℒ solve algebraically ℒ⁻¹ classical methods: hard

Step by step:

  1. Transform. Apply $\mathcal{L}$ to both sides. Use linearity, the derivative rule (which folds in $y(0)$ and $y'(0)$), and the table.
  2. Solve. What's left is an algebraic equation in $Y(s)$. Isolate it: $Y(s) = (\text{stuff involving } G(s) \text{ and initial conditions}) \,/\, P(s)$, where $P(s)$ is the characteristic polynomial.
  3. Invert. Decompose $Y(s)$ into pieces that match entries in your table — partial fractions for rationals, plus the shift theorems for any $e^{-as}$ factors. Look each piece up to read off $y(t)$.
The center of gravity is partial fractions

Almost every inverse Laplace boils down to: decompose a rational function of $s$ into simple fractions, then invert each fraction using the table. The art is in the decomposition, not the inversion. Practice partial fractions until they're automatic.

6. A complete worked example

Solve the initial-value problem

$$ y'' + y \;=\; u(t - \pi), \qquad y(0) = 0, \quad y'(0) = 0. $$

Physically: a unit-mass undamped oscillator, at rest, that gets a constant unit force switched on at time $t = \pi$. Classical methods can do it, but they require splitting into cases before and after $t = \pi$ and matching at the seam. Laplace handles the discontinuity without any of that.

Step 1 — Transform both sides

Use $\mathcal{L}\{y''\} = s^2 Y - s\,y(0) - y'(0)$ and $\mathcal{L}\{u(t - \pi)\} = e^{-\pi s}/s$. With zero initial conditions,

$$ s^2 Y(s) + Y(s) \;=\; \frac{e^{-\pi s}}{s}. $$

Step 2 — Solve for $Y(s)$

Factor and isolate:

$$ Y(s) \;=\; \frac{e^{-\pi s}}{s\,(s^2 + 1)}. $$

The job is now to invert this. The $e^{-\pi s}$ is a flag for the second shift theorem — set it aside, and focus on inverting $\dfrac{1}{s(s^2 + 1)}$.

Step 3 — Partial fractions

Decompose:

$$ \frac{1}{s(s^2 + 1)} \;=\; \frac{A}{s} + \frac{Bs + C}{s^2 + 1}. $$

Clearing denominators gives $1 = A(s^2 + 1) + (Bs + C)\,s$. Matching coefficients (constant: $A = 1$; $s^2$: $A + B = 0$, so $B = -1$; $s$: $C = 0$), we get

$$ \frac{1}{s(s^2 + 1)} \;=\; \frac{1}{s} - \frac{s}{s^2 + 1}. $$

Reading the table: this is the transform of $1 - \cos t$.

Step 4 — Apply the second shift

So $Y(s) = e^{-\pi s} \cdot \mathcal{L}\{1 - \cos t\}$. By the second shift theorem, multiplying a transform by $e^{-\pi s}$ corresponds to shifting the time-domain function right by $\pi$ and gating it with $u(t - \pi)$:

$$ y(t) \;=\; u(t - \pi)\,\big[\,1 - \cos(t - \pi)\,\big]. $$

Since $\cos(t - \pi) = -\cos t$, this simplifies beautifully:

$$ \boxed{\;y(t) \;=\; u(t - \pi)\,(1 + \cos t).\;} $$

Step 5 — Sanity check

For $t < \pi$, the step is zero, so $y \equiv 0$ — the oscillator hasn't been kicked yet. Check. At $t = \pi$, $1 + \cos\pi = 0$ and $y$ joins continuously. Check. For $t > \pi$, $y'' = -\cos t$ and $y = 1 + \cos t$, so $y'' + y = -\cos t + 1 + \cos t = 1$, matching the forcing $u(t - \pi) = 1$. Check.

One transform-solve-invert pass; no case-splitting, no matching conditions. That is the Laplace method.

7. Dirac delta and impulse response

The Dirac delta $\delta(t - a)$ is the limit of taller-and-narrower pulses of unit area concentrated at $t = a$. It is not a function in the classical sense — there is no rule of the form "$\delta(t)$ equals such-and-such number at each $t$" — but it behaves consistently inside integrals: for any continuous $f$,

$$ \int_0^\infty \delta(t - a)\, f(t)\,dt \;=\; f(a). $$

Plug that into the definition of the Laplace transform:

$$ \mathcal{L}\{\delta(t - a)\} \;=\; \int_0^\infty e^{-st}\,\delta(t - a)\,dt \;=\; e^{-as}. $$

That is why the transform of an impulse is the single simplest entry in the table.

For a linear system $a y'' + b y' + c y = g(t)$ at rest ($y(0) = y'(0) = 0$), the response to $g(t) = \delta(t)$ is called the impulse response $h(t)$, and its transform is

$$ H(s) \;=\; \mathcal{L}\{h(t)\} \;=\; \frac{1}{a s^2 + b s + c}. $$

By the convolution theorem, the response to any input $g(t)$ is

$$ y(t) \;=\; (h * g)(t) \;=\; \int_0^t h(\tau)\, g(t - \tau)\,d\tau. $$

In other words: one calculation — the impulse response — characterizes the system entirely. Hand it any input, and the output is a convolution away. This is the algebraic heart of linear-systems theory and the reason the Laplace transform underlies control engineering and signal processing.

8. Common pitfalls

Forgetting the initial conditions

The transform of a derivative is $sF(s) - f(0)$, not $sF(s)$. The whole appeal of Laplace is that initial conditions enter automatically — but only if you write the $-f(0)$ and $-f'(0)$ terms. Drop them and you've quietly switched problems.

Confusing the two shift theorems

First shift: $\mathcal{L}\{e^{at} f(t)\} = F(s - a)$ — a shift in $s$. Second shift: $\mathcal{L}\{u(t - a) f(t - a)\} = e^{-as} F(s)$ — a shift in $t$. When you see an $e^{-as}$ factor in $F(s)$, that's the second shift's signature; the inverse will have a $u(t - a)$ and a delayed argument.

Inverting without partial fractions

You almost never invert a rational $Y(s)$ in one move. Decompose into simple fractions whose denominators match table entries ($1/s$, $1/(s - a)$, $1/(s^2 + \omega^2)$, $s/(s^2 + \omega^2)$, etc.) first. Then each piece becomes a table lookup.

Treating $\delta$ as a function

The Dirac delta is a distribution, not a function with pointwise values. Don't try to plot it, don't ask for "$\delta(0)$," and don't differentiate it in the elementary-calculus sense. Use it only inside integrals or as an idealized input — that's where its rules are well defined.

Forgetting where Laplace doesn't apply

Laplace handles linear ODEs with constant coefficients beautifully. For variable coefficients (e.g. $t y'' + y = 0$) the transform usually produces a new differential equation in $s$, not an algebraic one — sometimes useful, often not. And Laplace assumes $f$ is defined for $t \geq 0$ and grows no faster than an exponential; outside that, the integral may not converge.

9. Worked examples

Try each one yourself first. The point is to internalize the workflow until "transform, solve, invert" runs on autopilot.

Example 1 · Solve $y' + 2y = 0$, $y(0) = 1$

Transform. Using $\mathcal{L}\{y'\} = sY - y(0) = sY - 1$:

$$ (sY - 1) + 2Y = 0 $$

Solve. $(s + 2)Y = 1$, so $Y(s) = \dfrac{1}{s + 2}$.

Invert. The table gives $\mathcal{L}^{-1}\{1/(s - a)\} = e^{at}$, here with $a = -2$:

$$ y(t) = e^{-2t}. $$
Example 2 · Solve $y'' + 4y = 0$, $y(0) = 1$, $y'(0) = 0$

Transform. Using $\mathcal{L}\{y''\} = s^2 Y - s\,y(0) - y'(0) = s^2 Y - s$:

$$ (s^2 Y - s) + 4Y = 0 $$

Solve. $(s^2 + 4)Y = s$, so $Y(s) = \dfrac{s}{s^2 + 4}$.

Invert. This matches $\mathcal{L}\{\cos(\omega t)\} = s/(s^2 + \omega^2)$ with $\omega = 2$:

$$ y(t) = \cos(2t). $$
Example 3 · Resonant forcing: $y'' + y = \sin t$, $y(0) = y'(0) = 0$

Transform. $(s^2 + 1) Y = \dfrac{1}{s^2 + 1}$, so $Y(s) = \dfrac{1}{(s^2 + 1)^2}$.

Invert. This isn't a table entry on its own. One route: recognize $\dfrac{1}{(s^2 + 1)^2}$ as $\mathcal{L}\{\sin t\} \cdot \mathcal{L}\{\sin t\}$, so by the convolution theorem $y = \sin t * \sin t$. Working through the convolution integral gives

$$ y(t) = \frac{\sin t - t\cos t}{2}. $$

The $t \cos t$ term is the trademark of resonance: forcing the oscillator at its own natural frequency makes the amplitude grow linearly in time.

Example 4 · Switched-on forcing: $y' + y = u(t - 2)$, $y(0) = 0$

Transform. $sY + Y = \dfrac{e^{-2s}}{s}$, so $Y(s) = \dfrac{e^{-2s}}{s(s + 1)}$.

Partial fractions on $\dfrac{1}{s(s+1)}$.

$$ \frac{1}{s(s+1)} = \frac{1}{s} - \frac{1}{s+1}. $$

So $Y(s) = e^{-2s}\!\left(\dfrac{1}{s} - \dfrac{1}{s+1}\right)$, which is the transform of $1 - e^{-t}$ multiplied by $e^{-2s}$.

Invert with second shift. Shift right by $2$, gate with $u(t - 2)$:

$$ y(t) = u(t - 2)\,\big[1 - e^{-(t - 2)}\big]. $$

Zero before the switch, then a saturating exponential approach to $y = 1$ afterwards.

Example 5 · Impulse response of $y'' + y$

Problem. $y'' + y = \delta(t)$, $y(0) = y'(0) = 0$.

Transform. $(s^2 + 1) Y = 1$, so $Y(s) = \dfrac{1}{s^2 + 1}$.

Invert. $y(t) = h(t) = \sin t$.

So the impulse response of the harmonic oscillator $y'' + y$ is just $\sin t$. By the convolution theorem, the response to any forcing $g(t)$ is $y(t) = \int_0^t \sin(\tau)\, g(t - \tau)\,d\tau$ — one integral encodes the system's reaction to anything.

Sources & further reading

The content above synthesizes the standard ODE-textbook treatment. The primary sources are where to go for proofs of convergence, more exotic transforms, and the connection to complex analysis (the Bromwich inversion integral).

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