Laplace Transform of Derivatives and Integrals

This is the section the whole chapter has been building toward. Transforming a derivative multiplies by $s$ and subtracts the initial value, so a differential equation becomes an algebraic equation in which the initial conditions are already built in. Integration works in reverse, dividing by $s$ .

Quick Reference

Item	Statement
First derivative	\mathscr{L}\left\{f'(t)\right\} = sF(s) - f(0)
Second derivative	\mathscr{L}\left\{f''(t)\right\} = s^{2}F(s) - sf(0) - f'(0)
$n$ th derivative	\mathscr{L}\left\{f^{(n)}(t)\right\} = s^{n}F(s) - s^{n-1}f(0) - s^{n-2}f'(0) - \cdots - f^{(n-1)}(0)
Integral	$ℒ {\int_{0}^{t} f (u) d u} = \frac{1}{s} F (s)$
Hypotheses (first derivative)	$f$ continuous and of exponential order on $t \geq 0$ ; f' piecewise continuous on every finite interval
Memory aid	Each derivative contributes one power of $s$ and strips off one initial value
Solving strategy	Transform, solve for $Y (s)$ , invert

The Laplace Transform of a Derivative

If $ℒ {f (t)} = F (s)$ , then

\mathscr{L}\left\{f'(t)\right\} = sF(s) - f(0) .

The formula holds if $f (t)$ is continuous for all $t \geq 0$ and of exponential order as $t \to \infty$ , and if f'(t) is piecewise continuous on every finite interval in $t \geq 0$ .

Recall that $f$ is of exponential order if there exist constants $K$ , $a$ , and $M$ with $| f (t) | \leq K e^{a t}$ for $t \geq M$ .

Proof

By definition,

\mathscr{L}\left\{f'(t)\right\} = \lim_{R\to\infty}\int_0^{R}e^{-st}f'(t)\,dt .

Apply integration by parts with $u = e^{- s t}$ and dv = f'(t)\,dt, so $d u = - s e^{- s t} d t$ and $v = f (t)$ :

\begin{aligned} \int_0^{R}e^{-st}f'(t)\,dt &= \left.e^{-st}f(t)\right|_0^{R} - \int_0^{R}f(t)\left(-se^{-st}\right)dt \\ &= e^{-sR}f(R) - f(0) + s\int_0^{R}e^{-st}f(t)\,dt . \end{aligned}

Now dispose of the boundary term at $R$ . Since $f$ is of exponential order, there exist $K$ , $a$ , $M$ with $| f (t) | \leq K e^{a t}$ for $t > M$ , so

| e^{- s R} f (R) | = e^{- s R} | f (R) | \leq K e^{- (s - a) R} for R > M .

As $R \to \infty$ , $e^{- (s - a) R} \to 0$ whenever $s > a$ . By the squeeze theorem, $e^{- s R} f (R) \to 0$ .

Therefore

\underbrace{\int_0^{\infty}e^{-st}f'(t)\,dt}_{\mathscr{L}\left\{f'(t)\right\}} = -f(0) + s\underbrace{\int_0^{\infty}e^{-st}f(t)\,dt}_{\mathscr{L}\left\{f(t)\right\}} = sF(s) - f(0) .

Read the formula carefully. The differentiation has become multiplication by $s$ , and the price paid is the extra term $- f (0)$ . That term is not a nuisance: it is exactly the initial condition, delivered into the algebra automatically. This is why the Laplace method never needs a general solution with arbitrary constants.

Higher Derivatives

If f' and f'' satisfy the conditions imposed on $f$ and f', then applying the theorem twice gives

\begin{aligned} \mathscr{L}\left\{f''(t)\right\} &= s\mathscr{L}\left\{f'(t)\right\} - f'(0) \\ &= s\left[s\mathscr{L}\left\{f(t)\right\} - f(0)\right] - f'(0) \\ &= s^{2}\mathscr{L}\left\{f(t)\right\} - sf(0) - f'(0) \\ &= s^{2}F(s) - sf(0) - f'(0) . \end{aligned}

The Laplace transforms of $f (t)$ and its higher derivatives are related by

\mathscr{L}\left\{f''(t)\right\} = s^{2}F(s) - sf(0) - f'(0)

and in general

\mathscr{L}\left\{f^{(n)}(t)\right\} = s^{n}F(s) - s^{n-1}f(0) - s^{n-2}f'(0) - \cdots - f^{(n-1)}(0) ,

where $F (s) = ℒ {f (t)}$ .

The formula holds if f(t), f'(t), \dots, f^{(n-1)}(t) are continuous for all $t \geq 0$ and of exponential order, and if $f^{(n)}$ is piecewise continuous on every finite interval in $t \geq 0$ .

The pattern is easy to remember: powers of $s$ count down from $n - 1$ while the order of the derivative in the initial value counts up from $0$ , and the exponents in each term always sum to $n - 1$ . The last term $f^{(n - 1)} (0)$ carries no $s$ at all.

Given $ℒ {1} = \frac{1}{s}$ , find $ℒ {t^{n}}$ using the derivative theorem.

Solution

Let $f (t) = t^{n}$ . All its derivatives are continuous and of exponential order. Note that

f(0) = f'(0) = \cdots = f^{(n-1)}(0) = 0 ,

since every derivative of order below $n$ still carries a positive power of $t$ , and

f^{(n)} (t) = n!,

a constant.

Substituting into the general formula, all the initial-value terms vanish:

\underset{n! ℒ {1}}{\underset{⏟}{ℒ {f^{(n)} (t)}}} = s^{n} ℒ {t^{n}} - 0 - 0 - \dots - 0

n! \cdot \frac{1}{s} = s^{n} ℒ {t^{n}} .

Therefore

ℒ {t^{n}} = \frac{n!}{s^{n + 1}},

recovering the power rule of Section 7.1 without any integration by parts.

The Laplace Transform of an Integral

Laplace Transform of Integrals. If $ℒ {f (t)} = F (s)$ , then

ℒ {\int_{0}^{t} f (u) d u} = \frac{1}{s} F (s) .

The formula holds if $f (t)$ is piecewise continuous on $[0, \infty)$ and of exponential order.

Proof

Let

g (t) = \int_{0}^{t} f (u) d u .

By the Fundamental Theorem of Calculus, g'(t) = f(t). Applying the derivative theorem to $g$ ,

\mathscr{L}\left\{\underbrace{g'(t)}_{f(t)}\right\} = s\,\mathscr{L}\left\{\underbrace{g(t)}_{\int_0^{t}f(u)\,du}\right\} - \underbrace{g(0)}_{\int_0^{0}f(u)\,du\,=\,0} .

The initial term vanishes because an integral over an interval of zero length is zero. Therefore

ℒ {\int_{0}^{t} f (u) d u} = \frac{1}{s} ℒ {f (t)} = \frac{1}{s} F (s) .

Integration divides by $s$ ; differentiation multiplies by it. Read in reverse, the rule is a useful inversion tool: whenever $F (s)$ has a factor $\frac{1}{s}$ , the inverse transform is an integral of the simpler inverse. For instance, $ℒ^{- 1} {\frac{1}{s (s^{2} + 1)}} = \int_{0}^{t} \sin u d u = 1 - \cos t$ .

Solving Initial Value Problems

The procedure.

Apply $ℒ$ to both sides of the equation, using the derivative formulas. Write $Y (s)$ for $ℒ {y (t)}$ and substitute the given values of $y (0)$ and y'(0).
Solve the resulting algebraic equation for $Y (s)$ .
Invert, usually by partial fractions, to obtain $y (t)$ .

No general solution and no arbitrary constants ever appear.

Solve y' + 3y = e^{-2t} with $y (0) = 1$ .

Solution

Transform. Using \mathscr{L}\left\{y'\right\} = sY(s) - y(0) = sY(s) - 1,

s Y (s) - 1 + 3 Y (s) = \frac{1}{s + 2} .

Solve for $Y$ .

(s + 3) Y (s) = 1 + \frac{1}{s + 2} = \frac{s + 3}{s + 2} ⟹ Y (s) = \frac{1}{s + 2} .

The factor $s + 3$ cancelled outright, which is unusually clean.

Invert.

y (t) = e^{- 2 t} .

Check. y' + 3y = -2e^{-2t} + 3e^{-2t} = e^{-2t} ✓, and $y (0) = 1$ ✓.

Solve y'' + 4y = 0 with $y (0) = 2$ and y'(0) = -1.

Solution

Transform. With \mathscr{L}\left\{y''\right\} = s^{2}Y(s) - sy(0) - y'(0) = s^{2}Y(s) - 2s + 1,

s^{2} Y (s) - 2 s + 1 + 4 Y (s) = 0 .

Solve.

(s^{2} + 4) Y (s) = 2 s - 1 ⟹ Y (s) = \frac{2 s - 1}{s^{2} + 4} = \frac{2 s}{s^{2} + 4} - \frac{1}{2} \cdot \frac{2}{s^{2} + 4} .

The second term was written with a $2$ in the numerator to match the pattern $\frac{k}{s^{2} + k^{2}}$ with $k = 2$ .

Invert.

y (t) = 2 \cos 2 t - \frac{1}{2} \sin 2 t .

Notice that no arbitrary constants ever appeared, and the initial conditions were used at the very first step rather than the last.

Solve y'' - 3y' + 2y = 4 with $y (0) = 0$ and y'(0) = 1.

Solution

Transform. Here \mathscr{L}\left\{y''\right\} = s^{2}Y - 0\cdot s - 1 = s^{2}Y - 1 and \mathscr{L}\left\{y'\right\} = sY - 0 = sY, so

s^{2} Y - 1 - 3 s Y + 2 Y = \frac{4}{s} .

Solve.

(s^{2} - 3 s + 2) Y = 1 + \frac{4}{s} = \frac{s + 4}{s} ⟹ Y (s) = \frac{s + 4}{s (s - 1) (s - 2)},

factoring $s^{2} - 3 s + 2 = (s - 1) (s - 2)$ .

Partial fractions.

\frac{s + 4}{s (s - 1) (s - 2)} = \frac{A}{s} + \frac{B}{s - 1} + \frac{C}{s - 2} .

Multiplying through: $s + 4 = A (s - 1) (s - 2) + B s (s - 2) + C s (s - 1)$ .

Setting $s = 0$ : $4 = 2 A$ , so $A = 2$ .
Setting $s = 1$ : $5 = - B$ , so $B = - 5$ .
Setting $s = 2$ : $6 = 2 C$ , so $C = 3$ .

Invert.

y (t) = 2 - 5 e^{t} + 3 e^{2 t} .

Check the initial conditions. $y (0) = 2 - 5 + 3 = 0$ ✓. And y'(t) = -5e^{t} + 6e^{2t}, so y'(0) = -5 + 6 = 1 ✓.

Note the structure of the answer: the constant $2$ is the particular solution, and $- 5 e^{t} + 3 e^{2 t}$ is the homogeneous part with its coefficients already determined. The Laplace method produced both at once.

Exercises

Solve y' - 2y = 0 with $y (0) = 5$ .

Solution

Transforming, $s Y - 5 - 2 Y = 0$ , so $Y (s) = \frac{5}{s - 2}$ and $y (t) = 5 e^{2 t}$ .

Solve y'' + y = \sin t with $y (0) = 0$ and y'(0) = 0.

Solution

Transforming, $s^{2} Y + Y = \frac{1}{s^{2} + 1}$ , so

Y (s) = \frac{1}{{(s^{2} + 1)}^{2}} .

From Section 7.4, $ℒ {t \sin k t} = \frac{2 k s}{{(s^{2} + k^{2})}^{2}}$ and $ℒ {t \cos k t} = \frac{s^{2} - k^{2}}{{(s^{2} + k^{2})}^{2}}$ . With $k = 1$ , subtracting the second from $\frac{1}{s^{2} + 1} = \frac{s^{2} + 1}{{(s^{2} + 1)}^{2}}$ gives

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

Therefore $\frac{1}{{(s^{2} + 1)}^{2}} = \frac{1}{2} [ℒ {\sin t} - ℒ {t \cos t}]$ , so

y (t) = \frac{1}{2} (\sin t - t \cos t) .

The unbounded factor $t$ signals resonance, since the forcing frequency matches the natural frequency, exactly as the Modification Rule predicted in Section 4.3.

Use the integral theorem to find $ℒ^{- 1} {\frac{1}{s (s - 3)}}$ .

Solution

Since $ℒ^{- 1} {\frac{1}{s - 3}} = e^{3 t}$ , the extra factor of $\frac{1}{s}$ corresponds to integrating:

ℒ^{- 1} {\frac{1}{s (s - 3)}} = \int_{0}^{t} e^{3 u} d u = \frac{e^{3 t} - 1}{3} .

Partial fractions gives the same answer: $\frac{1}{s (s - 3)} = \frac{1}{3} (\frac{1}{s - 3} - \frac{1}{s})$ .

Solve the system x' = y, y' = -x with $x (0) = 1$ , $y (0) = 0$ .

Solution

Transforming both equations,

s X - 1 = Y, s Y - 0 = - X .

Substituting the first into the second, $s (s X - 1) = - X$ , so $(s^{2} + 1) X = s$ and

X (s) = \frac{s}{s^{2} + 1} ⟹ x (t) = \cos t .

Then $Y = s X - 1 = \frac{s^{2}}{s^{2} + 1} - 1 = \frac{- 1}{s^{2} + 1}$ , so $y (t) = - \sin t$ .

Systems transform into simultaneous algebraic equations, solved by ordinary elimination.

Frequently Asked Questions

Why is \mathscr{L}\left\{f'(t)\right\} = sF(s) - f(0)?

It comes from integrating by parts: the boundary term at $t = 0$ contributes $- f (0)$ , the boundary term at infinity vanishes by exponential order, and what remains is $s$ times the original integral. The formula is the engine of the entire method.

Where do the initial conditions enter?

At the transform step. Each derivative formula contains the initial values explicitly, so substituting the given numbers puts them into the algebra immediately. This is a genuine advantage over the methods of Chapter 4, where a general solution is constructed first and the constants fitted afterward.

What if my initial conditions are given at

t = 3

instead of

t = 0

Shift the variable. Set $τ = t - 3$ and rewrite the problem in $τ$ , so the conditions are at $τ = 0$ . The unilateral Laplace transform only ever sees $t \geq 0$ , so this preprocessing is required.

Why do I need

f

to be continuous, not just piecewise continuous?

Because a jump in $f$ would make the integration by parts pick up extra boundary terms at each discontinuity, so the tidy formula would acquire correction terms. Note that f' is only required to be piecewise continuous, so corners in the graph of $f$ are perfectly acceptable.

How does this compare with the methods of Chapter 4?

Chapter 4 finds a general solution with arbitrary constants and fits them at the end. The Laplace method never forms a general solution: the answer emerges already satisfying the conditions. For a homogeneous equation with no data the older method is quicker; for a forced problem with given initial values, especially a discontinuous one, Laplace usually wins.

What does

\frac{1}{s} F (s)

correspond to?

Integrating $f$ from $0$ to $t$ . Multiplying a transform by $\frac{1}{s}$ is the $s$ -domain shadow of antidifferentiation, just as multiplying by $s$ shadows differentiation. This makes a stray factor of $\frac{1}{s}$ easy to handle during inversion.

Laplace Transform of Derivatives and Integrals

Quick Reference

The Laplace Transform of a Derivative

Higher Derivatives

The Laplace Transform of an Integral

Solving Initial Value Problems

Exercises

Frequently Asked Questions

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