Laplace of derivatives — key property for solving ODEs: Question bank

Before we start, a reminder of the only three formulas everything below leans on:

Here $F (s) = L {f} = \int_{0}^{\infty} e^{- s t} f (t) d t$ , so $F (s)$ means "the transform of the original function $f$ ", and $f (0)$ means "the value of $f$ at the starting time $t = 0$ ".

True or false — justify

The property is

L {f^{'}} = s F (s)

with no extra term.

False. It is

s F (s) - f (0)

; the

- f (0)

comes from evaluating the boundary term of integration by parts at the lower limit

t = 0

, and it is exactly what injects the initial condition.

L {f^{''}}

can be written using only

F (s)

and

f (0)

False. A second derivative needs two starting facts, so both

f (0)

and

f^{'} (0)

appear:

s^{2} F (s) - s f (0) - f^{'} (0)

f (0) = 0

, then

L {f^{'}}

really is just

s F (s)

True. The fee

- f (0)

becomes

- 0

, so the boundary term vanishes and only

s F (s)

survives — but this is a special case, not the general rule.

The Fourier transform of a derivative has no boundary term because it integrates over all of

R

, not from

0

True. With no lower endpoint at

t = 0

(and decay at both infinities) there is no surviving boundary value, which is exactly why the habit of dropping

- f (0)

is dangerous when you switch to Fourier and back.

L {f^{''}} = s^{2} F (s) - f (0) - s f^{'} (0)

is a valid way to write the second-derivative rule.

False — the powers are swapped. The rule pairs the higher power of

s

with the lower-order derivative:

s^{1}

multiplies

f (0)

s^{0}

multiplies

f^{'} (0)

, giving

s^{2} F (s) - s f (0) - f^{'} (0)

The

- f (0)

term makes Laplace worse than other methods because it adds an extra piece to track.

False. That term is the whole advantage: it carries the initial condition into the algebra automatically, so you never have to fit an arbitrary constant at the end like in classical IVP solving.

The derivative rule works for any function whatsoever.

False. It needs

f

to be differentiable and of exponential order so that

e^{- s t} f (t) \to 0

t \to \infty

; otherwise the boundary term at infinity does not vanish and the derivation breaks (see Exponential Order and Convergence of Integrals).

Because

L {f^{'}}

mixes

F (s)

and a constant, the transform is nonlinear.

False. It is still perfectly linear;

L {a f + b g} = a F + b G

always (see Linearity of the Laplace Transform). The

- f (0)

is a constant tied to

f

, not a nonlinear operation on it.

L {f^{(n)}}

the coefficient of

f^{(k)} (0)

- s^{k}

False. It is

- s^{n - 1 - k}

: as the derivative order

k

goes up, the power of

s

comes down, from

s^{n - 1}

(on

f (0)

) to

s^{0}

(on

f^{(n - 1)} (0)

Spot the error

L {y^{'}} + 3 L {y} = 0

gives

s Y + 3 Y = 0

, so

Y = 0

The step dropped

- y (0)

. With the initial value

y (0) = 2

it is

(s Y - 2) + 3 Y = 0

, giving

Y = \frac{2}{s + 3}

and

y = 2 e^{- 3 t}

— not the trivial zero solution.

"For

y^{''} - y = 0

I wrote

L {y^{''}} = s^{2} Y - y^{'} (0) - s y (0)

The two initial-value terms are in the wrong slots. Correct is

s^{2} Y - s y (0) - y^{'} (0)

; matching

s^{1}

with

y (0)

and

s^{0}

with

y^{'} (0)

"The problem gives

y (1) = 5

, so I set

f (0) = 5

in the formula."

You cannot: the derivative rule evaluates the boundary strictly at

t = 0

, so

f (0)

must be the value at

t = 0

. With an IC at

t = 1

you solve generally and fit, or shift the time variable so the condition sits at

0

F (s)

in the rule means the transform of

f^{'}

, since that's the function we're transforming."

No.

F (s) = L {f}

is the transform of the original undifferentiated

f

. The whole point is to express

L {f^{'}}

in terms of

F (s)

, the thing you already know.

L {f^{'''}} = s^{3} F - s^{2} f (0) - s f^{'} (0)

— three terms for third order."

One term is missing. A third derivative needs three initial values:

s^{3} F - s^{2} f (0) - s f^{'} (0) - f^{''} (0)

. The count of subtracted terms always equals the derivative order

n

"Since integration by parts moves the derivative onto

e^{- s t}

, and

\frac{d}{d t} e^{- s t} = - s e^{- s t}

, we get a

- s

, so the rule is

- s F (s)

The sign flips again. The by-parts formula has

- \int v d u = - \int f \cdot (- s e^{- s t}) d t = + s \int e^{- s t} f d t = + s F (s)

. Two minus signs cancel to give

+ s F (s)

; see Integration by Parts.

Why questions

Why does the boundary term at

t \to \infty

vanish?

Because

f

is of exponential order,

e^{- s t}

(for

s

large enough) crushes

f (t)

to zero faster than

f

can grow, so

e^{- s t} f (t) \to 0

. This decay is the "secret ingredient" that makes the integral converge at all.

Why does differentiation turn into multiplication by

s

Integration by parts hands the derivative from

f

to the factor

e^{- s t}

, and

\frac{d}{d t} e^{- s t} = - s e^{- s t}

pulls out exactly one factor of

s

— so each derivative removed costs one power of

s

gained.

Why is the "

- f (0)

" term convenient rather than annoying?

It plants the initial condition directly inside the algebraic equation, so solving the algebra already respects

y (0)

; no separate constant-fitting step is needed after inverting.

Why does the second-derivative rule need both

f (0)

and

f^{'} (0)

Applying the first rule twice generates two boundary terms — one from transforming

f^{''} = (f^{'})^{'}

(gives

- f^{'} (0)

) and one from transforming

f^{'}

inside it (gives

- s f (0)

). A second-order problem physically requires two starting facts anyway.

Why must the ODE's initial data live at

t = 0

specifically for this method to plug in directly?

The derivation evaluated the by-parts boundary term at the lower limit of the integral, which is

t = 0

. The formula literally contains "

f

evaluated at

0

", so only

t = 0

data drops straight in.

Why can we replace the leftover integral

\int_{0}^{\infty} e^{- s t} f (t) d t

F (s)

at the end of the derivation?

Because that leftover integral is, letter for letter, the definition of

F (s) = L {f}

. Recognising it is what closes the derivation into a formula (see Laplace Transform — Definition and Existence).

Edge cases

f (t) = c

(a constant), what is

L {f^{'}}

and does the rule agree?

f^{'} = 0

L {f^{'}} = 0

. The rule gives

s F (s) - f (0) = s \cdot \frac{c}{s} - c = c - c = 0

— consistent.

What if

f

has a jump discontinuity at

t = 0

, so

f (0^{-}) \neq = f (0^{+})

The rule uses the value the integral "sees", i.e.

f (0^{+})

(the limit from the right), since the transform integrates over

t \geq 0

. Using

f (0^{-})

would give a wrong boundary term.

Does the property still make sense if

f^{'} (0)

is required but

f

is only once differentiable?

No — to use

L {f^{''}}

you need

f^{''}

to exist and

f, f^{'}

to be of exponential order. If those fail, the derivation's boundary/decay assumptions collapse and the formula does not apply.

What happens to

L {f^{(n)}}

when all initial values

f (0), \dots, f^{(n - 1)} (0)

are zero?

Every subtracted term drops out and you get the clean

s^{n} F (s)

. This "rest" or zero-state case is exactly when Laplace looks like pure multiplication — but it is the exception, not the rule.

s

is chosen too small (below the exponential-order threshold), is the derivative rule still valid?

No. For small

s

the integral defining

F (s)

(and the boundary term at infinity) may not converge, so

F (s)

isn't defined there and the rule has no meaning; it holds only in the region of convergence.

Connections

Laplace Transform — Definition and Existence
Inverse Laplace Transform and Partial Fractions
Solving Initial Value Problems with Laplace
Integration by Parts
Exponential Order and Convergence of Integrals
Fourier Transform — comparison (no boundary term)
Linearity of the Laplace Transform