Solving ODEs with Laplace (including discontinuous forcing): Question bank

True or false — justify

True or false:

L {y^{''}} = s^{2} Y

for any twice-differentiable

y

False. The correct rule is

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

; dropping the boundary terms silently sets the initial conditions to zero, which is only right when

y (0) = y^{'} (0) = 0

True or false: the Laplace transform of a sum equals the sum of the transforms.

True. The transform is an integral, and integration is linear, so

L {a f + b g} = a F (s) + b G (s)

— this is exactly why we can transform an ODE term-by-term. (

F, G

are the transforms of

f, g

True or false:

L {f (t) g (t)} = F (s) G (s)

False. The transform of a product in time is not the product of transforms; it is

F (s) G (s)

that inverts to a convolution

\int_{0}^{t} f (τ) g (t - τ) d τ

(see Transfer Functions and Convolution).

True or false: the factor

e^{- a s}

multiplying

G (s)

means "multiply the answer by

e^{- a}

False.

e^{- a s}

is a time-delay operator, not a numerical scale: it inverts to

u (t - a) g (t - a)

, shifting

g

later by

a

and gating it off before

t = a

True or false:

L {e^{a t}} = \frac{1}{s - a}

holds for every real

s

False. The defining integral

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

only converges when

s > a

(region of convergence

Re (s) > a

); for

s \leq a

the integrand does not decay and the transform is undefined.

True or false: the pole of

F (s)

farthest to the right can lie strictly inside the region of convergence.

False. Poles are exactly the points where the integral blows up, so they sit on or to the left of the convergence boundary line

Re (s) = c

; the rightmost pole defines that boundary.

True or false:

L {t f (t)} = t F (s)

False. Multiplying by

t

in time becomes differentiating in

s

L {t f (t)} = - \frac{d F}{d s}

(the moment-generating property). E.g. from

L {e^{- t}} = \frac{1}{s + 1}

we get

L {t e^{- t}} = - \frac{d}{d s} \frac{1}{s + 1} = \frac{1}{( s + 1 ) ^{2}}

True or false:

L {f (2 t)} = F (2 s)

False. Scaling time by

a > 0

scales and shrinks the transform:

L {f (a t)} = \frac{1}{a} F (\frac{s}{a})

. Squeezing a signal in time stretches and lowers it in

s

— you cannot just substitute

2 s

True or false: because

δ (t - a)

is "infinitely tall," its Laplace transform must be infinite.

False.

L {δ (t - a)} = e^{- a s}

, a perfectly finite factor — the delta is defined by the sifting property

\int_{0}^{\infty} e^{- s t} δ (t - a) d t = e^{- a s}

, not by a literal height.

True or false: the Laplace method requires you to guess a form for the particular solution.

False. Unlike Method of Undetermined Coefficients, no guessing is needed: the algebra for

Y (s)

produces the whole answer automatically, forced part and homogeneous part together.

True or false: partial fractions is optional — you could always invert

Y (s)

directly.

Mostly false in practice. Tables only list simple pieces like

\frac{1}{s - a}

and

\frac{ω}{s ^{2} + ω ^{2}}

; partial fractions is the tool that ==breaks a complicated

Y (s)

into those recognisable pieces== (see Partial Fractions).

True or false: Laplace only works for equations with constant coefficients.

Essentially true for the clean algebra we use here. Variable coefficients like

t y^{'}

transform into derivatives in $s$ (via

L {t f} = - F^{'} (s)

), turning the algebraic problem back into a differential one — so Linear Constant-Coefficient ODEs are the natural home.

Spot the error

A student writes

L^{- 1} {e^{- 2 s} / (s + 1)} = e^{- (t - 2)}

. What is missing?

The gate. The correct inverse is

u (t - 2) e^{- (t - 2)}

; without the step

u (t - 2)

the solution would be nonzero (and even large) before the switch at

t = 2

, which is physically impossible.

For

y^{'} + y = u (t - 2), y (0) = 0

, a student transforms the left side as

s Y + Y - y (0) = s Y + Y

but then writes

Y = \frac{e ^{- 2 s}}{s + 1}

. Where did it go wrong?

They forgot that

L {u (t - 2)} = \frac{e ^{- 2 s}}{s}

has a

\frac{1}{s}

; dividing by

(s + 1)

gives

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

, not

\frac{e ^{- 2 s}}{s + 1}

To find

L {t u (t - 2)}

a student writes

e^{- 2 s} \cdot \frac{1}{s ^{2}}

. Why is this wrong?

The second shift needs

g (t - a)

, but here

g (t) = t \neq = (t - 2)

. Rewrite

t = (t - 2) + 2

, giving

L {t u (t - 2)} = e^{- 2 s} (\frac{1}{s ^{2}} + \frac{2}{s})

A student inverts

\frac{1}{( s + 1 ) ^{2}}

e^{- t}

, "the same as

\frac{1}{s + 1}

just squared." Fix it.

A repeated pole carries an extra power of

t

\frac{1}{( s + 1 ) ^{2}} \to t e^{- t}

. This is exactly the moment-generating property

L {t f} = - F^{'} (s)

read backwards, since

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

A student computes the response of

H (s) = \frac{1}{s + 1}

to input

f

by writing

y (t) = h (t) f (t)

in time. Spot the error.

Multiplying the transfer function by

F (s)

s

does not correspond to a product in time; it corresponds to the convolution

y (t) = \int_{0}^{t} h (τ) f (t - τ) d τ

(see Transfer Functions and Convolution).

A student inverts

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

\frac{1}{s} - \frac{1}{s + 1}

but then writes

h (t) = 1 - 1 = 0

. What went wrong?

They evaluated the transforms instead of inverting each term.

\frac{1}{s} \to 1

and

\frac{1}{s + 1} \to e^{- t}

, so

h (t) = 1 - e^{- t}

, which is

0

only at

t = 0

A student solving

y^{''} + y = 0, y (0) = 0, y^{'} (0) = 2

gets

Y = \frac{2}{s ^{2} + 1}

then writes

y = 2 cos t

. Fix it.

They matched the wrong pair.

\frac{ω}{s ^{2} + ω ^{2}} \to sin ω t

, so

\frac{2}{s ^{2} + 1} \to 2 sin t

; the cosine pair has an

s

in the numerator.

A student claims the boundary term

[e^{- s t} f (t)]_{0}^{\infty}

in the derivative rule is always

- f (0)

. When can this fail?

When

f

grows faster than

e^{s t}

decays, the upper limit does not vanish and the transform (and hence the rule) does not exist — this is the region-of-convergence/existence condition in Laplace Transform — Definition and Existence.

Why questions

Why does differentiation in

t

become multiplication by

s

Integration by parts moves the derivative off

f

and onto

e^{- s t}

, and

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

pulls out a factor

s

— that single fact is the engine of the whole method.

Why does multiplying by $t$ become differentiating in $s$ (the moment property)?

Differentiating

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

under the integral brings down a factor

- t

F^{'} (s) = \int_{0}^{\infty} (- t) e^{- s t} f d t = - L {t f}

, hence

L {t f} = - F^{'} (s)

Why does a time shift

a

produce an exponential factor

e^{- a s}

rather than, say, a polynomial?

Substituting

τ = t - a

in the integral pulls

e^{- s (τ + a)} = e^{- s a} e^{- s τ}

apart; the constant

e^{- s a}

factors out of the integral, so shifting always converts to this exponential stamp.

Why does scaling time by

a

produce the factor

\frac{1}{a}

L {f (a t)} = \frac{1}{a} F (s / a)

Substituting

τ = a t

replaces

d t

\frac{d τ}{a}

(the

\frac{1}{a}

) and turns

e^{- s t}

into

e^{- (s / a) τ}

(the

s / a

) — the Jacobian of the stretch is what rescales the transform.

Why does Laplace handle impulses (

δ

) so easily while classical methods struggle?

The impulse just adds a finite factor

e^{- a s}

to the algebra, whereas classical methods must patch the solution across the impulse by matching jump conditions in slope by hand.

Why do the initial conditions appear inside the algebra rather than being applied at the end?

The boundary term of the derivative rule injects

- f (0)

(and

- f^{'} (0)

) directly into the transformed equation, so ICs are baked into

Y (s)

from the start instead of fixing arbitrary constants afterward.

Why do we need partial fractions before reading off the answer?

The inversion table only knows a short list of standard shapes; partial fractions rewrites a messy rational

Y (s)

as a sum of exactly those shapes so each piece can be inverted term-by-term.

Why can we split

Y (s) = e^{- a s} H (s)

and invert

H

first, then delay?

Because

e^{- a s}

is a pure delay operator: whatever

H (s)

inverts to, the full inverse is that same function shifted and gated by

u (t - a)

— so you never have to invert the exponential itself.

Edge cases

What is

L {u (t - a) g (t - a)}

when

a = 0

It reduces to

e^{0} G (s) = G (s)

; with no delay the step

u (t) = 1

for

t \geq 0

and the gate does nothing, recovering the ordinary transform.

For

y^{'} + y = u (t - 2)

, what is

y

exactly at and just before

t = 2

For

t < 2

the forcing is off and

y \equiv 0

; at

t = 2

the solution starts from

0

and begins relaxing toward

1

1 - e^{- (t - 2)}

, so it is continuous with value

0

at the switch.

What long-term value does

y (t) = u (t - 2) (1 - e^{- (t - 2)})

approach, and why?

t \to \infty

e^{- (t - 2)} \to 0

, so

y \to 1

: the system settles at the constant input level, the steady state where

y^{'} = 0

balances

y = u = 1

y^{''} + 3 y^{'} + 2 y = u (t - 1)

with

H (s) = \frac{1}{s ( s + 1 ) ( s + 2 )}

, why does the residue at

s = 0

give the final value

\frac{1}{2}

The

\frac{1/2}{s}

term inverts to the constant

\frac{1}{2}

, and since the exponential terms

e^{- t}, e^{- 2 t}

decay,

\frac{1}{2}

is the steady state — consistent with

2 y = 1 \Rightarrow y = \frac{1}{2}

What happens to the method if the forcing turns on before

t = 0

, i.e.

a < 0

The one-sided transform only "sees"

t \geq 0

, so a switch at negative

a

is already fully on at

t = 0

; you must fold its effect into the initial conditions rather than into a

u (t - a)

term.

If a system's characteristic polynomial (denominator of

H

) has a repeated root, does the table still apply directly?

Not directly. A repeated factor like

\frac{1}{( s + 1 ) ^{2}}

inverts to

t e^{- t}

(moment property), so partial fractions must include the

\frac{1}{( s + 1 ) ^{2}}

term whose inverse carries an extra power of

t

Where in the

s

-plane must

Re (s)

lie for a decaying

y (t)

made of

e^{- t}

and

e^{- 2 t}

pieces?

The transform converges for

Re (s) > - 1

(the rightmost pole is at

s = - 1

); both poles sit in the left half-plane, which is exactly why those time-pieces decay rather than grow.

Recall One-line self-test

Can you state, without notes, both what $e^{- a s}$ does to a transform and the two things its inverse must include? Answer ::: $e^{- a s}$ delays a signal by $a$ in time; its inverse must include the shift $g (t - a)$ and the gate $u (t - a)$ .