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Differential Equations Laplace Transforms and Operational Methods Flashcards
50 question-and-answer cards covering Laplace Transforms and Operational Methods as it is examined in Differential Equations. 24 of them are printed below, taken from across the deck — no signup, no paywall on the preview.
24 sample cards from the Laplace Transforms and Operational Methods deck
Sampled from the end of the deck, so these are different cards from the ones shown on the syllabus page.
State the inverse form of the Second Translation Theorem.
$$\mathcal{L}^{-1}\{e^{-as}F(s)\} = f(t-a)\,\mathcal{U}(t-a).$$
State the theorem for Derivatives of a Transform (multiplication by $t^{n}$).
$$\mathcal{L}\{t^{n} f(t)\} = (-1)^{n} \frac{d^{n}}{ds^{n}} F(s), \quad n = 1, 2, 3, \ldots$$ In particular $\mathcal{L}\{t\, f(t)\} = -F'(s)$.
Using derivatives of transforms, compute $\mathcal{L}\{t \sin kt\}$.
$$\mathcal{L}\{t \sin kt\} = -\frac{d}{ds}\left(\frac{k}{s^{2}+k^{2}}\right) = \frac{2ks}{(s^{2}+k^{2})^{2}}.$$
What is the general procedure (the three steps) for solving an initial value problem using the Laplace transform?
(1) Take the Laplace transform of both sides of the ODE, using the derivative rules to insert initial conditions; (2) Solve algebraically for $Y(s) = \mathcal{L}\{y(t)\}$; (3) Take the inverse Laplace transform (often via partial fractions) to recover $y(t)$.
What is a key advantage of the Laplace method for solving IVPs compared with the classical (undetermined coefficients / variation of parameters) approach?
The initial conditions are incorporated automatically at the transform stage, so there is no need to solve for arbitrary constants afterward; it also handles discontinuous or impulsive forcing functions directly.
For the IVP $y'' + \omega^{2} y = 0$ with $y(0)=y_0$, $y'(0)=y_1$, what is $Y(s)$?
Transforming: $(s^{2}+\omega^{2})Y(s) - s y_0 - y_1 = 0$, so $$Y(s) = \frac{s\, y_0 + y_1}{s^{2}+\omega^{2}}.$$
How do you express a function that 'switches on' a term $g(t)$ at time $t=a$ using the Heaviside function?
Multiply by the unit step: $g(t)\,\mathcal{U}(t-a)$ equals $0$ for $t<a$ and $g(t)$ for $t \geq a$. A piecewise function can be assembled as a sum of such step-multiplied terms.
How is a two-piece function $f(t) = \begin{cases} g(t), & 0 \le t < a \\ h(t), & t \ge a \end{cases}$ written in terms of $\mathcal{U}(t-a)$?
$$f(t) = g(t) + [\,h(t) - g(t)\,]\,\mathcal{U}(t-a).$$
What is the definition of a periodic function $f(t)$ with period $T$, and why is periodicity useful for Laplace transforms?
$f(t+T) = f(t)$ for all $t \geq 0$. Periodicity lets us compute the transform from a single period only, via a closed-form formula, rather than integrating over $[0,\infty)$ directly.
State the formula for the Laplace transform of a periodic function of period $T$.
$$\mathcal{L}\{f(t)\} = \frac{1}{1 - e^{-sT}} \int_{0}^{T} e^{-st} f(t)\, dt.$$
What are the two defining (informal) properties of the Dirac delta function $\delta(t-a)$?
(1) $\delta(t-a) = 0$ for $t \neq a$ (and is 'infinite' at $t=a$), and (2) $\int_{-\infty}^{\infty} \delta(t-a)\, dt = 1$. It models an instantaneous unit impulse concentrated at $t=a$.
State the sifting property of the Dirac delta function.
$$\int_{-\infty}^{\infty} f(t)\,\delta(t-a)\, dt = f(a).$$ The delta 'sifts out' the value of $f$ at $t=a$.
What is the Laplace transform of the Dirac delta $\delta(t-a)$ for $a \geq 0$?
$$\mathcal{L}\{\delta(t-a)\} = e^{-as}.$$ In particular $\mathcal{L}\{\delta(t)\} = 1$.
What is the relationship between the Dirac delta function and the Heaviside unit step function?
The delta function is the derivative of the unit step: $$\frac{d}{dt}\,\mathcal{U}(t-a) = \delta(t-a),$$ and equivalently $\mathcal{U}(t-a) = \int_{-\infty}^{t} \delta(\tau - a)\, d\tau$.
How can $\delta(t-a)$ be viewed as a limit of ordinary functions?
As the limit of a unit-area pulse of shrinking width, e.g. $\delta_a(t) = \frac{1}{2\epsilon}$ on $[a-\epsilon, a+\epsilon]$ and $0$ elsewhere, taking $\epsilon \to 0^{+}$. The pulse keeps area $1$ while becoming infinitely tall and narrow.
Define the convolution $(f * g)(t)$ of two functions defined for $t \geq 0$.
$$(f * g)(t) = \int_{0}^{t} f(\tau)\, g(t-\tau)\, d\tau.$$
State the Convolution Theorem for Laplace transforms.
If $\mathcal{L}\{f\} = F(s)$ and $\mathcal{L}\{g\} = G(s)$, then $$\mathcal{L}\{f * g\} = F(s)\, G(s).$$ The transform of a convolution is the product of the transforms.
State the inverse form of the Convolution Theorem.
$$\mathcal{L}^{-1}\{F(s)\,G(s)\} = (f * g)(t) = \int_{0}^{t} f(\tau)\, g(t-\tau)\, d\tau.$$ Useful for inverting a product of transforms whose individual inverses are known.
List the key algebraic properties of convolution.
Convolution is commutative $f*g = g*f$, distributive $f*(g+h) = f*g + f*h$, associative $(f*g)*h = f*(g*h)$, and $f*0 = 0$. Note: in general $f*1 \neq f$.
Using the convolution theorem, find $\mathcal{L}^{-1}\left\{\frac{1}{s(s^{2}+1)}\right\}$.
Write it as $\frac{1}{s}\cdot\frac{1}{s^{2}+1}$, with inverses $1$ and $\sin t$. Then $$\mathcal{L}^{-1} = \int_{0}^{t} \sin\tau\, d\tau = 1 - \cos t.$$
What is the Laplace transform of the integral of a convolution / the special convolution $\mathcal{L}\left\{\int_0^t f(\tau)d\tau\right\}$ viewed as $f * 1$?
Since $\int_0^t f(\tau)\,d\tau = (f * 1)(t)$, by the convolution theorem $$\mathcal{L}\{f*1\} = F(s)\cdot\frac{1}{s} = \frac{F(s)}{s},$$ consistent with the transform-of-an-integral rule.
What type of equation can the convolution theorem be used to solve, and what form does it take?
A Volterra integral equation of the form $$f(t) = g(t) + \int_{0}^{t} f(\tau)\, h(t-\tau)\, d\tau.$$ Transforming gives $F(s) = G(s) + F(s)H(s)$, so $F(s) = \frac{G(s)}{1 - H(s)}$, then invert.
What is the transfer function and impulse response of a linear system solved by $\mathcal{L}\{y\} = H(s)\,\mathcal{L}\{\text{input}\}$?
$H(s)$ is the transfer function (transform of the response to a unit impulse with zero initial conditions); its inverse $h(t) = \mathcal{L}^{-1}\{H(s)\}$ is the impulse response (weight function). The zero-state output is the convolution $y(t) = (h * g)(t)$ of the impulse response with the input $g$.
Compare the Laplace transform method with Fourier methods for solving differential equations: what distinguishes their typical use?
The Laplace transform handles initial value problems on $t \geq 0$ (transient/causal signals, including exponentially growing ones) and naturally incorporates initial conditions and discontinuous/impulsive forcing. Fourier methods are suited to boundary value problems and steady-state analysis of functions defined on the whole line that are absolutely integrable or periodic, encoding frequency content rather than initial data.
What this deck covers
The Laplace Transforms and Operational Methods deck follows the Differential Equations Laplace Transforms and Operational Methods syllabus — 4 chapters and 12 topics — so questions land on material that is genuinely examinable rather than trivia around it. That works out to roughly 12.5 cards per chapter.
Answers are written to be recallable, not just readable — averaging about 167 characters, which is long enough to carry the reasoning and short enough to say out loud.
A deck like this earns its keep on the second and third pass. Read the syllabus first so you know the shape of the subject, then use the cards to find the specific facts that have not stuck.
Laplace Transforms and Operational Methods flashcards FAQ
How many Laplace Transforms and Operational Methods flashcards are in this Differential Equations deck?
50 cards. This page previews 24 of them, sampled evenly across the deck so you can judge the difficulty before installing anything.
Are these Differential Equations flashcards free?
Yes. The preview here is free to read with no signup, and the full 50-card deck is free inside the Examius app.
What do the Laplace Transforms and Operational Methods cards cover?
They follow the Differential Equations Laplace Transforms and Operational Methods syllabus — 4 chapters and 12 topics — so the questions track what is actually examinable.
How should I use these flashcards?
Read the syllabus first so you know the shape of the subject, then drill the deck. Examius schedules each card with spaced repetition, so cards you keep missing come back sooner and ones you know drift further apart.