Differential Equations Quiz: Convolution Theorem

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Convolution TheoremQuestion 1 of 18

Given that $F(s) = \mathcal{L}\{f(t)\}$ , which of the following expressions is equivalent to the Laplace transform of $\int_0^t e^{-a(t-\tau)} f(\tau) d\tau$ ?

\frac{F(s)}{s-a}

F(s+a)

\frac{F(s)}{s+a}

F(s-a)

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Differential Equations Quiz

Differential Equations Quiz: Convolution Theorem

Practice Convolution Theorem in Differential Equations with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Convolution Theorem, giving you a quick way to practice the rules, question types, and explanations that matter most for Differential Equations.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

Given that $F(s) = \mathcal{L}\{f(t)\}$ , which of the following expressions is equivalent to the Laplace transform of $\int_0^t e^{-a(t-\tau)} f(\tau) d\tau$ ?

$\frac{F(s)}{s-a}$
$F(s+a)$
$\frac{F(s)}{s+a}$ (correct answer)
$F(s-a)$

Explanation: When you encounter an integral like

\int_0^t e^{-a(t-\tau)} f(\tau) d\tau

, you're looking at a convolution. This is a fundamental pattern in Laplace transforms where two functions are "mixed" together through integration. The key insight is recognizing this as the convolution of

e^{-at}

and

f(t)

. The convolution theorem states that the Laplace transform of a convolution equals the product of the individual Laplace transforms:

\mathcal{L}\{g * h\} = G(s) \cdot H(s)

Here, we need the Laplace transforms of our two functions:

$\mathcal{L}\{f(t)\} = F(s)$ (given)
$\mathcal{L}\{e^{-at}\} = \frac{1}{s+a}$ (standard transform)

Therefore:

\mathcal{L}\left\{\int_0^t e^{-a(t-\tau)} f(\tau) d\tau\right\} = F(s) \cdot \frac{1}{s+a} = \frac{F(s)}{s+a}

This confirms answer C is correct. A is wrong because

\frac{F(s)}{s-a}

would correspond to the convolution with

e^{at}

, not

e^{-at}

. B is incorrect because

F(s+a)

represents the transform of

e^{-at}f(t)

(frequency shifting), not a convolution. D is wrong because

F(s-a)

represents the transform of

e^{at}f(t)

(frequency shifting in the opposite direction). Study tip: Whenever you see an integral from 0 to

t

with the integrand containing

(t-\tau)

, immediately think "convolution" and apply the convolution theorem. Master the basic transforms like

\mathcal{L}\{e^{±at}\} = \frac{1}{s∓a}

- they appear constantly in convolution problems.

Question 2

If $F(s) = \frac{1}{s^2+1}$ and $G(s) = \frac{s}{s^2+1}$ , then $\mathcal{L}^{-1}\{F(s) \cdot G(s)\}$ represents which convolution?

$(\sin(t) * \sin(t))(t)$
$(\cos(t) * \sin(t))(t)$
$(\sin(t) * \cos(t))(t)$ (correct answer)
$(\cos(t) * \cos(t))(t)$

Explanation: When you encounter a product of two Laplace transforms and need to find the inverse, you're dealing with the convolution theorem. This theorem states that

\mathcal{L}^{-1}\{F(s) \cdot G(s)\} = (f * g)(t)

, where

f(t) = \mathcal{L}^{-1}\{F(s)\}

and

g(t) = \mathcal{L}^{-1}\{G(s)\}

. First, find the individual inverse transforms. For

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

, you should recognize this as the standard form for

\mathcal{L}^{-1}\{\frac{1}{s^2+1}\} = \sin(t)

. For

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

, this is the standard form for

\mathcal{L}^{-1}\{\frac{s}{s^2+1}\} = \cos(t)

. Therefore,

\mathcal{L}^{-1}\{F(s) \cdot G(s)\} = (\sin(t) * \cos(t))(t)

, which is answer C. Looking at the wrong answers: A gives

(\sin(t) * \sin(t))(t)

, which would result from

F(s) = G(s) = \frac{1}{s^2+1}

. B gives

(\cos(t) * \sin(t))(t)

, which reverses the order—this would come from

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

and

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

. D gives

(\cos(t) * \cos(t))(t)

, which would result from both transforms being

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

. Remember: always identify the individual inverse transforms first, then apply the convolution theorem. The order in the convolution follows the order of the original transforms:

F(s) \cdot G(s)

becomes

(f * g)(t)

Question 3

If $h(t) = \int_0^t e^{2\tau} \cos(t-\tau) d\tau$ , then $\mathcal{L}\{h(t)\}$ equals:

$\frac{s}{(s+2)(s^2+1)}$
$\frac{s}{(s-2)(s^2+1)}$ (correct answer)
$\frac{1}{(s-2)(s^2+1)}$
$\frac{s+2}{s^2+1}$

Explanation: This problem tests your ability to find Laplace transforms of convolution integrals. When you see an integral of the form

\int_0^t f(\tau)g(t-\tau)d\tau

, this is a convolution, and the Laplace transform of a convolution equals the product of the individual Laplace transforms. Here,

h(t) = \int_0^t e^{2\tau} \cos(t-\tau) d\tau

is the convolution of

f(t) = e^{2t}

and

g(t) = \cos(t)

. Using the convolution theorem:

\mathcal{L}\{h(t)\} = \mathcal{L}\{f(t)\} \cdot \mathcal{L}\{g(t)\}

. The Laplace transforms are:

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

Therefore:

\mathcal{L}\{h(t)\} = \frac{1}{s-2} \cdot \frac{s}{s^2+1} = \frac{s}{(s-2)(s^2+1)}

This confirms answer B is correct. A is wrong because it has

(s+2)

instead of

(s-2)

in the denominator—this would correspond to

e^{-2t}

rather than

e^{2t}

. C is missing the

s

in the numerator, which would result from using

\mathcal{L}\{\sin(t)\} = \frac{1}{s^2+1}

instead of

\mathcal{L}\{\cos(t)\} = \frac{s}{s^2+1}

. D completely misses the convolution structure and doesn't include the factor from

e^{2t}

. Study tip: Always recognize convolution integrals by their

\int_0^t f(\tau)g(t-\tau)d\tau

form, then use the convolution theorem rather than attempting direct integration.

Question 4

Let $f(t) = u(t) - u(t-1)$ , where $u(t)$ is the Heaviside step function. What is the value of the self-convolution $h(t) = (f*f)(t)$ at $t=1.5$ ?

$1.5$
$0$
$1$
$0.5$ (correct answer)

Explanation: Convolution problems with piecewise functions require careful attention to the integration limits and where the functions are non-zero. The function

f(t) = u(t) - u(t-1)

represents a rectangular pulse that equals 1 for

0 \leq t < 1

and 0 elsewhere. For the self-convolution

(f*f)(t) = \int_{-\infty}^{\infty} f(\tau)f(t-\tau)d\tau

, you need to find where both

f(\tau)

and

f(t-\tau)

are simultaneously non-zero. Since

f(\tau) = 1

only when

0 \leq \tau \leq 1

, and

f(t-\tau) = 1

only when

0 \leq t-\tau \leq 1

(or equivalently

t-1 \leq \tau \leq t

), the product is non-zero only where these intervals overlap. At

t = 1.5

, we need the overlap of

[0,1]

and

[0.5, 1.5]

, which is

[0.5, 1]

. The convolution integral becomes

\int_{0.5}^{1} 1 \cdot 1 \, d\tau = 1 - 0.5 = 0.5

. Choice A (

1.5

) incorrectly uses the full value of

t

without considering the overlap constraint. Choice B (

0

) would suggest the functions don't overlap at all, which is wrong since there's a clear intersection. Choice C (

1

) represents the maximum possible overlap length but ignores that at

t = 1.5

, the overlap is only partial. When working with convolutions of piecewise functions, always sketch both functions and identify their overlap regions carefully. The convolution value equals the length of the overlap interval when convolving identical rectangular pulses.

Question 5

Which statement about the convolution theorem is FALSE when applied to solve differential equations using Laplace transforms?

If $\mathcal{L}\{y(t)\} = F(s) \cdot G(s)$ , then $y(t) = (f * g)(t)$ where $f = \mathcal{L}^{-1}\{F(s)\}$ and $g = \mathcal{L}^{-1}\{G(s)\}$
The convolution $(f * g)(t) = \int_0^t f(\tau)g(t-\tau)d\tau$ is always equal to $(g * f)(t) = \int_0^t g(\tau)f(t-\tau)d\tau$
For the equation $y'' + ay = f(t) * g(t)$ , the solution involves $\mathcal{L}\{y\} = \frac{F(s)G(s)}{s^2 + a}$ where $F(s) = \mathcal{L}\{f\}$ and $G(s) = \mathcal{L}\{g\}$
The convolution integral $(f * g)(0)$ always equals $f(0) \cdot g(0)$ regardless of the functions $f$ and $g$ (correct answer)

Explanation: Statement D is false. The convolution

(f * g)(0) = \int_0^0 f(\tau)g(-\tau)d\tau = 0

since the integration limits are the same, regardless of

f(0)

and

g(0)

. Statement A correctly describes the convolution theorem. Statement B correctly states that convolution is commutative. Statement C correctly applies the convolution theorem to a second-order differential equation (assuming zero initial conditions).

Question 6

The convolution $(t^2 * e^{-3t})(t)$ can be evaluated as $\mathcal{L}^{-1}\{F(s) \cdot G(s)\}$ where $F(s)$ and $G(s)$ are the Laplace transforms of the respective functions. What is the correct product $F(s) \cdot G(s)$ ?

$\frac{2}{s^3} \cdot \frac{1}{s+3} = \frac{2}{s^3(s+3)}$ (correct answer)
$\frac{2!}{s^3} \cdot \frac{1}{s-3} = \frac{2}{s^3(s-3)}$
$\frac{1}{s^2} \cdot \frac{1}{s+3} = \frac{1}{s^2(s+3)}$
$\frac{2}{s^2} \cdot \frac{1}{s+3} = \frac{2}{s^2(s+3)}$

Explanation: For the convolution theorem, we need

\mathcal{L}\{t^2\} = \frac{2!}{s^{2+1}} = \frac{2}{s^3}

and

\mathcal{L}\{e^{-3t}\} = \frac{1}{s+3}

(not

s-3

). The product gives

\frac{2}{s^3(s+3)}

. Choice B has wrong sign in exponential transform. Choice C uses

\mathcal{L}\{t\}

instead of

\mathcal{L}\{t^2\}

. Choice D incorrectly uses

s^2

instead of

s^3

for

t^2

Question 7

Let $h(t) = (f*g)(t)$ , where $f(t) = e^{2t}$ and $g(t) = 2e^{-2t}$ . Which of the following is an expression for $h(t)$ ?

$t e^{-4t}$
$e^{2t} - e^{-2t}$
$\sinh(2t)$ (correct answer)
$2$

Explanation: When you see the notation

(f*g)(t)

, this represents the convolution of two functions, which is a fundamental operation in differential equations and signal processing. The convolution integral is defined as

(f*g)(t) = \int_0^t f(t-\tau)g(\tau)d\tau

Let's compute this step by step with

f(t) = e^{2t}

and

g(t) = 2e^{-2t}

h(t) = \int_0^t e^{2(t-\tau)} \cdot 2e^{-2\tau} d\tau = \int_0^t 2e^{2t-2\tau}e^{-2\tau} d\tau = \int_0^t 2e^{2t-4\tau} d\tau

= 2e^{2t} \int_0^t e^{-4\tau} d\tau = 2e^{2t} \left[\frac{e^{-4\tau}}{-4}\right]_0^t = 2e^{2t} \cdot \frac{1-e^{-4t}}{4}

= \frac{e^{2t} - e^{2t}e^{-4t}}{2} = \frac{e^{2t} - e^{-2t}}{2}

Since

\sinh(2t) = \frac{e^{2t} - e^{-2t}}{2}

, the answer is C. Looking at the wrong answers: A

te^{-4t}

suggests confusion with multiplication rather than convolution. B

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

is missing the factor of

\frac{1}{2}

that comes from the convolution integral. D

2

would only result if the functions were constants, not exponentials. Study tip: Convolution problems often involve exponential functions because they're common solutions to differential equations. Remember that convolution isn't simple multiplication—it requires integration, and the result often involves hyperbolic functions when exponentials are involved.

Question 8

The Laplace transform of a function $h(t)$ is given by the expression $H(s) = \frac{s}{(s^2+9)(s-2)}$ . Which of the following integrals correctly represents $h(t)$ ?

$\int_{0}^{t} \cos(3\tau) e^{2(t-\tau)} d\tau$ (correct answer)
$\int_{0}^{t} \sin(3\tau) e^{2(t-\tau)} d\tau$
$\int_{0}^{t} \cos(3\tau) e^{2\tau} d\tau$
$\cos(3t) e^{2t}$

Explanation: According to the Convolution Theorem, the inverse Laplace transform of a product of two functions,

H(s) = F(s)G(s)

, is the convolution of their inverse transforms,

h(t) = (f*g)(t) = \int_{0}^{t} f(\tau)g(t-\tau)d\tau

. We can decompose

H(s)

into

F(s) = \frac{s}{s^2+9}

and

G(s) = \frac{1}{s-2}

. The inverse Laplace transforms are

f(t) = \mathcal{L}^{-1}\{\frac{s}{s^2+9}\} = \cos(3t)

and

g(t) = \mathcal{L}^{-1}\{\frac{1}{s-2}\} = e^{2t}

. Therefore,

h(t)

is the convolution of these two functions, which is correctly expressed as

\int_{0}^{t} \cos(3\tau) e^{2(t-\tau)} d\tau

Question 9

Consider the integro-differential equation $y(t) = 4t - \int_0^t y(\tau) \sin(t-\tau) d\tau$ . Find the Laplace transform $Y(s) = \mathcal{L}\{y(t)\}$ .

$\frac{4(s^2+1)}{s^2(s^2+2)}$ (correct answer)
$\frac{4(s^2+1)}{s^4}$
$\frac{4(s^2+1)}{s^2(s^2+s+1)}$
$\frac{4}{s^2} - \frac{Y(s)}{s^2+1}$

Explanation: Take the Laplace transform of the entire equation. The integral term is the convolution of

y(t)

and

\sin(t)

, denoted

(y*\sin)(t)

. By the Convolution Theorem, its Laplace transform is

\mathcal{L}\{(y*\sin)(t)\} = Y(s) \mathcal{L}\{\sin(t)\} = Y(s) \frac{1}{s^2+1}

. The equation in the s-domain becomes

Y(s) = \mathcal{L}\{4t\} - Y(s) \frac{1}{s^2+1}

, which is

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

. To solve for

Y(s)

, we gather terms:

Y(s) + Y(s) \frac{1}{s^2+1} = \frac{4}{s^2}

. Factoring out

Y(s)

gives

Y(s) \left(1 + \frac{1}{s^2+1}\right) = \frac{4}{s^2}

, which simplifies to

Y(s) \left(\frac{s^2+2}{s^2+1}\right) = \frac{4}{s^2}

. Isolating

Y(s)

yields

Y(s) = \frac{4(s^2+1)}{s^2(s^2+2)}

Question 10

Let $Y(s) = \frac{1}{s^2(s+1)}$ . According to the Convolution Theorem, which of the following integrals represents $y(t) = \mathcal{L}^{-1}\{Y(s)\}$ ?

$\int_0^t (t-\tau)e^{-\tau} d\tau$ (correct answer)
$\int_0^t \tau e^{-\tau} d\tau$
$\int_0^t t e^{-(t-\tau)} d\tau$
$t e^{-t}$

Explanation: We can write

Y(s)

as the product of two functions,

F(s) = \frac{1}{s^2}

and

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

. By the Convolution Theorem,

y(t)

is the convolution of their inverse Laplace transforms,

f(t)

and

g(t)

. We have

f(t) = \mathcal{L}^{-1}\{\frac{1}{s^2}\} = t

and

g(t) = \mathcal{L}^{-1}\{\frac{1}{s+1}\} = e^{-t}

. The convolution can be written as

(f*g)(t) = \int_0^t f(\tau)g(t-\tau) d\tau

(g*f)(t) = \int_0^t g(\tau)f(t-\tau) d\tau

. The second form gives

\int_0^t e^{-\tau} (t-\tau) d\tau

, which matches choice A. The first form would be

\int_0^t \tau e^{-(t-\tau)} d\tau

, which is also correct but not an option.

Question 11

Let $h(t) = \mathcal{L}^{-1}\{\frac{1}{s(s-a)}\}$ . Which statement correctly describes the limiting behavior of $h(t)$ as $t \to \infty$ for a constant $a > 0$ ?

$h(t)$ approaches a constant value of $-1/a$ .
$h(t)$ grows exponentially without bound. (correct answer)
$h(t)$ approaches 0.
$h(t)$ grows linearly without bound.

Explanation: When you encounter inverse Laplace transforms involving partial fractions, you need to decompose the expression and analyze the behavior of each component as time increases. To find

h(t) = \mathcal{L}^{-1}\left\{\frac{1}{s(s-a)}\right\}

, first use partial fraction decomposition:

\frac{1}{s(s-a)} = \frac{A}{s} + \frac{B}{s-a}

. Solving for the constants:

1 = A(s-a) + Bs

. Setting

s = 0

gives

A = -\frac{1}{a}

, and setting

s = a

gives

B = \frac{1}{a}

. Therefore:

\frac{1}{s(s-a)} = \frac{-1/a}{s} + \frac{1/a}{s-a}

Taking the inverse Laplace transform:

h(t) = -\frac{1}{a}\mathcal{L}^{-1}\left\{\frac{1}{s}\right\} + \frac{1}{a}\mathcal{L}^{-1}\left\{\frac{1}{s-a}\right\} = -\frac{1}{a} + \frac{1}{a}e^{at}

t \to \infty

with

a > 0

, the exponential term

e^{at}

grows without bound, making

h(t)

grow exponentially. This confirms answer B. Answer A incorrectly focuses only on the constant term, ignoring the exponential growth. Answer C would be correct if

a < 0

, but here

a > 0

makes the exponential term dominate. Answer D describes polynomial growth, not the exponential behavior we see here. Study tip: When analyzing limiting behavior of inverse Laplace transforms, always identify which term dominates as

t \to \infty

. Exponential terms with positive exponents will always dominate over constants or polynomial terms.

Question 12

Let $h(t) = \int_0^t (t-\tau)^2 e^{3\tau} d\tau$ . Using the Convolution Theorem, find the Laplace transform $H(s)$ .

$\frac{2}{s^3(s-3)}$ (correct answer)
$\frac{2}{(s-3)^3}$
$\frac{1}{s^3(s-3)}$
$\frac{2}{s^2(s-3)}$

Explanation: The integral is in the form of a convolution

(g*f)(t) = \int_0^t g(t-\tau) f(\tau) d\tau

. By comparing forms, we can identify

g(t-\tau) = (t-\tau)^2

and

f(\tau) = e^{3\tau}

. This means

g(t) = t^2

and

f(t) = e^{3t}

. The Convolution Theorem states that

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

. We find the individual Laplace transforms:

G(s) = \mathcal{L}\{t^2\} = \frac{2!}{s^{2+1}} = \frac{2}{s^3}

and

F(s) = \mathcal{L}\{e^{3t}\} = \frac{1}{s-3}

. Therefore,

H(s) = G(s)F(s) = \frac{2}{s^3} \cdot \frac{1}{s-3} = \frac{2}{s^3(s-3)}

Question 13

The solution to the initial value problem $y'' + \omega^2 y = g(t)$ , with $y(0)=0$ and $y'(0)=0$ , can be expressed as $y(t) = \int_0^t h(\tau) g(t-\tau) d\tau$ . What is the function $h(t)$ ?

$\frac{1}{\omega^2} \cos(\omega t)$
$\sin(\omega t)$
$\cos(\omega t)$
$\frac{1}{\omega} \sin(\omega t)$ (correct answer)

Explanation: When you see an equation like

y'' + \omega^2 y = g(t)

with zero initial conditions, you're dealing with a convolution integral where

h(t)

represents the impulse response function - the solution when the forcing function is a Dirac delta function. To find

h(t)

, we need to solve the initial value problem

h'' + \omega^2 h = \delta(t)

with

h(0) = 0

and

h'(0) = 0

. For

t > 0

, this becomes the homogeneous equation

h'' + \omega^2 h = 0

, which has the general solution

h(t) = A\cos(\omega t) + B\sin(\omega t)

. Applying the initial condition

h(0) = 0

gives us

A = 0

, so

h(t) = B\sin(\omega t)

. The condition

h'(0) = 0

might seem to give

B = 0

, but the delta function at

t = 0

creates a jump discontinuity in

h'(t)

. The proper analysis shows that

h'(0^+) - h'(0^-) = 1

, which gives us

B\omega = 1

, so

B = \frac{1}{\omega}

. Therefore,

h(t) = \frac{1}{\omega}\sin(\omega t)

, which is choice D. Choice A,

\frac{1}{\omega^2}\cos(\omega t)

, fails because it doesn't satisfy

h(0) = 0

. Choice B,

\sin(\omega t)

, has the wrong amplitude - it would give

h'(0^+) = \omega

instead of 1. Choice C,

\cos(\omega t)

, also violates

h(0) = 0

. Remember: For second-order differential equations with impulse forcing, the impulse response always involves the factor

\frac{1}{\omega}

multiplying the sine term to ensure proper normalization of the derivative jump.

Question 14

The function $h(t) = \int_0^t \sin(\tau) \cos(t-\tau) d\tau$ is the solution to which initial value problem?

$y'' + y = \sin(t)$ , with $y(0)=0, y'(0)=0$
$y'' + y = \cos(t)$ , with $y(0)=0, y'(0)=0$ (correct answer)
$y' + y = \cos(t)$ , with $y(0)=0$
$y'' - y = \cos(t)$ , with $y(0)=0, y'(0)=0$

Explanation: When you encounter an integral expression like

h(t) = \int_0^t \sin(\tau) \cos(t-\tau) d\tau

, this is likely a convolution integral that represents the solution to a differential equation. The key insight is recognizing this as the convolution of

\sin(t)

and

\cos(t)

, which appears in solutions to second-order linear differential equations. To find which equation this solves, you can either differentiate the integral twice or use convolution theory. Let's differentiate: Using the fundamental theorem of calculus and product rule,

h'(t) = \sin(t)\cos(0) + \int_0^t \sin(\tau)(-\sin(t-\tau)) d\tau = \sin(t) - \int_0^t \sin(\tau)\sin(t-\tau) d\tau

. Differentiating again:

h''(t) = \cos(t) - \sin(t)\sin(0) - \int_0^t \sin(\tau)\cos(t-\tau) d\tau = \cos(t) - h(t)

. Rearranging gives

h'' + h = \cos(t)

, and clearly

h(0) = 0

and

h'(0) = 0

. Option A is incorrect because the forcing function is

\sin(t)

, not

\cos(t)

. Option C represents a first-order equation, but our integral solution naturally satisfies a second-order equation. Option D has the wrong sign in the differential operator (minus instead of plus), which would lead to exponential rather than oscillatory behavior. Study tip: When you see convolution integrals involving trigonometric functions, think second-order constant coefficient equations. The forcing function in the differential equation typically appears as one of the functions in the convolution.

Question 15

Consider the differential equation $y'' + 4y = f(t) * g(t)$ with zero initial conditions, where $f(t) = e^t$ and $g(t) = t$ . Using the convolution theorem, what is $\mathcal{L}\{y(t)\}$ ?

$\frac{s^2+4}{(s-1)s^2}$
$\frac{1}{(s^2+4)(s-1)s}$
$\frac{1}{(s^2+4)(s-1)s^2}$ (correct answer)
$\frac{1}{(s^2+4)} \cdot \frac{1}{(s-1)} \cdot \frac{1}{s^2}$

Explanation: When you encounter a differential equation with a convolution on the right side, the Laplace transform and convolution theorem are your primary tools. The convolution theorem states that

\mathcal{L}\{f(t) * g(t)\} = \mathcal{L}\{f(t)\} \cdot \mathcal{L}\{g(t)\}

, which transforms the convolution into simple multiplication. Taking the Laplace transform of the entire equation

y'' + 4y = e^t * t

, you get:

\mathcal{L}\{y''\} + 4\mathcal{L}\{y\} = \mathcal{L}\{e^t * t\}

With zero initial conditions,

\mathcal{L}\{y''\} = s^2\mathcal{L}\{y\}

. Using the convolution theorem on the right side:

\mathcal{L}\{e^t * t\} = \mathcal{L}\{e^t\} \cdot \mathcal{L}\{t\} = \frac{1}{s-1} \cdot \frac{1}{s^2}

This gives you:

s^2\mathcal{L}\{y\} + 4\mathcal{L}\{y\} = \frac{1}{(s-1)s^2}

Factoring out

\mathcal{L}\{y\}

(s^2 + 4)\mathcal{L}\{y\} = \frac{1}{(s-1)s^2}

Therefore:

\mathcal{L}\{y\} = \frac{1}{(s^2+4)(s-1)s^2}

Choice A incorrectly places

(s^2+4)

in the numerator instead of denominator. Choice B has the wrong power of

s

in the denominator—it should be

s^2

, not

s

, since

\mathcal{L}\{t\} = \frac{1}{s^2}

. Choice D shows the individual transforms but fails to combine them properly after applying the convolution theorem. Remember: when using the convolution theorem, multiply the individual Laplace transforms, then solve algebraically for

\mathcal{L}\{y\}

by moving all terms to the appropriate side.

Question 16

Given that $\mathcal{L}\{h(t)\} = \frac{2}{(s+1)(s^2+4)}$ , which functions $f(t)$ and $g(t)$ satisfy $h(t) = (f * g)(t)$ when applying the convolution theorem?

$f(t) = 2e^{-t}$ and $g(t) = \sin(2t)$
$f(t) = e^{-t}$ and $g(t) = 2\sin(2t)$
$f(t) = 2e^{-t}$ and $g(t) = \cos(2t)$
$f(t) = e^{-t}$ and $g(t) = \sin(2t)$ (correct answer)

Explanation: By the convolution theorem, if

h(t) = (f * g)(t)

, then

\mathcal{L}\{h(t)\} = F(s) \cdot G(s)

. We need to factor

\frac{2}{(s+1)(s^2+4)}

as a product. Since

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

and

\mathcal{L}\{\sin(2t)\} = \frac{2}{s^2+4}

, we get

F(s) \cdot G(s) = \frac{1}{s+1} \cdot \frac{2}{s^2+4} = \frac{2}{(s+1)(s^2+4)}

. Choices A and C have incorrect coefficients, and choice B has the coefficient in the wrong function.

Question 17

If $f(t) = e^{2t}$ and $g(t) = \sin(3t)$ , which expression correctly represents $(f * g)(t)$ using the convolution theorem?

$\mathcal{L}^{-1}\left\{\frac{1}{s-2} \cdot \frac{3}{s^2+9}\right\}$ (correct answer)
$\mathcal{L}^{-1}\left\{\frac{1}{s+2} \cdot \frac{3}{s^2+9}\right\}$
$\mathcal{L}^{-1}\left\{\frac{1}{s-2} + \frac{3}{s^2+9}\right\}$
$\mathcal{L}^{-1}\left\{\frac{3}{(s-2)(s^2+9)}\right\}$

Explanation: The convolution theorem states that

\mathcal{L}\{(f * g)(t)\} = F(s) \cdot G(s)

, so

(f * g)(t) = \mathcal{L}^{-1}\{F(s) \cdot G(s)\}

. Here,

\mathcal{L}\{e^{2t}\} = \frac{1}{s-2}

and

\mathcal{L}\{\sin(3t)\} = \frac{3}{s^2+9}

. Choice B has incorrect sign in exponential transform. Choice C shows addition instead of multiplication (linearity property confusion). Choice D incorrectly combines the denominators.

Question 18

The function $f(t) = \mathcal{L}^{-1}\left\{\frac{4}{s^2(s^2+4)^2}\right\}$ can be expressed as a convolution. Which pair of functions, when convolved, produces $f(t)$ ?

$2t$ and $\frac{1}{2}t\sin(2t)$
$t$ and $t\sin(2t)$ (correct answer)
$t$ and $\frac{1}{2}t\sin(2t)$
$2t$ and $t\cos(2t)$

Explanation: When you encounter a Laplace transform that's a product of two simpler transforms in the denominator, convolution is often the key to finding the inverse transform. The convolution theorem states that

\mathcal{L}^{-1}\{F(s)G(s)\} = (f * g)(t)

, where

*

denotes convolution. Looking at

\frac{4}{s^2(s^2+4)^2}

, you can rewrite this as

\frac{1}{s^2} \cdot \frac{4}{(s^2+4)^2}

. Now you need to identify what functions have these Laplace transforms. From standard tables:

\mathcal{L}^{-1}\left\{\frac{1}{s^2}\right\} = t

and

\mathcal{L}^{-1}\left\{\frac{4}{(s^2+4)^2}\right\} = t\sin(2t)

. Therefore,

f(t) = t * t\sin(2t)

, making choice B correct. Let's examine why the other options fail. Choice A suggests

2t

and

\frac{1}{2}t\sin(2t)

, but

\mathcal{L}\{2t\} = \frac{2}{s^2}

, not

\frac{1}{s^2}

. Choice C proposes

t

and

\frac{1}{2}t\sin(2t)

, but

\mathcal{L}\left\{\frac{1}{2}t\sin(2t)\right\} = \frac{2}{(s^2+4)^2}

, not

\frac{4}{(s^2+4)^2}

. Choice D suggests

2t

and

t\cos(2t)

, but this gives the wrong first factor, and

\mathcal{L}\{t\cos(2t)\} = \frac{2(s^2-4)}{(s^2+4)^2}

, which doesn't match our second factor. Study tip: Master the standard Laplace transforms for polynomial and trigonometric functions, especially forms like

\frac{n!}{s^{n+1}}

and

\frac{2as}{(s^2+a^2)^2}

for

t^n

and

t\sin(at)

respectively.