Differential Equations Quiz: Laplace Of Derivatives

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Laplace Of DerivativesQuestion 1 of 13

Given that $f(0) = 3$ , $f'(0) = -2$ , and $f''(0) = 5$ , what is $\mathcal{L}\{f''(t) - 3f'(t) + 2f(t)\}$ ?

s^2F(s) - 3sF(s) + 2F(s) - 3s + 8

s^2F(s) - 3sF(s) + 2F(s) - 3s + 7

s^2F(s) - 3sF(s) + 2F(s) - 3s + 6

s^2F(s) - 3sF(s) + 2F(s) - 3s + 5

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Differential Equations Quiz

Differential Equations Quiz: Laplace Of Derivatives

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

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

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Question 1

Given that $f(0) = 3$ , $f'(0) = -2$ , and $f''(0) = 5$ , what is $\mathcal{L}\{f''(t) - 3f'(t) + 2f(t)\}$ ?

$s^2F(s) - 3sF(s) + 2F(s) - 3s + 8$
$s^2F(s) - 3sF(s) + 2F(s) - 3s + 7$ (correct answer)
$s^2F(s) - 3sF(s) + 2F(s) - 3s + 6$
$s^2F(s) - 3sF(s) + 2F(s) - 3s + 5$

Explanation: Using linearity of Laplace transforms and the derivative formulas:

\mathcal{L}\{f''(t)\} = s^2F(s) - sf(0) - f'(0) = s^2F(s) - 3s + 2

\mathcal{L}\{f'(t)\} = sF(s) - f(0) = sF(s) - 3

, and

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

. Therefore:

\mathcal{L}\{f''(t) - 3f'(t) + 2f(t)\} = (s^2F(s) - 3s + 2) - 3(sF(s) - 3) + 2F(s) = s^2F(s) - 3sF(s) + 2F(s) - 3s + 2 + 9 = s^2F(s) - 3sF(s) + 2F(s) - 3s + 7

. Choice A incorrectly adds 8 instead of 11-4=7. Choice C omits the +2 from the second derivative formula. Choice D uses only the constant term from

f''(0)

Question 2

Let $F(s) = \mathcal{L}\{f(t)\}$ and assume $f(0)=2$ . Which expression represents the Laplace transform of $f'(t) - \int_0^t f(\tau)d\tau$ ?

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

Explanation: When working with Laplace transforms of combined operations like derivatives and integrals, you need to apply the transform properties to each term separately, then combine the results algebraically. For the derivative

f'(t)

, use the derivative property:

\mathcal{L}\{f'(t)\} = sF(s) - f(0) = sF(s) - 2

. For the integral

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

, use the integration property:

\mathcal{L}\left\{\int_0^t f(\tau)d\tau\right\} = \frac{F(s)}{s}

. The expression

f'(t) - \int_0^t f(\tau)d\tau

becomes:

\mathcal{L}\{f'(t)\} - \mathcal{L}\left\{\int_0^t f(\tau)d\tau\right\} = (sF(s) - 2) - \frac{F(s)}{s} = sF(s) - \frac{F(s)}{s} - 2

. Factoring gives us

(s - \frac{1}{s})F(s) - 2

, which is answer D. Choice A omits the initial condition

f(0) = 2

when transforming the derivative, giving only

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

. Choice B incorrectly adds the initial condition instead of subtracting it in the derivative term. Choice C has the right factored form but wrong sign on the constant term—it adds 2 instead of subtracting it. Remember that the derivative property always involves the initial condition

f(0)

, and pay careful attention to plus and minus signs when combining terms. The linearity of Laplace transforms means you can work with each piece separately, but don't forget how the signs affect the final result.

Question 3

The Laplace transform of a function $y(t)$ is given by $Y(s) = \frac{2s - 7}{s^2 + 6s + 13}$ . What are the initial conditions $y(0)$ and $y'(0)$ ?

$y(0) = 2$ , $y'(0) = -7$
$y(0) = 2$ , $y'(0) = -13$
$y(0) = -7$ , $y'(0) = 2$
$y(0) = 2$ , $y'(0) = -19$ (correct answer)

Explanation: When you encounter a Laplace transform and need to find initial conditions, you'll use the Initial Value Theorem, which states that

\lim_{t \to 0^+} f(t) = \lim_{s \to \infty} sF(s)

for finding

f(0)

, and

\lim_{t \to 0^+} f'(t) = \lim_{s \to \infty} s^2F(s) - sf(0)

for finding

f'(0)

. First, let's find

y(0)

. Using the Initial Value Theorem:

y(0) = \lim_{s \to \infty} s \cdot \frac{2s - 7}{s^2 + 6s + 13} = \lim_{s \to \infty} \frac{2s^2 - 7s}{s^2 + 6s + 13}

Dividing numerator and denominator by

s^2

y(0) = \lim_{s \to \infty} \frac{2 - \frac{7}{s}}{1 + \frac{6}{s} + \frac{13}{s^2}} = \frac{2}{1} = 2

Next, for

y'(0)

y'(0) = \lim_{s \to \infty} \left(s^2 \cdot \frac{2s - 7}{s^2 + 6s + 13} - s \cdot 2\right) = \lim_{s \to \infty} \frac{2s^3 - 7s^2 - 2s^3 - 12s^2 - 26s}{s^2 + 6s + 13} = \lim_{s \to \infty} \frac{-19s^2 - 26s}{s^2 + 6s + 13} = -19

Therefore,

y(0) = 2

and

y'(0) = -19

, making D correct. Option A incorrectly uses -7 for

y'(0)

, likely from misapplying the coefficient. Option B uses -13, possibly confusing it with the constant term. Option C swaps the values entirely. Strategy tip: Always apply the Initial Value Theorem systematically—don't guess from coefficients. The limits as

s \to \infty

give you the true initial conditions.

Question 4

Let $Y(s) = \mathcal{L}\{y(t)\}$ . The expression for $\mathcal{L}\{y'''(t)\}$ is $s^3Y(s) - s^2y(0) - sy'(0) - y''(0)$ . If the initial condition $y(0)$ is changed from $c_1$ to $c_2$ , while $y'(0)$ and $y''(0)$ remain fixed, what is the resulting change in $\mathcal{L}\{y'''(t)\}$ ?

An additive change of $c_1 - c_2$
An additive change of $s(c_1 - c_2)$
An additive change of $s^2(c_2 - c_1)$
An additive change of $s^2(c_1 - c_2)$ (correct answer)

Explanation: Let

L_1

be the transform with

y(0)=c_1

and

L_2

be the transform with

y(0)=c_2

. The other initial conditions and

Y(s)

are unchanged.

L_1 = s^3Y(s) - s^2c_1 - sy'(0) - y''(0)

L_2 = s^3Y(s) - s^2c_2 - sy'(0) - y''(0)

. The change is

L_2 - L_1

L_2 - L_1 = (s^3Y(s) - s^2c_2 - sy'(0) - y''(0)) - (s^3Y(s) - s^2c_1 - sy'(0) - y''(0))

L_2 - L_1 = -s^2c_2 + s^2c_1 = s^2(c_1 - c_2)

. This is an additive change of

s^2(c_1 - c_2)

. The distractors test whether the student correctly associates the initial condition

y^{(k)}(0)

with the term

s^{n-1-k}

in the transform of

y^{(n)}(t)

. For

y'''(t)

n=3

. For

y(0)

k=0

. The associated term is

-s^{3-1-0}y(0) = -s^2y(0)

. The change is

-s^2c_2 - (-s^2c_1) = s^2(c_1-c_2)

Question 5

Consider the initial value problem $y'' + 2y' + 5y = 0$ with $y(0) = 1$ and $y'(0) = 3$ . If $Y(s) = \mathcal{L}\{y(t)\}$ , which of the following is the correct expression for $Y(s)$ ?

$Y(s) = \frac{s+3}{s^2+2s+5}$
$Y(s) = \frac{s+5}{s^2+2s+5}$ (correct answer)
$Y(s) = \frac{s-1}{s^2+2s+5}$
$Y(s) = \frac{s+2}{s^2+2s+5}$

Explanation: We take the Laplace transform of the entire differential equation.

\mathcal{L}\{y'' + 2y' + 5y\} = \mathcal{L}\{0\}

\mathcal{L}\{y''\} + 2\mathcal{L}\{y'\} + 5\mathcal{L}\{y\} = 0

Using the formulas for the transform of derivatives:

[s^2Y(s) - sy(0) - y'(0)] + 2[sY(s) - y(0)] + 5Y(s) = 0

Substitute the initial conditions

y(0)=1

and

y'(0)=3

[s^2Y(s) - s(1) - 3] + 2[sY(s) - 1] + 5Y(s) = 0

s^2Y(s) - s - 3 + 2sY(s) - 2 + 5Y(s) = 0

Group the terms with

Y(s)

and the polynomial terms separately:

Y(s)(s^2 + 2s + 5) - s - 5 = 0

Now, solve for

Y(s)

Y(s)(s^2 + 2s + 5) = s + 5

Y(s) = \frac{s+5}{s^2+2s+5}

Choice B is correct. Choice A results from a sign error in

y'(0)

[s^2Y-s+3] + 2[sY-1] + 5Y = 0 \implies Y(s^2+2s+5) = s-1

. This matches C. Choice A results from forgetting the

y(0)

term in

2\mathcal{L}\{y'\}

[s^2Y-s-3] + 2[sY] + 5Y = 0 \implies Y(s^2+2s+5) = s+3

. Choice C results from a sign error on

y'(0)

Question 6

The current $i(t)$ in an electrical circuit is modeled by the integro-differential equation $\frac{di}{dt} + 4i + 13\int_0^t i(\tau)d\tau = 26$ with the initial condition $i(0) = 5$ . Find the Laplace transform $I(s) = \mathcal{L}\{i(t)\}$ .

$I(s) = \frac{5s+26}{s^2+4s+13}$ (correct answer)
$I(s) = \frac{5s-26}{s^2+4s+13}$
$I(s) = \frac{26}{s^2+4s+13}$
$I(s) = \frac{5s}{s^2+4s+13}$

Explanation: We take the Laplace transform of each term in the equation.

\mathcal{L}\{\frac{di}{dt}\} + \mathcal{L}\{4i\} + \mathcal{L}\{13\int_0^t i(\tau)d\tau\} = \mathcal{L}\{26\}

Using the rules for transforms of derivatives and integrals:

[sI(s) - i(0)] + 4I(s) + 13\frac{I(s)}{s} = \frac{26}{s}

. Substitute the initial condition

i(0) = 5

[sI(s) - 5] + 4I(s) + \frac{13I(s)}{s} = \frac{26}{s}

. Group the terms containing

I(s)

I(s)(s + 4 + \frac{13}{s}) - 5 = \frac{26}{s}

. Move the constant term to the right side:

I(s)(s + 4 + \frac{13}{s}) = 5 + \frac{26}{s}

. Combine the terms on each side with a common denominator:

I(s)\left(\frac{s^2 + 4s + 13}{s}\right) = \frac{5s + 26}{s}

. Finally, solve for

I(s)

by multiplying both sides by

\frac{s}{s^2 + 4s + 13}

I(s) = \frac{5s+26}{s^2+4s+13}

. Choice A is correct. Choice C results from incorrectly assuming the initial condition

i(0)=0

. Choice D ignores the constant term on the right side of the equation. Choice B results from a sign error when moving the

-5

term.

Question 7

Let $F(s) = \mathcal{L}\{f(t)\}$ . If the Laplace transform of the second derivative of $f(t)$ is $\mathcal{L}\{f''(t)\} = \frac{1}{s^2-4}$ , and the initial conditions are $f(0) = -2$ and $f'(0) = 1$ , what is the expression for $F(s)$ ?

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

Explanation: The formula for the Laplace transform of a second derivative is

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

. We are given

\mathcal{L}\{f''(t)\}

f(0)

, and

f'(0)

, and we need to solve for

F(s)

. Substituting the given values into the formula:

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

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

Now, we rearrange the equation to isolate

F(s)

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

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

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

This matches choice A. B is incorrect due to sign errors when moving the terms

2s

and

-1

to the other side of the equation. C is incorrect as it represents

s^2F(s)

, not

F(s)

. D is incorrect because it confuses the roles of the transform and the function, and makes an algebraic error.

Question 8

A function $f(t)$ is defined as $f(t) = \begin{cases} 2 & 0 \le t < 1 \\ 0 & t \ge 1 \end{cases}$ . The function $f(t)$ has a jump discontinuity at $t=1$ . What is the Laplace transform of its derivative, $\mathcal{L}\{f'(t)\}$ ?

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

Explanation: The standard formula

\mathcal{L}\{f'(t)\} = sF(s) - f(0)

is only valid for continuous functions

f(t)

. For a function with a jump discontinuity, the derivative involves a Dirac delta function. We can write

f(t)

using the Heaviside step function

u(t)

as:

f(t) = 2(u(t-0) - u(t-1)) = 2 - 2u(t-1)

. The derivative in the sense of generalized functions is:

f'(t) = \frac{d}{dt}(2 - 2u(t-1)) = -2\delta(t-1)

, where

\delta(t)

is the Dirac delta function. The Laplace transform of the delta function shifted by

a

\mathcal{L}\{\delta(t-a)\} = e^{-as}

. Therefore,

\mathcal{L}\{f'(t)\} = \mathcal{L}\{-2\delta(t-1)\} = -2e^{-s}

. Alternatively, one can use the extended formula for discontinuous functions:

\mathcal{L}\{f'(t)\} = sF(s) - f(0) - \sum_k e^{-s t_k} [f(t_k^+) - f(t_k^-)]

. Here,

f(0)=2

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

. The jump at

t=1

f(1^+)-f(1^-) = 0-2 = -2

. The formula gives

s(\frac{2}{s} - \frac{2e^{-s}}{s}) - 2 - e^{-s}(-2) = (2 - 2e^{-s}) - 2 + 2e^{-s} = 0

. This is incorrect. The general formula includes all jumps. This method is tricky. The first method using the delta function is more direct. Choice A is

F(s)

, not

\mathcal{L}\{f'(t)\}

. Choice C is

sF(s)

. Choice D is incorrect and does not account for the behavior at

t=1

Question 9

The Laplace transform of the solution to $y''-k^2y=0$ with $y(0)=y_0$ and $y'(0)=v_0$ is $Y(s) = \frac{y_0 s + v_0}{s^2-k^2}$ . Let $g(t) = y'(t)$ . What is $G(s) = \mathcal{L}\{g(t)\}$ ?

$sY(s)$
$\frac{y_0 s^2 + v_0 s}{s^2-k^2}$
$\frac{v_0 s + y_0 k^2}{s^2-k^2}$ (correct answer)
$\frac{v_0 s - y_0 k^2}{s^2-k^2}$

Explanation: We are asked to find the Laplace transform of

g(t) = y'(t)

. Using the property for the transform of a derivative:

G(s) = \mathcal{L}\{y'(t)\} = sY(s) - y(0)

. We are given

Y(s) = \frac{y_0 s + v_0}{s^2-k^2}

and

y(0)=y_0

. Substituting these into the formula:

G(s) = s\left(\frac{y_0 s + v_0}{s^2-k^2}\right) - y_0

G(s) = \frac{y_0 s^2 + v_0 s}{s^2-k^2} - y_0

To combine these terms, we find a common denominator:

G(s) = \frac{y_0 s^2 + v_0 s}{s^2-k^2} - \frac{y_0(s^2-k^2)}{s^2-k^2}

G(s) = \frac{(y_0 s^2 + v_0 s) - (y_0 s^2 - y_0 k^2)}{s^2-k^2}

G(s) = \frac{y_0 s^2 + v_0 s - y_0 s^2 + y_0 k^2}{s^2-k^2}

G(s) = \frac{v_0 s + y_0 k^2}{s^2-k^2}

. This matches choice C. Choice A,

sY(s)

, is a common mistake that forgets to subtract the initial condition

y(0)

. Choice B is the intermediate term

sY(s)

before subtracting

y_0

. Choice D results from a sign error in the algebraic simplification.

Question 10

Let $F(s) = \mathcal{L}\{f(t)\}$ . Given that $f(0) = 4$ , which of the following expressions correctly represents $\mathcal{L}\{e^{3t}f'(t)\}$ ?

$sF(s-3) - 4$
$(s-3)F(s-3) - 4$ (correct answer)
$sF(s) - 4e^{-3s}$
$(s-3)F(s) - 4$

Explanation: This problem requires applying two Laplace transform properties: the transform of a derivative and the first shifting theorem (s-shift). First, find the Laplace transform of

f'(t)

. Let

g(t) = f'(t)

G(s) = \mathcal{L}\{f'(t)\} = sF(s) - f(0) = sF(s) - 4

. Next, apply the first shifting theorem to find

\mathcal{L}\{e^{3t}g(t)\}

. The theorem states that

\mathcal{L}\{e^{at}g(t)\} = G(s-a)

. In our case,

a=3

, so we replace every

s

G(s)

with

(s-3)

\mathcal{L}\{e^{3t}f'(t)\} = G(s-3) = (s-3)F(s-3) - 4

. Choice B is correct. Choice A,

sF(s-3) - 4

, incorrectly applies the shift only to

F(s)

and not to the factor

s

multiplying it. This would be the transform of

\frac{d}{dt}(e^{3t}f(t)) = 3e^{3t}f(t) + e^{3t}f'(t)

. Choice D does not shift the argument of

F(s)

. Choice C incorrectly mixes the s-shift and t-shift properties.

Question 11

Let $Y(s) = \mathcal{L}\{y(t)\} = \frac{3}{s^2+9}$ . If it is known that $y(0)=0$ , what must be the value of the initial condition $y'(0)$ such that $\mathcal{L}\{y''(t)\}$ is equal to $-9Y(s)$ ?

$0$
$1$
$3$ (correct answer)
$9$

Explanation: The formula for the Laplace transform of a second derivative is

\mathcal{L}\{y''(t)\} = s^2Y(s) - sy(0) - y'(0)

. We are given

Y(s) = \frac{3}{s^2+9}

and

y(0) = 0

. We are also given the condition that

\mathcal{L}\{y''(t)\} = -9Y(s)

. Let's substitute the known information into the derivative formula:

\mathcal{L}\{y''(t)\} = s^2\left(\frac{3}{s^2+9}\right) - s(0) - y'(0)

\mathcal{L}\{y''(t)\} = \frac{3s^2}{s^2+9} - y'(0)

. Now we set this equal to the given condition,

-9Y(s)

\frac{3s^2}{s^2+9} - y'(0) = -9\left(\frac{3}{s^2+9}\right) = \frac{-27}{s^2+9}

. Now, we solve for

y'(0)

y'(0) = \frac{3s^2}{s^2+9} + \frac{27}{s^2+9}

y'(0) = \frac{3s^2 + 27}{s^2+9}

y'(0) = \frac{3(s^2 + 9)}{s^2+9} = 3

. So, the initial condition must be

y'(0)=3

. (Note: The inverse transform of

Y(s) = \frac{3}{s^2+9}

y(t) = \sin(3t)

. This satisfies

y(0)=0

and

y'(0) = 3\cos(0) = 3

. The DE is

y''+9y=0

, which transforms to

\mathcal{L}\{y''\}=-9\mathcal{L}\{y\}=-9Y(s)

.) Distractors: A, B, and D are other plausible integer values that arise from miscalculation.

Question 12

Let $Y(s) = \mathcal{L}\{y(t)\}$ . Given the initial conditions $y(0)=2$ , $y'(0)=0$ , $y''(0)=-1$ , and $y'''(0)=5$ , find the Laplace transform of the fourth derivative, $y^{(4)}(t)$ .

$s^4Y(s) - 2s^3 - s - 5$
$s^4Y(s) - 2s^3 + s - 5$ (correct answer)
$s^4Y(s) - 5s^3 + s^2 - 2$
$s^4Y(s) - 2s^3 - s^2 + 5s$

Explanation: The formula for the Laplace transform of the fourth derivative is:

\mathcal{L}\{y^{(4)}(t)\} = s^4Y(s) - s^3y(0) - s^2y'(0) - sy''(0) - y'''(0)

. Substitute the given initial conditions into the formula:

= s^4Y(s) - s^3(2) - s^2(0) - s(-1) - (5)

= s^4Y(s) - 2s^3 - 0 + s - 5

= s^4Y(s) - 2s^3 + s - 5

. This corresponds to choice B. Choice A,

s^4Y(s) - 2s^3 - s - 5

, results from a sign error on the

sy''(0)

term. Choice C,

s^4Y(s) - 5s^3 + s^2 - 2

, results from reversing the order of the initial conditions in the formula. Choice D results from misaligning the powers of

s

with the corresponding initial conditions.

Question 13

Given that $\mathcal{L}\{f'(t)\} = sF(s) - 4$ and $\mathcal{L}\{f''(t)\} = s^2F(s) - 4s + 3$ , what is $f'(0)$ ?

$3$
$-3$ (correct answer)
$4$
$-4$

Explanation: From

\mathcal{L}\{f'(t)\} = sF(s) - 4

, we can identify that

\mathcal{L}\{f'(t)\} = sF(s) - f(0)

, so

f(0) = 4

. From

\mathcal{L}\{f''(t)\} = s^2F(s) - 4s + 3

, we use the formula

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

. Substituting

f(0) = 4

s^2F(s) - 4s - f'(0) = s^2F(s) - 4s + 3

. Therefore,

-f'(0) = 3

, which gives

f'(0) = -3

. Choice A confuses the sign. Choice C uses

f(0)

instead of

f'(0)

. Choice D uses

-f(0)