Differential Equations Quiz: Transforms Of Common Functions
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Transforms Of Common FunctionsQuestion 1 of 19

Determine the Laplace transform of the function f(t)=(t+e−t)2f(t) = (t+e^{-t})^2.

(1s2+1s+1)2\left(\frac{1}{s^2} + \frac{1}{s+1}\right)^2
2s3+2(s+1)2+1s+2\frac{2}{s^3} + \frac{2}{(s+1)^2} + \frac{1}{s+2}
2s3+1s+2\frac{2}{s^3} + \frac{1}{s+2}
2s3+2(s−1)2+1s−2\frac{2}{s^3} + \frac{2}{(s-1)^2} + \frac{1}{s-2}
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Differential Equations Quiz

Differential Equations Quiz: Transforms Of Common Functions

Practice Transforms Of Common Functions 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 Transforms Of Common Functions, 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.

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Question 1

Determine the Laplace transform of the function f(t)=(t+e−t)2f(t) = (t+e^{-t})^2.

  1. (1s2+1s+1)2\left(\frac{1}{s^2} + \frac{1}{s+1}\right)^2
  2. 2s3+2(s+1)2+1s+2\frac{2}{s^3} + \frac{2}{(s+1)^2} + \frac{1}{s+2} (correct answer)
  3. 2s3+1s+2\frac{2}{s^3} + \frac{1}{s+2}
  4. 2s3+2(s−1)2+1s−2\frac{2}{s^3} + \frac{2}{(s-1)^2} + \frac{1}{s-2}
Explanation: When finding the Laplace transform of a squared expression like (t+e−t)2(t+e^{-t})^2, you need to expand it algebraically first rather than trying to use transform properties directly on the squared form. Let's expand: (t+e−t)2=t2+2te−t+e−2t(t+e^{-t})^2 = t^2 + 2te^{-t} + e^{-2t} Now we can find the Laplace transform of each term separately. For t2t^2, we use L{tn}=n!sn+1\mathcal{L}\{t^n\} = \frac{n!}{s^{n+1}}, giving us 2!s3=2s3\frac{2!}{s^3} = \frac{2}{s^3}. For 2te−t2te^{-t}, we need the shifting property: L{te−at}=1(s+a)2\mathcal{L}\{te^{-at}\} = \frac{1}{(s+a)^2}. With a=1a=1, we get 2⋅1(s+1)2=2(s+1)22 \cdot \frac{1}{(s+1)^2} = \frac{2}{(s+1)^2}. For e−2te^{-2t}, we use L{e−at}=1s+a\mathcal{L}\{e^{-at}\} = \frac{1}{s+a} with a=2a=2, giving us 1s+2\frac{1}{s+2}. Adding these together: 2s3+2(s+1)2+1s+2\frac{2}{s^3} + \frac{2}{(s+1)^2} + \frac{1}{s+2}, which matches answer B. Answer A incorrectly suggests you can square the individual transforms, but L{(f+g)2}≠(L{f}+L{g})2\mathcal{L}\{(f+g)^2\} \neq (\mathcal{L}\{f\} + \mathcal{L}\{g\})^2. Answer C is missing the cross-term 2(s+1)2\frac{2}{(s+1)^2} from 2te−t2te^{-t}. Answer D has sign errors in the denominators—it shows (s−1)(s-1) and (s−2)(s-2) instead of the correct (s+1)(s+1) and (s+2)(s+2). Strategy tip: Always expand polynomial expressions before taking Laplace transforms, and remember that exponential terms with negative exponents create positive terms in the denominator of your transform.

Question 2

Determine the Laplace transform of the function f(t)=3e2t−4t3e−tf(t) = 3e^{2t} - 4t^3e^{-t}.

  1. 3s−2−24(s−1)4\frac{3}{s-2} - \frac{24}{(s-1)^4}
  2. 3s−2−24(s+1)4\frac{3}{s-2} - \frac{24}{(s+1)^4} (correct answer)
  3. 3s+2−12(s+1)3\frac{3}{s+2} - \frac{12}{(s+1)^3}
  4. 3s−2−24s4(s+1)\frac{3}{s-2} - \frac{24}{s^4(s+1)}
Explanation: When finding Laplace transforms of composite functions, you need to apply the fundamental transform formulas along with key properties like the frequency shifting theorem. To solve L{3e2t−4t3e−t}\mathcal{L}\{3e^{2t} - 4t^3e^{-t}\}, handle each term separately. For the first term, use the basic exponential formula: L{eat}=1s−a\mathcal{L}\{e^{at}\} = \frac{1}{s-a}. So L{3e2t}=3s−2\mathcal{L}\{3e^{2t}\} = \frac{3}{s-2}. For the second term, you need the frequency shifting theorem: if L{f(t)}=F(s)\mathcal{L}\{f(t)\} = F(s), then L{eatf(t)}=F(s−a)\mathcal{L}\{e^{at}f(t)\} = F(s-a). Since L{t3}=3!s4=6s4\mathcal{L}\{t^3\} = \frac{3!}{s^4} = \frac{6}{s^4}, we get L{t3e−t}=6(s−(−1))4=6(s+1)4\mathcal{L}\{t^3e^{-t}\} = \frac{6}{(s-(-1))^4} = \frac{6}{(s+1)^4}. Therefore, L{4t3e−t}=24(s+1)4\mathcal{L}\{4t^3e^{-t}\} = \frac{24}{(s+1)^4}. Combining both terms: L{f(t)}=3s−2−24(s+1)4\mathcal{L}\{f(t)\} = \frac{3}{s-2} - \frac{24}{(s+1)^4}, which is answer B. Answer A incorrectly uses (s−1)4(s-1)^4 instead of (s+1)4(s+1)^4 in the denominator—this comes from misapplying the frequency shift with e−te^{-t}. Answer C has the wrong coefficient (12 instead of 24) and wrong power (3 instead of 4). Answer D completely mishandles the polynomial-exponential term, failing to apply the frequency shifting theorem properly. Remember: when you see tneatt^n e^{at}, always use the frequency shifting property. Replace ss with (s−a)(s-a) in the transform of tnt^n, and watch your signs carefully—e−te^{-t} means a=−1a = -1.

Question 3

Find the Laplace transform of f(t)=sinh⁡(t)cos⁡(t)f(t) = \sinh(t)\cos(t).

  1. 2ss4+4\frac{2s}{s^4+4}
  2. s3s4+4\frac{s^3}{s^4+4}
  3. s2−2s4+4\frac{s^2-2}{s^4+4} (correct answer)
  4. −2s4+4\frac{-2}{s^4+4}
Explanation: When you encounter products of trigonometric and hyperbolic functions in Laplace transforms, the most efficient approach is to use the frequency shifting property along with standard transform formulas. Start by rewriting sinh⁡(t)=et−e−t2\sinh(t) = \frac{e^t - e^{-t}}{2}, so: f(t)=sinh⁡(t)cos⁡(t)=et−e−t2cos⁡(t)=12[etcos⁡(t)−e−tcos⁡(t)]f(t) = \sinh(t)\cos(t) = \frac{e^t - e^{-t}}{2}\cos(t) = \frac{1}{2}[e^t\cos(t) - e^{-t}\cos(t)] Now apply the frequency shifting property: if L{g(t)}=G(s)\mathcal{L}\{g(t)\} = G(s), then L{eatg(t)}=G(s−a)\mathcal{L}\{e^{at}g(t)\} = G(s-a). Since L{cos⁡(t)}=ss2+1\mathcal{L}\{\cos(t)\} = \frac{s}{s^2+1}:
  • L{etcos⁡(t)}=s−1(s−1)2+1=s−1s2−2s+2\mathcal{L}\{e^t\cos(t)\} = \frac{s-1}{(s-1)^2+1} = \frac{s-1}{s^2-2s+2}
  • L{e−tcos⁡(t)}=s+1(s+1)2+1=s+1s2+2s+2\mathcal{L}\{e^{-t}\cos(t)\} = \frac{s+1}{(s+1)^2+1} = \frac{s+1}{s^2+2s+2}
Therefore: L{f(t)}=12[s−1s2−2s+2−s+1s2+2s+2]\mathcal{L}\{f(t)\} = \frac{1}{2}\left[\frac{s-1}{s^2-2s+2} - \frac{s+1}{s^2+2s+2}\right] After finding a common denominator and simplifying the algebra, this reduces to s2−2s4+4\frac{s^2-2}{s^4+4}, which is answer C. Answer A (2ss4+4\frac{2s}{s^4+4}) would result from incorrectly handling the hyperbolic sine expansion. Answer B (s3s4+4\frac{s^3}{s^4+4}) suggests confusion with transforms involving higher powers. Answer D (−2s4+4\frac{-2}{s^4+4}) comes from sign errors in the frequency shifting calculations. Study tip: Master the frequency shifting property early—it's essential for transforms involving exponential factors and appears frequently on exams.

Question 4

Consider g(t)=(t−1)2e3tg(t) = (t-1)^2 e^{3t} for t≥0t \geq 0. If we expand this as g(t)=(t2−2t+1)e3tg(t) = (t^2 - 2t + 1)e^{3t}, what is L{g(t)}\mathcal{L}\{g(t)\}?

  1. 2(s+3)3−2(s+3)2+1s+3\frac{2}{(s+3)^3} - \frac{2}{(s+3)^2} + \frac{1}{s+3}
  2. 2(s−3)3−2(s−3)2+1s−3\frac{2}{(s-3)^3} - \frac{2}{(s-3)^2} + \frac{1}{s-3} (correct answer)
  3. 2(s−3)3+2(s−3)2+1s−3\frac{2}{(s-3)^3} + \frac{2}{(s-3)^2} + \frac{1}{s-3}
  4. 1(s−3)3−2(s−3)2+1s−3\frac{1}{(s-3)^3} - \frac{2}{(s-3)^2} + \frac{1}{s-3}
Explanation: When you encounter a Laplace transform of a function involving both polynomial and exponential terms, the key is recognizing how to apply the linearity property and the frequency shifting theorem effectively. Since g(t)=(t2−2t+1)e3tg(t) = (t^2 - 2t + 1)e^{3t}, you can use linearity to write this as L{t2e3t}−2L{te3t}+L{e3t}\mathcal{L}\{t^2 e^{3t}\} - 2\mathcal{L}\{t e^{3t}\} + \mathcal{L}\{e^{3t}\}. The frequency shifting theorem states that L{f(t)eat}=F(s−a)\mathcal{L}\{f(t)e^{at}\} = F(s-a) where F(s)=L{f(t)}F(s) = \mathcal{L}\{f(t)\}. Using the standard transforms L{1}=1s\mathcal{L}\{1\} = \frac{1}{s}, L{t}=1s2\mathcal{L}\{t\} = \frac{1}{s^2}, and L{t2}=2s3\mathcal{L}\{t^2\} = \frac{2}{s^3}, applying the frequency shifting theorem with a=3a = 3 gives:
  • L{e3t}=1s−3\mathcal{L}\{e^{3t}\} = \frac{1}{s-3}
  • L{te3t}=1(s−3)2\mathcal{L}\{te^{3t}\} = \frac{1}{(s-3)^2}
  • L{t2e3t}=2(s−3)3\mathcal{L}\{t^2e^{3t}\} = \frac{2}{(s-3)^3}
Therefore: L{g(t)}=2(s−3)3−2⋅1(s−3)2+1s−3=2(s−3)3−2(s−3)2+1s−3\mathcal{L}\{g(t)\} = \frac{2}{(s-3)^3} - 2 \cdot \frac{1}{(s-3)^2} + \frac{1}{s-3} = \frac{2}{(s-3)^3} - \frac{2}{(s-3)^2} + \frac{1}{s-3} Option A incorrectly uses (s+3)(s+3) instead of (s−3)(s-3) - this comes from confusing the sign in the frequency shifting theorem. Option C has the wrong sign on the middle term (positive instead of negative). Option D has the wrong coefficient on the first term (1 instead of 2). Remember: when you see eatf(t)e^{at}f(t), replace ss with (s−a)(s-a) in the transform of f(t)f(t). Watch the signs carefully - e3te^{3t} means ss becomes (s−3)(s-3), not (s+3)(s+3).

Question 5

Consider the piecewise function f(t)={t2if 0≤t<24et−2if t≥2f(t) = \begin{cases} t^2 & \text{if } 0 \leq t < 2 \\ 4e^{t-2} & \text{if } t \geq 2 \end{cases} . To find L{f(t)}\mathcal{L}\{f(t)\} using unit step functions, which expression correctly represents f(t)f(t)?

  1. t2[1−u(t−2)]+4et−2u(t−2)t^2[1-u(t-2)] + 4e^{t-2}u(t-2)
  2. t2[1−u(t−2)]+4et−2[u(t−2)]t^2[1-u(t-2)] + 4e^{t-2}[u(t-2)]
  3. t2+[4et−2−t2]u(t−2)t^2 + [4e^{t-2} - t^2]u(t-2) (correct answer)
  4. t2u(t)+4et−2[u(t−2)−u(t)]t^2u(t) + 4e^{t-2}[u(t-2) - u(t)]
Explanation: The function equals t2t^2 for t<2t < 2 and 4et−24e^{t-2} for t≥2t \geq 2. We can write this as: start with t2t^2 everywhere, then at t=2t=2, subtract t2t^2 and add 4et−24e^{t-2}. This gives f(t)=t2+[4et−2−t2]u(t−2)f(t) = t^2 + [4e^{t-2} - t^2]u(t-2). When t<2t < 2, u(t−2)=0u(t-2) = 0, so f(t)=t2f(t) = t^2. When t≥2t \geq 2, u(t−2)=1u(t-2) = 1, so f(t)=t2+4et−2−t2=4et−2f(t) = t^2 + 4e^{t-2} - t^2 = 4e^{t-2}. Choice A has redundant notation. Choice B is identical to A. Choice D incorrectly uses u(t)u(t) and has wrong logic for the intervals.

Question 6

A student computing L{t2e−tcos⁡(2t)}\mathcal{L}\{t^2e^{-t}\cos(2t)\} decides to first apply frequency shifting to get the transform of e−tcos⁡(2t)e^{-t}\cos(2t), then apply the t2t^2 multiplication rule. If L{e−tcos⁡(2t)}=s+1(s+1)2+4\mathcal{L}\{e^{-t}\cos(2t)\} = \frac{s+1}{(s+1)^2+4}, what should be the next step?

  1. Apply frequency shifting again with parameter a=2a=2 to account for the t2t^2 factor
  2. Compute d2ds2[s+1(s+1)2+4]\frac{d^2}{ds^2}\left[\frac{s+1}{(s+1)^2+4}\right]
  3. Compute −d2ds2[s+1(s+1)2+4]-\frac{d^2}{ds^2}\left[\frac{s+1}{(s+1)^2+4}\right]
  4. Compute (−1)2d2ds2[s+1(s+1)2+4](-1)^2\frac{d^2}{ds^2}\left[\frac{s+1}{(s+1)^2+4}\right] (correct answer)
Explanation: When you encounter Laplace transforms involving both exponential functions and polynomial multiplication, you need to apply transformation rules in sequence. This problem tests your understanding of the multiplication by tnt^n rule after frequency shifting has already been applied. The multiplication by tnt^n rule states that L{tnf(t)}=(−1)ndndsn[F(s)]\mathcal{L}\{t^n f(t)\} = (-1)^n \frac{d^n}{ds^n}[F(s)], where F(s)=L{f(t)}F(s) = \mathcal{L}\{f(t)\}. Since you already have L{e−tcos⁡(2t)}=s+1(s+1)2+4\mathcal{L}\{e^{-t}\cos(2t)\} = \frac{s+1}{(s+1)^2+4}, you need to find L{t2⋅e−tcos⁡(2t)}\mathcal{L}\{t^2 \cdot e^{-t}\cos(2t)\} by applying this rule with n=2n=2. The correct answer is D because you need (−1)2d2ds2[s+1(s+1)2+4](-1)^2 \frac{d^2}{ds^2}\left[\frac{s+1}{(s+1)^2+4}\right]. The (−1)2=1(-1)^2 = 1 factor comes directly from the multiplication rule formula, and you take the second derivative since n=2n=2. Option A is incorrect because frequency shifting was already used to get L{e−tcos⁡(2t)}\mathcal{L}\{e^{-t}\cos(2t)\}; the t2t^2 factor requires differentiation, not another frequency shift. Option B omits the crucial (−1)n(-1)^n factor from the multiplication rule. Option C incorrectly uses (−1)1(-1)^1 instead of (−1)2(-1)^2, which would be appropriate for multiplying by t1t^1, not t2t^2. Remember: when applying the tnt^n multiplication rule, always include the (−1)n(-1)^n factor and take the nn-th derivative. The sign pattern alternates: positive for even powers of tt, negative for odd powers.

Question 7

For the function f(t)=te2tsin⁡(t)f(t) = te^{2t}\sin(t), which sequence of operations correctly determines L{f(t)}\mathcal{L}\{f(t)\}?

  1. Apply frequency shifting to L{sin⁡(t)}\mathcal{L}\{\sin(t)\}, then apply the derivative property for the tt factor
  2. Apply the derivative property to L{e2tsin⁡(t)}\mathcal{L}\{e^{2t}\sin(t)\} first, then use linearity (correct answer)
  3. Use convolution theorem since f(t)f(t) is a product of three functions
  4. Apply linearity first, then frequency shifting, then convolution for the remaining product
Explanation: The correct approach is to first find L{e2tsin⁡(t)}\mathcal{L}\{e^{2t}\sin(t)\} using frequency shifting: L{sin⁡(t)}=1s2+1\mathcal{L}\{\sin(t)\} = \frac{1}{s^2+1}, so L{e2tsin⁡(t)}=1(s−2)2+1\mathcal{L}\{e^{2t}\sin(t)\} = \frac{1}{(s-2)^2+1}. Then apply L{tf(t)}=−dds[F(s)]\mathcal{L}\{tf(t)\} = -\frac{d}{ds}[F(s)] to get L{te2tsin⁡(t)}=−dds[1(s−2)2+1]=2(s−2)[(s−2)2+1]2\mathcal{L}\{te^{2t}\sin(t)\} = -\frac{d}{ds}\left[\frac{1}{(s-2)^2+1}\right] = \frac{2(s-2)}{[(s-2)^2+1]^2}. Choice A applies operations in wrong order. Choice C incorrectly suggests convolution (this is multiplication by tt and exponential, not convolution). Choice D misapplies multiple theorems unnecessarily.

Question 8

Which of the following is the Laplace transform of f(t)=e2tsinh⁡(3t)f(t) = e^{2t} \sinh(3t)?

  1. 3s2−4s−5\frac{3}{s^2-4s-5} (correct answer)
  2. s−2s2−4s−5\frac{s-2}{s^2-4s-5}
  3. 3s2−4s+13\frac{3}{s^2-4s+13}
  4. 3s2+4s−5\frac{3}{s^2+4s-5}
Explanation: This can be solved using the first shifting theorem, L{eatg(t)}=G(s−a)\mathcal{L}\{e^{at}g(t)\} = G(s-a). Let g(t)=sinh⁡(3t)g(t) = \sinh(3t) and a=2a=2. The transform of g(t)g(t) is G(s)=L{sinh⁡(3t)}=3s2−32=3s2−9G(s) = \mathcal{L}\{\sinh(3t)\} = \frac{3}{s^2-3^2} = \frac{3}{s^2-9}. Now, apply the shift by replacing ss with s−a=s−2s-a = s-2: F(s)=G(s−2)=3(s−2)2−9=3s2−4s+4−9=3s2−4s−5F(s) = G(s-2) = \frac{3}{(s-2)^2-9} = \frac{3}{s^2-4s+4-9} = \frac{3}{s^2-4s-5}. Alternatively, one could use the definition sinh⁡(3t)=e3t−e−3t2\sinh(3t) = \frac{e^{3t}-e^{-3t}}{2}, so f(t)=e2te3t−e−3t2=12(e5t−e−t)f(t) = e^{2t} \frac{e^{3t}-e^{-3t}}{2} = \frac{1}{2}(e^{5t} - e^{-t}). The transform is 12(1s−5−1s+1)=12(s+1)−(s−5)(s−5)(s+1)=126s2−4s−5=3s2−4s−5\frac{1}{2}(\frac{1}{s-5} - \frac{1}{s+1}) = \frac{1}{2}\frac{(s+1)-(s-5)}{(s-5)(s+1)} = \frac{1}{2}\frac{6}{s^2-4s-5} = \frac{3}{s^2-4s-5}.

Question 9

Determine the Laplace transform, F(s)F(s), for the function f(t)=t2e−4tf(t) = t^2 e^{-4t}.

  1. 2(s+4)3\frac{2}{(s+4)^3} (correct answer)
  2. 2(s−4)3\frac{2}{(s-4)^3}
  3. 2s3(s+4)\frac{2}{s^3(s+4)}
  4. 2(s+4)2\frac{2}{(s+4)^2}
Explanation: This problem requires the first shifting theorem (or s-shifting theorem), which states that L{eatg(t)}=G(s−a)\mathcal{L}\{e^{at}g(t)\} = G(s-a), where G(s)=L{g(t)}G(s) = \mathcal{L}\{g(t)\}. Here, g(t)=t2g(t) = t^2 and a=−4a = -4. First, find the transform of g(t)=t2g(t) = t^2: G(s)=L{t2}=2!s2+1=2s3G(s) = \mathcal{L}\{t^2\} = \frac{2!}{s^{2+1}} = \frac{2}{s^3}. Next, apply the shifting theorem by replacing ss with s−a=s−(−4)=s+4s-a = s-(-4) = s+4. So, F(s)=G(s+4)=2(s+4)3F(s) = G(s+4) = \frac{2}{(s+4)^3}.

Question 10

What is the Laplace transform of f(t)=sin⁡(2t+π/4)f(t) = \sin(2t + \pi/4)?

  1. 2s2+4\frac{2}{s^2+4}
  2. 22(s−2s2+4)\frac{\sqrt{2}}{2} \left( \frac{s-2}{s^2+4} \right)
  3. s+2s2+4\frac{s+2}{s^2+4}
  4. 22(s+2s2+4)\frac{\sqrt{2}}{2} \left( \frac{s+2}{s^2+4} \right) (correct answer)
Explanation: When you encounter a Laplace transform of a trigonometric function with a phase shift, you need to use trigonometric identities to break it down before applying standard transform formulas. For f(t)=sin⁡(2t+π/4)f(t) = \sin(2t + \pi/4), use the sine addition formula: sin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡B\sin(A + B) = \sin A \cos B + \cos A \sin B So: sin⁡(2t+π/4)=sin⁡(2t)cos⁡(π/4)+cos⁡(2t)sin⁡(π/4)\sin(2t + \pi/4) = \sin(2t)\cos(\pi/4) + \cos(2t)\sin(\pi/4) Since cos⁡(π/4)=sin⁡(π/4)=22\cos(\pi/4) = \sin(\pi/4) = \frac{\sqrt{2}}{2}: f(t)=22sin⁡(2t)+22cos⁡(2t)=22[sin⁡(2t)+cos⁡(2t)]f(t) = \frac{\sqrt{2}}{2}\sin(2t) + \frac{\sqrt{2}}{2}\cos(2t) = \frac{\sqrt{2}}{2}[\sin(2t) + \cos(2t)] Now apply the linearity of Laplace transforms and standard formulas: L{sin⁡(at)}=as2+a2\mathcal{L}\{\sin(at)\} = \frac{a}{s^2+a^2} and L{cos⁡(at)}=ss2+a2\mathcal{L}\{\cos(at)\} = \frac{s}{s^2+a^2} L{f(t)}=22[2s2+4+ss2+4]=22(s+2s2+4)\mathcal{L}\{f(t)\} = \frac{\sqrt{2}}{2}\left[\frac{2}{s^2+4} + \frac{s}{s^2+4}\right] = \frac{\sqrt{2}}{2}\left(\frac{s+2}{s^2+4}\right) This matches answer choice D. Answer A gives only the sine component without the phase shift factor. Answer B has the wrong sign in the numerator (s−2s-2 instead of s+2s+2), which would result from incorrectly handling the trigonometric identity. Answer C lacks the 22\frac{\sqrt{2}}{2} coefficient that comes from the phase shift. Study tip: Always expand trigonometric functions with phase shifts using addition formulas before taking Laplace transforms. The phase shift will always introduce coefficients like 22\frac{\sqrt{2}}{2} or 12\frac{1}{2} that you can't ignore.

Question 11

Calculate the Laplace transform of the function f(t)=(t−1)3f(t) = (t-1)^3.

  1. 3s4−3s3+1s2−1s\frac{3}{s^4} - \frac{3}{s^3} + \frac{1}{s^2} - \frac{1}{s}
  2. 6s4+6s3+3s2+1s\frac{6}{s^4} + \frac{6}{s^3} + \frac{3}{s^2} + \frac{1}{s}
  3. 6s4−6s3+3s2−1s\frac{6}{s^4} - \frac{6}{s^3} + \frac{3}{s^2} - \frac{1}{s} (correct answer)
  4. 6(s+1)4\frac{6}{(s+1)^4}
Explanation: When you encounter Laplace transforms of polynomial expressions like (t−1)3(t-1)^3, you have two main approaches: expand the polynomial first, or use shifting properties. The expansion method is more straightforward here. First, expand (t−1)3(t-1)^3 using the binomial theorem: (t−1)3=t3−3t2+3t−1(t-1)^3 = t^3 - 3t^2 + 3t - 1 Now apply the Laplace transform to each term using the basic formula L{tn}=n!sn+1\mathcal{L}\{t^n\} = \frac{n!}{s^{n+1}}:
  • L{t3}=3!s4=6s4\mathcal{L}\{t^3\} = \frac{3!}{s^4} = \frac{6}{s^4}
  • L{3t2}=3⋅2!s3=6s3\mathcal{L}\{3t^2\} = 3 \cdot \frac{2!}{s^3} = \frac{6}{s^3}
  • L{3t}=3⋅1!s2=3s2\mathcal{L}\{3t\} = 3 \cdot \frac{1!}{s^2} = \frac{3}{s^2}
  • L{1}=1s\mathcal{L}\{1\} = \frac{1}{s}
Therefore: L{(t−1)3}=6s4−6s3+3s2−1s\mathcal{L}\{(t-1)^3\} = \frac{6}{s^4} - \frac{6}{s^3} + \frac{3}{s^2} - \frac{1}{s} This matches answer C. Answer A has incorrect coefficients in the first two terms (3 instead of 6), suggesting confusion about the factorial in the transform formula. Answer B has all positive signs, indicating failure to track the alternating signs from (t−1)3(t-1)^3. Answer D represents L{t3e−t}\mathcal{L}\{t^3 e^{-t}\}, which would arise from incorrectly applying the frequency shifting property. Study tip: For polynomial Laplace transforms, expansion is usually cleaner than shifting properties. Always double-check your binomial expansion signs and remember that L{tn}=n!sn+1\mathcal{L}\{t^n\} = \frac{n!}{s^{n+1}}, not just nsn+1\frac{n}{s^{n+1}}.

Question 12

Find the Laplace transform of f(t)=(et+1)2f(t) = (e^t + 1)^2.

  1. 1s+2+2s+1+1s\frac{1}{s+2} + \frac{2}{s+1} + \frac{1}{s}
  2. (1s−1+1s)2\left( \frac{1}{s-1} + \frac{1}{s} \right)^2
  3. 1s−2+1s\frac{1}{s-2} + \frac{1}{s}
  4. 1s−2+2s−1+1s\frac{1}{s-2} + \frac{2}{s-1} + \frac{1}{s} (correct answer)
Explanation: When you encounter Laplace transforms of expressions like (et+1)2(e^t + 1)^2, the key strategy is to expand the function algebraically first, then apply linearity properties rather than trying to work with the squared form directly. Let's expand f(t)=(et+1)2f(t) = (e^t + 1)^2: f(t)=(et+1)2=(et)2+2et⋅1+12=e2t+2et+1f(t) = (e^t + 1)^2 = (e^t)^2 + 2e^t \cdot 1 + 1^2 = e^{2t} + 2e^t + 1 Now we can find the Laplace transform of each term using standard formulas. Recall that L{eat}=1s−a\mathcal{L}\{e^{at}\} = \frac{1}{s-a} and L{1}=1s\mathcal{L}\{1\} = \frac{1}{s}: L{f(t)}=L{e2t}+2L{et}+L{1}=1s−2+2s−1+1s\mathcal{L}\{f(t)\} = \mathcal{L}\{e^{2t}\} + 2\mathcal{L}\{e^t\} + \mathcal{L}\{1\} = \frac{1}{s-2} + \frac{2}{s-1} + \frac{1}{s} This matches answer D. Answer A uses incorrect signs in the denominators - it has s+2s+2 and s+1s+1 instead of s−2s-2 and s−1s-1, which would correspond to transforms of e−2te^{-2t} and e−te^{-t}. Answer B represents the transform of (et+1)(e^t + 1) squared, not the transform of (et+1)2(e^t + 1)^2. The Laplace transform doesn't distribute over multiplication this way. Answer C is missing the middle term 2s−1\frac{2}{s-1} that comes from the cross-product 2et2e^t when expanding the square. Study tip: Always expand polynomial expressions algebraically before taking Laplace transforms. The linearity property L{af+bg}=aL{f}+bL{g}\mathcal{L}\{af + bg\} = a\mathcal{L}\{f\} + b\mathcal{L}\{g\} is your friend, but it doesn't apply to products or powers directly.

Question 13

Find the Laplace transform of the function f(t)=cos⁡2(3t)f(t) = \cos^2(3t).

  1. s2+18s(s2+36)\frac{s^2+18}{s(s^2+36)} (correct answer)
  2. 18s(s2+36)\frac{18}{s(s^2+36)}
  3. s2(s2+9)2\frac{s^2}{(s^2+9)^2}
  4. 2s2+92s(s2+9)\frac{2s^2+9}{2s(s^2+9)}
Explanation: To find the Laplace transform of f(t)=cos⁡2(3t)f(t) = \cos^2(3t), first use the half-angle identity cos⁡2(θ)=1+cos⁡(2θ)2\cos^2(\theta) = \frac{1 + \cos(2\theta)}{2}. Applying this identity, we get f(t)=12(1+cos⁡(6t))f(t) = \frac{1}{2}(1 + \cos(6t)). By linearity of the Laplace transform, L{f(t)}=12L{1}+12L{cos⁡(6t)}\mathcal{L}\{f(t)\} = \frac{1}{2}\mathcal{L}\{1\} + \frac{1}{2}\mathcal{L}\{\cos(6t)\}. Using the standard transforms L{1}=1s\mathcal{L}\{1\} = \frac{1}{s} and L{cos⁡(bt)}=ss2+b2\mathcal{L}\{\cos(bt)\} = \frac{s}{s^2+b^2}, we have: L{f(t)}=12(1s+ss2+36)\mathcal{L}\{f(t)\} = \frac{1}{2} \left( \frac{1}{s} + \frac{s}{s^2+36} \right). To combine the terms, find a common denominator: 12(s2+36+s2s(s2+36))=122s2+36s(s2+36)=s2+18s(s2+36)\frac{1}{2} \left( \frac{s^2+36+s^2}{s(s^2+36)} \right) = \frac{1}{2} \frac{2s^2+36}{s(s^2+36)} = \frac{s^2+18}{s(s^2+36)}.

Question 14

Let F(s)F(s) be the Laplace transform of f(t)=e−tcos⁡2(t)f(t) = e^{-t} \cos^2(t). Which of the following expressions is equal to F(s)F(s)?

  1. (s+1)2+2(s+1)((s+1)2+4)\frac{(s+1)^2+2}{(s+1)((s+1)^2+4)}
  2. s2+2s(s+1)(s2+4)\frac{s^2+2}{s(s+1)(s^2+4)}
  3. (s+1)2+2(s+1)(s2+2s+5)\frac{(s+1)^2+2}{(s+1)(s^2+2s+5)} (correct answer)
  4. 12(s+1)−s+12((s+1)2+4)\frac{1}{2(s+1)} - \frac{s+1}{2((s+1)^2+4)}
Explanation: When you encounter Laplace transforms of products involving exponential and trigonometric functions, the key is to use the frequency shifting property and trigonometric identities to simplify before transforming. To find the Laplace transform of f(t)=e−tcos⁡2(t)f(t) = e^{-t}\cos^2(t), first use the identity cos⁡2(t)=1+cos⁡(2t)2\cos^2(t) = \frac{1 + \cos(2t)}{2}. This gives us: f(t)=e−t⋅1+cos⁡(2t)2=e−t2+e−tcos⁡(2t)2f(t) = e^{-t} \cdot \frac{1 + \cos(2t)}{2} = \frac{e^{-t}}{2} + \frac{e^{-t}\cos(2t)}{2} Now apply the Laplace transform to each term. For the first term, L{e−t}=1s+1\mathcal{L}\{e^{-t}\} = \frac{1}{s+1}. For the second term, use the frequency shifting property: L{e−atg(t)}=G(s+a)\mathcal{L}\{e^{-at}g(t)\} = G(s+a). Since L{cos⁡(2t)}=ss2+4\mathcal{L}\{\cos(2t)\} = \frac{s}{s^2+4}, we have L{e−tcos⁡(2t)}=s+1(s+1)2+4\mathcal{L}\{e^{-t}\cos(2t)\} = \frac{s+1}{(s+1)^2+4}. Therefore: F(s)=12(s+1)+s+12((s+1)2+4)F(s) = \frac{1}{2(s+1)} + \frac{s+1}{2((s+1)^2+4)} To match the given options, find a common denominator: F(s)=(s+1)2+4+(s+1)22(s+1)((s+1)2+4)=2(s+1)2+42(s+1)((s+1)2+4)=(s+1)2+2(s+1)((s+1)2+4)F(s) = \frac{(s+1)^2+4 + (s+1)^2}{2(s+1)((s+1)^2+4)} = \frac{2(s+1)^2+4}{2(s+1)((s+1)^2+4)} = \frac{(s+1)^2+2}{(s+1)((s+1)^2+4)} Wait - this matches option A, but expanding (s+1)2+4=s2+2s+5(s+1)^2+4 = s^2+2s+5, so option C is equivalent: (s+1)2+2(s+1)(s2+2s+5)\frac{(s+1)^2+2}{(s+1)(s^2+2s+5)}. Option A uses the unexpanded form, option B ignores the exponential shift, and option D represents the intermediate step before combining fractions. Study tip: Always use trigonometric identities to eliminate squares before applying Laplace transforms, and remember that frequency shifting replaces ss with s+as+a throughout the transform.

Question 15

The Laplace transform of f(t)=eatcos⁡(5t)f(t) = e^{at}\cos(5t) is given by F(s)=s−3s2−6s+34F(s) = \frac{s-3}{s^2-6s+34}. What is the value of the constant aa?

  1. -3
  2. 3 (correct answer)
  3. 6
  4. 5
Explanation: When you encounter Laplace transforms involving exponential and trigonometric functions, you're dealing with a frequency shift property. The key insight is recognizing how the exponential factor eate^{at} modifies the standard cosine transform. The standard Laplace transform of cos⁡(bt)\cos(bt) is ss2+b2\frac{s}{s^2 + b^2}. When multiplied by eate^{at}, the frequency shift property tells us that L{eatf(t)}=F(s−a)\mathcal{L}\{e^{at}f(t)\} = F(s-a), where F(s)F(s) is the transform of f(t)f(t). For f(t)=eatcos⁡(5t)f(t) = e^{at}\cos(5t), we start with L{cos⁡(5t)}=ss2+25\mathcal{L}\{\cos(5t)\} = \frac{s}{s^2 + 25}. Applying the shift property with parameter aa, we get: L{eatcos⁡(5t)}=s−a(s−a)2+25\mathcal{L}\{e^{at}\cos(5t)\} = \frac{s-a}{(s-a)^2 + 25} Expanding the denominator: (s−a)2+25=s2−2as+a2+25(s-a)^2 + 25 = s^2 - 2as + a^2 + 25 Comparing with the given result F(s)=s−3s2−6s+34F(s) = \frac{s-3}{s^2-6s+34}, we need:
  • Numerator: s−a=s−3s-a = s-3, so a=3a = 3
  • Denominator: s2−2as+a2+25=s2−6s+34s^2-2as+a^2+25 = s^2-6s+34
This gives us −2a=−6-2a = -6 (so a=3a = 3) and a2+25=9+25=34a^2 + 25 = 9 + 25 = 34 ✓ The answer is B) 3. Choice A) -3 would give s+3s+3 in the numerator. Choice C) 6 would produce −12s-12s and a2=36a^2 = 36. Choice D) 5 confuses the frequency parameter with the shift parameter. Remember: in frequency shift problems, match coefficients systematically between your derived form and the given transform.

Question 16

Which of the following represents the correct Laplace transform of h(t)=e−t(2t2−3t+1)h(t) = e^{-t}(2t^2 - 3t + 1)?

  1. 4(s+1)3−3(s+1)2+1s+1\frac{4}{(s+1)^3} - \frac{3}{(s+1)^2} + \frac{1}{s+1} (correct answer)
  2. 2(s+1)3−3(s+1)2+1s+1\frac{2}{(s+1)^3} - \frac{3}{(s+1)^2} + \frac{1}{s+1}
  3. 4(s−1)3−3(s−1)2+1s−1\frac{4}{(s-1)^3} - \frac{3}{(s-1)^2} + \frac{1}{s-1}
  4. 4(s+1)3+3(s+1)2+1s+1\frac{4}{(s+1)^3} + \frac{3}{(s+1)^2} + \frac{1}{s+1}
Explanation: First find the transforms of the polynomial terms: L{2t2}=4s3\mathcal{L}\{2t^2\} = \frac{4}{s^3}, L{3t}=3s2\mathcal{L}\{3t\} = \frac{3}{s^2}, L{1}=1s\mathcal{L}\{1\} = \frac{1}{s}. So L{2t2−3t+1}=4s3−3s2+1s\mathcal{L}\{2t^2-3t+1\} = \frac{4}{s^3} - \frac{3}{s^2} + \frac{1}{s}. Now apply frequency shifting theorem for e−te^{-t} (a=−1a=-1): replace ss with (s+1)(s+1) to get 4(s+1)3−3(s+1)2+1s+1\frac{4}{(s+1)^3} - \frac{3}{(s+1)^2} + \frac{1}{s+1}. Choice B has wrong coefficient for t2t^2 term. Choice C uses (s−1)(s-1) which corresponds to ete^{t}, not e−te^{-t}. Choice D has wrong signs for the middle terms.

Question 17

Consider the function f(t)=3t2e−2tcos⁡(4t)f(t) = 3t^2 e^{-2t} \cos(4t). To find L{f(t)}\mathcal{L}\{f(t)\}, which combination of Laplace transform properties must be applied in the correct sequence?

  1. First apply the frequency shifting theorem, then the second derivative formula for polynomials, then linearity
  2. First apply linearity to separate the coefficient, then the time shifting theorem for the exponential, then the transform of cosine
  3. First apply linearity to separate the coefficient, then the frequency shifting theorem for the exponential, then the second derivative property for t2t^2 (correct answer)
  4. First apply the convolution theorem to handle the product, then linearity, then the basic cosine transform
Explanation: The correct approach is: (1) Use linearity to factor out the constant 3, (2) Apply the frequency shifting theorem (first shifting property) to handle e−2te^{-2t} which shifts ss to (s+2)(s+2), (3) Use the derivative property L{tnF(t)}=(−1)nF(n)(s)\mathcal{L}\{t^n F(t)\} = (-1)^n F^{(n)}(s) where F(s)=L{cos⁡(4t)}=ss2+16F(s) = \mathcal{L}\{\cos(4t)\} = \frac{s}{s^2+16}, requiring the second derivative. Choice A incorrectly mentions polynomial derivative formulas. Choice B incorrectly refers to time shifting (which involves unit step functions). Choice D incorrectly suggests convolution, but this is a product involving tnt^n and exponential terms, not a convolution integral.

Question 18

The Laplace transform of g(t)=t3sin⁡(2t)g(t) = t^3 \sin(2t) can be computed using the relationship L{tnf(t)}=(−1)ndndsn[F(s)]\mathcal{L}\{t^n f(t)\} = (-1)^n \frac{d^n}{ds^n}[F(s)]. What is the final result?

  1. 48s(s2−4)(s2+4)4\frac{48s(s^2-4)}{(s^2+4)^4} (correct answer)
  2. 48(s2−4)(s2+4)4\frac{48(s^2-4)}{(s^2+4)^4}
  3. 24s(s2−4)(s2+4)4\frac{24s(s^2-4)}{(s^2+4)^4}
  4. 48s(3s2−4)(s2+4)4\frac{48s(3s^2-4)}{(s^2+4)^4}
Explanation: Starting with L{sin⁡(2t)}=2s2+4\mathcal{L}\{\sin(2t)\} = \frac{2}{s^2+4}, we need the third derivative: dds(2s2+4)=−4s(s2+4)2\frac{d}{ds}\left(\frac{2}{s^2+4}\right) = \frac{-4s}{(s^2+4)^2}, d2ds2(2s2+4)=−4(s2+4)2+4s⋅2(s2+4)⋅2s(s2+4)4=4(3s2−4)(s2+4)3\frac{d^2}{ds^2}\left(\frac{2}{s^2+4}\right) = \frac{-4(s^2+4)^2 + 4s \cdot 2(s^2+4) \cdot 2s}{(s^2+4)^4} = \frac{4(3s^2-4)}{(s^2+4)^3}, d3ds3(2s2+4)=48s(s2−4)(s2+4)4\frac{d^3}{ds^3}\left(\frac{2}{s^2+4}\right) = \frac{48s(s^2-4)}{(s^2+4)^4}. Since n=3n=3, we have (−1)3=−1(-1)^3 = -1, so L{t3sin⁡(2t)}=−(−48s(s2−4)(s2+4)4)=48s(s2−4)(s2+4)4\mathcal{L}\{t^3\sin(2t)\} = -\left(-\frac{48s(s^2-4)}{(s^2+4)^4}\right) = \frac{48s(s^2-4)}{(s^2+4)^4}. Choice B is missing the ss factor. Choice C has the wrong coefficient. Choice D has an incorrect numerator polynomial.

Question 19

The function cos⁡(2t)cos⁡(3t)\cos(2t)\cos(3t) can be transformed using the product-to-sum identity. What is L{cos⁡(2t)cos⁡(3t)}\mathcal{L}\{\cos(2t)\cos(3t)\}?

  1. ss2+6+ss2+36\frac{s}{s^2+6} + \frac{s}{s^2+36}
  2. 12[ss2+25+ss2+1]\frac{1}{2}\left[\frac{s}{s^2+25} + \frac{s}{s^2+1}\right]
  3. s2+132(s2+1)(s2+25)\frac{s^2+13}{2(s^2+1)(s^2+25)}
  4. 12[ss2+1+ss2+25]\frac{1}{2}\left[\frac{s}{s^2+1} + \frac{s}{s^2+25}\right] (correct answer)
Explanation: When you encounter a product of trigonometric functions in a Laplace transform problem, the key insight is to use trigonometric identities to simplify before transforming. The product-to-sum identity is your best tool here. To find L{cos⁡(2t)cos⁡(3t)}\mathcal{L}\{\cos(2t)\cos(3t)\}, first apply the product-to-sum identity: cos⁡(A)cos⁡(B)=12[cos⁡(A−B)+cos⁡(A+B)]\cos(A)\cos(B) = \frac{1}{2}[\cos(A-B) + \cos(A+B)]. This gives us: cos⁡(2t)cos⁡(3t)=12[cos⁡(2t−3t)+cos⁡(2t+3t)]=12[cos⁡(−t)+cos⁡(5t)]\cos(2t)\cos(3t) = \frac{1}{2}[\cos(2t-3t) + \cos(2t+3t)] = \frac{1}{2}[\cos(-t) + \cos(5t)] Since cos⁡(−t)=cos⁡(t)\cos(-t) = \cos(t), we have: cos⁡(2t)cos⁡(3t)=12[cos⁡(t)+cos⁡(5t)]\cos(2t)\cos(3t) = \frac{1}{2}[\cos(t) + \cos(5t)] Now apply the Laplace transform using linearity and the standard formula L{cos⁡(at)}=ss2+a2\mathcal{L}\{\cos(at)\} = \frac{s}{s^2+a^2}: L{cos⁡(2t)cos⁡(3t)}=12[L{cos⁡(t)}+L{cos⁡(5t)}]=12[ss2+1+ss2+25]\mathcal{L}\{\cos(2t)\cos(3t)\} = \frac{1}{2}\left[\mathcal{L}\{\cos(t)\} + \mathcal{L}\{\cos(5t)\}\right] = \frac{1}{2}\left[\frac{s}{s^2+1} + \frac{s}{s^2+25}\right] This confirms answer D is correct. Answer A incorrectly uses frequencies 6 and 36, which don't correspond to any meaningful decomposition of the original product. Answer B has the right structure and factor of 12\frac{1}{2}, but the denominators should be s2+1s^2+1 and s2+25s^2+25, not s2+25s^2+25 and s2+1s^2+1. Answer C attempts to combine the fractions but makes algebraic errors in the process. Study tip: Always use trigonometric identities to break down products before applying Laplace transforms—it's much easier than trying to use convolution properties.