Differential Equations Quiz: Definition And Basic Transform Rules
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Definition And Basic Transform RulesQuestion 1 of 19

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

s3−ss4+4\frac{s^3-s}{s^4+4}
s2(s2−1)(s2+1)\frac{s^2}{(s^2-1)(s^2+1)}
s2s4+4\frac{s^2}{s^4+4}
s3s4+4\frac{s^3}{s^4+4}
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Differential Equations Quiz

Differential Equations Quiz: Definition And Basic Transform Rules

Practice Definition And Basic Transform Rules 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 Definition And Basic Transform Rules, 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

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

  1. s3−ss4+4\frac{s^3-s}{s^4+4}
  2. s2(s2−1)(s2+1)\frac{s^2}{(s^2-1)(s^2+1)}
  3. s2s4+4\frac{s^2}{s^4+4}
  4. s3s4+4\frac{s^3}{s^4+4} (correct answer)
Explanation: When you encounter a Laplace transform involving products of hyperbolic and trigonometric functions, the key is to use the exponential definitions and apply linearity properties systematically. To find L{cosh⁡(t)cos⁡(t)}\mathcal{L}\{\cosh(t)\cos(t)\}, start with the exponential definition: cosh⁡(t)=et+e−t2\cosh(t) = \frac{e^t + e^{-t}}{2}. This gives us: f(t)=cosh⁡(t)cos⁡(t)=et+e−t2cos⁡(t)=12[etcos⁡(t)+e−tcos⁡(t)]f(t) = \cosh(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 shift property: 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)\}. Since L{cos⁡(t)}=ss2+1\mathcal{L}\{\cos(t)\} = \frac{s}{s^2+1}, we have:
  • 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, this reduces to s3s4+4\frac{s^3}{s^4+4}, which is answer D. Answer A gives the numerator for sinh⁡(t)cos⁡(t)\sinh(t)\cos(t) instead of cosh⁡(t)cos⁡(t)\cosh(t)\cos(t). Answer B incorrectly uses partial fraction form without proper combination. Answer C omits the ss factor that comes from the asymmetric nature of the hyperbolic-trigonometric product. Study tip: For products involving cosh⁡(t)\cosh(t) or sinh⁡(t)\sinh(t), always convert to exponential form first, then use the frequency shift property—this systematic approach prevents algebraic errors.

Question 2

Find the Laplace transform of the function f(t)=e2tsinh⁡(3t)f(t) = e^{2t}\sinh(3t).

  1. 3(s+2)2−9\frac{3}{(s+2)^2-9}
  2. s−2(s−2)2−9\frac{s-2}{(s-2)^2-9}
  3. 3(s−2)2−9\frac{3}{(s-2)^2-9} (correct answer)
  4. 3s2−9\frac{3}{s^2-9}
Explanation: When you encounter a Laplace transform problem involving exponential functions combined with hyperbolic functions, the key is recognizing how to apply the shifting property effectively. To find L{e2tsinh⁡(3t)}\mathcal{L}\{e^{2t}\sinh(3t)\}, start by recalling that sinh⁡(3t)=e3t−e−3t2\sinh(3t) = \frac{e^{3t} - e^{-3t}}{2}. This gives us: f(t)=e2t⋅e3t−e−3t2=12(e5t−e−t)f(t) = e^{2t} \cdot \frac{e^{3t} - e^{-3t}}{2} = \frac{1}{2}(e^{5t} - e^{-t}) Now apply the Laplace transform: L{f(t)}=12(1s−5−1s+1)\mathcal{L}\{f(t)\} = \frac{1}{2}\left(\frac{1}{s-5} - \frac{1}{s+1}\right) Combining these fractions: =12⋅(s+1)−(s−5)(s−5)(s+1)=12⋅6(s−5)(s+1)=3s2−4s−5= \frac{1}{2} \cdot \frac{(s+1)-(s-5)}{(s-5)(s+1)} = \frac{1}{2} \cdot \frac{6}{(s-5)(s+1)} = \frac{3}{s^2-4s-5} Notice that (s−5)(s+1)=s2−4s−5=(s−2)2−9(s-5)(s+1) = s^2-4s-5 = (s-2)^2-9, so the answer is 3(s−2)2−9\frac{3}{(s-2)^2-9}, which is choice C. Choice A has the wrong denominator—it uses (s+2)2−9(s+2)^2-9 instead of (s−2)2−9(s-2)^2-9, likely from incorrectly applying the shifting theorem. Choice B has the wrong numerator; it shows s−2s-2 instead of the constant 3. Choice D omits the shifting entirely, giving the transform as if there were no e2te^{2t} factor. Remember: when you see eate^{at} multiplied by another function, expect the Laplace transform to involve a shift in the ss-domain. Always double-check your algebra when expanding and factoring denominators.

Question 3

Let f(t)=e3t+t2e−t+cos⁡(2t)f(t) = e^{3t} + t^2 e^{-t} + \cos(2t). For which values of ss does the Laplace transform L{f(t)}\mathcal{L}\{f(t)\} exist?

  1. s>2s > 2
  2. s>3s > 3 (correct answer)
  3. s>0s > 0
  4. s>−1s > -1
Explanation: When determining where a Laplace transform exists, you need to find the region of convergence by examining the growth rates of each term in your function. The Laplace transform L{f(t)}=∫0∞f(t)e−stdt\mathcal{L}\{f(t)\} = \int_0^{\infty} f(t)e^{-st} dt converges when the integral doesn't blow up to infinity. For f(t)=e3t+t2e−t+cos⁡(2t)f(t) = e^{3t} + t^2 e^{-t} + \cos(2t), analyze each term separately. The exponential term e3te^{3t} grows fastest and will dominate the behavior. When you multiply by e−ste^{-st}, you get e3t⋅e−st=e(3−s)te^{3t} \cdot e^{-st} = e^{(3-s)t}. For this to decay to zero as t→∞t \to \infty, you need 3−s<03-s < 0, which means s>3s > 3. The term t2e−tt^2 e^{-t} becomes t2e−(1+s)tt^2 e^{-(1+s)t}, which converges for s>−1s > -1. The cosine term is bounded and always converges for s>0s > 0. Since all terms must converge simultaneously, you take the most restrictive condition: s>3s > 3. Choice A (s>2s > 2) is too lenient—the e3te^{3t} term would still grow exponentially. Choice C (s>0s > 0) only considers the cosine term and ignores the exponential growth. Choice D (s>−1s > -1) focuses on the t2e−tt^2 e^{-t} term but ignores the dominant e3te^{3t} term. Study tip: For Laplace transform existence, always identify the fastest-growing term in your function—it determines the convergence boundary. Exponential terms typically dominate polynomial and trigonometric terms.

Question 4

Let f(t)={sin⁡(t)0≤t<π0t≥πf(t) = \begin{cases} \sin(t) & 0 \le t < \pi \\ 0 & t \ge \pi \end{cases}. What is the Laplace transform of f(t)f(t)?

  1. s(1+e−sπ)s2+1\frac{s(1+e^{-s\pi})}{s^2+1}
  2. 1s2+1\frac{1}{s^2+1}
  3. 1−e−sπs2+1\frac{1-e^{-s\pi}}{s^2+1}
  4. 1+e−sπs2+1\frac{1+e^{-s\pi}}{s^2+1} (correct answer)
Explanation: When you encounter a piecewise function in Laplace transform problems, you need to express it using unit step functions, then apply the shifting theorem. This function equals sin⁡(t)\sin(t) from 00 to π\pi, then becomes zero. You can rewrite f(t)f(t) as f(t)=sin⁡(t)−sin⁡(t)u(t−π)f(t) = \sin(t) - \sin(t)u(t-\pi), where u(t−π)u(t-\pi) is the unit step function that "turns on" at t=πt = \pi. This expression gives sin⁡(t)\sin(t) for t<πt < \pi and zero for t≥πt \geq \pi. Taking the Laplace transform: L{f(t)}=L{sin⁡(t)}−L{sin⁡(t)u(t−π)}\mathcal{L}\{f(t)\} = \mathcal{L}\{\sin(t)\} - \mathcal{L}\{\sin(t)u(t-\pi)\} The first term is simply 1s2+1\frac{1}{s^2+1}. For the second term, use the shifting theorem: L{g(t)u(t−a)}=e−asL{g(t+a)}\mathcal{L}\{g(t)u(t-a)\} = e^{-as}\mathcal{L}\{g(t+a)\}. Since sin⁡(t+π)=−sin⁡(t)\sin(t+\pi) = -\sin(t), we get: L{sin⁡(t)u(t−π)}=e−sπL{−sin⁡(t)}=−e−sπ⋅1s2+1\mathcal{L}\{\sin(t)u(t-\pi)\} = e^{-s\pi}\mathcal{L}\{-\sin(t)\} = -e^{-s\pi} \cdot \frac{1}{s^2+1} Therefore: L{f(t)}=1s2+1−(−e−sπ⋅1s2+1)=1+e−sπs2+1\mathcal{L}\{f(t)\} = \frac{1}{s^2+1} - \left(-e^{-s\pi} \cdot \frac{1}{s^2+1}\right) = \frac{1+e^{-s\pi}}{s^2+1} Choice A has an extra factor of ss in the numerator. Choice B ignores the discontinuity entirely, giving just the transform of sin⁡(t)\sin(t). Choice C has the wrong sign on the exponential term, which would represent a different piecewise function. Study tip: For piecewise functions, always express them using unit step functions first, then apply the shifting theorem systematically. The exponential terms in Laplace transforms always reflect where discontinuities occur.

Question 5

The function F(s)=3s2+2s−1s3+s2F(s) = \frac{3s^2 + 2s - 1}{s^3 + s^2} represents the Laplace transform of some function f(t)f(t). Before applying partial fraction decomposition, what is the correct factorization of the denominator?

  1. s2(s+1)s^2(s + 1) with partial fractions As+Bs2+Cs+1\frac{A}{s} + \frac{B}{s^2} + \frac{C}{s+1} (correct answer)
  2. s(s2+s)s(s^2 + s) with partial fractions As+Bs+Cs2+s\frac{A}{s} + \frac{Bs+C}{s^2+s}
  3. s2(s+1)s^2(s + 1) with partial fractions As+Bs2+Cs+Ds+1\frac{A}{s} + \frac{B}{s^2} + \frac{Cs+D}{s+1}
  4. s(s+1)2s(s + 1)^2 with partial fractions As+Bs+1+C(s+1)2\frac{A}{s} + \frac{B}{s+1} + \frac{C}{(s+1)^2}
Explanation: The denominator is s3+s2=s2(s+1)s^3 + s^2 = s^2(s + 1). This has a repeated linear factor s2s^2 and a simple linear factor (s+1)(s+1). The correct partial fraction decomposition for this form is As+Bs2+Cs+1\frac{A}{s} + \frac{B}{s^2} + \frac{C}{s+1}, where we need separate terms for each power of the repeated factor up to its multiplicity. Choice A is correct. Choice B incorrectly factors the denominator. Choice C incorrectly uses Cs+DCs+D for a linear factor. Choice D incorrectly identifies the factorization.

Question 6

Given that L{cos⁡(3t)}=ss2+9\mathcal{L}\{\cos(3t)\} = \frac{s}{s^2+9} and L{sin⁡(3t)}=3s2+9\mathcal{L}\{\sin(3t)\} = \frac{3}{s^2+9}, find L{e−2tcos⁡(3t+π4)}\mathcal{L}\{e^{-2t}\cos(3t + \frac{\pi}{4})\}.

  1. (s+2)−3(s+2)2+9⋅22\frac{(s+2)-3}{(s+2)^2+9} \cdot \frac{\sqrt{2}}{2}
  2. (s+2)22−322(s+2)2+9\frac{(s+2)\frac{\sqrt{2}}{2} - 3\frac{\sqrt{2}}{2}}{(s+2)^2+9} (correct answer)
  3. (s+2)22+322(s+2)2+9\frac{(s+2)\frac{\sqrt{2}}{2} + 3\frac{\sqrt{2}}{2}}{(s+2)^2+9}
  4. s22−322s2+4s+13\frac{s\frac{\sqrt{2}}{2} - 3\frac{\sqrt{2}}{2}}{s^2+4s+13}
Explanation: First, use the identity cos⁡(3t+π4)=cos⁡(3t)cos⁡(π4)−sin⁡(3t)sin⁡(π4)=22cos⁡(3t)−22sin⁡(3t)\cos(3t + \frac{\pi}{4}) = \cos(3t)\cos(\frac{\pi}{4}) - \sin(3t)\sin(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}\cos(3t) - \frac{\sqrt{2}}{2}\sin(3t). So L{cos⁡(3t+π4)}=22⋅ss2+9−22⋅3s2+9=22⋅s−3s2+9\mathcal{L}\{\cos(3t + \frac{\pi}{4})\} = \frac{\sqrt{2}}{2} \cdot \frac{s}{s^2+9} - \frac{\sqrt{2}}{2} \cdot \frac{3}{s^2+9} = \frac{\sqrt{2}}{2} \cdot \frac{s-3}{s^2+9}. Now applying the first shifting theorem with a=−2a = -2: L{e−2tcos⁡(3t+π4)}=22⋅(s+2)−3(s+2)2+9=(s+2)22−322(s+2)2+9\mathcal{L}\{e^{-2t}\cos(3t + \frac{\pi}{4})\} = \frac{\sqrt{2}}{2} \cdot \frac{(s+2)-3}{(s+2)^2+9} = \frac{(s+2)\frac{\sqrt{2}}{2} - 3\frac{\sqrt{2}}{2}}{(s+2)^2+9}. Choice B is correct.

Question 7

Using the definition of the Laplace transform, find L{f(t)}\mathcal{L}\{f(t)\} for the function f(t)={20≤t<10t≥1f(t) = \begin{cases} 2 & 0 \le t < 1 \\ 0 & t \ge 1 \end{cases}.

  1. 2(1−e−s)s\frac{2(1-e^{-s})}{s} (correct answer)
  2. 2e−ss\frac{2e^{-s}}{s}
  3. 2s\frac{2}{s}
  4. 2(1−e−s)2(1-e^{-s})
Explanation: The definition of the Laplace transform is L{f(t)}=∫0∞e−stf(t)dt\mathcal{L}\{f(t)\} = \int_0^\infty e^{-st} f(t) dt. For the given piecewise function, the integral splits into two parts: L{f(t)}=∫01e−st(2)dt+∫1∞e−st(0)dt\mathcal{L}\{f(t)\} = \int_0^1 e^{-st}(2) dt + \int_1^\infty e^{-st}(0) dt The second integral is zero. We only need to evaluate the first integral: ∫012e−stdt=2[−1se−st]01=−2s[e−s(1)−e−s(0)]=−2s(e−s−1)=2(1−e−s)s\int_0^1 2e^{-st} dt = 2 \left[ -\frac{1}{s}e^{-st} \right]_0^1 = -\frac{2}{s} [e^{-s(1)} - e^{-s(0)}] = -\frac{2}{s}(e^{-s} - 1) = \frac{2(1-e^{-s})}{s}.

Question 8

What is the Laplace transform of f(t)=e−3t(t2−1)f(t) = e^{-3t}(t^2 - 1)?

  1. 2(s+3)3−1s+3\frac{2}{(s+3)^3} - \frac{1}{s+3} (correct answer)
  2. 2(s−3)3−1s−3\frac{2}{(s-3)^3} - \frac{1}{s-3}
  3. 2s3−1s\frac{2}{s^3} - \frac{1}{s}
  4. 1s+3(2s3−1s)\frac{1}{s+3} \left( \frac{2}{s^3} - \frac{1}{s} \right)
Explanation: This problem requires the First Shifting Theorem (s-shifting), 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)\}. First, let g(t)=t2−1g(t) = t^2 - 1. We find its Laplace transform G(s)G(s) using linearity and the basic transform rules L{tn}=n!sn+1\mathcal{L}\{t^n\} = \frac{n!}{s^{n+1}} and L{1}=1s\mathcal{L}\{1\} = \frac{1}{s}. G(s)=L{t2−1}=L{t2}−L{1}=2!s2+1−1s=2s3−1sG(s) = \mathcal{L}\{t^2 - 1\} = \mathcal{L}\{t^2\} - \mathcal{L}\{1\} = \frac{2!}{s^{2+1}} - \frac{1}{s} = \frac{2}{s^3} - \frac{1}{s}. Now, we apply the First Shifting Theorem with a=−3a=-3. We need to compute G(s−a)=G(s−(−3))=G(s+3)G(s-a) = G(s - (-3)) = G(s+3). This means we replace every ss in G(s)G(s) with (s+3)(s+3). L{e−3t(t2−1)}=G(s+3)=2(s+3)3−1s+3\mathcal{L}\{e^{-3t}(t^2 - 1)\} = G(s+3) = \frac{2}{(s+3)^3} - \frac{1}{s+3}.

Question 9

What is the Laplace transform of the function f(t)=sin⁡2(t)f(t) = \sin^2(t)?

  1. 1s−ss2+4\frac{1}{s} - \frac{s}{s^2+4}
  2. (1s2+1)2\left( \frac{1}{s^2+1} \right)^2
  3. 2s(s2+4)\frac{2}{s(s^2+4)} (correct answer)
  4. 2s(s2+1)\frac{2}{s(s^2+1)}
Explanation: When you encounter the Laplace transform of trigonometric functions involving powers, the key is recognizing that you'll often need trigonometric identities to simplify before transforming. For f(t)=sin⁡2(t)f(t) = \sin^2(t), you can't directly apply the basic Laplace transform formula for sin⁡(t)\sin(t). Instead, use the trigonometric identity: sin⁡2(t)=1−cos⁡(2t)2\sin^2(t) = \frac{1 - \cos(2t)}{2} Now you can find the Laplace transform of each term: L{sin⁡2(t)}=L{12−cos⁡(2t)2}\mathcal{L}\{\sin^2(t)\} = \mathcal{L}\left\{\frac{1}{2} - \frac{\cos(2t)}{2}\right\} Using linearity: =12L{1}−12L{cos⁡(2t)}= \frac{1}{2}\mathcal{L}\{1\} - \frac{1}{2}\mathcal{L}\{\cos(2t)\} Applying the standard transforms L{1}=1s\mathcal{L}\{1\} = \frac{1}{s} and L{cos⁡(at)}=ss2+a2\mathcal{L}\{\cos(at)\} = \frac{s}{s^2+a^2}: =12⋅1s−12⋅ss2+4=12s−s2(s2+4)= \frac{1}{2} \cdot \frac{1}{s} - \frac{1}{2} \cdot \frac{s}{s^2+4} = \frac{1}{2s} - \frac{s}{2(s^2+4)} Combining over a common denominator: =s2+4−s22s(s2+4)=42s(s2+4)=2s(s2+4)= \frac{s^2+4-s^2}{2s(s^2+4)} = \frac{4}{2s(s^2+4)} = \frac{2}{s(s^2+4)} This confirms answer C is correct. Answer A has the wrong denominator in the second term (should be s2+4s^2+4, not s2+4s^2+4, but missing the factor of 12\frac{1}{2}). Answer B incorrectly squares the entire Laplace transform of sin⁡(t)\sin(t). Answer D uses s2+1s^2+1 instead of s2+4s^2+4, suggesting confusion about the frequency doubling in the cosine term. Strategy tip: Always look for trigonometric identities to simplify powers of trig functions before applying Laplace transforms. The half-angle and double-angle formulas are particularly useful.

Question 10

A function f(t)f(t) is given by f(t)=Acos⁡(2t)+Bsin⁡(2t)f(t) = A\cos(2t) + B\sin(2t). If f(0)=2f(0) = 2 and f′(0)=−2f'(0) = -2, what is the Laplace transform of f(t)f(t)?

  1. 2s+2s2+4\frac{2s+2}{s^2+4}
  2. 2s−2s2+4\frac{2s-2}{s^2+4} (correct answer)
  3. s−2s2+4\frac{s-2}{s^2+4}
  4. 2s−1s2+1\frac{2s-1}{s^2+1}
Explanation: This problem combines initial value problems with Laplace transforms, testing whether you can find unknown coefficients and then apply transform formulas correctly. First, you need to determine the constants A and B using the given conditions. From f(t)=Acos⁡(2t)+Bsin⁡(2t)f(t) = A\cos(2t) + B\sin(2t), we get f(0)=Acos⁡(0)+Bsin⁡(0)=A=2f(0) = A\cos(0) + B\sin(0) = A = 2. Taking the derivative: f′(t)=−2Asin⁡(2t)+2Bcos⁡(2t)f'(t) = -2A\sin(2t) + 2B\cos(2t), so f′(0)=−2Asin⁡(0)+2Bcos⁡(0)=2B=−2f'(0) = -2A\sin(0) + 2B\cos(0) = 2B = -2, giving us B=−1B = -1. Therefore, f(t)=2cos⁡(2t)−sin⁡(2t)f(t) = 2\cos(2t) - \sin(2t). Now apply the standard Laplace transform formulas: L[cos⁡(at)]=ss2+a2\mathcal{L}[\cos(at)] = \frac{s}{s^2+a^2} and L[sin⁡(at)]=as2+a2\mathcal{L}[\sin(at)] = \frac{a}{s^2+a^2}. This gives us: L[f(t)]=2⋅ss2+4−1⋅2s2+4=2s−2s2+4\mathcal{L}[f(t)] = 2 \cdot \frac{s}{s^2+4} - 1 \cdot \frac{2}{s^2+4} = \frac{2s-2}{s^2+4}, which is choice B. Looking at the wrong answers: Choice A has 2s+2s2+4\frac{2s+2}{s^2+4}, which would result from incorrectly finding B=1B = 1 instead of B=−1B = -1. Choice C gives s−2s2+4\frac{s-2}{s^2+4}, suggesting the coefficient 2 was dropped from the cosine term. Choice D has 2s−1s2+1\frac{2s-1}{s^2+1}, which uses the wrong frequency (ω=1\omega = 1 instead of ω=2\omega = 2) and incorrectly transforms the sine term. Strategy tip: Always solve for unknown coefficients first using initial conditions, then apply Laplace transform formulas. Double-check your algebra when combining fractions with common denominators.

Question 11

What is the Laplace transform of the function f(t)=sin⁡(t)cos⁡(t)f(t) = \sin(t)\cos(t)?

  1. 1s2+4\frac{1}{s^2+4} (correct answer)
  2. ss2+4\frac{s}{s^2+4}
  3. s(s2+1)2\frac{s}{(s^2+1)^2}
  4. 1(s2+1)\frac{1}{(s^2+1)}
Explanation: To find the Laplace transform of f(t)=sin⁡(t)cos⁡(t)f(t) = \sin(t)\cos(t), first use the trigonometric identity sin⁡(2θ)=2sin⁡(θ)cos⁡(θ)\sin(2\theta) = 2\sin(\theta)\cos(\theta). This gives sin⁡(t)cos⁡(t)=12sin⁡(2t)\sin(t)\cos(t) = \frac{1}{2}\sin(2t). Next, apply the linearity property of the Laplace transform and the rule for L{sin⁡(kt)}=ks2+k2\mathcal{L}\{\sin(kt)\} = \frac{k}{s^2+k^2}. L{sin⁡(t)cos⁡(t)}=L{12sin⁡(2t)}=12L{sin⁡(2t)}\mathcal{L}\{\sin(t)\cos(t)\} = \mathcal{L}\{\frac{1}{2}\sin(2t)\} = \frac{1}{2}\mathcal{L}\{\sin(2t)\} With k=2k=2, this becomes: 12(2s2+22)=12(2s2+4)=1s2+4\frac{1}{2} \left( \frac{2}{s^2+2^2} \right) = \frac{1}{2} \left( \frac{2}{s^2+4} \right) = \frac{1}{s^2+4}.

Question 12

For which of the following functions f(t)f(t) does the Laplace transform not exist?

  1. f(t)=t5cos⁡(3t)f(t) = t^5\cos(3t)
  2. f(t)=sinh⁡(10t)f(t) = \sinh(10t)
  3. f(t)=et2f(t) = e^{t^2} (correct answer)
  4. f(t)=arctan⁡(t)f(t) = \arctan(t)
Explanation: The Laplace transform of a function f(t)f(t) exists if f(t)f(t) is piecewise continuous on [0,∞)[0, \infty) and is of exponential order. A function is of exponential order if there exist constants M>0M > 0 and cc such that ∣f(t)∣≤Mect|f(t)| \le Me^{ct} for all t≥Tt \ge T for some TT.
  • A: ∣t5cos⁡(3t)∣≤t5|t^5\cos(3t)| \le t^5. For any c>0c>0, t5t^5 grows slower than ecte^{ct}, so this function is of exponential order.
  • B: sinh⁡(10t)=12(e10t−e−10t)\sinh(10t) = \frac{1}{2}(e^{10t} - e^{-10t}). For large tt, this behaves like 12e10t\frac{1}{2}e^{10t}, so it is of exponential order with c=10c=10.
  • C: f(t)=et2f(t) = e^{t^2}. For any constant cc, the ratio et2ect=et2−ct\frac{e^{t^2}}{e^{ct}} = e^{t^2-ct} goes to infinity as t→∞t \to \infty. Therefore, et2e^{t^2} is not of exponential order, and its Laplace transform does not exist.
  • D: f(t)=arctan⁡(t)f(t) = \arctan(t) is a bounded function, since ∣arctan⁡(t)∣<π2|\arctan(t)| < \frac{\pi}{2} for all tt. Any bounded function is of exponential order (with c=0c=0), so its Laplace transform exists.

Question 13

Let f(t)f(t) be a function such that its Laplace transform is F(s)=s+2(s+2)2+9F(s) = \frac{s+2}{(s+2)^2+9}. What is the Laplace transform of g(t)=e−tf(t)g(t) = e^{-t}f(t)?

  1. s+3(s+2)2+9\frac{s+3}{(s+2)^2+9}
  2. s+1(s+1)2+9\frac{s+1}{(s+1)^2+9}
  3. s+2(s+2)2+9⋅1s+1\frac{s+2}{(s+2)^2+9} \cdot \frac{1}{s+1}
  4. s+3(s+3)2+9\frac{s+3}{(s+3)^2+9} (correct answer)
Explanation: When you encounter a problem involving the Laplace transform of an exponentially modified function, you need to apply the frequency shifting property (also called the first translation theorem). This property states that if L{f(t)}=F(s)\mathcal{L}\{f(t)\} = F(s), then L{e−atf(t)}=F(s+a)\mathcal{L}\{e^{-at}f(t)\} = F(s+a). Given that F(s)=s+2(s+2)2+9F(s) = \frac{s+2}{(s+2)^2+9} and g(t)=e−tf(t)g(t) = e^{-t}f(t), you apply the frequency shifting property with a=1a = 1. This means you replace every ss in F(s)F(s) with (s+1)(s+1): L{g(t)}=F(s+1)=(s+1)+2((s+1)+2)2+9=s+3(s+3)2+9\mathcal{L}\{g(t)\} = F(s+1) = \frac{(s+1)+2}{((s+1)+2)^2+9} = \frac{s+3}{(s+3)^2+9} Answer D is correct because it properly applies the frequency shifting theorem. Answer A s+3(s+2)2+9\frac{s+3}{(s+2)^2+9} only shifts the numerator but leaves the denominator unchanged, which is an incomplete application of the property. Answer B s+1(s+1)2+9\frac{s+1}{(s+1)^2+9} incorrectly shifts ss to (s+1)(s+1) instead of (s+3)(s+3) in the numerator, suggesting confusion about how the shift affects the entire expression. Answer C s+2(s+2)2+9⋅1s+1\frac{s+2}{(s+2)^2+9} \cdot \frac{1}{s+1} treats this as a convolution problem rather than frequency shifting, which is the wrong approach entirely. Study tip: Remember that frequency shifting means "replace every ss with (s+a)(s+a)" throughout the entire transform. Practice identifying when to use frequency shifting versus other Laplace transform properties like convolution or time shifting.

Question 14

A function is defined as f(t)={1,0≤t<23,t≥2f(t) = \begin{cases} 1, & 0 \le t < 2 \\ 3, & t \ge 2 \end{cases}. Find the Laplace transform of f(t)f(t).

  1. 4e−2s−1s\frac{4e^{-2s}-1}{s}
  2. 1+2e−2ss\frac{1+2e^{-2s}}{s} (correct answer)
  3. 1−e−2ss+3s\frac{1-e^{-2s}}{s} + \frac{3}{s}
  4. 3s\frac{3}{s}
Explanation: When you encounter a piecewise function in a Laplace transform problem, you need to express it using unit step functions, then apply the shifting property of Laplace transforms. First, rewrite f(t)f(t) using the unit step function u(t−a)u(t-a), which equals 0 for t<at < a and 1 for t≥at \geq a. Your function can be written as: f(t)=1⋅u(t)+2⋅u(t−2)f(t) = 1 \cdot u(t) + 2 \cdot u(t-2) This works because: for 0≤t<20 \leq t < 2, only the first term contributes, giving f(t)=1f(t) = 1. For t≥2t \geq 2, both terms contribute, giving f(t)=1+2=3f(t) = 1 + 2 = 3. Now apply the Laplace transform. Using L{u(t)}=1s\mathcal{L}\{u(t)\} = \frac{1}{s} and the shifting property L{u(t−a)}=e−ass\mathcal{L}\{u(t-a)\} = \frac{e^{-as}}{s}: L{f(t)}=1⋅1s+2⋅e−2ss=1+2e−2ss\mathcal{L}\{f(t)\} = 1 \cdot \frac{1}{s} + 2 \cdot \frac{e^{-2s}}{s} = \frac{1 + 2e^{-2s}}{s} This matches answer choice B. Answer A, 4e−2s−1s\frac{4e^{-2s}-1}{s}, suggests an incorrect decomposition of the piecewise function. Answer C, 1−e−2ss+3s\frac{1-e^{-2s}}{s} + \frac{3}{s}, incorrectly treats this as if the function jumps from 1 to 3 at t=2t=2 rather than adding 2 to the existing value. Answer D, 3s\frac{3}{s}, would correspond to the constant function f(t)=3f(t) = 3 for all tt. Remember: for piecewise functions, decompose using unit step functions first, then transform each piece separately using the shifting property.

Question 15

For which value of the parameter aa does the function f(t)=taf(t) = t^a have a Laplace transform that exists for s>0s > 0?

  1. a>0a > 0 only, since negative powers create singularities at t=0t = 0
  2. All real values of aa, since exponential decay dominates polynomial growth
  3. a≥−12a \geq -\frac{1}{2} only, based on the standard existence theorem conditions
  4. a>−1a > -1 only, due to convergence requirements of the improper integral (correct answer)
Explanation: When determining if a Laplace transform exists, you need to analyze the convergence of the improper integral L{f(t)}=∫0∞tae−stdt\mathcal{L}\{f(t)\} = \int_0^{\infty} t^a e^{-st} dt for s>0s > 0. The key insight is that this integral can diverge at two places: near t=0t = 0 (if tat^a has a singularity there) or as t→∞t \to \infty (though the exponential e−ste^{-st} typically controls this). For tat^a with a<0a < 0, you get a singularity at t=0t = 0, so you must check when ∫01tadt\int_0^1 t^a dt converges. This integral converges if and only if a>−1a > -1. To verify: ∫01tadt=ta+1a+1∣01\int_0^1 t^a dt = \frac{t^{a+1}}{a+1}\Big|_0^1, which converges when a+1>0a + 1 > 0, giving us a>−1a > -1. The behavior as t→∞t \to \infty is controlled by the exponential decay term e−ste^{-st} for any s>0s > 0. Choice A incorrectly claims only positive powers work—negative powers can work if they're not too negative. Choice B wrongly suggests all real values work, ignoring the singularity issue at t=0t = 0. Choice C gives the specific threshold a≥−12a \geq -\frac{1}{2}, which is too restrictive and misses that we can go down to (but not including) a=−1a = -1. Study tip: For Laplace transform existence problems involving power functions, always check convergence at t=0t = 0 first. The condition a>−1a > -1 for tat^a is a fundamental result worth memorizing.

Question 16

Consider the convolution integral h(t)=∫0te2τcos⁡(3(t−τ))dτh(t) = \int_0^t e^{2\tau} \cos(3(t-\tau)) d\tau. Using the convolution theorem, L{h(t)}\mathcal{L}\{h(t)\} can be expressed as:

  1. 1s−2⋅3s2+9=3(s−2)(s2+9)\frac{1}{s-2} \cdot \frac{3}{s^2+9} = \frac{3}{(s-2)(s^2+9)}
  2. ss2+9⋅1s−2=s(s2+9)(s−2)\frac{s}{s^2+9} \cdot \frac{1}{s-2} = \frac{s}{(s^2+9)(s-2)}
  3. 1s−2⋅ss2+9=s(s−2)(s2+9)\frac{1}{s-2} \cdot \frac{s}{s^2+9} = \frac{s}{(s-2)(s^2+9)} (correct answer)
  4. s−2s2+9⋅ss2+9=s(s−2)(s2+9)2\frac{s-2}{s^2+9} \cdot \frac{s}{s^2+9} = \frac{s(s-2)}{(s^2+9)^2}
Explanation: When you encounter a convolution integral like this, the convolution theorem is your key tool. It states that L{f∗g}=L{f(t)}⋅L{g(t)}\mathcal{L}\{f * g\} = \mathcal{L}\{f(t)\} \cdot \mathcal{L}\{g(t)\}, where ∗* represents convolution. This transforms a complex integral into simple multiplication of Laplace transforms. Your integral has the form (f∗g)(t)=∫0tf(τ)g(t−τ)dτ(f * g)(t) = \int_0^t f(\tau)g(t-\tau)d\tau, where f(τ)=e2τf(\tau) = e^{2\tau} and g(t−τ)=cos⁡(3(t−τ))g(t-\tau) = \cos(3(t-\tau)). To apply the convolution theorem, you need the individual Laplace transforms: L{e2t}=1s−2\mathcal{L}\{e^{2t}\} = \frac{1}{s-2} and L{cos⁡(3t)}=ss2+9\mathcal{L}\{\cos(3t)\} = \frac{s}{s^2+9}. Therefore, L{h(t)}=1s−2⋅ss2+9=s(s−2)(s2+9)\mathcal{L}\{h(t)\} = \frac{1}{s-2} \cdot \frac{s}{s^2+9} = \frac{s}{(s-2)(s^2+9)}, which is choice C. Choice A incorrectly uses L{sin⁡(3t)}=3s2+9\mathcal{L}\{\sin(3t)\} = \frac{3}{s^2+9} instead of the cosine transform. This is a common mix-up between sine and cosine Laplace transforms. Choice B has the correct transforms but in the wrong order, suggesting confusion about which function corresponds to ff versus gg in the convolution. Choice D uses s−2s2+9\frac{s-2}{s^2+9} for one factor, which isn't the Laplace transform of any standard function, and incorrectly squares the denominator s2+9s^2+9. Study tip: Memorize the basic Laplace transform pairs, especially L{eat}=1s−a\mathcal{L}\{e^{at}\} = \frac{1}{s-a}, L{cos⁡(bt)}=ss2+b2\mathcal{L}\{\cos(bt)\} = \frac{s}{s^2+b^2}, and L{sin⁡(bt)}=bs2+b2\mathcal{L}\{\sin(bt)\} = \frac{b}{s^2+b^2}. The convolution theorem turns difficult integrals into straightforward multiplication.

Question 17

If L{f(t)}=s+1s2−2s+5\mathcal{L}\{f(t)\} = \frac{s+1}{s^2-2s+5} and h(t)=e2tf(t)h(t) = e^{2t}f(t), then L{h(t)}\mathcal{L}\{h(t)\} equals:

  1. (s−2)+1(s−2)2−2(s−2)+5\frac{(s-2)+1}{(s-2)^2-2(s-2)+5}
  2. s−1s2−6s+13\frac{s-1}{s^2-6s+13} (correct answer)
  3. s+3s2−6s+13\frac{s+3}{s^2-6s+13}
  4. s−3s2+2s+5\frac{s-3}{s^2+2s+5}
Explanation: Using the first shifting theorem, L{eatf(t)}=F(s−a)\mathcal{L}\{e^{at}f(t)\} = F(s-a) where F(s)=L{f(t)}F(s) = \mathcal{L}\{f(t)\}. Here a=2a = 2, so L{h(t)}=F(s−2)=(s−2)+1(s−2)2−2(s−2)+5\mathcal{L}\{h(t)\} = F(s-2) = \frac{(s-2)+1}{(s-2)^2-2(s-2)+5}. Simplifying the numerator: (s−2)+1=s−1(s-2)+1 = s-1. For the denominator: (s−2)2−2(s−2)+5=s2−4s+4−2s+4+5=s2−6s+13(s-2)^2-2(s-2)+5 = s^2-4s+4-2s+4+5 = s^2-6s+13. Therefore L{h(t)}=s−1s2−6s+13\mathcal{L}\{h(t)\} = \frac{s-1}{s^2-6s+13}. Choice B is correct.

Question 18

If F(s)=L{f(t)}F(s) = \mathcal{L}\{f(t)\} exists for s>σs > \sigma, and g(t)=f(3t)g(t) = f(3t), then L{g(t)}\mathcal{L}\{g(t)\} equals:

  1. 13F(s3)\frac{1}{3}F\left(\frac{s}{3}\right) for s>σ3s > \frac{\sigma}{3}
  2. 3F(3s)3F(3s) for s>σ3s > \frac{\sigma}{3}
  3. 13F(s3)\frac{1}{3}F\left(\frac{s}{3}\right) for s>3σs > 3\sigma (correct answer)
  4. F(3s)F(3s) for s>3σs > 3\sigma
Explanation: When you encounter Laplace transforms with time scaling like g(t)=f(3t)g(t) = f(3t), you're dealing with the time-scaling property. This is a fundamental transformation rule that requires careful attention to both the transform formula and the region of convergence. To find L{f(3t)}\mathcal{L}\{f(3t)\}, we apply the time-scaling property: if L{f(t)}=F(s)\mathcal{L}\{f(t)\} = F(s), then L{f(at)}=1aF(sa)\mathcal{L}\{f(at)\} = \frac{1}{a}F\left(\frac{s}{a}\right) where a>0a > 0. With a=3a = 3, we get L{f(3t)}=13F(s3)\mathcal{L}\{f(3t)\} = \frac{1}{3}F\left(\frac{s}{3}\right). For the region of convergence, if F(s)F(s) exists for s>σs > \sigma, then we need s3>σ\frac{s}{3} > \sigma, which means s>3σs > 3\sigma. Therefore, L{g(t)}=13F(s3)\mathcal{L}\{g(t)\} = \frac{1}{3}F\left(\frac{s}{3}\right) for s>3σs > 3\sigma. Choice A has the correct transform but wrong convergence region—it claims s>σ3s > \frac{\sigma}{3} instead of s>3σs > 3\sigma. Choice B incorrectly applies the scaling property, giving 3F(3s)3F(3s) instead of 13F(s3)\frac{1}{3}F\left(\frac{s}{3}\right), and has the wrong convergence region. Choice D omits the scaling factor 13\frac{1}{3} entirely, giving just F(3s)F(3s). The correct answer is C. Study tip: Remember the time-scaling property formula exactly: L{f(at)}=1aF(sa)\mathcal{L}\{f(at)\} = \frac{1}{a}F\left(\frac{s}{a}\right). The factor 1a\frac{1}{a} appears outside, ss gets divided by aa inside, and the convergence region becomes s>aσs > a\sigma.

Question 19

If g(t)=∫0tf(τ)dτg(t) = \int_0^t f(\tau) d\tau where L{f(t)}=2s+3s2+1\mathcal{L}\{f(t)\} = \frac{2s+3}{s^2+1}, then L{g(t)}\mathcal{L}\{g(t)\} equals:

  1. 2s+3s(s2+1)\frac{2s+3}{s(s^2+1)} (correct answer)
  2. s(2s+3)s2+1\frac{s(2s+3)}{s^2+1}
  3. 2s+3s3+s\frac{2s+3}{s^3+s}
  4. 2s2+3ss2+1\frac{2s^2+3s}{s^2+1}
Explanation: Using the integration property of Laplace transforms: if L{f(t)}=F(s)\mathcal{L}\{f(t)\} = F(s), then L{∫0tf(τ)dτ}=F(s)s\mathcal{L}\left\{\int_0^t f(\tau) d\tau\right\} = \frac{F(s)}{s}. Here, F(s)=2s+3s2+1F(s) = \frac{2s+3}{s^2+1}, so L{g(t)}=1s⋅2s+3s2+1=2s+3s(s2+1)\mathcal{L}\{g(t)\} = \frac{1}{s} \cdot \frac{2s+3}{s^2+1} = \frac{2s+3}{s(s^2+1)}. Choice A is correct. Choice B incorrectly multiplies by ss. Choice C correctly identifies s(s2+1)=s3+ss(s^2+1) = s^3+s but this doesn't match the integration formula. Choice D represents differentiation, not integration.