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Differential Equations Quiz

Differential Equations Quiz: Inverse Laplace Via Partial Fractions

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

Question 1 / 12

0 of 12 answered

Find the inverse Laplace transform of F(s)=1s3+sF(s) = \frac{1}{s^3+s}F(s)=s3+s1​.

Select an answer to continue

What this quiz covers

This quiz focuses on Inverse Laplace Via Partial Fractions, 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

Find the inverse Laplace transform of F(s)=1s3+sF(s) = \frac{1}{s^3+s}F(s)=s3+s1​.

  1. 1−sin⁡(t)1 - \sin(t)1−sin(t)
  2. 12t2−sin⁡(t)\frac{1}{2}t^2 - \sin(t)21​t2−sin(t)
  3. 1−cos⁡(t)1 - \cos(t)1−cos(t) (correct answer)
  4. 1+cos⁡(t)1 + \cos(t)1+cos(t)

Explanation: First, factor the denominator: s3+s=s(s2+1)s^3+s = s(s^2+1)s3+s=s(s2+1). The function is F(s)=1s(s2+1)F(s) = \frac{1}{s(s^2+1)}F(s)=s(s2+1)1​. We use partial fraction decomposition: F(s)=As+Bs+Cs2+1F(s) = \frac{A}{s} + \frac{Bs+C}{s^2+1}F(s)=sA​+s2+1Bs+C​. Multiplying by the denominator gives 1=A(s2+1)+(Bs+C)s1 = A(s^2+1) + (Bs+C)s1=A(s2+1)+(Bs+C)s. Let s=0s=0s=0: 1=A(1)  ⟹  A=11 = A(1) \implies A=11=A(1)⟹A=1. Equating coefficients of s2s^2s2: 0=A+B  ⟹  B=−A=−10 = A+B \implies B = -A = -10=A+B⟹B=−A=−1. Equating coefficients of sss: 0=C0 = C0=C. So, F(s)=1s−ss2+1F(s) = \frac{1}{s} - \frac{s}{s^2+1}F(s)=s1​−s2+1s​. Taking the inverse Laplace transform term by term: L−1{1s}=1\mathcal{L}^{-1}\{\frac{1}{s}\} = 1L−1{s1​}=1 L−1{ss2+1}=cos⁡(t)\mathcal{L}^{-1}\{\frac{s}{s^2+1}\} = \cos(t)L−1{s2+1s​}=cos(t) Therefore, f(t)=1−cos⁡(t)f(t) = 1 - \cos(t)f(t)=1−cos(t).

Question 2

Find the inverse Laplace transform, f(t)f(t)f(t), of the function F(s)=3s(s−1)(s2+2s+5)F(s) = \frac{3s}{(s-1)(s^2+2s+5)}F(s)=(s−1)(s2+2s+5)3s​.

  1. f(t)=38et−38e−tcos⁡(2t)−98e−tsin⁡(2t)f(t) = \frac{3}{8}e^t - \frac{3}{8}e^{-t}\cos(2t) - \frac{9}{8}e^{-t}\sin(2t)f(t)=83​et−83​e−tcos(2t)−89​e−tsin(2t)
  2. f(t)=38et+38e−tcos⁡(2t)−158e−tsin⁡(2t)f(t) = \frac{3}{8}e^t + \frac{3}{8}e^{-t}\cos(2t) - \frac{15}{8}e^{-t}\sin(2t)f(t)=83​et+83​e−tcos(2t)−815​e−tsin(2t)
  3. f(t)=38et−38e−tcos⁡(2t)+98e−tsin⁡(2t)f(t) = \frac{3}{8}e^t - \frac{3}{8}e^{-t}\cos(2t) + \frac{9}{8}e^{-t}\sin(2t)f(t)=83​et−83​e−tcos(2t)+89​e−tsin(2t) (correct answer)
  4. f(t)=38e−t−38etcos⁡(2t)+98etsin⁡(2t)f(t) = \frac{3}{8}e^{-t} - \frac{3}{8}e^{t}\cos(2t) + \frac{9}{8}e^{t}\sin(2t)f(t)=83​e−t−83​etcos(2t)+89​etsin(2t)

Explanation: The partial fraction decomposition is F(s)=As−1+Bs+Cs2+2s+5F(s) = \frac{A}{s-1} + \frac{Bs+C}{s^2+2s+5}F(s)=s−1A​+s2+2s+5Bs+C​. Using the cover-up method for AAA: A=3ss2+2s+5∣s=1=3(1)1+2+5=38A = \left. \frac{3s}{s^2+2s+5} \right|_{s=1} = \frac{3(1)}{1+2+5} = \frac{3}{8}A=s2+2s+53s​​s=1​=1+2+53(1)​=83​. Multiply by the denominator: 3s=A(s2+2s+5)+(Bs+C)(s−1)3s = A(s^2+2s+5) + (Bs+C)(s-1)3s=A(s2+2s+5)+(Bs+C)(s−1). Equating coefficients of s2s^2s2: 0=A+B  ⟹  B=−A=−380 = A+B \implies B = -A = -\frac{3}{8}0=A+B⟹B=−A=−83​. Equating constant terms: 0=5A−C  ⟹  C=5A=1580 = 5A-C \implies C = 5A = \frac{15}{8}0=5A−C⟹C=5A=815​. So, F(s)=3/8s−1+−3/8s+15/8s2+2s+5F(s) = \frac{3/8}{s-1} + \frac{-3/8 s + 15/8}{s^2+2s+5}F(s)=s−13/8​+s2+2s+5−3/8s+15/8​. Complete the square for the quadratic term: s2+2s+5=(s+1)2+4=(s+1)2+22s^2+2s+5 = (s+1)^2+4 = (s+1)^2+2^2s2+2s+5=(s+1)2+4=(s+1)2+22. Rewrite the second term's numerator to match the form for sine and cosine transforms: −3/8s+15/8(s+1)2+4=−3/8(s+1)+3/8+15/8(s+1)2+4=−3/8(s+1)+18/8(s+1)2+4\frac{-3/8 s + 15/8}{(s+1)^2+4} = \frac{-3/8(s+1) + 3/8 + 15/8}{(s+1)^2+4} = \frac{-3/8(s+1) + 18/8}{(s+1)^2+4}(s+1)2+4−3/8s+15/8​=(s+1)2+4−3/8(s+1)+3/8+15/8​=(s+1)2+4−3/8(s+1)+18/8​. Split this into two terms: −38s+1(s+1)2+22+941(s+1)2+22=−38s+1(s+1)2+22+982(s+1)2+22-\frac{3}{8}\frac{s+1}{(s+1)^2+2^2} + \frac{9}{4}\frac{1}{(s+1)^2+2^2} = -\frac{3}{8}\frac{s+1}{(s+1)^2+2^2} + \frac{9}{8}\frac{2}{(s+1)^2+2^2}−83​(s+1)2+22s+1​+49​(s+1)2+221​=−83​(s+1)2+22s+1​+89​(s+1)2+222​. Taking the inverse Laplace transform gives: f(t)=38et−38e−tcos⁡(2t)+98e−tsin⁡(2t)f(t) = \frac{3}{8}e^t - \frac{3}{8}e^{-t}\cos(2t) + \frac{9}{8}e^{-t}\sin(2t)f(t)=83​et−83​e−tcos(2t)+89​e−tsin(2t).

Question 3

The Laplace transform of a function y(t)y(t)y(t) is given by Y(s)=s3−2s+1s(s−3)2(s2+4)Y(s) = \frac{s^3 - 2s + 1}{s(s-3)^2(s^2+4)}Y(s)=s(s−3)2(s2+4)s3−2s+1​. Without computing the coefficients, what is the general form of y(t)y(t)y(t) for t>0t > 0t>0?

  1. y(t)=C1+C2e3t+C3e−3t+C4cos⁡(2t)+C5sin⁡(2t)y(t) = C_1 + C_2 e^{3t} + C_3 e^{-3t} + C_4 \cos(2t) + C_5 \sin(2t)y(t)=C1​+C2​e3t+C3​e−3t+C4​cos(2t)+C5​sin(2t)
  2. y(t)=C1+C2e3t+C3te3t+C4e2t+C5e−2ty(t) = C_1 + C_2 e^{3t} + C_3 t e^{3t} + C_4 e^{2t} + C_5 e^{-2t}y(t)=C1​+C2​e3t+C3​te3t+C4​e2t+C5​e−2t
  3. y(t)=C1+C2e3t+C3te3t+C4cos⁡(2t)+C5sin⁡(2t)y(t) = C_1 + C_2 e^{3t} + C_3 t e^{3t} + C_4 \cos(2t) + C_5 \sin(2t)y(t)=C1​+C2​e3t+C3​te3t+C4​cos(2t)+C5​sin(2t) (correct answer)
  4. y(t)=C1+C2e3t+C3cos⁡(4t)+C4sin⁡(4t)y(t) = C_1 + C_2 e^{3t} + C_3 \cos(4t) + C_4 \sin(4t)y(t)=C1​+C2​e3t+C3​cos(4t)+C4​sin(4t)

Explanation: The form of the solution y(t)y(t)y(t) is determined by the poles of Y(s)Y(s)Y(s), which are the roots of the denominator. The denominator is s(s−3)2(s2+4)s(s-3)^2(s^2+4)s(s−3)2(s2+4). The poles are:

  • s=0s=0s=0: a simple pole, which corresponds to a constant term C1C_1C1​ in y(t)y(t)y(t).
  • s=3s=3s=3: a repeated pole of order 2, which corresponds to terms C2e3t+C3te3tC_2 e^{3t} + C_3 t e^{3t}C2​e3t+C3​te3t.
  • s2+4=0  ⟹  s=±2is^2+4=0 \implies s = \pm 2is2+4=0⟹s=±2i: a pair of complex conjugate poles, which corresponds to terms C4cos⁡(2t)+C5sin⁡(2t)C_4 \cos(2t) + C_5 \sin(2t)C4​cos(2t)+C5​sin(2t). Combining these gives the general form y(t)=C1+C2e3t+C3te3t+C4cos⁡(2t)+C5sin⁡(2t)y(t) = C_1 + C_2 e^{3t} + C_3 t e^{3t} + C_4 \cos(2t) + C_5 \sin(2t)y(t)=C1​+C2​e3t+C3​te3t+C4​cos(2t)+C5​sin(2t).

Question 4

The function F(s)=s2+1s(s−1)(s+2)2F(s) = \frac{s^2+1}{s(s-1)(s+2)^2}F(s)=s(s−1)(s+2)2s2+1​ has the partial fraction decomposition As+Bs−1+Cs+2+D(s+2)2\frac{A}{s} + \frac{B}{s-1} + \frac{C}{s+2} + \frac{D}{(s+2)^2}sA​+s−1B​+s+2C​+(s+2)2D​. What is the value of the coefficient CCC?

  1. −14-\frac{1}{4}−41​
  2. 56\frac{5}{6}65​
  3. 136\frac{1}{36}361​ (correct answer)
  4. 29\frac{2}{9}92​

Explanation: To find CCC, we cannot use the simple cover-up method. One method is to find the other coefficients first and then equate coefficients of the highest power of sss. A=s2+1(s−1)(s+2)2∣s=0=1(−1)(4)=−14A = \left. \frac{s^2+1}{(s-1)(s+2)^2} \right|_{s=0} = \frac{1}{(-1)(4)} = -\frac{1}{4}A=(s−1)(s+2)2s2+1​​s=0​=(−1)(4)1​=−41​. B=s2+1s(s+2)2∣s=1=21(9)=29B = \left. \frac{s^2+1}{s(s+2)^2} \right|_{s=1} = \frac{2}{1(9)} = \frac{2}{9}B=s(s+2)2s2+1​​s=1​=1(9)2​=92​. The original expression is s2+1s(s−1)(s2+4s+4)=s2+1s(s3+3s2−4)=s2+1s4+3s3−4s\frac{s^2+1}{s(s-1)(s^2+4s+4)} = \frac{s^2+1}{s(s^3+3s^2-4)} = \frac{s^2+1}{s^4+3s^3-4s}s(s−1)(s2+4s+4)s2+1​=s(s3+3s2−4)s2+1​=s4+3s3−4ss2+1​. The decomposition is As+Bs−1+Cs+2+D(s+2)2\frac{A}{s} + \frac{B}{s-1} + \frac{C}{s+2} + \frac{D}{(s+2)^2}sA​+s−1B​+s+2C​+(s+2)2D​. When combined, the numerator is A(s−1)(s+2)2+Bs(s+2)2+Cs(s−1)(s+2)+Ds(s−1)A(s-1)(s+2)^2 + Bs(s+2)^2 + Cs(s-1)(s+2) + Ds(s-1)A(s−1)(s+2)2+Bs(s+2)2+Cs(s−1)(s+2)+Ds(s−1). The s3s^3s3 term in the numerator comes from As3+Bs3+Cs3As^3 + Bs^3 + Cs^3As3+Bs3+Cs3. The coefficient of s3s^3s3 in the original numerator is 0. So, A+B+C=0A+B+C=0A+B+C=0. We can find CCC from AAA and BBB. C=−A−B=−(−14)−29=14−29=9−836=136C = -A - B = -(-\frac{1}{4}) - \frac{2}{9} = \frac{1}{4} - \frac{2}{9} = \frac{9-8}{36} = \frac{1}{36}C=−A−B=−(−41​)−92​=41​−92​=369−8​=361​. Distractor B is the value of D, and D is the value of B.

Question 5

A system is governed by the differential equation y′′−3y′+2y=u(t−1)y'' - 3y' + 2y = u(t-1)y′′−3y′+2y=u(t−1) with initial conditions y(0)=0y(0)=0y(0)=0 and y′(0)=0y'(0)=0y′(0)=0, where u(t−1)u(t-1)u(t−1) is the unit step function. Find the solution y(t)y(t)y(t).

  1. y(t)=u(t−1)(12−et+12e2t)y(t) = u(t-1) \left( \frac{1}{2} - e^{t} + \frac{1}{2}e^{2t} \right)y(t)=u(t−1)(21​−et+21​e2t)
  2. y(t)=u(t−1)(12−et−1+12e2(t−1))y(t) = u(t-1) \left( \frac{1}{2} - e^{t-1} + \frac{1}{2}e^{2(t-1)} \right)y(t)=u(t−1)(21​−et−1+21​e2(t−1)) (correct answer)
  3. y(t)=u(t)(12−et−1+12e2(t−1))y(t) = u(t) \left( \frac{1}{2} - e^{t-1} + \frac{1}{2}e^{2(t-1)} \right)y(t)=u(t)(21​−et−1+21​e2(t−1))
  4. y(t)=u(t−1)(−12+et−1−12e2(t−1))y(t) = u(t-1) \left( -\frac{1}{2} + e^{t-1} - \frac{1}{2}e^{2(t-1)} \right)y(t)=u(t−1)(−21​+et−1−21​e2(t−1))

Explanation: Take the Laplace transform of the equation: (s2−3s+2)Y(s)=e−ss(s^2 - 3s + 2)Y(s) = \frac{e^{-s}}{s}(s2−3s+2)Y(s)=se−s​. Y(s)=e−ss(s2−3s+2)=e−s1s(s−1)(s−2)Y(s) = \frac{e^{-s}}{s(s^2 - 3s + 2)} = e^{-s} \frac{1}{s(s-1)(s-2)}Y(s)=s(s2−3s+2)e−s​=e−ss(s−1)(s−2)1​. Let G(s)=1s(s−1)(s−2)G(s) = \frac{1}{s(s-1)(s-2)}G(s)=s(s−1)(s−2)1​. We find its inverse transform g(t)g(t)g(t) first. The partial fraction decomposition is G(s)=As+Bs−1+Cs−2G(s) = \frac{A}{s} + \frac{B}{s-1} + \frac{C}{s-2}G(s)=sA​+s−1B​+s−2C​. A=1(s−1)(s−2)∣s=0=12A = \left. \frac{1}{(s-1)(s-2)} \right|_{s=0} = \frac{1}{2}A=(s−1)(s−2)1​​s=0​=21​. B=1s(s−2)∣s=1=−1B = \left. \frac{1}{s(s-2)} \right|_{s=1} = -1B=s(s−2)1​​s=1​=−1. C=1s(s−1)∣s=2=12C = \left. \frac{1}{s(s-1)} \right|_{s=2} = \frac{1}{2}C=s(s−1)1​​s=2​=21​. So G(s)=1/2s−1s−1+1/2s−2G(s) = \frac{1/2}{s} - \frac{1}{s-1} + \frac{1/2}{s-2}G(s)=s1/2​−s−11​+s−21/2​. Then g(t)=L−1{G(s)}=12−et+12e2tg(t) = \mathcal{L}^{-1}\{G(s)\} = \frac{1}{2} - e^t + \frac{1}{2}e^{2t}g(t)=L−1{G(s)}=21​−et+21​e2t. Finally, we find y(t)y(t)y(t) using the second shifting theorem: y(t)=L−1{e−sG(s)}=u(t−1)g(t−1)y(t) = \mathcal{L}^{-1}\{e^{-s}G(s)\} = u(t-1)g(t-1)y(t)=L−1{e−sG(s)}=u(t−1)g(t−1). y(t)=u(t−1)(12−et−1+12e2(t−1))y(t) = u(t-1) \left( \frac{1}{2} - e^{t-1} + \frac{1}{2}e^{2(t-1)} \right)y(t)=u(t−1)(21​−et−1+21​e2(t−1)).

Question 6

The Laplace transform of the solution to an initial value problem is Y(s)=1s4−1Y(s) = \frac{1}{s^4 - 1}Y(s)=s4−11​. Find the solution y(t)y(t)y(t).

  1. y(t)=12cosh⁡(t)−12cos⁡(t)y(t) = \frac{1}{2}\cosh(t) - \frac{1}{2}\cos(t)y(t)=21​cosh(t)−21​cos(t)
  2. y(t)=12sinh⁡(t)−12sin⁡(t)y(t) = \frac{1}{2}\sinh(t) - \frac{1}{2}\sin(t)y(t)=21​sinh(t)−21​sin(t) (correct answer)
  3. y(t)=14cosh⁡(t)−12sin⁡(t)y(t) = \frac{1}{4}\cosh(t) - \frac{1}{2}\sin(t)y(t)=41​cosh(t)−21​sin(t)
  4. y(t)=14et−14e−t+12cos⁡(t)y(t) = \frac{1}{4}e^t - \frac{1}{4}e^{-t} + \frac{1}{2}\cos(t)y(t)=41​et−41​e−t+21​cos(t)

Explanation: First, factor the denominator: s4−1=(s2−1)(s2+1)=(s−1)(s+1)(s2+1)s^4 - 1 = (s^2-1)(s^2+1) = (s-1)(s+1)(s^2+1)s4−1=(s2−1)(s2+1)=(s−1)(s+1)(s2+1). The partial fraction decomposition is Y(s)=As−1+Bs+1+Cs+Ds2+1Y(s) = \frac{A}{s-1} + \frac{B}{s+1} + \frac{Cs+D}{s^2+1}Y(s)=s−1A​+s+1B​+s2+1Cs+D​. A=1(s+1)(s2+1)∣s=1=1(2)(2)=14A = \left. \frac{1}{(s+1)(s^2+1)} \right|_{s=1} = \frac{1}{(2)(2)} = \frac{1}{4}A=(s+1)(s2+1)1​​s=1​=(2)(2)1​=41​. B=1(s−1)(s2+1)∣s=−1=1(−2)(2)=−14B = \left. \frac{1}{(s-1)(s^2+1)} \right|_{s=-1} = \frac{1}{(-2)(2)} = -\frac{1}{4}B=(s−1)(s2+1)1​​s=−1​=(−2)(2)1​=−41​. To find C and D, we multiply by the denominator: 1=A(s+1)(s2+1)+B(s−1)(s2+1)+(Cs+D)(s2−1)1 = A(s+1)(s^2+1) + B(s-1)(s^2+1) + (Cs+D)(s^2-1)1=A(s+1)(s2+1)+B(s−1)(s2+1)+(Cs+D)(s2−1). Equating coefficients for s3s^3s3: 0=A+B+C=14−14+C  ⟹  C=00 = A+B+C = \frac{1}{4} - \frac{1}{4} + C \implies C=00=A+B+C=41​−41​+C⟹C=0. Equating constant terms: 1=A−B−D=14−(−14)−D=12−D  ⟹  D=−121 = A-B-D = \frac{1}{4} - (-\frac{1}{4}) - D = \frac{1}{2} - D \implies D = -\frac{1}{2}1=A−B−D=41​−(−41​)−D=21​−D⟹D=−21​. So, Y(s)=1/4s−1−1/4s+1−1/2s2+1Y(s) = \frac{1/4}{s-1} - \frac{1/4}{s+1} - \frac{1/2}{s^2+1}Y(s)=s−11/4​−s+11/4​−s2+11/2​. Taking the inverse Laplace transform: y(t)=14et−14e−t−12sin⁡(t)y(t) = \frac{1}{4}e^t - \frac{1}{4}e^{-t} - \frac{1}{2}\sin(t)y(t)=41​et−41​e−t−21​sin(t). We can write this in terms of hyperbolic functions: y(t)=12(et−e−t2)−12sin⁡(t)=12sinh⁡(t)−12sin⁡(t)y(t) = \frac{1}{2}\left(\frac{e^t-e^{-t}}{2}\right) - \frac{1}{2}\sin(t) = \frac{1}{2}\sinh(t) - \frac{1}{2}\sin(t)y(t)=21​(2et−e−t​)−21​sin(t)=21​sinh(t)−21​sin(t).

Question 7

Find the inverse Laplace transform of F(s)=s2−s−6(s+1)(s2−4)F(s) = \frac{s^2-s-6}{(s+1)(s^2-4)}F(s)=(s+1)(s2−4)s2−s−6​.

  1. 43e−t−13e2t\frac{4}{3}e^{-t} - \frac{1}{3}e^{2t}34​e−t−31​e2t (correct answer)
  2. 43e−t−13e2t−23e−2t\frac{4}{3}e^{-t} - \frac{1}{3}e^{2t} - \frac{2}{3}e^{-2t}34​e−t−31​e2t−32​e−2t
  3. 53e−t+13e2t\frac{5}{3}e^{-t} + \frac{1}{3}e^{2t}35​e−t+31​e2t
  4. e−t+e2t−e−2te^{-t} + e^{2t} - e^{-2t}e−t+e2t−e−2t

Explanation: Before applying partial fractions, it is crucial to check for common factors between the numerator and denominator. Factor the numerator: s2−s−6=(s−3)(s+2)s^2-s-6 = (s-3)(s+2)s2−s−6=(s−3)(s+2). Factor the denominator: (s+1)(s2−4)=(s+1)(s−2)(s+2)(s+1)(s^2-4) = (s+1)(s-2)(s+2)(s+1)(s2−4)=(s+1)(s−2)(s+2). So, F(s)=(s−3)(s+2)(s+1)(s−2)(s+2)F(s) = \frac{(s-3)(s+2)}{(s+1)(s-2)(s+2)}F(s)=(s+1)(s−2)(s+2)(s−3)(s+2)​. The term (s+2)(s+2)(s+2) cancels out, simplifying the expression to F(s)=s−3(s+1)(s−2)F(s) = \frac{s-3}{(s+1)(s-2)}F(s)=(s+1)(s−2)s−3​. Now, we use partial fraction decomposition: s−3(s+1)(s−2)=As+1+Bs−2\frac{s-3}{(s+1)(s-2)} = \frac{A}{s+1} + \frac{B}{s-2}(s+1)(s−2)s−3​=s+1A​+s−2B​. Using the cover-up method: A=s−3s−2∣s=−1=−1−3−1−2=−4−3=43A = \left. \frac{s-3}{s-2} \right|_{s=-1} = \frac{-1-3}{-1-2} = \frac{-4}{-3} = \frac{4}{3}A=s−2s−3​​s=−1​=−1−2−1−3​=−3−4​=34​. B=s−3s+1∣s=2=2−32+1=−13B = \left. \frac{s-3}{s+1} \right|_{s=2} = \frac{2-3}{2+1} = -\frac{1}{3}B=s+1s−3​​s=2​=2+12−3​=−31​. Thus, F(s)=4/3s+1−1/3s−2F(s) = \frac{4/3}{s+1} - \frac{1/3}{s-2}F(s)=s+14/3​−s−21/3​. Taking the inverse Laplace transform gives f(t)=43e−t−13e2tf(t) = \frac{4}{3}e^{-t} - \frac{1}{3}e^{2t}f(t)=34​e−t−31​e2t.

Question 8

The Laplace transform Y(s)Y(s)Y(s) of a function y(t)y(t)y(t) satisfies the equation sY(s)−y(0)=1s(s−2)sY(s) - y(0) = \frac{1}{s(s-2)}sY(s)−y(0)=s(s−2)1​. If y(0)=1y(0) = 1y(0)=1, what is y(t)y(t)y(t)?

  1. y(t)=34−12t+14e2ty(t) = \frac{3}{4} - \frac{1}{2}t + \frac{1}{4}e^{2t}y(t)=43​−21​t+41​e2t (correct answer)
  2. y(t)=−14−12t+14e2ty(t) = -\frac{1}{4} - \frac{1}{2}t + \frac{1}{4}e^{2t}y(t)=−41​−21​t+41​e2t
  3. y(t)=34−12+14e2ty(t) = \frac{3}{4} - \frac{1}{2} + \frac{1}{4}e^{2t}y(t)=43​−21​+41​e2t
  4. y(t)=1−12t+12e2ty(t) = 1 - \frac{1}{2}t + \frac{1}{2}e^{2t}y(t)=1−21​t+21​e2t

Explanation: First, we must solve for Y(s)Y(s)Y(s). Given sY(s)−y(0)=1s(s−2)sY(s) - y(0) = \frac{1}{s(s-2)}sY(s)−y(0)=s(s−2)1​ and y(0)=1y(0)=1y(0)=1, we substitute y(0)y(0)y(0): sY(s)−1=1s(s−2)sY(s) - 1 = \frac{1}{s(s-2)}sY(s)−1=s(s−2)1​. sY(s)=1+1s(s−2)=s(s−2)+1s(s−2)=s2−2s+1s(s−2)=(s−1)2s(s−2)sY(s) = 1 + \frac{1}{s(s-2)} = \frac{s(s-2)+1}{s(s-2)} = \frac{s^2-2s+1}{s(s-2)} = \frac{(s-1)^2}{s(s-2)}sY(s)=1+s(s−2)1​=s(s−2)s(s−2)+1​=s(s−2)s2−2s+1​=s(s−2)(s−1)2​. Y(s)=(s−1)2s2(s−2)Y(s) = \frac{(s-1)^2}{s^2(s-2)}Y(s)=s2(s−2)(s−1)2​. Now we find the inverse transform using partial fractions. The form is Y(s)=As+Bs2+Cs−2Y(s) = \frac{A}{s} + \frac{B}{s^2} + \frac{C}{s-2}Y(s)=sA​+s2B​+s−2C​. Using the cover-up method for BBB and CCC: B=(s−1)2s−2∣s=0=(−1)2−2=−12B = \left. \frac{(s-1)^2}{s-2} \right|_{s=0} = \frac{(-1)^2}{-2} = -\frac{1}{2}B=s−2(s−1)2​​s=0​=−2(−1)2​=−21​. C=(s−1)2s2∣s=2=(2−1)222=14C = \left. \frac{(s-1)^2}{s^2} \right|_{s=2} = \frac{(2-1)^2}{2^2} = \frac{1}{4}C=s2(s−1)2​​s=2​=22(2−1)2​=41​. To find AAA, we can equate coefficients. The equation is (s−1)2=As(s−2)+B(s−2)+Cs2(s-1)^2 = As(s-2) + B(s-2) + Cs^2(s−1)2=As(s−2)+B(s−2)+Cs2. Equating coefficients of s2s^2s2: 1=A+C1 = A+C1=A+C. Since C=1/4C=1/4C=1/4, A=1−C=1−1/4=3/4A = 1 - C = 1 - 1/4 = 3/4A=1−C=1−1/4=3/4. So, Y(s)=3/4s−1/2s2+1/4s−2Y(s) = \frac{3/4}{s} - \frac{1/2}{s^2} + \frac{1/4}{s-2}Y(s)=s3/4​−s21/2​+s−21/4​. Taking the inverse Laplace transform: y(t)=34−12t+14e2ty(t) = \frac{3}{4} - \frac{1}{2}t + \frac{1}{4}e^{2t}y(t)=43​−21​t+41​e2t.

Question 9

The transfer function of a linear time-invariant system is given by H(s)=1s2+2s+kH(s) = \frac{1}{s^2+2s+k}H(s)=s2+2s+k1​. For which of the following values of the parameter kkk will the impulse response h(t)=L−1{H(s)}h(t) = \mathcal{L}^{-1}\{H(s)\}h(t)=L−1{H(s)} be a product of a decaying exponential and a sinusoidal function?

  1. k=1k = 1k=1
  2. k=−3k = -3k=−3
  3. k=0.5k = 0.5k=0.5
  4. k=5k = 5k=5 (correct answer)

Explanation: The form of the impulse response h(t)h(t)h(t) depends on the roots of the denominator of H(s)H(s)H(s), which is the characteristic equation s2+2s+k=0s^2+2s+k=0s2+2s+k=0. The roots are given by the quadratic formula: s=−2±4−4k2=−1±1−ks = \frac{-2 \pm \sqrt{4-4k}}{2} = -1 \pm \sqrt{1-k}s=2−2±4−4k​​=−1±1−k​. For the solution to be a product of a decaying exponential and a sinusoidal function, the roots must be a complex conjugate pair. This occurs when the discriminant is negative. 1−k<0  ⟹  k>11-k < 0 \implies k > 11−k<0⟹k>1. We examine the answer choices: (A) k=1k=1k=1: 1−k=01-k=01−k=0. This gives a repeated real root, leading to critical damping (h(t)h(t)h(t) has a te−tte^{-t}te−t term). (B) k=−3k=-3k=−3: 1−k=4>01-k = 4 > 01−k=4>0. This gives two distinct real roots, leading to an overdamped response. (C) k=0.5k=0.5k=0.5: 1−k=0.5>01-k = 0.5 > 01−k=0.5>0. This also gives two distinct real roots, leading to an overdamped response. (D) k=5k=5k=5: 1−k=−4<01-k = -4 < 01−k=−4<0. This gives complex roots s=−1±2is = -1 \pm 2is=−1±2i, leading to an underdamped response of the form h(t)=e−t(Acos⁡(2t)+Bsin⁡(2t))h(t) = e^{-t}(A\cos(2t)+B\sin(2t))h(t)=e−t(Acos(2t)+Bsin(2t)). This matches the description.

Question 10

Let f(t)=L−1{s+1s(s−1)2}f(t) = \mathcal{L}^{-1}\left\{\frac{s+1}{s(s-1)^2}\right\}f(t)=L−1{s(s−1)2s+1​}. Which of the following is f(t)f(t)f(t)?

  1. f(t)=1+et+2tetf(t) = 1 + e^t + 2te^tf(t)=1+et+2tet
  2. f(t)=1−et+2tetf(t) = 1 - e^t + 2te^tf(t)=1−et+2tet (correct answer)
  3. f(t)=1+etf(t) = 1 + e^tf(t)=1+et
  4. f(t)=1−et+2etf(t) = 1 - e^t + 2e^tf(t)=1−et+2et

Explanation: The function F(s)=s+1s(s−1)2F(s) = \frac{s+1}{s(s-1)^2}F(s)=s(s−1)2s+1​ has a partial fraction decomposition of the form As+Bs−1+C(s−1)2\frac{A}{s} + \frac{B}{s-1} + \frac{C}{(s-1)^2}sA​+s−1B​+(s−1)2C​. To find the coefficients: A=s+1(s−1)2∣s=0=1(−1)2=1A = \left. \frac{s+1}{(s-1)^2} \right|_{s=0} = \frac{1}{(-1)^2} = 1A=(s−1)2s+1​​s=0​=(−1)21​=1. C=s+1s∣s=1=21=2C = \left. \frac{s+1}{s} \right|_{s=1} = \frac{2}{1} = 2C=ss+1​​s=1​=12​=2. To find BBB, we can use the equation s+1=A(s−1)2+Bs(s−1)+Css+1 = A(s-1)^2 + Bs(s-1) + Css+1=A(s−1)2+Bs(s−1)+Cs. Expanding and equating coefficients of s2s^2s2: 0=A+B0 = A+B0=A+B. Since A=1A=1A=1, we have B=−1B=-1B=−1. So, F(s)=1s−1s−1+2(s−1)2F(s) = \frac{1}{s} - \frac{1}{s-1} + \frac{2}{(s-1)^2}F(s)=s1​−s−11​+(s−1)22​. Taking the inverse Laplace transform of each term yields: f(t)=L−1{1s}−L−1{1s−1}+2L−1{1(s−1)2}=1−et+2tetf(t) = \mathcal{L}^{-1}\{\frac{1}{s}\} - \mathcal{L}^{-1}\{\frac{1}{s-1}\} + 2\mathcal{L}^{-1}\{\frac{1}{(s-1)^2}\} = 1 - e^t + 2te^tf(t)=L−1{s1​}−L−1{s−11​}+2L−1{(s−1)21​}=1−et+2tet.

Question 11

Determine the inverse Laplace transform of F(s)=s+3s2+s+1F(s) = \frac{s+3}{s^2+s+1}F(s)=s2+s+1s+3​.

  1. e−t/2(cos⁡(32t)+533sin⁡(32t))e^{-t/2}\left(\cos\left(\frac{\sqrt{3}}{2}t\right) + \frac{5\sqrt{3}}{3}\sin\left(\frac{\sqrt{3}}{2}t\right)\right)e−t/2(cos(23​​t)+353​​sin(23​​t)) (correct answer)
  2. et/2(cos⁡(32t)+533sin⁡(32t))e^{t/2}\left(\cos\left(\frac{\sqrt{3}}{2}t\right) + \frac{5\sqrt{3}}{3}\sin\left(\frac{\sqrt{3}}{2}t\right)\right)et/2(cos(23​​t)+353​​sin(23​​t))
  3. e−t/2(cos⁡(32t)+52sin⁡(32t))e^{-t/2}\left(\cos\left(\frac{\sqrt{3}}{2}t\right) + \frac{5}{2}\sin\left(\frac{\sqrt{3}}{2}t\right)\right)e−t/2(cos(23​​t)+25​sin(23​​t))
  4. e−t(cos⁡(t)+3sin⁡(t))e^{-t}\left(\cos(t) + 3\sin(t)\right)e−t(cos(t)+3sin(t))

Explanation: The denominator s2+s+1s^2+s+1s2+s+1 is an irreducible quadratic. We complete the square: s2+s+1=(s2+s+14)−14+1=(s+12)2+34s^2+s+1 = (s^2+s+\frac{1}{4}) - \frac{1}{4} + 1 = (s+\frac{1}{2})^2 + \frac{3}{4}s2+s+1=(s2+s+41​)−41​+1=(s+21​)2+43​. So F(s)=s+3(s+12)2+(32)2F(s) = \frac{s+3}{(s+\frac{1}{2})^2 + (\frac{\sqrt{3}}{2})^2}F(s)=(s+21​)2+(23​​)2s+3​. To match the transform formulas for sine and cosine with a shift, we rewrite the numerator in terms of (s+12)(s+\frac{1}{2})(s+21​): s+3=(s+12)+52s+3 = (s+\frac{1}{2}) + \frac{5}{2}s+3=(s+21​)+25​. So, F(s)=(s+12)+52(s+12)2+(32)2=s+12(s+12)2+(32)2+52(s+12)2+(32)2F(s) = \frac{(s+\frac{1}{2}) + \frac{5}{2}}{(s+\frac{1}{2})^2 + (\frac{\sqrt{3}}{2})^2} = \frac{s+\frac{1}{2}}{(s+\frac{1}{2})^2 + (\frac{\sqrt{3}}{2})^2} + \frac{\frac{5}{2}}{(s+\frac{1}{2})^2 + (\frac{\sqrt{3}}{2})^2}F(s)=(s+21​)2+(23​​)2(s+21​)+25​​=(s+21​)2+(23​​)2s+21​​+(s+21​)2+(23​​)225​​. For the second term, we need 32\frac{\sqrt{3}}{2}23​​ in the numerator for the sine transform. We multiply and divide by it: F(s)=s+12(s+12)2+(32)2+5/23/2⋅3/2(s+12)2+(32)2=s+12(s+12)2+(32)2+53⋅3/2(s+12)2+(32)2F(s) = \frac{s+\frac{1}{2}}{(s+\frac{1}{2})^2 + (\frac{\sqrt{3}}{2})^2} + \frac{5/2}{\sqrt{3}/2} \cdot \frac{\sqrt{3}/2}{(s+\frac{1}{2})^2 + (\frac{\sqrt{3}}{2})^2} = \frac{s+\frac{1}{2}}{(s+\frac{1}{2})^2 + (\frac{\sqrt{3}}{2})^2} + \frac{5}{\sqrt{3}} \cdot \frac{\sqrt{3}/2}{(s+\frac{1}{2})^2 + (\frac{\sqrt{3}}{2})^2}F(s)=(s+21​)2+(23​​)2s+21​​+3​/25/2​⋅(s+21​)2+(23​​)23​/2​=(s+21​)2+(23​​)2s+21​​+3​5​⋅(s+21​)2+(23​​)23​/2​. Taking the inverse transform gives: f(t)=e−t/2cos⁡(32t)+53e−t/2sin⁡(32t)f(t) = e^{-t/2}\cos\left(\frac{\sqrt{3}}{2}t\right) + \frac{5}{\sqrt{3}}e^{-t/2}\sin\left(\frac{\sqrt{3}}{2}t\right)f(t)=e−t/2cos(23​​t)+3​5​e−t/2sin(23​​t). Rationalizing the coefficient 53=533\frac{5}{\sqrt{3}} = \frac{5\sqrt{3}}{3}3​5​=353​​ gives the final answer.

Question 12

Solve the initial value problem y′′+y=4ety'' + y = 4e^ty′′+y=4et with initial conditions y(0)=1y(0)=1y(0)=1 and y′(0)=−1y'(0)=-1y′(0)=−1.

  1. y(t)=2et−cos⁡(t)−3sin⁡(t)y(t) = 2e^t - \cos(t) - 3\sin(t)y(t)=2et−cos(t)−3sin(t) (correct answer)
  2. y(t)=2et−cos⁡(t)+sin⁡(t)y(t) = 2e^t - \cos(t) + \sin(t)y(t)=2et−cos(t)+sin(t)
  3. y(t)=2et−3cos⁡(t)−3sin⁡(t)y(t) = 2e^t - 3\cos(t) - 3\sin(t)y(t)=2et−3cos(t)−3sin(t)
  4. y(t)=2et+cos⁡(t)+3sin⁡(t)y(t) = 2e^t + \cos(t) + 3\sin(t)y(t)=2et+cos(t)+3sin(t)

Explanation: Taking the Laplace transform of the differential equation gives: L{y′′}+L{y}=L{4et}\mathcal{L}\{y''\} + \mathcal{L}\{y\} = \mathcal{L}\{4e^t\}L{y′′}+L{y}=L{4et} s2Y(s)−sy(0)−y′(0)+Y(s)=4s−1s^2Y(s) - sy(0) - y'(0) + Y(s) = \frac{4}{s-1}s2Y(s)−sy(0)−y′(0)+Y(s)=s−14​. Substituting the initial conditions y(0)=1y(0)=1y(0)=1 and y′(0)=−1y'(0)=-1y′(0)=−1: s2Y(s)−s−(−1)+Y(s)=4s−1s^2Y(s) - s - (-1) + Y(s) = \frac{4}{s-1}s2Y(s)−s−(−1)+Y(s)=s−14​ (s2+1)Y(s)−s+1=4s−1(s^2+1)Y(s) - s + 1 = \frac{4}{s-1}(s2+1)Y(s)−s+1=s−14​ Y(s)=4(s−1)(s2+1)+s−1s2+1Y(s) = \frac{4}{(s-1)(s^2+1)} + \frac{s-1}{s^2+1}Y(s)=(s−1)(s2+1)4​+s2+1s−1​. We apply partial fractions to the first term: 4(s−1)(s2+1)=As−1+Bs+Cs2+1\frac{4}{(s-1)(s^2+1)} = \frac{A}{s-1} + \frac{Bs+C}{s^2+1}(s−1)(s2+1)4​=s−1A​+s2+1Bs+C​. Solving for coefficients gives A=2,B=−2,C=−2A=2, B=-2, C=-2A=2,B=−2,C=−2. So, Y(s)=(2s−1−2s+2s2+1)+s−1s2+1=2s−1−2ss2+1−2s2+1+ss2+1−1s2+1Y(s) = \left(\frac{2}{s-1} - \frac{2s+2}{s^2+1}\right) + \frac{s-1}{s^2+1} = \frac{2}{s-1} - \frac{2s}{s^2+1} - \frac{2}{s^2+1} + \frac{s}{s^2+1} - \frac{1}{s^2+1}Y(s)=(s−12​−s2+12s+2​)+s2+1s−1​=s−12​−s2+12s​−s2+12​+s2+1s​−s2+11​. Combining terms: Y(s)=2s−1−ss2+1−3s2+1Y(s) = \frac{2}{s-1} - \frac{s}{s^2+1} - \frac{3}{s^2+1}Y(s)=s−12​−s2+1s​−s2+13​. Taking the inverse Laplace transform: y(t)=2et−cos⁡(t)−3sin⁡(t)y(t) = 2e^t - \cos(t) - 3\sin(t)y(t)=2et−cos(t)−3sin(t).