Skip to main content
Sign in Create a free account
Varsity Tutors
Sign in

Inverse Laplace via Partial Fractions - Differential Equations

  1. 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)}.
    Answer: 43e−t−13e2t\frac{4}{3}e^{-t} - \frac{1}{3}e^{2t} 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). Factor the denominator: (s+1)(s2−4)=(s+1)(s−2)(s+2)(s+1)(s^2-4) = (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)}. The term (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)}. 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}. 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}. 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}. Thus, F(s)=4/3s+1−1/3s−2F(s) = \frac{4/3}{s+1} - \frac{1/3}{s-2}. Taking the inverse Laplace transform gives f(t)=43e−t−13e2tf(t) = \frac{4}{3}e^{-t} - \frac{1}{3}e^{2t}.
  2. Solve the initial value problem y′′+y=4ety'' + y = 4e^t with initial conditions y(0)=1y(0)=1 and y′(0)=−1y'(0)=-1.
    Answer: y(t)=2et−cos⁡(t)−3sin⁡(t)y(t) = 2e^t - \cos(t) - 3\sin(t) 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\} s2Y(s)−sy(0)−y′(0)+Y(s)=4s−1s^2Y(s) - sy(0) - y'(0) + Y(s) = \frac{4}{s-1}. Substituting the initial conditions y(0)=1y(0)=1 and y′(0)=−1y'(0)=-1: s2Y(s)−s−(−1)+Y(s)=4s−1s^2Y(s) - s - (-1) + Y(s) = \frac{4}{s-1} (s2+1)Y(s)−s+1=4s−1(s^2+1)Y(s) - s + 1 = \frac{4}{s-1} Y(s)=4(s−1)(s2+1)+s−1s2+1Y(s) = \frac{4}{(s-1)(s^2+1)} + \frac{s-1}{s^2+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}. Solving for coefficients gives A=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}. 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}. Taking the inverse Laplace transform: y(t)=2et−cos⁡(t)−3sin⁡(t)y(t) = 2e^t - \cos(t) - 3\sin(t).
  3. Find the inverse Laplace transform of F(s)=1s3+sF(s) = \frac{1}{s^3+s}.
    Answer: 1−cos⁡(t)1 - \cos(t) First, factor the denominator: s3+s=s(s2+1)s^3+s = s(s^2+1). The function is F(s)=1s(s2+1)F(s) = \frac{1}{s(s^2+1)}. We use partial fraction decomposition: F(s)=As+Bs+Cs2+1F(s) = \frac{A}{s} + \frac{Bs+C}{s^2+1}. Multiplying by the denominator gives 1=A(s2+1)+(Bs+C)s1 = A(s^2+1) + (Bs+C)s. Let s=0s=0: 1=A(1)  ⟹  A=11 = A(1) \implies A=1. Equating coefficients of s2s^2: 0=A+B  ⟹  B=−A=−10 = A+B \implies B = -A = -1. Equating coefficients of ss: 0=C0 = C. So, F(s)=1s−ss2+1F(s) = \frac{1}{s} - \frac{s}{s^2+1}. Taking the inverse Laplace transform term by term: L−1{1s}=1\mathcal{L}^{-1}\{\frac{1}{s}\} = 1 L−1{ss2+1}=cos⁡(t)\mathcal{L}^{-1}\{\frac{s}{s^2+1}\} = \cos(t) Therefore, f(t)=1−cos⁡(t)f(t) = 1 - \cos(t).
  4. Determine the inverse Laplace transform of F(s)=s+3s2+s+1F(s) = \frac{s+3}{s^2+s+1}.
    Answer: 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) The denominator s2+s+1s^2+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}. So F(s)=s+3(s+12)2+(32)2F(s) = \frac{s+3}{(s+\frac{1}{2})^2 + (\frac{\sqrt{3}}{2})^2}. 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+3=(s+12)+52s+3 = (s+\frac{1}{2}) + \frac{5}{2}. 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}. For the second term, we need 32\frac{\sqrt{3}}{2} 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}. 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). Rationalizing the coefficient 53=533\frac{5}{\sqrt{3}} = \frac{5\sqrt{3}}{3} gives the final answer.
  5. Find the inverse Laplace transform, 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)}.
    Answer: 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) 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}. Using the cover-up method for AA: 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}. Multiply by the denominator: 3s=A(s2+2s+5)+(Bs+C)(s−1)3s = A(s^2+2s+5) + (Bs+C)(s-1). Equating coefficients of s2s^2: 0=A+B  ⟹  B=−A=−380 = A+B \implies B = -A = -\frac{3}{8}. Equating constant terms: 0=5A−C  ⟹  C=5A=1580 = 5A-C \implies C = 5A = \frac{15}{8}. 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}. 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+2^2. 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}. 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}. 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).
  6. Let f(t)=L−1{s+1s(s−1)2}f(t) = \mathcal{L}^{-1}\left\{\frac{s+1}{s(s-1)^2}\right\}. Which of the following is f(t)f(t)?
    Answer: f(t)=1−et+2tetf(t) = 1 - e^t + 2te^t The function F(s)=s+1s(s−1)2F(s) = \frac{s+1}{s(s-1)^2} 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}. 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} = 1. C=s+1s∣s=1=21=2C = \left. \frac{s+1}{s} \right|_{s=1} = \frac{2}{1} = 2. To find BB, we can use the equation s+1=A(s−1)2+Bs(s−1)+Css+1 = A(s-1)^2 + Bs(s-1) + Cs. Expanding and equating coefficients of s2s^2: 0=A+B0 = A+B. Since A=1A=1, we have B=−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}. 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^t.
  7. The transfer function of a linear time-invariant system is given by H(s)=1s2+2s+kH(s) = \frac{1}{s^2+2s+k}. For which of the following values of the parameter kk will the impulse response h(t)=L−1{H(s)}h(t) = \mathcal{L}^{-1}\{H(s)\} be a product of a decaying exponential and a sinusoidal function?
    Answer: k=5k = 5 The form of the impulse response h(t)h(t) depends on the roots of the denominator of H(s)H(s), which is the characteristic equation s2+2s+k=0s^2+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}. 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 > 1. We examine the answer choices: (A) k=1k=1: 1−k=01-k=0. This gives a repeated real root, leading to critical damping (h(t)h(t) has a te−tte^{-t} term). (B) k=−3k=-3: 1−k=4>01-k = 4 > 0. This gives two distinct real roots, leading to an overdamped response. (C) k=0.5k=0.5: 1−k=0.5>01-k = 0.5 > 0. This also gives two distinct real roots, leading to an overdamped response. (D) k=5k=5: 1−k=−4<01-k = -4 < 0. This gives complex roots s=−1±2is = -1 \pm 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)). This matches the description.
  8. The Laplace transform of the solution to an initial value problem is Y(s)=1s4−1Y(s) = \frac{1}{s^4 - 1}. Find the solution y(t)y(t).
    Answer: y(t)=12sinh⁡(t)−12sin⁡(t)y(t) = \frac{1}{2}\sinh(t) - \frac{1}{2}\sin(t) 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). 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}. 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}. 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}. 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). Equating coefficients for s3s^3: 0=A+B+C=14−14+C  ⟹  C=00 = A+B+C = \frac{1}{4} - \frac{1}{4} + C \implies 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}. 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}. 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). 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).
  9. A system is governed by the differential equation y′′−3y′+2y=u(t−1)y'' - 3y' + 2y = u(t-1) with initial conditions y(0)=0y(0)=0 and y′(0)=0y'(0)=0, where u(t−1)u(t-1) is the unit step function. Find the solution y(t)y(t).
    Answer: 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) Take the Laplace transform of the equation: (s2−3s+2)Y(s)=e−ss(s^2 - 3s + 2)Y(s) = \frac{e^{-s}}{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)}. Let G(s)=1s(s−1)(s−2)G(s) = \frac{1}{s(s-1)(s-2)}. We find its inverse transform 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}. A=1(s−1)(s−2)∣s=0=12A = \left. \frac{1}{(s-1)(s-2)} \right|_{s=0} = \frac{1}{2}. B=1s(s−2)∣s=1=−1B = \left. \frac{1}{s(s-2)} \right|_{s=1} = -1. C=1s(s−1)∣s=2=12C = \left. \frac{1}{s(s-1)} \right|_{s=2} = \frac{1}{2}. So G(s)=1/2s−1s−1+1/2s−2G(s) = \frac{1/2}{s} - \frac{1}{s-1} + \frac{1/2}{s-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}. Finally, we find 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)=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).
  10. The Laplace transform Y(s)Y(s) of a function y(t)y(t) satisfies the equation sY(s)−y(0)=1s(s−2)sY(s) - y(0) = \frac{1}{s(s-2)}. If y(0)=1y(0) = 1, what is y(t)y(t)?
    Answer: y(t)=34−12t+14e2ty(t) = \frac{3}{4} - \frac{1}{2}t + \frac{1}{4}e^{2t} First, we must solve for Y(s)Y(s). Given sY(s)−y(0)=1s(s−2)sY(s) - y(0) = \frac{1}{s(s-2)} and y(0)=1y(0)=1, we substitute y(0)y(0): sY(s)−1=1s(s−2)sY(s) - 1 = \frac{1}{s(s-2)}. 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)}. Y(s)=(s−1)2s2(s−2)Y(s) = \frac{(s-1)^2}{s^2(s-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}. Using the cover-up method for BB and CC: 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}. 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}. To find AA, 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. Equating coefficients of s2s^2: 1=A+C1 = A+C. Since C=1/4C=1/4, A=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}. Taking the inverse Laplace transform: y(t)=34−12t+14e2ty(t) = \frac{3}{4} - \frac{1}{2}t + \frac{1}{4}e^{2t}.
  11. The Laplace transform of a function 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)}. Without computing the coefficients, what is the general form of y(t)y(t) for t>0t > 0?
    Answer: 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) The form of the solution y(t)y(t) is determined by the poles of 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). The poles are: - s=0s=0: a simple pole, which corresponds to a constant term C1C_1 in y(t)y(t). - s=3s=3: a repeated pole of order 2, which corresponds to terms C2e3t+C3te3tC_2 e^{3t} + C_3 t e^{3t}. - s2+4=0  ⟹  s=±2is^2+4=0 \implies s = \pm 2i: a pair of complex conjugate poles, which corresponds to terms C4cos⁡(2t)+C5sin⁡(2t)C_4 \cos(2t) + C_5 \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).
  12. The function F(s)=s2+1s(s−1)(s+2)2F(s) = \frac{s^2+1}{s(s-1)(s+2)^2} 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}. What is the value of the coefficient CC?
    Answer: 136\frac{1}{36} To find CC, 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 ss. 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}. 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}. 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}. 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}. 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). The s3s^3 term in the numerator comes from As3+Bs3+Cs3As^3 + Bs^3 + Cs^3. The coefficient of s3s^3 in the original numerator is 0. So, A+B+C=0A+B+C=0. We can find CC from AA and BB. 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}. Distractor B is the value of D, and D is the value of B.

Start this quiz