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Heaviside Step Functions - Differential Equations

  1. Find the Laplace transform of the function f(t)=cos⁡(2t)(uπ/2(t)−u3π/2(t))f(t) = \cos(2t)(u_{\pi/2}(t) - u_{3\pi/2}(t)).
    Answer: ss2+4(e−3πs/2−e−πs/2)\frac{s}{s^2+4}(e^{-3\pi s/2} - e^{-\pi s/2}) We need to compute L{cos⁡(2t)uπ/2(t)}−L{cos⁡(2t)u3π/2(t)}\mathcal{L}\{\cos(2t)u_{\pi/2}(t)\} - \mathcal{L}\{\cos(2t)u_{3\pi/2}(t)\}. For the first term, with c=π/2c=\pi/2, we must write cos⁡(2t)\cos(2t) as a function of (t−π/2)(t-\pi/2). 2t=2(t−π/2)+π2t = 2(t-\pi/2) + \pi. Using the identity cos⁡(x+π)=−cos⁡(x)\cos(x+\pi) = -\cos(x), we have cos⁡(2t)=cos⁡(2(t−π/2)+π)=−cos⁡(2(t−π/2))\cos(2t) = \cos(2(t-\pi/2)+\pi) = -\cos(2(t-\pi/2)). So, L{cos⁡(2t)uπ/2(t)}=L{−cos⁡(2(t−π/2))uπ/2(t)}=−e−πs/2L{cos⁡(2t)}=−e−πs/2ss2+4\mathcal{L}\{\cos(2t)u_{\pi/2}(t)\} = \mathcal{L}\{-\cos(2(t-\pi/2))u_{\pi/2}(t)\} = -e^{-\pi s/2}\mathcal{L}\{\cos(2t)\} = -e^{-\pi s/2}\frac{s}{s^2+4}. For the second term, with c=3π/2c=3\pi/2, we write cos⁡(2t)\cos(2t) as a function of (t−3π/2)(t-3\pi/2). 2t=2(t−3π/2)+3π2t = 2(t-3\pi/2) + 3\pi. Using the identity cos⁡(x+3π)=−cos⁡(x)\cos(x+3\pi) = -\cos(x), we have cos⁡(2t)=cos⁡(2(t−3π/2)+3π)=−cos⁡(2(t−3π/2))\cos(2t) = \cos(2(t-3\pi/2)+3\pi) = -\cos(2(t-3\pi/2)). So, L{cos⁡(2t)u3π/2(t)}=−e−3πs/2L{cos⁡(2t)}=−e−3πs/2ss2+4\mathcal{L}\{\cos(2t)u_{3\pi/2}(t)\} = -e^{-3\pi s/2}\mathcal{L}\{\cos(2t)\} = -e^{-3\pi s/2}\frac{s}{s^2+4}. Combining the results: F(s)=−e−πs/2ss2+4−(−e−3πs/2ss2+4)=ss2+4(e−3πs/2−e−πs/2)F(s) = -e^{-\pi s/2}\frac{s}{s^2+4} - \left(-e^{-3\pi s/2}\frac{s}{s^2+4}\right) = \frac{s}{s^2+4}(e^{-3\pi s/2} - e^{-\pi s/2}).
  2. What is the Laplace transform of the function f(t)=(t2+1)u2(t)f(t) = (t^2+1)u_2(t)?
    Answer: e−2s(2s3+4s2+5s)e^{-2s}\left( \frac{2}{s^3} + \frac{4}{s^2} + \frac{5}{s} \right) To find the Laplace transform of f(t)=g(t)uc(t)f(t)=g(t)u_c(t), we must use the second shifting theorem, L{g(t)uc(t)}=e−csL{g(t+c)}\mathcal{L}\{g(t)u_c(t)\} = e^{-cs}\mathcal{L}\{g(t+c)\}. In this case, g(t)=t2+1g(t)=t^2+1 and c=2c=2. We need to find L{g(t+2)}\mathcal{L}\{g(t+2)\}. g(t+2)=(t+2)2+1=(t2+4t+4)+1=t2+4t+5g(t+2) = (t+2)^2 + 1 = (t^2 + 4t + 4) + 1 = t^2 + 4t + 5. Now, we take the Laplace transform of this new polynomial: L{t2+4t+5}=L{t2}+4L{t}+L{5}=2s3+4s2+5s\mathcal{L}\{t^2 + 4t + 5\} = \mathcal{L}\{t^2\} + 4\mathcal{L}\{t\} + \mathcal{L}\{5\} = \frac{2}{s^3} + \frac{4}{s^2} + \frac{5}{s}. Finally, we multiply by e−cs=e−2se^{-cs} = e^{-2s} to get the final answer: F(s)=e−2s(2s3+4s2+5s)F(s) = e^{-2s}\left( \frac{2}{s^3} + \frac{4}{s^2} + \frac{5}{s} \right). Distractor A represents the common mistake of applying e−csL{g(t)}e^{-cs}\mathcal{L}\{g(t)\} instead of e−csL{g(t+c)}e^{-cs}\mathcal{L}\{g(t+c)\}.
  3. Using the provided solution to the initial value problem, what is the value of y(2)y(2)?
    Answer: 3e−4+52−52e−23e^{-4} + \frac{5}{2} - \frac{5}{2}e^{-2} The solution is given by y(t)=3e−2t+52(1−e−2(t−1))u1(t)y(t) = 3e^{-2t} + \frac{5}{2}(1 - e^{-2(t-1)})u_1(t). To find y(2)y(2), we substitute t=2t=2 into the expression. Since 2>12 > 1, the Heaviside function u1(2)u_1(2) is equal to 1. So, we evaluate the entire expression at t=2t=2: y(2)=3e−2(2)+52(1−e−2(2−1))u1(2)y(2) = 3e^{-2(2)} + \frac{5}{2}(1 - e^{-2(2-1)})u_1(2) y(2)=3e−4+52(1−e−2(1))(1)y(2) = 3e^{-4} + \frac{5}{2}(1 - e^{-2(1)})(1) y(2)=3e−4+52(1−e−2)y(2) = 3e^{-4} + \frac{5}{2}(1 - e^{-2}) y(2)=3e−4+52−52e−2y(2) = 3e^{-4} + \frac{5}{2} - \frac{5}{2}e^{-2}. Distractor A results from ignoring the forcing term (evaluating only the homogeneous part). Distractor B results from incorrectly using the unshifted function g(t)g(t) instead of g(t−1)g(t-1). Distractor D results from an algebraic error in the exponent.
  4. A function f(t)f(t) is defined piecewise as f(t)={t20≤t<111≤t<34−tt≥3f(t) = \begin{cases} t^2 & 0 \le t < 1 \\ 1 & 1 \le t < 3 \\ 4-t & t \ge 3 \end{cases}. Which of the following expressions correctly represents f(t)f(t) using Heaviside step functions uc(t)u_c(t)?
    Answer: t2+(1−t2)u1(t)+(3−t)u3(t)t^2 + (1-t^2)u_1(t) + (3-t)u_3(t) A standard way to write a piecewise function f(t)={g0(t)0≤t<c1g1(t)c1≤t<c2g2(t)t≥c2f(t) = \begin{cases} g_0(t) & 0 \le t < c_1 \\ g_1(t) & c_1 \le t < c_2 \\ g_2(t) & t \ge c_2 \end{cases} is f(t)=g0(t)+[g1(t)−g0(t)]uc1(t)+[g2(t)−g1(t)]uc2(t)f(t) = g_0(t) + [g_1(t) - g_0(t)]u_{c_1}(t) + [g_2(t) - g_1(t)]u_{c_2}(t). In this case, g0(t)=t2g_0(t) = t^2, g1(t)=1g_1(t) = 1, and g2(t)=4−tg_2(t) = 4-t. The points of discontinuity are c1=1c_1=1 and c2=3c_2=3. Applying the formula: f(t)=t2+(1−t2)u1(t)+((4−t)−1)u3(t)f(t) = t^2 + (1-t^2)u_1(t) + ((4-t) - 1)u_3(t) f(t)=t2+(1−t2)u1(t)+(3−t)u3(t)f(t) = t^2 + (1-t^2)u_1(t) + (3-t)u_3(t). Let's verify: For 0≤t<10 \le t < 1, u1=0,u3=0u_1=0, u_3=0, so f(t)=t2f(t)=t^2. Correct. For 1≤t<31 \le t < 3, u1=1,u3=0u_1=1, u_3=0, so f(t)=t2+(1−t2)=1f(t)=t^2+(1-t^2)=1. Correct. For t≥3t \ge 3, u1=1,u3=1u_1=1, u_3=1, so f(t)=t2+(1−t2)+(3−t)=1+3−t=4−tf(t)=t^2+(1-t^2)+(3-t)=1+3-t=4-t. Correct.
  5. Given that Y(s)=(s+3)e−πss(s2+1)Y(s) = \frac{(s+3)e^{-\pi s}}{s(s^2+1)}, find the inverse Laplace transform y(t)y(t).
    Answer: (3+3cos⁡(t)−sin⁡(t))uπ(t)(3 + 3\cos(t) - \sin(t))u_\pi(t) This problem requires using the second shifting theorem, L−1{e−csG(s)}=g(t−c)uc(t)\mathcal{L}^{-1}\{e^{-cs}G(s)\} = g(t-c)u_c(t). Here, c=πc=\pi and G(s)=s+3s(s2+1)G(s) = \frac{s+3}{s(s^2+1)}. First, we find g(t)=L−1{G(s)}g(t) = \mathcal{L}^{-1}\{G(s)\} using partial fraction decomposition. s+3s(s2+1)=As+Bs+Cs2+1\frac{s+3}{s(s^2+1)} = \frac{A}{s} + \frac{Bs+C}{s^2+1}. Solving for the coefficients gives A(s2+1)+(Bs+C)s=s+3A(s^2+1) + (Bs+C)s = s+3, which leads to A=3,C=1,B=−3A=3, C=1, B=-3. So, G(s)=3s−3ss2+1+1s2+1G(s) = \frac{3}{s} - \frac{3s}{s^2+1} + \frac{1}{s^2+1}. The inverse transform is g(t)=3−3cos⁡(t)+sin⁡(t)g(t) = 3 - 3\cos(t) + \sin(t). Now, we must find g(t−π)g(t-\pi). g(t−π)=3−3cos⁡(t−π)+sin⁡(t−π)g(t-\pi) = 3 - 3\cos(t-\pi) + \sin(t-\pi). Using the identities cos⁡(t−π)=−cos⁡(t)\cos(t-\pi) = -\cos(t) and sin⁡(t−π)=−sin⁡(t)\sin(t-\pi) = -\sin(t), we get: g(t−π)=3−3(−cos⁡(t))+(−sin⁡(t))=3+3cos⁡(t)−sin⁡(t)g(t-\pi) = 3 - 3(-\cos(t)) + (-\sin(t)) = 3 + 3\cos(t) - \sin(t). Finally, the solution is y(t)=g(t−π)uπ(t)=(3+3cos⁡(t)−sin⁡(t))uπ(t)y(t) = g(t-\pi)u_\pi(t) = (3 + 3\cos(t) - \sin(t))u_\pi(t).
  6. Find the inverse Laplace transform y(t)y(t) of the function Y(s)=2s3+(s−1)e−4ss2+9Y(s) = \frac{2}{s^3} + \frac{(s-1)e^{-4s}}{s^2+9}.
    Answer: t2+(cos⁡(3(t−4))−13sin⁡(3(t−4)))u4(t)t^2 + (\cos(3(t-4)) - \frac{1}{3}\sin(3(t-4)))u_4(t) We find the inverse transform of each term separately. For the first term: L−1{2s3}=L−1{2!s2+1}=t2\mathcal{L}^{-1}\left\{\frac{2}{s^3}\right\} = \mathcal{L}^{-1}\left\{\frac{2!}{s^{2+1}}\right\} = t^2. For the second term, we use the second shifting theorem, L−1{e−csF(s)}=f(t−c)uc(t)\mathcal{L}^{-1}\{e^{-cs}F(s)\} = f(t-c)u_c(t). Here, c=4c=4 and F(s)=s−1s2+9F(s) = \frac{s-1}{s^2+9}. We find f(t)=L−1{F(s)}f(t) = \mathcal{L}^{-1}\{F(s)\} first. F(s)=ss2+9−1s2+9=ss2+32−133s2+32F(s) = \frac{s}{s^2+9} - \frac{1}{s^2+9} = \frac{s}{s^2+3^2} - \frac{1}{3}\frac{3}{s^2+3^2}. f(t)=cos⁡(3t)−13sin⁡(3t)f(t) = \cos(3t) - \frac{1}{3}\sin(3t). Now, we apply the shift: f(t−4)u4(t)=(cos⁡(3(t−4))−13sin⁡(3(t−4)))u4(t)f(t-4)u_4(t) = \left(\cos(3(t-4)) - \frac{1}{3}\sin(3(t-4))\right)u_4(t). Combining the two parts, the complete inverse transform is y(t)=t2+(cos⁡(3(t−4))−13sin⁡(3(t−4)))u4(t)y(t) = t^2 + \left(\cos(3(t-4)) - \frac{1}{3}\sin(3(t-4))\right)u_4(t).
  7. Solve the initial value problem y′+2y=f(t)y' + 2y = f(t), with y(0)=3y(0)=3 and f(t)={00≤t<15t≥1f(t) = \begin{cases} 0 & 0 \le t < 1 \\ 5 & t \ge 1 \end{cases}.
    Answer: y(t)=3e−2t+52(1−e−2(t−1))u1(t)y(t) = 3e^{-2t} + \frac{5}{2}(1 - e^{-2(t-1)})u_1(t) First, write the forcing function as f(t)=5u1(t)f(t) = 5u_1(t). Take the Laplace transform of the entire equation: L{y′}+2L{y}=L{5u1(t)}\mathcal{L}\{y'\} + 2\mathcal{L}\{y\} = \mathcal{L}\{5u_1(t)\} [sY(s)−y(0)]+2Y(s)=5e−ss[sY(s) - y(0)] + 2Y(s) = \frac{5e^{-s}}{s} Substitute y(0)=3y(0)=3: (s+2)Y(s)−3=5e−ss(s+2)Y(s) - 3 = \frac{5e^{-s}}{s} Y(s)=3s+2+5e−ss(s+2)Y(s) = \frac{3}{s+2} + \frac{5e^{-s}}{s(s+2)}. To find the inverse transform of the second term, use partial fraction decomposition for 5s(s+2)=As+Bs+2\frac{5}{s(s+2)} = \frac{A}{s} + \frac{B}{s+2}. This gives A=5/2A=5/2 and B=−5/2B=-5/2. So, 5s(s+2)=52(1s−1s+2)\frac{5}{s(s+2)} = \frac{5}{2}\left(\frac{1}{s} - \frac{1}{s+2}\right). Let G(s)=52(1s−1s+2)G(s) = \frac{5}{2}\left(\frac{1}{s} - \frac{1}{s+2}\right), then g(t)=L−1{G(s)}=52(1−e−2t)g(t) = \mathcal{L}^{-1}\{G(s)\} = \frac{5}{2}(1 - e^{-2t}). The second term in Y(s)Y(s) is e−sG(s)e^{-s}G(s), so its inverse transform is g(t−1)u1(t)=52(1−e−2(t−1))u1(t)g(t-1)u_1(t) = \frac{5}{2}(1 - e^{-2(t-1)})u_1(t). The inverse transform of the first term is L−1{3s+2}=3e−2t\mathcal{L}^{-1}\left\{\frac{3}{s+2}\right\} = 3e^{-2t}. Combining the parts, the solution is y(t)=3e−2t+52(1−e−2(t−1))u1(t)y(t) = 3e^{-2t} + \frac{5}{2}(1 - e^{-2(t-1)})u_1(t).
  8. Let f(t)f(t) be a function defined as f(t)={cos⁡(t)0≤t<π−1t≥πf(t) = \begin{cases} \cos(t) & 0 \le t < \pi \\ -1 & t \ge \pi \end{cases}. Find the Laplace transform, F(s)=L{f(t)}F(s) = \mathcal{L}\{f(t)\}.
    Answer: ss2+1+e−πs(ss2+1−1s)\frac{s}{s^2+1} + e^{-\pi s} \left( \frac{s}{s^2+1} - \frac{1}{s} \right) First, represent f(t)f(t) using the Heaviside step function uπ(t)u_\pi(t). f(t)=cos⁡(t)(1−uπ(t))−1⋅uπ(t)=cos⁡(t)−(cos⁡(t)+1)uπ(t)f(t) = \cos(t)(1 - u_\pi(t)) - 1 \cdot u_\pi(t) = \cos(t) - (\cos(t) + 1)u_\pi(t). To find the Laplace transform of the second term using the second shifting theorem, L{g(t−c)uc(t)}=e−csG(s)\mathcal{L}\{g(t-c)u_c(t)\} = e^{-cs}G(s), we must express cos⁡(t)+1\cos(t)+1 as a function of (t−π)(t-\pi). Let g(t)=cos⁡(t)+1g(t) = \cos(t)+1. We need to find L{g(t)uπ(t)}=e−πsL{g(t+π)}\mathcal{L}\{g(t)u_\pi(t)\} = e^{-\pi s}\mathcal{L}\{g(t+\pi)\}. g(t+π)=cos⁡(t+π)+1=−cos⁡(t)+1g(t+\pi) = \cos(t+\pi)+1 = -\cos(t)+1. So, L{g(t+π)}=L{−cos⁡(t)+1}=−ss2+1+1s\mathcal{L}\{g(t+\pi)\} = \mathcal{L}\{-\cos(t)+1\} = -\frac{s}{s^2+1} + \frac{1}{s}. Therefore, L{(cos⁡(t)+1)uπ(t)}=e−πs(1s−ss2+1)\mathcal{L}\{(\cos(t)+1)u_\pi(t)\} = e^{-\pi s} \left(\frac{1}{s} - \frac{s}{s^2+1}\right). The complete transform is: F(s)=L{cos⁡(t)}−L{(cos⁡(t)+1)uπ(t)}=ss2+1−e−πs(1s−ss2+1)=ss2+1+e−πs(ss2+1−1s)F(s) = \mathcal{L}\{\cos(t)\} - \mathcal{L}\{(\cos(t)+1)u_\pi(t)\} = \frac{s}{s^2+1} - e^{-\pi s} \left(\frac{1}{s} - \frac{s}{s^2+1}\right) = \frac{s}{s^2+1} + e^{-\pi s} \left(\frac{s}{s^2+1} - \frac{1}{s}\right).
  9. The staircase function, defined as f(t)=⌊t⌋f(t) = \lfloor t \rfloor for t≥0t \ge 0, can be expressed as an infinite sum of Heaviside functions: f(t)=∑n=1∞un(t)f(t) = \sum_{n=1}^{\infty} u_n(t). What is the Laplace transform of f(t)f(t)?
    Answer: 1s(es−1)\frac{1}{s(e^s-1)} The Laplace transform of the infinite sum is the sum of the Laplace transforms: F(s)=L{∑n=1∞un(t)}=∑n=1∞L{un(t)}F(s) = \mathcal{L}\left\{\sum_{n=1}^{\infty} u_n(t)\right\} = \sum_{n=1}^{\infty} \mathcal{L}\{u_n(t)\}. The Laplace transform of a single Heaviside function is L{un(t)}=e−nss\mathcal{L}\{u_n(t)\} = \frac{e^{-ns}}{s}. So, F(s)=∑n=1∞e−nss=1s∑n=1∞e−nsF(s) = \sum_{n=1}^{\infty} \frac{e^{-ns}}{s} = \frac{1}{s} \sum_{n=1}^{\infty} e^{-ns}. The sum is a geometric series with first term a=e−sa = e^{-s} and common ratio r=e−sr = e^{-s}. Since ∣r∣=e−s<1|r| = e^{-s} < 1 for s>0s>0, the sum converges to a1−r=e−s1−e−s\frac{a}{1-r} = \frac{e^{-s}}{1-e^{-s}}. Therefore, F(s)=1s(e−s1−e−s)F(s) = \frac{1}{s} \left( \frac{e^{-s}}{1-e^{-s}} \right). To match the answer choices, we can multiply the numerator and denominator by ese^s: F(s)=1s(e−ses(1−e−s)es)=1s(1es−1)=1s(es−1)F(s) = \frac{1}{s} \left( \frac{e^{-s}e^s}{(1-e^{-s})e^s} \right) = \frac{1}{s} \left( \frac{1}{e^s-1} \right) = \frac{1}{s(e^s-1)}.
  10. The convolution integral ∫0te−(t−τ)u2(τ)dτ\int_0^t e^{-(t-\tau)} u_2(\tau) d\tau can be evaluated using properties of Heaviside functions. What is the result for t=5t = 5?
    Answer: 1−e−31 - e^{-3} For t=5t = 5, we evaluate ∫05e−(5−τ)u2(τ)dτ\int_0^5 e^{-(5-\tau)} u_2(\tau) d\tau. Since u2(τ)=0u_2(\tau) = 0 for τ<2\tau < 2 and u2(τ)=1u_2(\tau) = 1 for τ≥2\tau \geq 2, the integral becomes ∫25e−(5−τ)dτ=∫25eτ−5dτ=e−5∫25eτdτ=e−5[eτ]25=e−5(e5−e2)=1−e−3\int_2^5 e^{-(5-\tau)} d\tau = \int_2^5 e^{\tau-5} d\tau = e^{-5}\int_2^5 e^{\tau} d\tau = e^{-5}[e^{\tau}]_2^5 = e^{-5}(e^5 - e^2) = 1 - e^{-3}. Choice A incorrectly includes extra exponential terms that don't arise from the integration. Choice C gives the wrong limits and signs. Choice D adds an incorrect e−5e^{-5} term that would come from mishandling the exponential factor.
  11. Consider the initial value problem representing an undamped harmonic oscillator with a finite-duration force: y′′+4y=8sin⁡(t)(1−uπ(t))y'' + 4y = 8\sin(t)(1 - u_\pi(t)), with initial conditions y(0)=0y(0)=0 and y′(0)=0y'(0)=0. What is the solution y(t)y(t) for t≥πt \ge \pi?
    Answer: −83sin⁡(2t)-\frac{8}{3}\sin(2t) First, take the Laplace transform of the IVP. The forcing function is f(t)=8sin⁡(t)−8sin⁡(t)uπ(t)f(t) = 8\sin(t) - 8\sin(t)u_\pi(t). We rewrite the second term: sin⁡(t)=sin⁡((t−π)+π)=−sin⁡(t−π)\sin(t) = \sin((t-\pi)+\pi) = -\sin(t-\pi). So f(t)=8sin⁡(t)+8sin⁡(t−π)uπ(t)f(t) = 8\sin(t) + 8\sin(t-\pi)u_\pi(t). \{L}\{f(t)\} = \frac{8}{s^2+1} + \frac{8e^{-\pi s}}{s^2+1}. The transformed DE is (s2+4)Y(s)=8s2+1(1+e−πs)(s^2+4)Y(s) = \frac{8}{s^2+1}(1+e^{-\pi s}). Y(s)=8(s2+1)(s2+4)(1+e−πs)Y(s) = \frac{8}{(s^2+1)(s^2+4)}(1+e^{-\pi s}). Using partial fractions, 8(s2+1)(s2+4)=8/3s2+1−8/3s2+4\frac{8}{(s^2+1)(s^2+4)} = \frac{8/3}{s^2+1} - \frac{8/3}{s^2+4}. Let G(s)=8/3s2+1−8/3s2+4G(s) = \frac{8/3}{s^2+1} - \frac{8/3}{s^2+4}. Its inverse transform is g(t)=83sin⁡(t)−43sin⁡(2t)g(t) = \frac{8}{3}\sin(t) - \frac{4}{3}\sin(2t). Then y(t)=g(t)+g(t−π)uπ(t)y(t) = g(t) + g(t-\pi)u_\pi(t). g(t−π)=83sin⁡(t−π)−43sin⁡(2(t−π))=−83sin⁡(t)−43sin⁡(2t)g(t-\pi) = \frac{8}{3}\sin(t-\pi) - \frac{4}{3}\sin(2(t-\pi)) = -\frac{8}{3}\sin(t) - \frac{4}{3}\sin(2t). For t≥πt \ge \pi, uπ(t)=1u_\pi(t)=1, so the solution is y(t)=g(t)+g(t−π)y(t) = g(t) + g(t-\pi). y(t)=(83sin⁡(t)−43sin⁡(2t))+(−83sin⁡(t)−43sin⁡(2t))=−83sin⁡(2t)y(t) = \left(\frac{8}{3}\sin(t) - \frac{4}{3}\sin(2t)\right) + \left(-\frac{8}{3}\sin(t) - \frac{4}{3}\sin(2t)\right) = -\frac{8}{3}\sin(2t).
  12. A system is governed by dydt+2y=g(t)\frac{dy}{dt} + 2y = g(t) where g(t)=∑n=1∞(−1)n+1un(t)g(t) = \sum_{n=1}^{\infty} (-1)^{n+1} u_n(t) and y(0)=0y(0) = 0. What is the behavior of y(t)y(t) as t→∞t \to \infty?
    Answer: y(t)y(t) oscillates between 12\frac{1}{2} and 00 with period approaching 1 The forcing function g(t)=u1(t)−u2(t)+u3(t)−u4(t)+…g(t) = u_1(t) - u_2(t) + u_3(t) - u_4(t) + \ldots creates a square wave that alternates between 1 and 0. For t∈[n,n+1)t \in [n, n+1), we have g(t)=(−1)n+1g(t) = (-1)^{n+1}. The solution to y′+2y=g(t)y' + 2y = g(t) with y(0)=0y(0) = 0 is found using integrating factors. For each interval [n,n+1)[n, n+1), the solution is y(t)=(−1)n+12(1−e−2(t−n))+y(n)e−2(t−n)y(t) = \frac{(-1)^{n+1}}{2}(1 - e^{-2(t-n)}) + y(n)e^{-2(t-n)}. As tt increases through each interval, y(t)y(t) exponentially approaches (−1)n+12\frac{(-1)^{n+1}}{2} from its value at the start of the interval. Since the forcing alternates sign every unit of time, and the exponential decay is rapid (time constant 12\frac{1}{2}), the solution oscillates between approximately 12\frac{1}{2} and 00, spending most of each interval close to the steady-state value 12\frac{1}{2} or −12-\frac{1}{2} corresponding to the current forcing level. However, since the forcing switches between 1 and 0 (not 1 and -1), the oscillation is between 12\frac{1}{2} and 00. Choice A incorrectly assumes symmetric oscillation about zero. Choice B misses the periodic switching. Choice D ignores the stabilizing effect of the decay term.

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