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Linearity and Shifting Theorems - Differential Equations

  1. Let F(s)=L{f(t)}F(s) = \mathcal{L}\{f(t)\}. Which of the following is equivalent to L{2f(t)−e3tf(t)}\mathcal{L}\{2f(t) - e^{3t}f(t)\}?
    Answer: 2F(s)−F(s−3)2F(s) - F(s-3) This question requires applying both the linearity property and the first shifting theorem. First, use the linearity property: L{2f(t)−e3tf(t)}=L{2f(t)}−L{e3tf(t)}=2L{f(t)}−L{e3tf(t)}\mathcal{L}\{2f(t) - e^{3t}f(t)\} = \mathcal{L}\{2f(t)\} - \mathcal{L}\{e^{3t}f(t)\} = 2\mathcal{L}\{f(t)\} - \mathcal{L}\{e^{3t}f(t)\} The first term is 2F(s)2F(s). For the second term, apply the first shifting theorem, L{eatf(t)}=F(s−a)\mathcal{L}\{e^{at}f(t)\} = F(s-a). Here, a=3a=3, so L{e3tf(t)}=F(s−3)\mathcal{L}\{e^{3t}f(t)\} = F(s-3). Combining the two parts, the expression is 2F(s)−F(s−3)2F(s) - F(s-3).
  2. Consider the system of functions where L{u(t)}=1s\mathcal{L}\{u(t)\} = \frac{1}{s} and L{e−at}=1s+a\mathcal{L}\{e^{-at}\} = \frac{1}{s+a}. Using linearity and shifting theorems, which expression correctly represents L{∫0t−3e−2τdτ⋅u(t−3)}\mathcal{L}\left\{\int_0^{t-3} e^{-2\tau} d\tau \cdot u(t-3)\right\}?
    Answer: e−3ss(s+2)\frac{e^{-3s}}{s(s+2)} First, let's evaluate the integral: ∫0t−3e−2τdτ=[−12e−2τ]0t−3=−12e−2(t−3)+12=12(1−e−2(t−3))\int_0^{t-3} e^{-2\tau} d\tau = \left[-\frac{1}{2}e^{-2\tau}\right]_0^{t-3} = -\frac{1}{2}e^{-2(t-3)} + \frac{1}{2} = \frac{1}{2}(1 - e^{-2(t-3)}). So our function becomes 12(1−e−2(t−3))u(t−3)\frac{1}{2}(1 - e^{-2(t-3)})u(t-3). This can be written as 12u(t−3)−12e−2(t−3)u(t−3)\frac{1}{2}u(t-3) - \frac{1}{2}e^{-2(t-3)}u(t-3). Using the second shifting theorem: L{u(t−3)}=e−3ss\mathcal{L}\{u(t-3)\} = \frac{e^{-3s}}{s} and L{e−2(t−3)u(t−3)}=e−3s⋅1s+2\mathcal{L}\{e^{-2(t-3)}u(t-3)\} = e^{-3s} \cdot \frac{1}{s+2}. Therefore, L{12(1−e−2(t−3))u(t−3)}=12⋅e−3ss−12⋅e−3ss+2=e−3s2(1s−1s+2)=e−3s2⋅(s+2)−ss(s+2)=e−3s2⋅2s(s+2)=e−3ss(s+2)\mathcal{L}\left\{\frac{1}{2}(1 - e^{-2(t-3)})u(t-3)\right\} = \frac{1}{2} \cdot \frac{e^{-3s}}{s} - \frac{1}{2} \cdot \frac{e^{-3s}}{s+2} = \frac{e^{-3s}}{2}\left(\frac{1}{s} - \frac{1}{s+2}\right) = \frac{e^{-3s}}{2} \cdot \frac{(s+2)-s}{s(s+2)} = \frac{e^{-3s}}{2} \cdot \frac{2}{s(s+2)} = \frac{e^{-3s}}{s(s+2)}.
  3. Let F(s)=L{f(t)}F(s) = \mathcal{L}\{f(t)\} and G(s)=L{g(t)}G(s) = \mathcal{L}\{g(t)\}. The function g(t)g(t) is defined as g(t)={f(t−a),t≥a0,t<ag(t) = \begin{cases} f(t-a), & t \ge a \\ 0, & t < a \end{cases} for a constant a>0a > 0. Which statement correctly relates G(s)G(s) and F(s)F(s)?
    Answer: G(s)=e−asF(s)G(s) = e^{-as}F(s) The piecewise definition of g(t)g(t) is the standard way to represent a time-shifted function that is zero before the shift. This definition is equivalent to writing g(t)=f(t−a)u(t−a)g(t) = f(t-a)u(t-a), where u(t−a)u(t-a) is the Heaviside step function. The Laplace transform of this expression is given by the second shifting theorem (or t-shifting theorem), which states that L{f(t−a)u(t−a)}=e−asF(s)\mathcal{L}\{f(t-a)u(t-a)\} = e^{-as}F(s), where F(s)=L{f(t)}F(s) = \mathcal{L}\{f(t)\}. Therefore, the correct relationship is G(s)=e−asF(s)G(s) = e^{-as}F(s).
  4. Let F(s)=L{f(t)}F(s) = \mathcal{L}\{f(t)\}. A second function is defined as g(t)=3f(t−2)u(t−2)+e−tf(t)g(t) = 3f(t-2)u(t-2) + e^{-t}f(t). Find L{g(t)}\mathcal{L}\{g(t)\} in terms of F(s)F(s).
    Answer: 3e−2sF(s)+F(s+1)3e^{-2s}F(s) + F(s+1) We use the linearity of the Laplace transform to transform each term separately. L{g(t)}=L{3f(t−2)u(t−2)}+L{e−tf(t)}\mathcal{L}\{g(t)\} = \mathcal{L}\{3f(t-2)u(t-2)\} + \mathcal{L}\{e^{-t}f(t)\} For the first term, we use the property of linearity with constants and the second shifting theorem, L{h(t−a)u(t−a)}=e−asH(s)\mathcal{L}\{h(t-a)u(t-a)\} = e^{-as}H(s). Here, a=2a=2 and h(t)=f(t)h(t)=f(t). So, L{3f(t−2)u(t−2)}=3e−2sF(s)\mathcal{L}\{3f(t-2)u(t-2)\} = 3e^{-2s}F(s). For the second term, we use the first shifting theorem, L{eath(t)}=H(s−a)\mathcal{L}\{e^{at}h(t)\} = H(s-a). Here, a=−1a=-1 and h(t)=f(t)h(t)=f(t). So, L{e−tf(t)}=F(s−(−1))=F(s+1)\mathcal{L}\{e^{-t}f(t)\} = F(s-(-1)) = F(s+1). Combining the results gives L{g(t)}=3e−2sF(s)+F(s+1)\mathcal{L}\{g(t)\} = 3e^{-2s}F(s) + F(s+1).
  5. Find the inverse Laplace transform of the function F(s)=s+3s2+2s+5F(s) = \frac{s+3}{s^2+2s+5}.
    Answer: e−t(cos⁡(2t)+sin⁡(2t))e^{-t}(\cos(2t) + \sin(2t)) To find the inverse Laplace transform, first complete the square in the denominator: s2+2s+5=(s2+2s+1)+4=(s+1)2+22s^2+2s+5 = (s^2+2s+1) + 4 = (s+1)^2 + 2^2. This suggests using the first shifting theorem with a=−1a=-1. Rewrite the numerator in terms of (s+1)(s+1): s+3=(s+1)+2s+3 = (s+1) + 2. Now, split the fraction: F(s)=s+1(s+1)2+22+2(s+1)2+22F(s) = \frac{s+1}{(s+1)^2+2^2} + \frac{2}{(s+1)^2+2^2} The inverse transform is found using linearity and the first shifting theorem, L−1{G(s−a)}=eatg(t)\mathcal{L}^{-1}\{G(s-a)\} = e^{at}g(t). The first term corresponds to the shifted cosine transform: L−1{s+1(s+1)2+22}=e−tcos⁡(2t)\mathcal{L}^{-1}\left\{\frac{s+1}{(s+1)^2+2^2}\right\} = e^{-t}\cos(2t). The second term corresponds to the shifted sine transform: L−1{2(s+1)2+22}=e−tsin⁡(2t)\mathcal{L}^{-1}\left\{\frac{2}{(s+1)^2+2^2}\right\} = e^{-t}\sin(2t). Summing these results gives e−t(cos⁡(2t)+sin⁡(2t))e^{-t}(\cos(2t) + \sin(2t)).
  6. What is the Laplace transform of f(t)=e2tsin⁡2(t)f(t) = e^{2t}\sin^2(t)?
    Answer: 2(s−2)((s−2)2+4)\frac{2}{(s-2)((s-2)^2+4)} This problem requires a trigonometric identity followed by the first shifting theorem. First, find the Laplace transform of g(t)=sin⁡2(t)g(t) = \sin^2(t). Use the identity sin⁡2(t)=12(1−cos⁡(2t))\sin^2(t) = \frac{1}{2}(1 - \cos(2t)). Using linearity: G(s)=L{sin⁡2(t)}=L{12−12cos⁡(2t)}=12s−12ss2+4G(s) = \mathcal{L}\{\sin^2(t)\} = \mathcal{L}\{\frac{1}{2} - \frac{1}{2}\cos(2t)\} = \frac{1}{2s} - \frac{1}{2}\frac{s}{s^2+4} Combining the terms gives: G(s)=s2+4−s22s(s2+4)=42s(s2+4)=2s(s2+4)G(s) = \frac{s^2+4-s^2}{2s(s^2+4)} = \frac{4}{2s(s^2+4)} = \frac{2}{s(s^2+4)} Now, apply the first shifting theorem, L{eatg(t)}=G(s−a)\mathcal{L}\{e^{at}g(t)\} = G(s-a), with a=2a=2. We replace every ss in G(s)G(s) with (s−2)(s-2): L{e2tsin⁡2(t)}=G(s−2)=2(s−2)((s−2)2+4)\mathcal{L}\{e^{2t}\sin^2(t)\} = G(s-2) = \frac{2}{(s-2)((s-2)^2+4)}
  7. Find the Laplace transform of the function g(t)=t2u(t−2)g(t) = t^2 u(t-2), where u(t)u(t) is the Heaviside step function.
    Answer: e−2s(2s3+4s2+4s)e^{-2s}\left(\frac{2}{s^3} + \frac{4}{s^2} + \frac{4}{s}\right) This problem requires the second shifting theorem, L{f(t−a)u(t−a)}=e−asF(s)\mathcal{L}\{f(t-a)u(t-a)\} = e^{-as}F(s). The given function is g(t)=t2u(t−2)g(t) = t^2 u(t-2). We must express the function t2t^2 in terms of (t−2)(t-2). Let τ=t−2\tau = t-2, which means t=τ+2t = \tau+2. Then t2=(τ+2)2=τ2+4τ+4t^2 = (\tau+2)^2 = \tau^2+4\tau+4. So, g(t)=((t−2)2+4(t−2)+4)u(t−2)g(t) = ((t-2)^2 + 4(t-2) + 4)u(t-2). This is in the form f(t−2)u(t−2)f(t-2)u(t-2) where f(t)=t2+4t+4f(t) = t^2+4t+4. First, we find the Laplace transform of f(t)f(t): F(s)=L{t2+4t+4}=2!s3+4(1!)s2+4s=2s3+4s2+4sF(s) = \mathcal{L}\{t^2+4t+4\} = \frac{2!}{s^3} + \frac{4(1!)}{s^2} + \frac{4}{s} = \frac{2}{s^3} + \frac{4}{s^2} + \frac{4}{s} Now, applying the second shifting theorem with a=2a=2, we get L{g(t)}=e−2sF(s)=e−2s(2s3+4s2+4s)\mathcal{L}\{g(t)\} = e^{-2s}F(s) = e^{-2s}\left(\frac{2}{s^3} + \frac{4}{s^2} + \frac{4}{s}\right).
  8. Find the inverse Laplace transform of G(s)=2s−5s2−6s+13G(s) = \frac{2s-5}{s^2-6s+13}.
    Answer: e3t(2cos⁡(2t)+12sin⁡(2t))e^{3t}(2\cos(2t) + \frac{1}{2}\sin(2t)) The denominator s2−6s+13s^2-6s+13 suggests completing the square. s2−6s+13=(s2−6s+9)+4=(s−3)2+22s^2-6s+13 = (s^2-6s+9) + 4 = (s-3)^2 + 2^2. This indicates a shift of a=3a=3. We must rewrite the numerator, 2s−52s-5, in terms of (s−3)(s-3): 2s−5=2(s−3)+6−5=2(s−3)+12s-5 = 2(s-3) + 6 - 5 = 2(s-3) + 1 Now, substitute this back into the fraction and split it: G(s)=2(s−3)+1(s−3)2+22=2(s−3)(s−3)2+22+1(s−3)2+22G(s) = \frac{2(s-3) + 1}{(s-3)^2+2^2} = \frac{2(s-3)}{(s-3)^2+2^2} + \frac{1}{(s-3)^2+2^2} We find the inverse transform of each part using linearity and the first shifting theorem. For the first term: L−1{2(s−3)(s−3)2+22}=2e3tcos⁡(2t)\mathcal{L}^{-1}\left\{ \frac{2(s-3)}{(s-3)^2+2^2} \right\} = 2e^{3t}\cos(2t) For the second term, we need a 2 in the numerator for the sine transform: L−1{1(s−3)2+22}=12L−1{2(s−3)2+22}=12e3tsin⁡(2t)\mathcal{L}^{-1}\left\{ \frac{1}{(s-3)^2+2^2} \right\} = \frac{1}{2}\mathcal{L}^{-1}\left\{ \frac{2}{(s-3)^2+2^2} \right\} = \frac{1}{2}e^{3t}\sin(2t) Combining these gives the final answer: g(t)=2e3tcos⁡(2t)+12e3tsin⁡(2t)=e3t(2cos⁡(2t)+12sin⁡(2t))g(t) = 2e^{3t}\cos(2t) + \frac{1}{2}e^{3t}\sin(2t) = e^{3t}(2\cos(2t) + \frac{1}{2}\sin(2t)).
  9. Find the inverse Laplace transform of F(s)=se−2ss2+4s+13F(s) = \frac{se^{-2s}}{s^2+4s+13}.
    Answer: e−2(t−2)(cos⁡(3(t−2))−23sin⁡(3(t−2)))u(t−2)e^{-2(t-2)}(\cos(3(t-2)) - \frac{2}{3}\sin(3(t-2)))u(t-2) When you encounter an inverse Laplace transform with an exponential factor like e−2se^{-2s}, this signals a time-shifting problem. The exponential e−ase^{-as} in the Laplace domain corresponds to shifting the function by aa units in the time domain, creating a unit step function u(t−a)u(t-a). To solve this, first work with ss2+4s+13\frac{s}{s^2+4s+13} by completing the square in the denominator: s2+4s+13=(s+2)2+9s^2+4s+13 = (s+2)^2+9. Rewrite the numerator to match this form: s=(s+2)−2s = (s+2)-2. This gives us ss2+4s+13=(s+2)−2(s+2)2+9=s+2(s+2)2+9−2(s+2)2+9\frac{s}{s^2+4s+13} = \frac{(s+2)-2}{(s+2)^2+9} = \frac{s+2}{(s+2)^2+9} - \frac{2}{(s+2)^2+9}. These correspond to standard forms: s+2(s+2)2+9→e−2tcos⁡(3t)\frac{s+2}{(s+2)^2+9} \rightarrow e^{-2t}\cos(3t) and 2(s+2)2+9→23e−2tsin⁡(3t)\frac{2}{(s+2)^2+9} \rightarrow \frac{2}{3}e^{-2t}\sin(3t). So the inverse of ss2+4s+13\frac{s}{s^2+4s+13} is e−2t(cos⁡(3t)−23sin⁡(3t))e^{-2t}(\cos(3t) - \frac{2}{3}\sin(3t)). The e−2se^{-2s} factor shifts this by 2 units: replace tt with (t−2)(t-2) and multiply by u(t−2)u(t-2), giving answer A. Answer B incorrectly keeps the original time variable tt in the exponential and trigonometric functions instead of shifting to (t−2)(t-2). Answer C has the wrong coefficient −2-2 instead of −23-\frac{2}{3} for the sine term. Answer D incorrectly uses e2(t−2)e^{2(t-2)} instead of e−2(t−2)e^{-2(t-2)}, missing the negative sign. Remember: e−ase^{-as} in the Laplace domain means shift everything by aa units in time and add u(t−a)u(t-a).
  10. Find the inverse Laplace transform of F(s)=1(s+3)3F(s) = \frac{1}{(s+3)^3}.
    Answer: 12t2e−3t\frac{1}{2}t^2e^{-3t} When you encounter inverse Laplace transforms with powers in the denominator like (s+a)n(s+a)^n, you're dealing with a combination of shifting and polynomial multiplication patterns. The key is recognizing that this involves both the shifting property and the transform of polynomial functions. To find L−1{1(s+3)3}\mathcal{L}^{-1}\left\{\frac{1}{(s+3)^3}\right\}, start with the basic transform L−1{1s3}=t22!=t22\mathcal{L}^{-1}\left\{\frac{1}{s^3}\right\} = \frac{t^2}{2!} = \frac{t^2}{2}. Then apply the first shifting theorem: if L−1{F(s)}=f(t)\mathcal{L}^{-1}\{F(s)\} = f(t), then L−1{F(s+a)}=e−atf(t)\mathcal{L}^{-1}\{F(s+a)\} = e^{-at}f(t). Since 1(s+3)3=F(s+3)\frac{1}{(s+3)^3} = F(s+3) where F(s)=1s3F(s) = \frac{1}{s^3}, we get: L−1{1(s+3)3}=e−3t⋅t22=12t2e−3t\mathcal{L}^{-1}\left\{\frac{1}{(s+3)^3}\right\} = e^{-3t} \cdot \frac{t^2}{2} = \frac{1}{2}t^2e^{-3t} This confirms answer A is correct. Answer B (t2e−3tt^2e^{-3t}) omits the factorial correction factor 12!\frac{1}{2!} that comes from the 1s3\frac{1}{s^3} transform. Answer C (16t3e−3t\frac{1}{6}t^3e^{-3t}) incorrectly uses the transform for 1s4\frac{1}{s^4}, which would give t33!\frac{t^3}{3!}. Answer D (12t2e3t\frac{1}{2}t^2e^{3t}) has the wrong sign in the exponential—confusing the shifting direction. Study tip: Memorize that L−1{1sn}=tn−1(n−1)!\mathcal{L}^{-1}\left\{\frac{1}{s^n}\right\} = \frac{t^{n-1}}{(n-1)!} and remember the first shifting theorem always produces e−ate^{-at} when you see (s+a)(s+a) terms.
  11. Find the Laplace transform of the function f(t)f(t) defined as: f(t)={2,0≤t<1t,1≤t<30,t≥3f(t) = \begin{cases} 2, & 0 \le t < 1 \\ t, & 1 \le t < 3 \\ 0, & t \ge 3 \end{cases}
    Answer: 2s−e−s1s+e−s1s2−e−3s3s−e−3s1s2\frac{2}{s} - e^{-s}\frac{1}{s} + e^{-s}\frac{1}{s^2} - e^{-3s}\frac{3}{s} - e^{-3s}\frac{1}{s^2} First, express the piecewise function f(t)f(t) using Heaviside step functions u(t−a)u(t-a). f(t)=2[u(t)−u(t−1)]+t[u(t−1)−u(t−3)]f(t) = 2[u(t)-u(t-1)] + t[u(t-1)-u(t-3)] f(t)=2−2u(t−1)+tu(t−1)−tu(t−3)f(t) = 2 - 2u(t-1) + tu(t-1) - tu(t-3) f(t)=2+(t−2)u(t−1)−tu(t−3)f(t) = 2 + (t-2)u(t-1) - tu(t-3) To use the second shifting theorem, L{g(t−a)u(t−a)}=e−asG(s)\mathcal{L}\{g(t-a)u(t-a)\} = e^{-as}G(s), we rewrite the terms: For (t−2)u(t−1)(t-2)u(t-1), let t−1=τt-1 = \tau, so t−2=τ−1t-2 = \tau - 1. This term becomes ((t−1)−1)u(t−1)( (t-1) - 1 )u(t-1). Here g(t)=t−1g(t) = t-1, so G(s)=1s2−1sG(s) = \frac{1}{s^2}-\frac{1}{s}. The transform is e−s(1s2−1s)e^{-s}(\frac{1}{s^2}-\frac{1}{s}). For −tu(t−3)-tu(t-3), let t−3=τt-3 = \tau, so −t=−(τ+3)-t = -(\tau+3). This term is −((t−3)+3)u(t−3)-((t-3)+3)u(t-3). Here g(t)=−(t+3)g(t) = -(t+3), so G(s)=−(1s2+3s)G(s) = -(\frac{1}{s^2}+\frac{3}{s}). The transform is −e−3s(1s2+3s)-e^{-3s}(\frac{1}{s^2}+\frac{3}{s}). Combining all parts: L{f(t)}=L{2}+L{(t−2)u(t−1)}+L{−tu(t−3)}=2s+e−s(1s2−1s)−e−3s(1s2+3s)\mathcal{L}\{f(t)\} = \mathcal{L}\{2\} + \mathcal{L}\{(t-2)u(t-1)\} + \mathcal{L}\{-tu(t-3)\} = \frac{2}{s} + e^{-s}(\frac{1}{s^2}-\frac{1}{s}) - e^{-3s}(\frac{1}{s^2}+\frac{3}{s}). Distributing gives 2s−e−ss+e−ss2−3e−3ss−e−3ss2\frac{2}{s} - \frac{e^{-s}}{s} + \frac{e^{-s}}{s^2} - \frac{3e^{-3s}}{s} - \frac{e^{-3s}}{s^2}. This corresponds to choice B. Choice C had a sign error in the setup.
  12. Find the inverse Laplace transform of G(s)=e−3ss(s+2)G(s) = \frac{e^{-3s}}{s(s+2)}.
    Answer: (12−12e−2(t−3))u(t−3)\left(\frac{1}{2} - \frac{1}{2}e^{-2(t-3)}\right)u(t-3) This problem uses the second shifting theorem for inverse transforms: L−1{e−asF(s)}=f(t−a)u(t−a)\mathcal{L}^{-1}\{e^{-as}F(s)\} = f(t-a)u(t-a). First, identify F(s)=1s(s+2)F(s) = \frac{1}{s(s+2)} and a=3a=3. We need to find f(t)=L−1{F(s)}f(t) = \mathcal{L}^{-1}\{F(s)\}. Using partial fraction decomposition on F(s)F(s): 1s(s+2)=As+Bs+2  ⟹  1=A(s+2)+Bs\frac{1}{s(s+2)} = \frac{A}{s} + \frac{B}{s+2} \implies 1 = A(s+2) + Bs Setting s=0s=0 gives A=1/2A=1/2. Setting s=−2s=-2 gives B=−1/2B=-1/2. So, F(s)=1/2s−1/2s+2F(s) = \frac{1/2}{s} - \frac{1/2}{s+2}. The inverse transform is f(t)=12−12e−2tf(t) = \frac{1}{2} - \frac{1}{2}e^{-2t}. Now, apply the shifting theorem: L−1{e−3sF(s)}=f(t−3)u(t−3)\mathcal{L}^{-1}\{e^{-3s}F(s)\} = f(t-3)u(t-3). Replace every tt in f(t)f(t) with (t−3)(t-3): g(t)=(12−12e−2(t−3))u(t−3)g(t) = \left(\frac{1}{2} - \frac{1}{2}e^{-2(t-3)}\right)u(t-3)
  13. Given the initial value problem y′′+4y′+5y=δ(t−2)+e−tu(t−1)y'' + 4y' + 5y = \delta(t-2) + e^{-t}u(t-1) with y(0)=1y(0) = 1 and y′(0)=0y'(0) = 0, where δ(t−2)\delta(t-2) is the Dirac delta function, which expression correctly represents the Laplace transform of the right-hand side using linearity and shifting theorems?
    Answer: e−2s+e−se−1s+1e^{-2s} + \frac{e^{-s}e^{-1}}{s+1} For the Dirac delta function δ(t−2)\delta(t-2), we use the property L{δ(t−a)}=e−as\mathcal{L}\{\delta(t-a)\} = e^{-as}, so L{δ(t−2)}=e−2s\mathcal{L}\{\delta(t-2)\} = e^{-2s}. For the term e−tu(t−1)e^{-t}u(t-1), we need to apply the second shifting theorem carefully. We have e−tu(t−1)=e−(t−1+1)u(t−1)=e−1e−(t−1)u(t−1)e^{-t}u(t-1) = e^{-(t-1+1)}u(t-1) = e^{-1}e^{-(t-1)}u(t-1). Using the second shifting theorem, L{g(t−a)u(t−a)}=e−asG(s)\mathcal{L}\{g(t-a)u(t-a)\} = e^{-as}G(s), where g(t)=e−tg(t) = e^{-t}, so G(s)=1s+1G(s) = \frac{1}{s+1}. Therefore, L{e−(t−1)u(t−1)}=e−s⋅1s+1=e−ss+1\mathcal{L}\{e^{-(t-1)}u(t-1)\} = e^{-s} \cdot \frac{1}{s+1} = \frac{e^{-s}}{s+1}. The factor e−1e^{-1} remains, giving us L{e−tu(t−1)}=e−1⋅e−ss+1=e−se−1s+1\mathcal{L}\{e^{-t}u(t-1)\} = e^{-1} \cdot \frac{e^{-s}}{s+1} = \frac{e^{-s}e^{-1}}{s+1}. Therefore, the total transform of the right-hand side is e−2s+e−se−1s+1e^{-2s} + \frac{e^{-s}e^{-1}}{s+1}.

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