Substitutions…Again!

Big Idea

The exponential function in the construction of the Laplace transform yields the dual shift theorems:

$\begin{aligned} \mathcal L\{e^{at}f(t) \} &= F(s-a), \\ \mathcal L^{-1} \{ e^{-as}F(s) \} &= f(t-a)U(t-a). \end{aligned}$

Questions

Question 1. Evaluate $\mathcal L\{te^{2t} + 2e^{3t} - 4e^{2t}\cos(2t)\}$ .

(Click for Solution)

Solution. Factorising $e^{2t}$ ,

$\begin{aligned} te^{2t} + 2e^{3t} - 4e^{2t}\cos(2t) &= e^{2t}\cdot \underbrace{(t + 2e^{t} - 4\cos(2t))}_{f(t)}. \end{aligned}$

By the shift theorems, $\mathcal L\{ e^{2t} f(t) \} = F(s-2)$ . It suffices to evaluate $F(s) = \mathcal L\{f(t)\}$ . By the linearity of Laplace transforms,

$\begin{aligned} F(s) &= \mathcal L\{ t + 2e^{t} - 4\cos(2t) \} \\ &= \mathcal L\{ t \} + 2 \cdot \mathcal L\{ e^t \} - 4 \cdot \mathcal L\{ \cos(2t) \} \\ &= \frac{1!}{s^{1+1}} + 2 \cdot \frac 1{s-1} - 4 \cdot \frac{s}{s^2 + 2^2} \\ &= \frac 1{s^2} + \frac 2{s-1} - \frac{4s}{s^2 + 4}. \end{aligned}$

Hence,

$\displaystyle \begin{aligned} \mathcal L\{e^{2t} f(t)\} &= \frac 1{(s-2)^2} + \frac 2{(s-2)-1} - \frac{4(s-2)}{(s-2)^2 + 4} \\ &= \frac 1{(s-2)^2} + \frac 2{s-3} - \frac{4s-8}{(s-2)^2 + 4}. \end{aligned}$

Question 2. Evaluate $\displaystyle \mathcal L^{-1}\left\{\frac 1{{(s-1)}^3} - \frac 1{s^2 - 2s + 5} + \frac{s-1}{s^2 -2s + 10}\right\}$ .

(Click for Solution)

Solution. Motivated by the first term $1/(s-1)^3$ , define

$\displaystyle F(s-1) := \frac 1{{(s-1)}^3} - \frac 1{s^2 - 2s + 5} + \frac{s-1}{s^2 -2s + 10}.$

Replacing $s$ with $s+1$ in each term,

$\begin{aligned} \frac 1{{(s+1-1)}^3} &= \frac 1{s^3}, \\ \frac 1{(s+1)^2 - 2(s+1) + 5} &= \frac{1}{(s^2 + 2s + 1) - (2s + 2) + 5} = \frac{1}{s^2 + 4}, \\ \frac{(s+1)-1}{(s+1)^2 -2(s+1) + 10} &= \frac{s}{(s^2 + 2s + 1) - (2s + 2) + 10} = \frac{s}{s^2 + 9}. \end{aligned}$

Hence, $\displaystyle F(s) = \frac 1{s^3} - \frac{1}{s^2 + 4} + \frac{s}{s^2 + 9}$ . By the shift theorems,

$\mathcal L^{-1} \{F(s-1)\} = e^{t} f(t).$

It remains to evaluate $f(t) = \mathcal L^{-1}(F(s))$ :

$\begin{aligned} f(t) = \mathcal L^{-1}(F(s)) &= \mathcal L^{-1} \left\{ \frac 1{s^3} - \frac{1}{s^2 + 4} + \frac{s}{s^2 + 9} \right\} \\ &= \mathcal L^{-1} \left\{ \frac 12 \cdot \frac {2!}{s^{2+1}} - \frac 12 \cdot \frac{2}{s^2 + 2^2} + \frac{s}{s^2 + 3^2} \right\} \\ &= \frac 12 t^2 - \frac 12 \sin(2t) + \cos(3t). \end{aligned}$

Therefore, $\mathcal L\{F(s-1)\} = \frac 12 e^{t} (t^2 - \sin(2t) + 2\cos(3t))$ .

Question 3. Evaluate $\mathcal L\{(t-1 + e^{t-2}) \cdot U(t-2)\}$ .

(Click for Solution)

Solution. Motivated by the term $U(t-2)$ , define

$f(t-2) := t-1 + e^{t-2}.$

By the shift theorems,

$\mathcal L\{ f(t-2) U(t-2) \} = e^{-2s} F(s).$

Replacing $t$ with $t+2$ ,

$f(t) = f((t+2) - 2) = (t+2)-1 + e^{(t+2)-2} = t + 1 + e^t.$

Taking Laplace transforms,

$\begin{aligned} F(s) = \mathcal L\{ f(t) \} &= \mathcal L\{ t + 1 + e^t \} \\ &= \frac{1!}{s^{1+1}} + \frac 1s + \frac 1{s-1} \\ &= \frac 1{s^2} + \frac 1s + \frac 1{s-1}. \end{aligned}$

Hence,

$\displaystyle \mathcal L\{ f(t-2) U(t-2) \} = e^{-2s} F(s) = e^{-2s} \cdot \left( \frac 1{s^2} + \frac 1s + \frac 1{s-1} \right).$

Question 4. Evaluate $\displaystyle \mathcal L^{-1} \left\{e^{-3s} \left( \frac 6{s^2+9} -\frac {s+2}{s(s-2)} \right)\right\}$ .

(Click for Solution)

Solution. Define $\displaystyle F(s) = \frac 6{s^2+9} -\frac {s+2}{s(s-2)}$ . By the shift theorems,

$\mathcal L^{-1} \{ e^{-3s} F(s) \} = f(t-3) U(t-3).$

Simplifying $F(s)$ using partial fractions,

$\begin{aligned} F(s) &= \frac 6{s^2+9} -\frac {s+2}{s(s-2)} = \frac 6{s^2+9} + \frac 1s - \frac 2{s-2}. \end{aligned}$

Taking inverse Laplace transforms,

$\begin{aligned} f(t) = \mathcal L^{-1}\{ F(s)\} &= \mathcal L^{-1} \left\{ 2 \cdot \frac 3{s^2+ 3^2} + \frac 1s - 2 \cdot \frac 1{s-2} \right\} \\ &= 2 \sin(3t) + 1 - 2e^{2t}. \end{aligned}$

Therefore,

$\begin{aligned} \mathcal L^{-1} \{ e^{-3s} F(s) \} &= f(t-3) U(t-3) \\ &= (2 \sin(3(t-3)) + 1 - 2e^{2(t-3)}) \cdot U(t-3).\end{aligned}$

—Joel Kindiak, 19 Apr 25, 0107H

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