Question

Discuss how you would fix up each of the following functions so that the theorem $7.3.2$could be used directly to and the given Laplace Transform. Check your answers using $\left(16\right)$of this section.

$$\begin{gathered} \left(a\right)+\left\{\right(2t11\left)8\right(t21\left) \\ \right(b)+\{et8\left(t25\right)\left\} \\ \right(c)+\{cost8\left(t2\right)\left\} \\ \right(d)+\{\left(t223t\right)8\left(t22\right)\} \end{gathered}$$

Short answer

The Laplace transform equations are,

$$\begin{gathered} \left(a\right)\mathcal{L}\left\{\right(2t+1\left)U\right(t-1\left)\right\}=e^{-s}\left(2\frac{1}{s^2}+\frac{3}{s}\right) \\ \left(b\right)\mathcal{L}\left\{e^{t}U(t-5)\right\}=e^{5s}{e}^5\frac{1}{s-1} \\ \left(c\right)\mathcal{L}\left\{\cos tU\right(t-\pi \left)\right\}=-e^{-\pi s}\frac{s}{s^2+1} \\ \left(d\right)\mathcal{L}\left\{\left(t^2-3t\right)U(t-2)\right\}=e^{-2s}\left(\frac{2}{s^3}+\frac{1}{s^2}-2\frac{1}{s}\right) \end{gathered}$$

Step-by-step solution

Define Heaviside Step Function:

• The Heaviside step function, also known as the unit step function, is a step function named after Oliver Heaviside (1850–1925), whose value is zero for negative arguments and one for positive arguments.
• It is generally denoted by H or (but sometimes u, 1 or 1)

(a) Find the Laplace Transform

From theorem 7.3.2, if$F\left(s\right)=\mathcal{L}\left\{f\right(t\left)\right\}anda>0,then$

$\mathcal{L}\left\{f\right(t-a\left)U\right(t-a\left)\right\}=e^{-as}F\left(s\right)$

Thus,

$$\begin{gathered} \mathcal{L}\left\{\right(2(\text{t}-1)+3\left)U\right(\text{t}-1\left)\right\}=\mathcal{L}\left\{2\right(\text{t}-1\left)\text{U}\right(\text{t}-1)+3\text{U}(\text{t}-1\left)\right\} \\ =2\mathcal{L}\left\{\right(\text{t}-1\left)\text{U}\right(\text{t}-1\left)\right\}+3\mathcal{L}\left\{\text{U}\right(\text{t}-1\left)\right\} \\ =2{\text{c}}^{-8}\frac{1}{{\text{s}}^2}+{\text{e}}^{-8}\frac{3}{\text{s}} \\ ={\text{e}}^{\text{s}}\left(2\frac{1}{{\text{s}}^2}+\frac{3}{\text{s}}\right) \end{gathered}$$

Now, using

$$\begin{gathered} \mathcal{L}\left\{g\right(t\left)U\right(t-a\left)\right\}=e^{-a8}\mathcal{L}\left\{g\right(t+a\left)\right\} \\ {e}^{-s}\mathcal{L}\left\{2\right(t+1)+1\}=e^{-s}\mathcal{L}\{2t+3\} \\ =e^{-s}\left(2\frac{1}{s^2}+\frac{3}{s}\right) \end{gathered}$$

This matches what we found above.

Hence, the Laplace transform equations is,

localid="1668581482293" $\mathcal{L}\left\{\right(2t+1\left)U\right(t-1\left)\right\}=e^{-s}\left(2\frac{1}{s^2}+\frac{3}{s}\right)$

(b) Determine the Laplace transform equation

From theorem 7.3.2, if$F\left(s\right)=\mathcal{L}\left\{f\right(t\left)\right\}$andlocalid="1663856645039" $a>0$, then

.$\mathcal{L}\left\{f\right(t-a\left)U\right(t-a\left)\right\}=e^{-as}F\left(s\right)$

Thus,

$$\begin{gathered} \mathcal{L}\left\{e^{l-5}{e}^{5}U(t-5)\right\}=e^5\mathcal{L}\left\{e^{t-5}U(t-5)\right\} \\ =e^5{e}^{-5s}\frac{1}{s-1} \end{gathered}$$

Now, using

$$\begin{gathered} \mathcal{L}\left\{g\right(l)\cup (l-a\left)\right\}=e^{-a\pi }\mathcal{L}\left\{g\right(l+a\left)\right\} \\ {e}^{-5s}\mathcal{L}\left\{e^{t+5}\right\}=e^{-5s}\mathcal{L}\left\{e^t{e}^5\right\} \\ =e^{-5s}{e}^5\frac{1}{s-1} \end{gathered}$$

This matches what we found above.

Hence, the Laplace transform equations is,

$\mathcal{L}\left\{e^{t}U(t-5)\right\}=e^{5s}{e}^5\frac{1}{s-1}$

(c) Find the Laplace transform using the theorem

From theorem 7.3.2, if$F\left(s\right)=\mathcal{L}\left\{f\right(t\left)\right\}$androle="math" localid="1663856805916" $a>0$, then

$\mathcal{L}\left\{f\right(t-a\left)U\right(t-a\left)\right\}=e^{-as}F\left(s\right)$

Thus,

$$\begin{gathered} \mathcal{L}\left\{\right(-\cos (t-\pi )\left)U\right(t-\pi \left)\right\}=-\mathcal{L}\left\{\cos \right(t-\pi \left)U\right(t-\pi \left)\right\} \\ =-c^{\pi s}\frac{s}{s^2+1} \end{gathered}$$

Now, using

$$\begin{gathered} \mathcal{L}\left\{g\right(t\left)U\right(t-a\left)\right\}=e^{-as}\mathcal{L}\left\{g\right(t+a\left)\right\} \\ {e}^{-\pi s}\mathcal{L}\left\{\cos \right(t+\pi \left)\right\}=e^{-\pi s}\mathcal{L}\{-\cos t\} \\ =-e^{-\pi s}\frac{s}{s^2+1} \end{gathered}$$

Hence, the Laplace transform equations is,

$\mathcal{L}\left\{\cos tU\right(t-\pi \left)\right\}=-e^{-\pi s}\frac{s}{s^2+1}$

(d) Find the Laplace transform equation

From theorem 7.3.2, if $F\left(s\right)=\mathcal{L}\left\{f\right(t\left)\right\}$ and $a>0$, then

$\mathcal{L}\left\{f\right(t-a\left)U\right(t-a\left)\right\}=e^{-as}F\left(s\right)$

Thus,

$$\begin{gathered} \mathcal{L}\left\{\left({(t-2)}^2+(t-2)-2\right)U(t-2)\right\}=\mathcal{L}\left\{{(t-2)}^{2}U(t-2)+(t-2)U(t-2)-2U(t-2)\right\} \\ =\mathcal{L}\left\{{(t-2)}^{2}U(t-2)\right\}+\mathcal{L}\left\{\right(t-2\left)U\right(t-2\left)\right\}-2\mathcal{L}\left\{U\right(t-2\left)\right\} \\ =e^{-2}\frac{2}{s^3}+e^{-2}\frac{1}{s^2}-2e^{-2s}\frac{1}{s} \end{gathered}$$

Now, using

$$\begin{gathered} \mathcal{L}\left\{g\right(t\left)U\right(t-a\left)\right\}=e^{-an}\mathcal{L}\left\{g\right(t+a\left)\right\} \\ {e}^{-2s}\mathcal{L}\left\{\left({(t+2)}^2-3(t+2)\right)\right\}=e^{-2t}\mathcal{L}\left\{t^2+4t+4-3t-6\right\} \\ =e^{-2s}\mathcal{L}\left\{t^2+t-2\right\} \\ =e^{-2v}\left(\frac{2}{s^3}+\frac{1}{s^2}-2\frac{1}{s}\right) \end{gathered}$$

Hence, the Laplace transform equations are:

$\mathcal{L}\left\{\left(t^2-3t\right)U(t-2)\right\}=e^{-2s}\left(\frac{2}{s^3}+\frac{1}{s^2}-2\frac{1}{s}\right)$