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\input spec_frac
\begin{document}
\title{Solutions, 316-XVIII}
\maketitle
Notation:
\[ u(t-c) \equiv u_c(t) := \left\{
\begin{array}{lcl}
0&,&t\le c \\
1&,&c < t 
\end{array} \right.
\]
7.6<1,2,3*,4,5*,6,15*,16,29,30,31*,35*,36>\\
CAUTION: there may be errors!!!
\section{Problem 7.6.1}
Sketch the graph and determine the Laplace transform.
\[ (t-1)^2 u(t-1) \ .\]
{\bf Solution:}\\
\begin{eqnarray*}
{\cal L}\left\{ t^2  \right\} 
                & = & \frac{2}{s^3} \rightarrow \\
{\cal L}\left\{ (t-1)^2 u(t-1)  \right\} 
                & = & e^{-s} \frac{2}{s^3}
\end{eqnarray*}

\begin{center}
\psfig{figure=para.ps,width=4.in}
\end{center}


\newpage

\section{Problem 7.6.3}
Sketch the graph and determine the Laplace transform.
\[ t^2 u(t-2) \ .\]
{\bf Solution:}\\
(hint: $t = (t-2)+2$ ).

\newpage

\section{Problem 7.6.4}
Sketch the graph and determine the Laplace transform.
\[ t u(t-1) \ .\]
{\bf Solution:}\\
(hint: $t = (t-1)+1$ is required so that the function is converted to
the canonical form $f(t-a)u(t-a)$ ).\\
Since
\[ t u(t-1) = \left[ (t-1) + 1\right] u(t-1) \]
we have
\begin{eqnarray*}
{\cal L}\left\{ t+1  \right\} 
                & = & \frac{1+s}{s^2} \rightarrow \\
{\cal L}\left\{ \left[(t-1)+1\right] u(t-1)  \right\} 
                & = & e^{-s} \frac{1+s}{s^2}
\end{eqnarray*}

\newpage

\section{Problem 7.6.5}
Express the given function using unit step functions and compute Laplace 
transform.\\
\[ g(t) = \left\{
\begin{array}{lcl}
0&,&0 \le t \le 1 \\
2&,&1 < t \le 2 \\
1&,&2 < t \le 3 \\
3&,&3 \le t 
\end{array} \right.
\]
{\bf Solution:}\\

\newpage

\section{Problem 7.6.5}
Express the given function using unit step functions and compute Laplace
transform.\\
\[ g(t) = \left\{
\begin{array}{lcl}
0&,&0 \le t \le 2 \\
t+1&,&2 \le t 
\end{array} \right.
\]
\begin{eqnarray*}
{\bf Solution:}\\
 g(t) & = & \left\{
\begin{array}{lcl}
0&,&0 \le t \le 2 \\
t+1&,&2 \le t 
\end{array} \right.  \rightarrow\\
g(t) & = & (t+1) u(t-2) = \left[(t-2) + 3 \right] u(t-2) \Rightarrow\\
{\cal L}\left\{ t+3  \right\} 
                & = & \frac{1+3s}{s^2} \rightarrow \\
{\cal L}\left\{ \left[(t-2)+3\right] u(t-2)  \right\} 
                & = & e^{-2s} \frac{1+3s}{s^2}
\end{eqnarray*}

\newpage

\section{Problem 7.7.15}
Determine an inverse Laplace transform for the function
\[ \frac{s e^{-3s}}{s^2 + 4s + 5} \ .\]
{\bf Solution:}\\

\newpage

\section{Problem 7.7.16}
Determine an inverse Laplace transform for the function
\[ \frac{e^{-s}}{s^2 + 4} \ .\]
{\bf Solution:}\\
\begin{eqnarray*}
{\cal L}^{-1}\left\{ \frac{1}{s^2 + 4} \right\} 
& = & \sin{2t} 
\Rightarrow \\
{\cal L}^{-1}\left\{ \frac{e^{-s}}{s^2 + 4} \right\} 
& = & \sin{2(t-1)}\ u(t-1) 
\end{eqnarray*}
\newpage

\section{Problem 7.7.30}
Solve the given IVP with Laplace transforms; graph the solution.
\[ w^{\prime \prime} + w = u(t-2) - u(t-4) \ ;\ w(0)=1\ ,\ w^\prime(0) =0\ .\]
{\bf Solution:}\\
\begin{eqnarray*}
 w^{\prime \prime} + w&  = & u(t-2) - u(t-4) \ ;\ w(0)=1\ ,\ w^\prime(0) =0\\ 
{\cal L}\left\{w^{\prime \prime} + w\right\} & = & 
{\cal L}\left\{ u(t-2) - u(t-4) \right\} \\
\left(s^2+1\right) W(s) - s & = & \frac{e^{-2s} - e^{-4s}}{s} \\
W(s) & = & \frac{s}{1+s^2} + \frac{e^{-2s} - e^{-4s}}{s} \frac{1}{1+s^2} 
\end{eqnarray*}
where for the last expression we need
\begin{eqnarray*}
\frac{1}{s(1+s^2)} & = & \frac{A}{s}+\frac{Bs+C}{1+s^2} \\
 A & = & \lim_{s\rightarrow 0} s\frac{1}{s(1+s^2)} = 1 \\
\frac{Bs+C}{1+s^2} & = & \frac{1}{s(1+s^2)} - \frac{1}{s} \\
                   & = & \frac{1 - (s^2+1)}{s(1+s^2)} = -\frac{s}{1+s^2}\\
\frac{1}{s(1+s^2)} & = & \frac{1}{s}-\frac{s}{1+s^2} 
\end{eqnarray*}
so that
\begin{eqnarray*}
W(s) & = & \frac{s}{1+s^2} + \left(e^{-2s} - e^{-4s}\right)
 \left( \frac{1}{s}-\frac{s}{1+s^2} \right) \\
y(t) & = & \cos{t} + \left(1-\cos{(t-2)}\right) u(t-2) - 
\left(1 - \cos{(t-4)}\right) u(t-4)
\end{eqnarray*}


\newpage

\section{Problem 7.7.31}
Solve the given IVP with Laplace transforms; graph the solution.
\[ y^{\prime \prime} + y = t - (t-4) u(t-2) \ ;\ y(0)=0\ ,\ y^\prime(0) =1\ .\]
{\bf Solution:}\\

\newpage

\section{Problem 7.7.35}
Solve the IVP using Laplace transforms.
\[ z^{\prime \prime} + 3z^\prime + 2z  = e^{-3t} u(t-2) \ ;\ z(0)=2\ ,
\ z^\prime(0) =-3\ .\]
{\bf Solution:}\\
	\newpage
\input laptab.tex
\end{document}

