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\begin{document}
\title{Solutions, 316-XVI}
\maketitle

\section{Problem 7.4.7}
Determine the inverse Laplace transform
\[ {\cal L}^{-1}\left\{ \frac{2s+16}{s^2 + 4s + 13} \right\} \]
{\bf Solution:}\\
\begin{eqnarray*}
 \frac{2s+16}{s^2 + 4s + 13} & = & \frac{2s+16}{(s^2+4s + 4)-4 + 13} \\
  & = & \frac{2(s+2)-4+16}{(s+2)^2 + 9} \\
  & = & \frac{2(s+2)+4 (3)}{(s+2)^2 + 3^2} 
\end{eqnarray*}
so that
\begin{eqnarray*}
{\cal L}^{-1}\left\{ \frac{2s+16}{s^2 + 4s + 13} \right\} 
& = & 2 {\cal L}^{-1}\left\{ \frac{(s+2)}{(s+2)^2 + 3^2} \right\} 
    + 4 {\cal L}^{-1}\left\{ \frac{ 3}{(s+2)^2 + 3^2} \right\} \\
& = & 2 e^{-2t} \cos{3t} + 4 e^{-2t} \sin{3t}
\end{eqnarray*}

\section{Problem 7.4.9}
Determine the inverse Laplace transform
\[ {\cal L}^{-1}\left\{ \frac{3s-15}{2s^2 - 4s + 10} \right\} \]
{\bf Solution:}\\


\section{Problem 7.4.11}
Determine the partial fractions expansion for
\[ F(s) = \frac{s^2 - 26s - 47}{(s-1)(s+2)(s+5)} \ . \]
{\bf Solution:}\\

\section{Problem 7.4.15}
Determine the partial fractions expansion for
\[ F(s) = \frac{-2 s^2 + 8 s - 14}{(s+1)(s^2 -2s+5)} \ . \]
{\bf Solution:}\\
\begin{eqnarray*}
F(s) & = & \frac{-2 s^2 + 8 s - 14}{(s+1)(s^2 -2s+5)} \\
     & = & \frac{A}{s+1} + \frac{B s + C}{s^2 -2s+1+4} \\ 
     & = & \frac{A}{s+1} + \frac{B s + C}{(s-1)^2+4} 
\end{eqnarray*}
Then
\begin{eqnarray*}
(s+1)F(s) & = & \frac{-2 s^2 + 8 s - 14}{(s^2 -2s+5)} \\
     & = & A + (s+1)\frac{(B s + C)}{s^2 -2s+1+4} \Rightarrow \\ 
\lim_{s\rightarrow -1} (s+1)F(s)     
     & = & \left. \frac{-2 s^2 + 8 s - 14}{(s^2 -2s+5)}\right|_{s=-1} = 
     \frac{-2 - 8 - 14}{(1 +2+5)} = \frac{-24}{8} = -3 = A 
\end{eqnarray*}
Subtracting:
\begin{eqnarray*}
\frac{B s + C}{(s-1)^2+4}
 & = & \frac{-2 s^2 + 8 s - 14}{(s+1)(s^2 -2s+5)} - \frac{-3}{s+1}  \\
 & = & \frac{-2 s^2 + 8 s - 14}{(s+1)(s^2 -2s+5)} + 
\frac{3(s^2-2s+5)}{(s+1)(s^2 -2s+5)}  \\
 & = & \frac{-2 s^2 + 8 s - 14 + 3(s^2-2s+5)}{(s+1)(s^2 -2s+5)}  \\
 & = & \frac{s^2 + 2 s + 1 }{(s+1)(s^2 -2s+5)}  \\
 & = & \frac{(s + 1)^2 }{(s+1)(s^2 -2s+5)}  \\
 & = & \frac{s + 1 }{(s^2 -2s+5)}  \\
 & = & \frac{(s-1) + 2 }{(s-1)^2+4}  \\
 & = & \frac{(s-1)}{(s-1)^2+4}  + \frac{2}{(s-1)^2+4} 
\end{eqnarray*}
so that
\[ F(s) = -3\frac{1}{s+1} + \frac{(s-1)}{(s-1)^2+4} + \frac{2}{(s-1)^2+4} \]
and
\[ {\cal L}^{-1}\left\{ F(s) \right\} =
 -3 e^{-t} + e^t \cos{2t} + e^t \sin{2t} \ . \] 

\section{Problem 7.4.23}
Determine the inverse Laplace transform
\[ F(s) = \frac{5s^2 + 34 s + 53}{(s+3)^2(s+1)} \ . \]
{\bf Solution:}\\

\section{Problem 7.4.26}
Determine the inverse Laplace transform
\[ F(s) = \frac{7s^3 -2 s^2 - 3 s + 6}{s^3 (s-2)} \ . \]
{\bf Solution:}\\
\begin{eqnarray*}
F(s) & = & \frac{7s^3 -2 s^2 - 3 s + 6}{s^3 (s-2)} \\
     & = & \frac{A}{s^3} + \frac{B}{s^2} + \frac{C}{s} + \frac{D}{s-2} 
\end{eqnarray*}
Then
\[ A =  \lim_{s\rightarrow 0} s^3 F(s) = 
\left. \frac{7s^3 -2 s^2 - 3 s + 6}{s-2}\right|_{s=0} = \frac{6}{-2} = -3\]
and
\[ D = \lim_{s\rightarrow 2} (s-2) F(s) =
\left. \frac{7s^3 -2 s^2 - 3 s + 6}{s^3} \right|_{s=2} = \frac{48}{8} = 6 \]
while
\[ B = \lim_{s\rightarrow 0} \frac{d}{ds} \left( s^3 F(s) \right) =
\left( \frac{21s^2 -4 s - 3 }{s-2} 
- \frac{7s^3 -2 s^2 - 3 s + 6}{(s-2)^2} 
\right)_{s=0} = \frac{3}{2}-\frac{6}{4} = 0 \]
and
\begin{eqnarray*}
 C & = & 
\lim_{s\rightarrow 0} \frac{1}{2}\frac{d^2}{ds^2} \left( s^3 F(s) \right) \\
   & = &
\frac{1}{2} \frac{d}{ds}\left( \frac{21s^2 -4 s - 3 }{s-2} 
- \frac{7s^3 -2 s^2 - 3 s + 6}{(s-2)^2} 
\right)_{s=0} \\
& = & \frac{1}{2}\left( \frac{42 s -4 }{s-2}-\frac{21s^2 -4 s - 3 }{(s-2)^2} 
- \frac{21 s^2 -4 s - 3 }{(s-2)^2} + 2 \frac{7s^3 -2 s^2 - 3 s + 6}{(s-2)^3}
\right)_{s=0} \\
& = &  \frac{1}{2}\left(\frac{-4 }{-2}-\frac{- 3 }{( -2)^2} - 
\frac{ - 3 }{(-2)^2}+ 2 \frac{ 6}{(-2)^3} = 2 + \frac{3}{4} + \frac{3}{4}
- \frac{3}{2}\right) = 1
\end{eqnarray*}
Then
\begin{eqnarray*}
F(s) & = & \frac{7s^3 -2 s^2 - 3 s + 6}{s^3 (s-2)} \\
     & = & \frac{-3}{s^3} + \frac{1}{s} + \frac{6}{s-2} 
\end{eqnarray*}
and
\[ f(t) = -\frac{3}{2} t^2 + 1 + 6 e^{2t} \ .\]

\newpage

\section{Problem 7.5.9}
Solve the given IVP using Laplace transforms
\[ z^{\prime \prime} + 5 z^\prime - 6z = 21 e^t \ ,\  z(0) = -1\ ,
\ z^\prime(0) = 9 \ . \]
{\bf Solution:}\\
\newpage

\section{Problem 7.5.13}
Solve the given IVP using Laplace transforms
\[ y^{\prime \prime} - y^\prime - 2 y = -8 \cos{t} - 2 \sin{t} \ ;
\ y(\pi/2)= 1 \ ,\  y^\prime(\pi/2) = 0 \ .\]
\newpage

\section{Problem 7.5.27}
Solve the given 3rd order IVP using Laplace transforms
\[ y^{\prime \prime\prime} + 3 y^{\prime \prime} + 3 y^\prime + y = 0 \ ;
\ y(0) = -4\ ,\ y^\prime(0) = 4\ ,\ y^{\prime \prime}(0)=-2 \ .\]
\newpage
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\end{document}

