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Assignment 16 (due 4.1)  6.2: 19, 20, 22, 23; 6.3:  5, 8, 10;
{6.3:  9, 13, 15, 24, 25, 28, 29}; 6.4 ?




Find the inverse LaPlace transform of ${5s + 21 \over s^2 + 3s + 4}$

Look at the denominator first to determine if you should factor and use partial fractions

$s^2 + 3s + 4$:  $b^2 - 4ac = 3^2 - 4(1)(4) = 9 - 16 < 0$

Hence $s^2 + 3s + 4$ does not factor over the reals.  Hence to avoid complex numbers, we won't
factor it.


$s^2 + 3s + 4$ is not an $s^2 - a^2$ or an $s^2 + a^2$, so it must be an $(s-a)^2 + b^2$.

Hence we will complete the square:

$s^2 + 3s + \underline{\hskip 0.2in} - \underline{\hskip 0.2in} + 4 = (s + \underline{\hskip
0.2in})^2- \underline{\hskip 0.2in} + 4$


Hence ${5s + 21 \over s^2 + 3s + 4}$ = ${5s + 21 \over (s + {3 \over 2})^2 + {7 \over 4}}$

\eject

Must now consider the numerator.  We need it to look like $s - a = s + {3 \over 2}$ or $b =
\sqrt{7 \over 4}$ in order to use ${\cal L}^{-1}({s-a\over (s -a)^2 + b^2}) = e^{at} cos bt$ 
\hfil 
\break
and/or ${\cal L}^{-1}({b \over (s -a)^2 + b^2}) = e^{at} sin bt$

$5s + 21 = 5(s + {3 \over 2}) - {15 \over 2} + 21 = 5(s + {3 \over 2}) - {27 \over 2}$

$ = 5(s +
{3 \over 2}) - [{27 \over 2}\sqrt{4 \over 7}] \sqrt{7 \over 4} = 5(s + {3 \over 2}) - [{27
\over \sqrt{7}}] \sqrt{7 \over 4} $

Hence ${5s + 21 \over s^2 + 3s + 4}$ = 
${5(s + {3 \over 2}) - [{27 \over\sqrt{7}}] \sqrt{7
\over 4} \over (s + {3 \over 2})^2 + {7 \over 4}}$ 

\hskip 0.95in= $5[{s + {3 \over 2} \over (s + {3 \over 2})^2 + {7 \over 4}}] 
- {27 \over\sqrt{7}}[{\sqrt{7 \over 4} \over  (s + {3 \over 2})^2 + {7 \over 4}}]$


Thus ${\cal L}^{-1}({5s + 21 \over s^2 + 3s + 4}) = 5e^{-{3 \over 2}t} cos
\sqrt{7 \over 4}t - {27 \over\sqrt{7}}e^{-{3 \over 2}t} sin \sqrt{7 \over 4}t

6.3:  Step functions.

$u_c(t) = \cases{0 & $t < c$ \cr 1 & $t \geq c$}$


1.)  Graph $u_c(t)$:
\eject

2.)  Given $f$, graph $u_c(t) f(t-c)$:
\vskip 0.8in

3.)  Calculate ${\cal L}(u_c(t) f(t-c))$ in terms of ${\cal L}(f(t))$:
\vfill

Example: 
 Find the LaPlace transform of 

1.)  
\centerline{$g(t) = \cases{0 & $t < 3$ \cr e^{t-3} & $t \geq 3$}$}

2.)  
\centerline{$f(t) = \cases{0 & $t < 3$ \cr 5 & $3 \leq t < 4$ \cr t 
- 5
& $t \geq 4$}$}
 

Example:  Find the inverse Laplace transform of ${e^{-8s} \over s^3 }

4.)  Calculate ${\cal L}(e^{ct} f(t))$ in terms of $F(s) = {\cal L}(f(t))$
\vfill

Example: Use formula 6 (p. 304)
to find the inverse LaPlace transform of ${s - c \over (s-c)^2 + a^2}$.


\eject
\end