\magnification 1200
\parindent 0pt
\parskip 10pt
\pageno=1
\hsize 7truein
\vsize 9.2truein
\def\u{\vskip -10pt}
\def\v{\vskip -5pt}


Note the following review problems DO NOT cover all problem types which may appear on the 
final.  

6.3 preliminaries:

1a.)  Suppose $f(t) = t^2$, then $f(t-2) = \underline{(t-2)^2}$

1b.)  Suppose $f(t) = t^2 + 3t + 4$, then $f(t-2) = \underline{(t-2)^2 + 3(t-2) + 4}$

1c.)  Suppose $f(t) = sin(t) + e^{8t}$, then $f(t-2) =  \underline{sin(t-2) + e^{8(t-2)}}$

2a.)  Suppose $f(t-2) = (t-2)^2$, then $f(t) = \underline{ t^2}$

2b.)  Suppose $f(t-2) = (t-2)^2 + 3(t-2) + 4$, then $f(t) = \underline{t^2 + 3t + 4}$

2c.)  Suppose $f(t-2) = sin(t-2) + e^{8(t-2)}$, then $f(t) =  \underline{sin(t) + e^{8t}}$

3a.)  Suppose $f(t-2) = t^2 + 2t + 5$, then $f(t) = \underline{t^2 + 6t + 13}$

$t^2 + 2t + 5 = (t-2)^2 + 4t - 4 + 2t + 5 = (t-2)^2 + 6t + 1 = (t-2)^2 + 6(t-2) + 12 + 1 $

\u
$=(t-2)^2 + 6(t-2) + 13$

Check:  $f(t-2) = (t-2)^2 + 6(t-2) + 13 = t^2 - 4t + 4 + 6t - 12 + 13 = t^2 + 2t + 5$

3b.)  Suppose $f(t-2) = 3t^2 + 8t + 1$, then $f(t) = \underline{3t^2 + 20t + 29}$

$ 3t^2 + 8t + 1 = 3(t-2)^2 - 3(-4t + 4) + 8t + 1 = 3(t-2)^2 + 12t -12 + 8t + 1$
\u
$ = 
3(t-2)^2 + 20t - 11 = 
$
\u
$3(t-2)^2 + 20(t-2) + 40 - 11 = 
3(t-2)^2 + 20(t-2) + 29$ 


Check:  $f(t-2) = 3(t-2)^2 + 20(t-2) + 29 = 3(t^2 - 4t + 4 ) + 20t - 40 + 29 = 3t^2 - 12t + 12 + 20t - 11
= 3t^2 + 8t + 1$


3c.)  Suppose $f(t-2) = cos(t) + e^{8t}$, then $f(t) =  \underline{cos(t+2) + e^{8t + 16}} $

$cos(t) + 4^{8t} = cos(t-2 + 2) + e^{8(t-2) + 16}$

Check:  $f(t-2) = cos(t-2+2) + e^{8(t-2) + 16} = cos(t) + e^{8t-16 + 16} = cos(t) + e^{8t}$ 

Chapter 6:



4.) Find the LaPlace transform of the following:  (used ${\cal L}(u_c(t)f(t-c)) = 
e^{-cs}{\cal L}(f(t))$) 

4a.)  ${\cal L}(u_3(t^2 - 2t + 1)) =  \underline{
e^{-3s}({2 \over s^3} + 4{1 \over s^2} + {4 \over s})} $

${\cal L}(u_3(t^2 - 2t + 1)) 
= {\cal L}(u_3((t-3)^2 + 6t - 9 - 2t + 1)) 
= {\cal L}(u_3((t-3)^2 + 4t - 8)) 
= {\cal L}(u_3 ( (t-3)^2 + 4(t-3) + 12 - 8 ) )
= {\cal L}(u_3 ( (t-3)^2 + 4(t-3) + 4 ) )
= e^{-3s}{\cal L}( t^2 + 4t + 4 ) ) 
= e^{-3s}({2 \over s^3} + 4{1 \over s^2} + {4 \over s})$ 

4b.)  ${\cal L}(u_4(e^{-8t})) =  \underline{e^{-4s-32}{1 \over s + 8}} $

${\cal L}(u_4e^{-8t} ) 
= {\cal L}(u_4e^{-8(t-4) - 32})
= e^{-4s}{\cal L}(e^{-8t - 32} )
= e^{-4s}e^{-32}{\cal L}(e^{-8t} )
= e^{-4s-32}{1 \over s + 8}$


4c.)  $ {\cal L}(u_2(t^2e^{3t})) =  \underline{e^{-2s+ 6}({2 \over (s - 3)^3} + {4 \over (s - 
3)^2} + {4 \over (s - 3)} )
} $

$ {\cal L}(u_2(t^2e^{3t}))
=   {\cal L}(u_2([(t-2)^2 + 4t - 4]e^{3(t-2) + 6}))
=  {\cal L}(u_2([(t-2)^2 + 4(t-2) + 8 - 4]e^{3(t-2) + 6}))
=  {\cal L}(u_2([(t-2)^2 + 4(t-2) + 4]e^{3(t-2) + 6}))
=  e^{-2s}{\cal L}([t^2 + 4t + 4]e^{3t + 6}))
=  e^{-2s}e^6{\cal L}([t^2 + 4t + 4]e^{3t}))
=  e^{-2s}e^6{\cal L}(t^2e^{3t} + 4te^{3t} + 4e^{3t}))
=  e^{-2s+ 6}({2 \over (s - 3)^3} + 4{1 \over (s - 3)^2} + {4 \over (s - 3)} )$




5.)  Find the inverse LaPlace transform of the following: (usually used 
$u_c(t)f(t-c) = {\cal L}^{-1}(e^{-cs}{\cal L}(f(t))) 

5a.)  $ {\cal L}^{-1}(e^{-8s}{1 \over s - 3}) =  \underline{u_8(t)e^{3(t-8)}} $

${\cal L}^{-1}(e^{-8s}{1 \over s - 3}) = u_8f(t-8)$ where

${\cal L}(f(t)) = {1 \over s - 3})$.  Hence $f(t) = {\cal L}^{-1}({1 \over s - 3}) = e^{3t}$

5b.)  ${\cal L}^{-1}(e^{4s}{1 \over s^2 - 3}) =  \underline{u_{-4}(t)sinh(\sqrt{3}(t + 4))} $

${\cal L}^{-1}(e^{4s}{1 \over s^2 - 3}) = u_{-4}(t)f(t+4)$ where

${\cal L}(f(t)) = {1 \over s^2 - 3})$.  Hence $f(t) 
= {1 \over \sqrt{3}}{\cal L}^{-1}({\sqrt(3) \over s^2 - 3}) = sinh(\sqrt{3}t)$


5c.)  ${\cal L}^{-1}(e^{s}{1 \over (s - 3)^2 + 4}) =  
\underline{{1 \over 2}u_{-1}(t)e^{3(t+1)}sin(2(t+ 
1))} $

${\cal L}^{-1}(e^{s}{1 \over (s - 3)^2 + 4}) = u_{-1}(t)f(t+1)$ where

${\cal L}(f(t)) = {1 \over (s - 3)^2 + 4})$.  Hence 
$f(t) = {1 \over 2}{\cal L}^{-1}({2 \over (s - 3)^2 + 4}) = {1 \over 2}e^{3t}sin(2t)$



5d.)  ${\cal L}^{-1}(e^{-s}{5 \over (s - 3)^4}) =  \underline{u_{1}(t){1 \over 
6}(t-1)^3e^{3(t-1)}} $

${\cal L}^{-1}(e^{-s}{5 \over (s - 3)^4}) 
= u_{1}(t)f(t-1)$ where

${\cal L}(f(t)) = {1 \over (s - 3)^4})$.  Hence 
$f(t) = {1 \over 6}{\cal L}^{-1}({6 \over (s - 3)^4}) = {1 \over 6}t^3e^{3t}$


5e.) ${\cal L}^{-1}(e^{s} \over 4s) =  \underline{{1 \over 4} u_{-1}(t)} $

${\cal L}^{-1}(e^{s} \over 4s) $
$= {1 \over 4}{\cal L}^{-1}(e^{s} \over s)$
$= {1 \over 4} u_{-1}(t)f(t+1)$ where

${\cal L}(f(t)) = {1 \over s}$.  Hence 
$f(t) = 1$.  Thus $f(t+1) = 1$

5f.)  ${\cal L}^{-1}(e^{s}) =  \underline{\delta(t + 1)} $





6.)  Use the definition and not the table to find the LaPlace transform of the following:

6a.)  ${\cal L}(t^2) =  \underline{{ 2 \over s^3}} $

$\int_0^\infty e^{-st}t^2 dt $
$= t^2 {e^{-st} \over -s} |_0^\infty - \int_0^\infty 2t{e^{-st} \over -s}$
$= lim_{t \rightarrow \infty}  t^2 {e^{-st} \over -s} -  0^2 {e^{0} \over -s} 
- [2t{e^{-st} \over s^2}|_0^\infty - \int_0^\infty 2{e^{-st} \over s^2}]$

\v
$= 0 -  0 
- [lim_{t \rightarrow \infty}2t{e^{-st} \over s^2} - 2(0){e^{0} \over s^2} - 
2{e^{-st} \over -s^3}|_0^\infty ]$
$= - [0 - 0 - 
(lim_{t \rightarrow \infty}2{e^{-st} \over -s^3} - 2{e^{0} \over -s^3}) ]$
$= 
(0 - {2 \over -s^3}) ]$
$= 
{ 2 \over s^3} $



Let $u = t^2, ~  dv = e^{-st}$
\v

~~~~$du = 2t, ~v = {e^{-st} \over -s}$
\v

~~~~$d^2u = 2, ~ \int v = {e^{-st} \over s^2}$




6a.)  ${\cal L}(cos(t)) =  \underline{\hskip 2in} $




7.)  Find the inverse LaPlace transform of the following.  Leave your answer in terms of a 
convolution integral:

7a.)  ${\cal L}^{-1}({1 \over (s-2)(s^2 + 4)}) =  \underline{\hskip 2in} $

7b.)  ${\cal L}^{-1}({1 \over (s-2)(s^2 - 4s)}) =  \underline{\hskip 2in} $

7c.)  ${\cal L}^{-1}({ 2s \over (s-2)(s^2 - 4)s}) =  \underline{\hskip 2in} $

8.) Find $f*g$

8a.)  $4t*5t^4 =  \underline{{ 2 \over 3}t^6 } $

$\int_0^t 4(t-s) 5s^4 ds = \int_0^t 20(ts^4- s^5) ds = 4ts^5 - {20 \over 6}s^6 |_0^t
= 4t^6 - {10 \over 3}t^6 = {2 \over 3}t^6 



8b.)   $5t^4*4t =  \underline{{2 \over 3}t^6 } $

$5t^4*4t =  4t*5t^4 = {2 \over 3}t^6 $ 


8c.)  $sin(t)*e^t =  \underline{\hskip 2in} $

$ \int_0^t e^{t-s}sin(s)ds =  \int_0^t e^{t}e^{-s}sin(s)ds = $
$e^t\int_0^t e^{-s}sin(s)ds = e^t[-e^{-s}sin(s)|_0^t - \int_0^t -e^{-s}cos(s)ds] =
 e^t[-e^{-t}sin(t) - -e^0 sin(0)  - \{e^{-s}cos(s)|_0^t - \int_0^t -e^{-s}sin(s)ds\}]
= e^t[-e^{-t}sin(t)  - \{(e^{-t}cos(t) - e^0 cos(0)) - \int_0^t -e^{-s}sin(s)ds\}]
= e^t[-e^{-t}sin(t)  - \{(e^{-t}cos(t) - 1) - \int_0^t -e^{-s}sin(s)ds\}]
= -e^t e^{-t}sin(t)  - \{(e^t e^{-t}cos(t) - e^t) - e^t \int_0^t -e^{-s}sin(s)ds\}]
= -sin(t)  - cos(t) + e^t) + e^t \int_0^t -e^{-s}sin(s)ds\}]
$
 

Let $u = sin(s), ~  dv = e^{-s}$

~~~~$du = cos(s), ~v = -e^{-s}$

~~~~$d^2u = -sin(s), ~ \int v = e^{-s}$






Make sure you can also solve a quick differential equation using the LaPlace transform and use any 
of the formulas on p. 304.

Chapter 3:

9.)  Solve the following initial problems:

9a.)  $y'' + 6y' + 8y = 0, ~y(0) = 0, ~y'(0) = 0 $

Suppose $y = e^{rt}$.  Then $y' = re^{rt}, ~y'' = r^2e^{rt}$

$r^2e^{rt} + 6re^{rt} + 8e^{rt} = 0$
Hence $r^2 + 6r + 8 = 0$.  Thus, (r + 2)(r + 4) = 0$.  Hence $r = -2, -4$

Hence general solution is $y = c_1e^{-2t} + c_2e^{-4t}$

$y(0) = 0:  0 = c_1 + c_2$.  Thus $c_2 = -c_1$

$y'(0) = 0:  y' = -2c_1e^{-2t} - 4c_2e^{-4t}$

$0 = -2c_1 - 4c_2 = 2c_2 - 4c_2 = -2c_2$  Thus $c_2 = 0, c_1 = 0$

Thus, $y = 0$


9b.)  $y'' + 6y' + 9y = 0, ~y(0) = 0, ~y'(0) = 0 $

9c.)  $y'' + 6y' + 10y = 0, ~y(0) = 0, ~y'(0) = 0$

9a.)  $y'' + 6y' + 8y = cos(t), ~y(0) = 0, ~y'(0) = 0 $

9b.)  $y'' + 6y' + 9y = cos(t), ~y(0) = 0, ~y'(0) = 0 $

9c.)  $y'' + 6y' + 10y = cos(t), ~y(0) = 0, ~y'(0) = 0$

3.8: 1-5, 7, 11, 14, 3.9:  1 - 8

Make sure you understand sections 3.8, 3.9

10.)  Solve the following initial problems:

10a.)  $y' + 3y + 1 = 0, ~y(0) = 0$

10b.)  $   , ~y(0) = 0$

*10c.)  $cos(t)y'- sin(t)y = {1 \over t^2}, ~y(0) = 0$

10d.)  $y' = {3x^2 - 2 \over xy - xy^2}, ~y(0) = 0$


Chapter 1:

11.)  For each of the following, draw the direction field for the given differential equation.  
Based on
the direction field, determine the behavior of $y$ as $t \rightarrow \infty$.  If this behavior 
depends on the initial value of $y$ at $t=o$, describe this dependency.


11a.)  $y' = y$

11a.)  $y' = 1$

11a.)  $y' = y(y + 4)$

Chapter 7:

12.)  Transform the given equation into a system of first order equations:

12a.)  $x''' - 2x'' + 3x' - 4x = t^2$

12b.)  $x'''' - 2x'' + 3x' - 4x = t^2$






Make sure you also study exam 1 and 2 as well as everything else.  Remember the above list is 
INCOMPLETE.


* means optional type problem.  If a problem like 9c appeared on the final, it would be in the 
"choose" section.


\end