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


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){5 \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

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


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

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




6b.)  ${\cal L}(cos(t)) =  \underline{{s \over 1 + s^2} } $

$\int_0^\infty e^{-st}cos(t) dt 
= e^{-st}sin(t)|_0^\infty - \int_0^\infty -se^{-st}sin(t)dt$

$= lim_{t \rightarrow \infty}  e^{-st}sin(t) - e^0sin(0) - [se^{-st}cos(t)|_0^\infty - 
\int_0^\infty -s^2e^{-st}cos(t)dt] $

$= 0 - 0 - [lim_{t \rightarrow \infty} se^{-st}cos(t) - se^0cos(0) + s^2 \int_0^\infty
e^{-st}cos(t)dt] $

$= -[0 - s + s^2 \int_0^\infty
e^{-st}cos(t)dt] $

$=  s - s^2 \int_0^\infty e^{-st}cos(t)dt $

Hence $\int_0^\infty e^{-st}cos(t) dt 
=  s - s^2 \int_0^\infty e^{-st}cos(t)dt $

Thus $\int_0^\infty e^{-st}cos(t) dt +  s^2 \int_0^\infty e^{-st}cos(t)dt 
=  s $

Thus $(1 + s^2)\int_0^\infty e^{-st}cos(t) dt 
=  s $

Thus $\int_0^\infty e^{-st}cos(t) dt 
=  {s \over 1 + s^2} $



Let $u = e^{-st}, ~dv = cos (t)$

~~~~$du = -se^{-st}, ~ v = sin(t)$
  
~~~$d^2u = s^2e^{-st}, \int v = -cos(t)$




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{{1\over 2}\int_0^t e^{2(t- s)} sin(2s)ds} $

 ${\cal L}^{-1}({1 \over (s-2)(s^2 + 4)}) = 
 {\cal L}^{-1}({1 \over (s-2)}) * {\cal L}^{-1}({1 \over (s^2 + 4)}) 
= e^{2t} * {1 \over 2}sin(2t) = {1\over 2}\int_0^t e^{2(t- s)} sin(2s)ds$


7b.)  ${\cal L}^{-1}({1 \over (s-2)(s^2 - 4s + 5)}) =  \underline{\int_0^t e^{2(t-s} e^{2s}sin(s)ds
} $

${\cal L}^{-1}({1 \over (s-2)})*{\cal L}^{-1}({1 \over (s^2 - 4s + 5)}) $
= ${\cal L}^{-1}({1 \over (s-2)})*{\cal L}^{-1}({1 \over (s-2)^2 + 1)})$ 
= $e^{2t} * e^{2t}sin(t) = \int_0^t e^{2(t-s)} e^{2s}sin(s)ds$

Note we can easily calculate this integral:

$= \int_0^t e^{2t}e^{-2s} e^{2s}sin(s)ds
= \int_0^t e^{2t}sin(s)ds
= e^{2t}\int_0^t  sin(s)ds
= -e^{2t}cos(s)|_0^t 
= -e^{2t}cos(t) + e^{2t}$
 

7c.)  ${\cal L}^{-1}({ 2s \over (s-2)(s^2 - 4s + 5)}) =  \underline{= 2\int_0^t e^{2(t-s)}
 (e^{2s}cos(s) + 2e^{2s}sin(s)) ds} $

${\cal L}^{-1}({ 2s \over (s-2)(s^2 - 4s + 5)}) $
$= 2{\cal L}^{-1}({ 1 \over (s-2)} \cdot {s -2 + 2\over ((s-2)^2 + 1)}) $
$= 2{\cal L}^{-1}({ 1 \over (s-2)} \cdot [{s -2 \over ((s-2)^2 + 1)} + {2 \over ((s-2)^2 + 
1)}]) $
$= 2e^{2t} * (e^{2t}cos(t) + 2e^{2t}sin(t))$
$= 2\int_0^t e^{2(t-s)} (e^{2s}cos(s) + 2e^{2s}sin(s)) ds$

Note we can easily calculate this integral:

$= 2e^{2t}\int_0^t e^{-2s} (e^{2s}cos(s) + 2e^{2s}sin(s)) ds$
$= 2e^{2t}\int_0^t cos(s) + 2sin(s)) ds$
$= 2e^{2t}[sin(s) - 2cos(s)]|_0^t$
$= 2e^{2t}[sin(t) - 2cos(t) - ( 0 - 2)] $
$= 2e^{2t}[sin(t) - 2cos(t) + 2] $

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{{1 \over 2}(-sin(t)  - cos(t) + 
e^t)} $

$ \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] 

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

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

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

$= -sin(t)  - cos(t) + e^t - e^t\int_0^t e^{-s}sin(s)ds$


Hence $e^t\int_0^t e^{-s}sin(s)ds =-sin(t)  - cos(t) + e^t - e^t\int_0^t e^{-s}sin(s)ds$ 

Hence $2e^t\int_0^t e^{-s}sin(s)ds =-sin(t)  - cos(t) + e^t$ 

Hence $e^t\int_0^t e^{-s}sin(s)ds ={1 \over 2}(-sin(t)  - cos(t) + e^t)$ 

 

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 $y(t) = c_1e^{-2t} + c_2e^{-4t}$

$y(0) = 0:  0 = c_1 + c_2$.  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, $y(t) = 0$ is the solution to the initial value problem.

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

$r^2 + 6r + 9 = (r + 3)(r +3) = 0$

Hence general solution is $y(t) = c_1e^{-3t} + c_2te^{-3t}$

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

$y = c_2te^{-3t}$

$y' = c_2[e^{-3t} -3t e^{-3t}]$

$y'(0) = 0:  0 =  c_2[e^{0} -3(0) e^{0}] $.  Thus $c_2 = 0$.

Thus, $y(t) = 0$ is the solution to the initial value problem.




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

$r^2 + 6r + 10 = 0$, $r = {-6 \pm \sqrt{36 - 4(1)(10)} \over 2}  
= {-6 \pm 2i \over 2} = -3 \pm i$

Hence general solution is  $y(t) = c_1e^{-3t}sin(t) + c_2e^{-3t}cos(t)$

$y(0) = 0:  0 = c_1(0) + c_2(1)$.  Thus $c_2 = 0$

$y = c_1 e^{-3t}sin(t) $

$y' = c_1[-3 e^{-3t}sin(t) +  e^{-3t}cos(t)] 

$y'(0) = 0:  0 =  c_1[0 + 1] $.  Thus $c_1 = 0$.

Thus, $y(t) = 0$ is the solution to the initial value problem.

Quicker method to solve 9a, b, c:

Note that $y = 0$ was the obvious solution to the initial value problems in 9a, b, c.

Clearly the constant function $y(t) = 0$ satisfies $ay'' + by' + cy = 0$ for any 
$a,~ b,~ c$.  It also satisfies $y(0) = 0, ~y'(0) = 0$.  Thus this constant function 
is a solution to IVP of this type.  Since  IVP of this type have unique solutions, $y(t) 
= 0$ is the only solution.
 
9d.)  $y'' + 6y' + 8y = cos(t), ~y(0) = 0, ~y'(0) = 0 $

Guess $y = Acos(t) + Bsin(t) $ is a solution.

Then $y' = -Asin(t) + Bcos(t)$ and $y'' = -Acos(t) - Bsin(t)$ 

$-Acos(t) - Bsin(t) + 6(-Asin(t) + Bcos(t)) + 8(Acos(t) + Bsin(t)) = cos(t)$

$(-A + 6B + 8A) cos(t) + (-B -6A + 8B) sin(t) = cos(t)$

$(6B + 7A) cos(t) + (-6A + 7B) sin(t) = cos(t) + 0 sin(t)$

$cos(t)$ and $sin(t)$ are linearly independent functions, 

Hence $ (-6A + 7B) sin(t) =  0 sin(t)$.  Thus $-6A + 7B = 0$  so $A = {7 \over 6}B $

And $(6B + 7A) cos(t)  = cos(t)$.  Thus $6B + 7A = 1$, so $6B + 7({7 \over 6}B ) = 1$.  
${36 + 49 \over 6}B = {85 \over 6}B = 1$

Thus 
$B = {6 \over 85} $ and $A = {7 \over 85} $.

Check: $-{7 \over 85} cos(t) - {6 \over 85} sin(t) + 6(-{7 \over 85} sin(t) + 
{6 \over 85} cos(t)) + 8({7 \over 85} cos(t) + {6 \over 85} sin(t)) = 
(-{7 \over 85}  + {36 \over 85} + {56 \over 85})cos(t) + 
(- {6 \over 85} - {42 \over 85} + {48 \over 85} )sin(t) = cos(t)$



Thus general solution is 
$y(t) = c_1e^{-2t} + c_2e^{-4t} +  {7 \over 85} cos(t) + {6 \over 85} sin(t) $

Use initial values to find $c_1, ~c_2$:

$0 = c_1 + c_2 + {7 \over 85} + 0$.  Hence $c_1 = -c_2 - {7 \over 85}$ 

$y'(t) = -2c_1e^{-2t} + -4c_2e^{-4t} -  {7 \over 85} sin(t) + {6 \over 85} cos(t) $

$0 = -2c_1 - 4c_2 - 0 + {6 \over 85} 
= -2(-c_2 - {7 \over 85} )- 4c_2 + {6 \over 85} 
= 2c_2 + {14 \over 85} - 4c_2 + {6 \over 85} 
= -2c_2 + {20 \over 85}$ 

Thus $2c_2 = {20 \over 85}$ or $c_2 = {10 \over 85} $ and 
$c_1 = -{10 \over 85} - {7 \over 85} = -{17 \over 85}$ 

Note these are NOT the same values for $c_1$ and $c_2$ for the homogeneous case.

The the solution to the initial value problem is
$y(t) = -{17 \over 85}e^{-2t} + {10 \over 85} e^{-4t} 
+  {7 \over 85} cos(t) + {6 \over 85} sin(t) $


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

Guess $y = Acos(t) + Bsin(t) $ is a solution.

Then $y' = -Asin(t) + Bcos(t)$ and $y'' = -Acos(t) - Bsin(t)$ 

$-Acos(t) - Bsin(t) + 6(-Asin(t) + Bcos(t)) + 9(Acos(t) + Bsin(t)) = cos(t)$

$(-A + 6B + 9A) cos(t) + (-B -6A + 9B) sin(t) = cos(t)$

$(6B + 8A) cos(t) + (-6A + 8B) sin(t) = cos(t) + 0 sin(t)$

$cos(t)$ and $sin(t)$ are linearly independent functions, 

Hence $ (-6A + 8B) sin(t) =  0 sin(t)$.  Thus $-6A + 8B = 0$  so $A = {8 \over 6}B =
{4 \over 3}B  $

And $(6B + 8A) cos(t)  = cos(t)$.  Thus $6B + 8A = 1$, so $6B + 8({4 \over 3}B ) = 1$.  
${18 + 32 \over 3}B = {50 \over 3}B = 1$

Thus 
$B = {3 \over 50} $ and $A = {8 \over 100}  = {2 \over 25}$.


Thus general solution is 
$y(t) = c_1e^{-3t} + c_2te^{-3t} +  {2 \over 25} cos(t) + {3 \over 50} sin(t) $


Use initial values to find $c_1, ~c_2$:

$0 = c_1 + c_2(0) + {2 \over 25} + 0$.  Hence $c_1 = - {2 \over 25}$ 

$y'(t) = -3c_1e^{-3t} + c_2e^{-3t} - 3c_2te^{-3t}  -  {2 \over 25} sin(t) 
+ {3 \over 50} cos(t) $

$0 = -3c_1 + c_2 - 0 - 0 + {3 \over 50} 
= -3(- {2 \over 25} ) + c_2 + {3 \over 50} 
= {6 \over 25}  + c_2 + {3 \over 50} 
= c_2 + {15 \over 50}$ 

Thus $c_2 = -{15 \over 50} $ and 
$c_1 = -{2 \over 25} $ 


Note these are NOT the same values for $c_1$ and $c_2$ for the homogeneous case.

The the solution to the initial value problem is

$y(t) = -{2 \over 25} e^{-3t}  -{15 \over 50} te^{-3t} + 
 {2 \over 25} cos(t) + {3 \over 50} sin(t) $

Check: $y(0) = -{2 \over 25} + 0 + {2 \over 25} = 0$

$y'(t) = {6 \over 25} e^{-3t} + {45 \over 50} te^{-3t} -{15 \over 50} e^{-3t}  
 - {2 \over 25} sin(t) + {3 \over 50} cos(t) $

$y'(0) = {6 \over 25}  + 0  -{15 \over 50}   
 - 0 + {3 \over 50} = {12 - 15 + 3 \over 50} = 0 $

Thus initial conditions are satisfied.  To fully check that this is the solution, 
should check if $y'' + 6y' + 9y = cos(t)$ 



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

Hence general solution to the homogeneous equation is  
$y(t) = c_1e^{-3t}sin(t) + c_2e^{-3t}cos(t)$

Thus $y = Acos(t) + Bsin(t) $ is not a solution to the homogeneous equation.

Guess $y = Acos(t) + Bsin(t) $ is a solution.

Then $y' = -Asin(t) + Bcos(t)$ and $y'' = -Acos(t) - Bsin(t)$ 

$-Acos(t) - Bsin(t) + 6(-Asin(t) + Bcos(t)) + 10(Acos(t) + Bsin(t)) = cos(t)$

$(-A + 6B + 10A) cos(t) + (-B -6A + 10B) sin(t) = cos(t)$

$(6B + 9A) cos(t) + (-6A + 9B) sin(t) = cos(t) + 0 sin(t)$

$cos(t)$ and $sin(t)$ are linearly independent functions, 

Hence $ (-6A + 9B) sin(t) =  0 sin(t)$.  Thus $-6A + 9B = 0$  so $A = {9 \over 6}B =
{3 \over 2}B  $

And $(6B + 9A) cos(t)  = cos(t)$.  Thus $6B + 9A = 1$, so $6B + 9({3 \over 2}B ) = 1$.  
${12 + 27 \over 2}B = {39 \over 2}B = 1$

Thus 
$B = {2 \over 39} $ and $A = {3 \over 39}  = {1 \over 13}$.


Thus general solution is 
$y(t) = c_1e^{-3t}sin(t) + c_2e^{-3t}cos(t) +  {2 \over 39} cos(t) + {3 \over 39} sin(t) $


Use initial values to find $c_1, ~c_2$:

$0 = c_1(0) + c_2(1) + {2 \over 39} + 0$.  Hence $c_2 = - {2 \over 39} $ 

$y'(t) =   -3c_1e^{-3t}sin(t) + c_1e^{-3t}cos(t) 
 -3c_2e^{-3t}cos(t) - c_2e^{-3t}sin(t)   
-  {2 \over 39} sin(t) 
+ {3 \over 39} cos(t) $

$0 = 0 + c_1 -3c_2 + 0 - 0 + {3 \over 39} $
Thus $c_1 = 3c_2-{3 \over 39}  = 3(- {2 \over 39}) -{3 \over 39} = {-12 \over 39}
{-4 \over 13} $  
 
Note these are NOT the same values for $c_1$ and $c_2$ for the homogeneous case.

The the solution to the initial value problem is

$y(t) = {-4 \over 13}e^{-3t}sin(t) + - {2 \over 39} e^{-3t}cos(t) +  
{2 \over 39} cos(t) + {3 \over 39} sin(t) $

Check: $y(0) = 0 - {2 \over 39}  +  
{2 \over 39}  + 0 = 0$

To fully check that this is the solution, 
should check if $y'' + 6y' + 9y = cos(t)$ and if $y'(0) = 0$.



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

Method 1:  separate variable

${dy \over dx} = -3y-1$

${dy \over -3y-1} = dx$

$\int {dy \over -3y-1} = \int dx$

Let $u = -3y - 1$, $du = -3dy$

$-{1 \over 3} \int {du \over u} = \int dx$

$-{1 \over 3} ln(u) = x + C$

$ln(-3y - 1) = -3x + C_1$

$-3y - 1 = e^{-3x + C_1}$

$-3y = e^{-3x}e^{C_1} + 1$

$y = {1 \over 3}e^{C_1}e^{-3x} - {1 \over 3}$

$y = Ke^{-3x} - {1 \over 3}$

$y(0) = 0:  0 = K - {1 \over 3}$.  Hence $K = {1 \over 3}$

Thus $y = {1 \over 3}e^{-3x} - {1 \over 3}$


Method 2:  integrating factor:

$y' + 3y = -1$


Let $u = e^{\int 3 dx} = e^{3x}$

$e^{3x}y' + 3e^{3x}y = -e^{3x}$

$(e^{3x}y)' = -e^{3x}$

$\int(e^{3x}y)' = -\int e^{3x}$

$e^{3x}y = -{1 \over 3}e^{3x} + C$

$y = -{1 \over 3} + Ce^{-3x}$

$y(0) = 0:  0 = -{1 \over 3} + C$.  Hence $C = {1 \over 3}$

Thus $y = {1 \over 3}e^{-3x} - {1 \over 3}$


Method 3:  LaPlace transform:

${\cal L}(y') + 3{\cal L}(y) = -{\cal L}(1)$

$s{\cal L}(y) + y(0) + 3{\cal L}(y) = -{1 \over s}$

$(s + 3){\cal L}(y) = -{1 \over s}$

${\cal L}(y) = -{1 \over s(s + 3) } = -{1 \over 3}[{1 \over s} - {1 \over s + 3}]$

$y = -{1 \over 3}[1 - e^{-3x}]$




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

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

Note $cos(t)y'- sin(t)y = (cos(t)y)'$

$\int (cos(t)y)' = \int {1 \over t^2} dt$

$cos(t)y = -t^{-1} + C = {-1 + Ct \over t}$


$y(0) = 0$:  $0 = -1 + C$.  Hence $C = 1$.

$y = {-1 + Ct \over tcos(t)}$




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

${dy \over dx} = {3x^2 - 2 \over x(y - y^2)}$

${(y - y^2)dy} = {3x^2 - 2 \over x}dx$

$int{(y - y^2)dy} = \int[3x - {2 \over x}]dx$

${1 \over 2}y^2 - {1 \over 3}y^3 = {3 \over 2}x^2 - 2ln(x) + C$


$y(0) = 0$: $0 = {3 \over 2}e^2 - 2ln(e) + C$


$C = 2 ln(e) -  {3 \over 2}e^2 = 2 -  {3 \over 2}e^2$

Answer:  ${1 \over 2}y^2 - {1 \over 3}y^3 = {3 \over 2}x^2 - 2ln(x) + 2 -  {3 \over 2}e^2$




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

~~~~~~~$x_1 = x$

$x_1' = x_2 = x'$

$x_2' = x_3 = x''$

$x_3' = x'''$


Answer:  $x_1' = x_2$, $x_2' = x_3$, $x_3' - 2x_3 + 3x_2 - 4x_1 = t^2$


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

~~~~~~~$x_1 = x$

$x_1' = x_2 = x'$

$x_2' = x_3 = x''$

$x_3' = x_4 = x'''$

$x_4' = x''''$

Answer:  $x_1' = x_2$, $x_2' = x_3$, $x_3' = x_4$, $x_4' - 2x_3 + 3x_2 - 4x_1 = t^2$






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.


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