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${\cal L}$ is a linear function.


Ex: ${\cal L}(5e^2e^{it}e^{3t}) = $

\eject

Solve (i.e., solve for $y$):


$y^{iv} = 0$, ~~$y(0) = 0, ~y'(0) = 0, ~y''(0) = 0, ~y'''(0) = 10~$


${\cal L}(y^{iv}) = {\cal L}(0)$

$s^4{\cal L}(y) - s^310y(0) - s^2y'(0) - sy''(0) - y'''(0) = 0$

$s^4{\cal L}(y)  - 10 = 0$

$s^4{\cal L}(y)  = 10$

${\cal L}(y)  = {10 \over s^4}$

${\cal L}^{-1}({\cal L}(y))  ={\cal L}^{-1}( {10 \over s^4})$

$y  ={\cal L}^{-1}( {10 \over s^4})$   ~~~~~~~~~~Formula 3:  ${\cal L}^{-1}( {n! \over s^{n+1}}) =t^n $ 

$y  ={\cal L}^{-1}(  {10 \over 3!}\cdot {3! \over s^4}) ={10 \over 3!}{\cal L}^{-1}(  \cdot {3! \over s^4}) = {10 \over 3!}t^3 = {5 \over 3}t^3$ 
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\eject

Solve (i.e., solve for $y$):


$y'' + 2y' + 5y = 0$, ~~$y(0) = 1, ~y'(0) = 7$

${\cal L}(y'' + 2y' + 5y) = {\cal L}(0)$

${\cal L}(y'') + 2{\cal L}(y') + 5{\cal L}(y) = 0$



$s^2{\cal L}(y)  - sy(0)  - y'(0)$
\v
~~~~~~~~~~$+ 2[s{\cal L}(y) - y(0)]$
\v
~~~~~~~~~~~~~~~~~~~~~~~~$ + 5{\cal L}(y) = 0$

$s^2{\cal L}(y)  - s - 7$
\v
~~~~~~~~~~$+ 2[s{\cal L}(y) - 1]$
\v
~~~~~~~~~~~~~~~~~~~~~~~~$ + 5{\cal L}(y) = 0$

$s^2{\cal L}(y)  - s  - 7$
\v
~~~~~~~~~~$+ 2s{\cal L}(y) -2$
\v
~~~~~~~~~~~~~~~~~~~~~~~~$ + 5{\cal L}(y) = 0$




$(s^2 + 2s + 5){\cal L}(y) =   s  + 7 + 2$

${\cal L}(y) =   {s  + 9 \over s^2 + 2s + 5}$

${\cal L}^{-1}({\cal L}(y)) =   {\cal L}^{-1}({s  + 9 \over s^2 + 2s + 5})$

$y =   {\cal L}^{-1}({s  + 9 \over s^2 + 2s + 5})$

\eject

$y =   {\cal L}^{-1}({s  + 9 \over s^2 + 2s +1 -1+ 5})$

$y =   {\cal L}^{-1}({s  + 9 \over (s+1)^2 -1+ 5})$

$y =   {\cal L}^{-1}({s  + 9 \over (s+1)^2 + 4})$

$y =   {\cal L}^{-1}({s  + 1 + 8 \over (s+1)^2 + 4})$


$y =   {\cal L}^{-1}({s  + 1  \over (s+1)^2 + 4} + {8 \over (s+1)^2 + 4})$

$y =   {\cal L}^{-1}({s  + 1  \over (s+1)^2 + 4}) + {\cal L}^{-1}({8 \over (s+1)^2 + 4})$

$y =   {\cal L}^{-1}({s  + 1  \over (s+1)^2 + 4}) + 4{\cal L}^{-1}({2 \over (s+1)^2 + 4})$


$y = e^{-t}cos(2t) + 4e^{-t}sin(2t)$
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Partial Check:  Plug $y = e^{rt}$ into $y'' + 2y' + 5y = 0$:  
\v
Characteristic eqn:  $r^2 + 2r + 5 = 0$  implies 
\v
~~~~~~~~~~~~{$(r + 1)^2 + 4 = 0$ and thus $(r + 1)^2  = -4$.  }
\v
Hence $r + 1 = \sqrt{-4} = \pm 2i$.~~~ Thus $r = -1 \pm 2i$  
\v
and the general homogeneous soln is 
\v
~~~~~~~~~~~~{$y = c_1e^{-t}cos(2t) + c_2e^{-t}sin(2t)$}

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{
\small{Partial Check:} 
\v\v
$y(0) = 1$ for IVP soln 
$y = e^{-t}cos(2t) + 4e^{-t}sin(2t)$}
\v

$1 = e^{0}cos(0) + 4e^{0}sin(0)$ where equality does hold
\v
\small{Partial Check:}  $y'(0) = 7$ - \small{too much work due to product rule.}

\end