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The LaPlace Transform is a method to change a differential equation to a linear equation.

Example:  Solve $2y'' + 3y' + 4y = 0$, $y(0) = 5$, $y'(0) = 6$


1.) Take the LaPlace Transform of both sides of the equation:

${\cal L}(2y'' + 3y' + 4y) = {\cal L}(0)$

2.)  Use the fact that the LaPlace Transform is linear:


${\cal L}(y'') + 3{\cal L}(y') + 4{\cal L}(y) = 0$


3.)  Use thm to change this equation into an algebraic equation:

$s^2{\cal L}(y) - sy(0) - y'(0) + 3[s{\cal L}(y) - y(0)] + 4{\cal L}(y) = 0$


3.5)  Substitute in the initial values:

$s^2{\cal L}(y) - 5s - 6 + 3[s{\cal L}(y) - 5] + 4{\cal L}(y) = 0$

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4.)  Solve the algebraic equation for ${\cal L}(y)$

$s^2{\cal L}(y) - 5s - 6 + 3s{\cal L}(y) - 15 + 4{\cal L}(y) = 0$

$[s^2 + 3s + 4]{\cal L}(y) = 5s + 21$

${\cal L}(y) = {5s + 21 \over s^2 + 3s + 4}$

Some algebra implies ${\cal L}(y)  = {5s + 21 \over s^2 + 3s + 4}$

5.)  Solve for $y$ by taking the inverse LaPlace transform of both sides (use a table):

${\cal L}^{-1}({\cal L}(y))  = {\cal L}^{-1}({5s + 21 \over s^2 + 3s + 4})$

$y  = {\cal L}^{-1}({5s + 21 \over s^2 + 3s + 4})$


Use partial fractions or completing the square to find ${\cal L}^{-1}({5s 
+ 21 \over s^2 + 3s + 4})$:

$b^2 - 4ac = 3^2 - 4(1)(4) < 0$


${5s + 21 \over s^2 + 3s + 4} =$

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To find the inverse LaPlace transform, you may need to use that the inverse LaPlace transform in linear.  
You may also need to use partial fractions or other methods in order to right the righthand side of (*) as a 
sum of functions whose inverse LaPlace transform is known.

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Calculus pre-requisites you must know.

Derivative = slope of tangent line = rate.

Integral = area between curve and x-axis (where area can be negative).

Integration by parts:

Derivative of a product:  $(uv)' = uv' + vu'$

\centerline{$uv' = (uv)' - vu'$}

\centerline{$\int uv' = \int (uv)' - \int vu'$}

\centerline{$\int uv' = (uv) - \int vu'$}


Example:

$\int e^{2x} sin(3x)$

Let $u = sin(3x)$, $dv = e^{2x}$

then $du = 3cos(3x)$, $v = {1 \over 2}e^{2x}$

then $d^2u = -9sin(3x)$, $\int v = {1 \over 4}e^{2x}$

$\int e^{2x} sin(3x) = {1 \over 2}sin(3x)e^{2x} - \int {3 \over 2}e^{2x}cos(3x)$

\hskip 0.4in $= {1 \over 2}sin(3x)e^{2x} - [{3 \over 4}cos(3x)e^{2x} - \int {-9 \over 4}sin(3x)e^{2x}$


$\int e^{2x} sin(3x) = {1 \over 2}sin(3x)e^{2x} - {3 \over 4}cos(3x)e^{2x} - {9 \over 4}\int sin(3x)e^{2x}$


${13 \over 4} \int e^{2x} sin(3x) = {1 \over 2}sin(3x)e^{2x} - {3 \over 4}cos(3x)e^{2x}$


$\int e^{2x} sin(3x) = {4 \over 13}[{1 \over 2}sin(3x)e^{2x} - {3 \over 4}cos(3x)e^{2x}]$

Exercise:  Calculate $\int e^{x} cos(2x)$


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