\magnification 2200
\parindent 0pt
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\hoffset -0.35truein
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\def\u{\vskip -5pt}
\def\v{\vskip 10pt}



Linear algebra pre-requisites you must know.


${\bf b_1, ..., b_n}$ are linearly independent if 
\u

$c_1{\bf b_1} + c_2{\bf b_2} + ... + c_n{\bf b_n}  = 
d_1{\bf b_1} + d_2{\bf b_2} + ... + d_n{\bf b_n}$ 

\centerline{implies
$c_1 = d_1$, $c_2 = d_2$..., $c_n = d_n$.}

or equivalently,
 
${\bf b_1, ..., b_n}$ are linearly independent if
\u

$c_1{\bf b_1} + c_2{\bf b_2} + ... + c_n{\bf b_n}  = 0$ 
implies $c_1 = c_2 = ... c_n$.

\v\v

Example 1:
${\bf b_1} = (1, 0, 0)$, ${\bf b_2} = (0, 1, 0)$, ${\bf b_3} = (0, 0, 1)$.
\v

\centerline{$(1, 2, 3) \not= (1, 2, 4)$.}

\v
\centerline{If $(a, b, c) = (1, 2, 3)$ then $a = 1$, $b = 2$, $c = 3$.}

\v\v
Example 2:
${\bf b_1} = 1$, ${\bf b_2} = t$, ${\bf b_3} = t^2$.
\v
\centerline{$1 + 2t +  3t^2 \not= 1 + 2t +  4t^2$.}
\v

\centerline{If $a + bt +  ct^2 = 1 + 2t + 3t^2$ then $a = 1$, $b = 2$, $c = 3$.}

\eject
Application:  Partial Fractions

${ 4\over (x^2 + 1)(x - 3)} = {Ax + B \over x^2 + 1} + {C \over x - 3}$


$\hskip .7in = {(Ax + B)(x-3) + C(x^2 + 1) \over (x^2 + 1)(x - 3)}$

Hence ${ 4\over (x^2 + 1)(x - 3)} = 
{(Ax + B)(x-3) + C(x^2 + 1)  \over (x^2 + 1)(x - 3)}$
\v
\centerline{Thus $4 = (Ax + B)(x-3) + C(x^2 + 1)$}
\v
\centerline{$ {4 = Ax^2 + Bx - 3Ax - 3B + Cx^2 + C}$}
\v
\centerline{${4 = (A + C)x^2 + (B - 3A)x - 3B + C }$}



I.e., $0x^2 + 0x + 4 = {(A + C)x^2 + (B - 3A)x - 3B + C }$

Thus $0 = A + C$, $0 = B - 3A$, $4 = -3B + C$.  

$C = -A$, $B = 3A$,\hfil \break
 $4 = -3(3A) + -A$ implies $4 = -10A$. \hfil \break  Hence $A = -{2 
\over 5}$, $B = 3(-{2 
\over 5}) = -{6 \over 5}$, $C = {2 \over 5}$.
\v
\centerline{Thus,
${ 4\over (x^2 + 1)(x - 3)} 
= {-{2 \over 5}x  - {6 \over 5} \over x^2 + 1} + {{2 \over 5} \over 
x - 3}$}

\hskip 1.65in = ${-{2}x - {6} \over 5(x^2 + 1)} + {{2} \over 
5(x - 3)}$}

\eject

Linear Functions

A function $f$ is linear if $f(a{\bf x} + b{\bf y}) = af({\bf x}) + 
bf({\bf y})$

Or equivalently $f$ is linear if\hfil \break
 1.)  $f(a{\bf x}) = af({\bf x})$ and 2.) $f({\bf x} + {\bf y}) = f({\bf 
x}) + f({\bf y})$


Theorem:  If $f$ is linear, then $f({\bf 0}) = {\bf 0}$

Proof:  $f({\bf 0}) = f(0 \cdot {\bf 0}) = 0 \cdot f({\bf 0}) = {\bf 0}$


Example 1.)  $f:R \rightarrow R$, $f(x) = 2x$

Proof:\hfil \break
  $f(ax + by) = 2(ax + by) = 2ax + 2by = af(x) + bf(y)$

Example 2.) $f:R^2 \rightarrow R^2$, 

\centerline{$f((x_1,x_2)) = (2x_1, x_1 + x_2)$}

Proof:  Let ${\bf x} = (x_1, x_2)$, ${\bf y} = (y_1, y_2)

$a{\bf x} + b{\bf y} = a(x_1, x_2) + b(y_1, y_2) = (ax_1, ax_2) + (by_1, by_2) 
= (ax_1 + by_1, ax_2 + by_2)$

\eject
$f(ax_1 + by_1, ax_2 + by_2) $

\hskip 0.4in $= (2(ax_1 + by_1), ax_1 + by_1 + ax_2 + by_2) $

\hskip 0.4in $= (2ax_1 + 2by_1, ax_1 + ax_2 + by_1 + by_2)   $

\hskip 0.4in $=(2ax_1, ax_1 + ax_2) + (2by_1, by_1 + by_2) $

\hskip 0.4in $= a(2x_1, x_1 + x_2) + b(2y_1, y_1 + y_2)$

\hskip 0.4in $= af((x_1, x_2)) + bf((y_1, y_2))$


Example 3.) $D:$ {set of all differential functions} $\rightarrow$ {set of all functions}, 
$D(f) = f'$

Proof: \hfil \break
 $D(af + bg) = (af + bg)' = af' + bg' = aD(f) + bD(g)$

Example 4.) Given $a, b$ real numbers,\hfil \break
 $I:$ {set of all integrable functions on [a, b]} $\rightarrow R$ , 
$I(f) = \int_a^b f$

Proof:  $I(sf + tg) = \int_a^b sf + tg = s \int_a^b f + t \int_a^b g = sI(f) + tI(g)$ 
\eject

Example 5.)  The inverse of a linear function is linear (when the inverse exists).

Suppose $f^{-1}(x) = c$, $f^{-1}(y) = d$. 

Then $f(c) = x$ and $f(d) = y$ and \hfil \break
$f(ac + bd) = af(c) + bf(d) = ax + by$.

 Hence $f^{-1}(ax + by) = ac + bd = a f^{-1}(x) + bf^{-1}(y)$.

Example 6.)  $D:$ {set of all twice differential functions} $\rightarrow$ {set of all functions}, 
$L(f) = af'' + bf' + cf$

Proof: \hfil \break
$L(sf + tg) = a(sf + tg)'' + b(sf + tg)' + c(sf + tg) $

\hskip 0.67in $= saf'' + tag'' + sbf' + tbg' + scf + tcg $

\hskip 0.67in $= s(af'' + bf' + cf) +t(ag'' + bg' + cg) $

\hskip 0.67in $= sL(f) + tL(g)$

\eject

Consequence 1:  If $\psi_1, \psi_2$ are  solutions to $af'' + bf' + cf = 0$, 
then $3 \psi_1 + 5 \psi_2$ is also a solution to \break
$af'' + bf' + cf = 0$,

Proof: Since $\psi_1, \psi_2$ are  solutions to $af'' + bf' + cf = 0$, 
$L(\psi_1) = 0$ and $L(\psi_2) = 0$.

Hence $L(3 \psi_1 + 5 \psi_2) = 3 L(\psi_1) + 5 L(\psi_2)$
\u
\hskip 1.24in$ = 3(0) + 5(0) 
= 
0$.  
\u
Thus $3 \psi_1 + 5 \psi_2$ is also a solution to $af'' + bf' + cf = 
0$

Consequence 2: \hfil \break
 If $\psi_1$ is a solution to $af'' + bf' + cf = h$ \hfil \break
and $\psi_2$ is a solution to $af'' + bf' + cf = k$, \hfil \break
then $3 \psi_1 + 5 \psi_2$ is  a solution to $af'' + bf' + cf = 3h + 5k$,

Since $\psi_1$ is a solution to $af'' + bf' + cf = h$, $L(\psi_1) = h$.
 
Since $\psi_2$ is a solution to $af'' + bf' + cf = k$, $L(\psi_2) = k$.

Hence $L(3 \psi_1 + 5 \psi_2) = 3 L(\psi_1) + 5 L(\psi_2)$
\u
\hskip 1.27in$ = 3h + 5k$.  
\u
Thus $3 \psi_1 + 5 \psi_2$ is also a solution to 

\centerline{$af'' + bf' + cf = 
3h + 5k$}


Example 7.)  The LaPlace transform ${\cal L}$: is linear

Theorem:  Suppose $f$, $f'$, ..., $f^{(n-1)}$ are continuous and $f^{(n)}$
is piecewise continuous on $0 \leq t \leq A$.  Suppose there exists
constance $K$, $a$, and $M$ such that $|f(t)| \leq Ke^{at}$, $|f'(t)| \leq
Ke^{at}$, ..., $|f^{(n-1)}(t)| \leq Ke^{at}$ for $t \geq M$.  Then ${\cal
L}(f^{(n)})$ exists for $s > a$ and is given by

${\cal L}(f^{(n)})$ 

\rightline{$ = s^n{\cal L}(f) - s^{n-1}f(0) - ... 
- sf^{(n-2)}(0) - f^{(n-1)}(0)$}}

LaPlace Transform

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

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

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

\vskip 1in

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

\vskip 0.4in

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

${\cal L}(f^{(n)})$ 

\rightline{$ = s^n{\cal L}(f) - s^{n-1}f(0) - ...
- sf^{(n-2)}(0) - f^{(n-1)}(0)$}}


\vskip 1in

3.5)  Substitute in the initial values.

\vskip 0.5in
4.)  Solve the algebraic equation for ${\cal L}(y)$

\vskip 0.2in

Some algebra implies ${\cal L}(y) = $

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

\vskip 1in


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.

\v
\hrule

Calculus pre-requisites you must know.

Derivative = slope of tangent line = rate.

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

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


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


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


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

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

\end