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\def\u{\vskip -5pt}
\def\s{\vskip -3pt}
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HW 5:  Look at comments before submitting HW 6 as HW 6 will be graded more rigourously.  

In class quizzes are available again.  While I recommend doing them on each class day, the  unofficial due date is Sunday (note there is no late penalty to allow schedule flexibility). 


\hr
\vskip -2pt
\hr
\vskip 5pt

%\eject




Linear matrix equation:~~$Ax = b$

Linear differential equation:  $$y^{(n)} + p_1(t)y^{(n-1)} + ... + p_{n-1}(t)y' + p_n(t)y = g(t)$$

\hr

Linear combination of vectors:  $c_1{\bf v_1} +...+  c_n{\bf v_n}$

Linear combination of functions:  $c_1{\bf \phi_1} +...+  c_n{\bf \phi_n}$

\hr

\vskip -2pt
\hr
\vskip 5pt

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 \hb
~~~ 1.)  $f(a{\bf x}) = af({\bf x})$ and  \hb
~~~ 2.) $f({\bf x} + {\bf y}) = f({\bf 
x}) + f({\bf y})$

\hr

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

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


\hr

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

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

\hr

Example 1b.) $f :R \rightarrow R$, $f(x) = 2x + 3$ is NOT linear.
\s
Proof:  $f(2 \cdot 0) = f(0) = 3$, but $2f(0) = 2 \cdot 3 = 6$. 
\vskip -8pt
Hence $f(2 \cdot 0)  \not= 2f(0)$

Alternate Proof:  $f(0 + 1) = f(1) = 5$, but \hfil \break
$f(0) + f(1) = 3 + 5 = 
8$.
Hence $f(0 + 1) \not= f(0) + f(1)$

Note confusing notation:  Most lines, $f(x) = mx + b$ are not linear 
functions. 

Question:  When is a line,  $f(x) = mx + b$, a linear function?
\eject
~
\vfill

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

\centerline{$f((x_1,x_2)) = (2x_1, x_1 + x_2)$}
\s
Proof:  Let ${\bf x} = (x_1, x_2)$, ${\bf y} = (y_1, y_2)$
\s
$a{\bf x} + b{\bf y} = a(x_1, x_2) + b(y_1, y_2) = (ax_1, ax_2) + (by_1, by_2) 
= (ax_1 + by_1, ax_2 + by_2)$

$f(ax_1 + by_1, ax_2 + by_2) $
\s
\hskip 0.4in $= (2(ax_1 + by_1), ax_1 + by_1 + ax_2 + by_2) $
\s
\hskip 0.4in $= (2ax_1 + 2by_1, ax_1 + ax_2 + by_1 + by_2)   $
\s
\hskip 0.4in $=(2ax_1, ax_1 + ax_2) + (2by_1, by_1 + by_2) $
\s
\hskip 0.4in $= a(2x_1, x_1 + x_2) + b(2y_1, y_1 + y_2)$
\s
\hskip 0.4in $= af((x_1, x_2)) + bf((y_1, y_2))$

\hr

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

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

\eject

Example 4.) Given $a, b$ real numbers,\hfil \break
 $I:$ {set of all integrable functions on [a, b]} $\rightarrow R$ , 
$I(f) = \int_a^b f$
\s
Proof:  $I(sf + tg) = \int_a^b sf + tg = s \int_a^b f + t \int_a^b g = sI(f) + tI(g)$ 

\hr

Example 5.)  The inverse of a linear function is linear (when the inverse exists).
\s
Suppose $f^{-1}(x) = c$, $f^{-1}(y) = d$. 
\t
Then $f(c) = x$ and $f(d) = y$ and \hfil \break
$f(ac + bd) = af(c) + bf(d) = ax + by$.
\t
 Hence $f^{-1}(ax + by) = ac + bd = a f^{-1}(x) + bf^{-1}(y)$.

\hr

Example 6.)  $D:$ {set of all twice differential functions} $\rightarrow$ {set of all functions}, 
$L(f) = af'' + bf' + cf$
\s
Proof: \hfil \break
$L(sf + tg) = a(sf + tg)'' + b(sf + tg)' + c(sf + tg) $
\t
\hskip 0.67in $= saf'' + tag'' + sbf' + tbg' + scf + tcg $
\t
\hskip 0.67in $= s(af'' + bf' + cf) +t(ag'' + bg' + cg) $
\t
\hskip 0.67in $= sL(f) + tL(g)$

\eject

Consequence 1:  If $\phi_1, \phi_2$ are  solutions to $af'' + bf' + cf = 0$, 
then $3 \phi_1 + 5 \phi_2$ is also a solution to \break
$af'' + bf' + cf = 0$,


\vfill\vfill\vfill\vfill

\hr

Thm 3.2.2:  If $\phi_1$ and $\phi_2$ are two solutions to a \underbar{homogeneous} linear differential 
equation 
\v\u
 \centerline{$y'' + p(t)y' + q(t)y = 0$}
\u
then 
$c_1\phi_1 + c_2\phi_2$ is also a solution to this linear differential equation.
\vskip 1in
\eject



Consequence 1:  If $\phi_1, \phi_2$ are  solutions to $af'' + bf' + cf = 0$, 
then $3 \phi_1 + 5 \phi_2$ is also a solution to \break
$af'' + bf' + cf = 0$,

Proof: Since $\phi_1, \phi_2$ are  solutions to $af'' + bf' + cf = 0$, 
$L(\phi_1) = 0$ and $L(\phi_2) = 0$.

Hence $L(3 \phi_1 + 5 \phi_2) = 3 L(\phi_1) + 5 L(\phi_2)$
\u
\hskip 1.24in$ = 3(0) + 5(0) 
= 
0$.  
\u
Thus $3 \phi_1 + 5 \phi_2$ is also a solution to $af'' + bf' + cf = 
0$


\hr

Thm 3.2.2:  If $\phi_1$ and $\phi_2$ are two solutions to a \underbar{homogeneous} linear DE
\v\u
 \centerline{$y'' + p(t)y' + q(t)y = 0$  ~~(*)}\u
$c_1\phi_1 + c_2\phi_2$ is also a solution to this linear DE.

Proof:  $L(y) = y'' + p(t)y' + q(t)y$ is a linear function.

The function  $y = h(t)$ is a solution to (*) iff $L(h) = 0$.  

Since $\phi_i$  are solutions to (*), $L(\phi_i)= 0$ for $i = 1, 2$.
\v
\centerline{$L(c_1\phi_1 + c_2\phi_2) = c_1L(\phi_1) + c_2L(\phi_2) = c_1(0) + c_2(0) = 0$ }


Thus $y = c_1\phi_1 + c_2\phi_2$ is also a solution to (*).




\eject

~
\vfill \vfill\vfill\vfill\vfill

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

\eject

Section 3.5:  Solving linear non-homogeneous DE.

Example:  Solve $y'' - 4y' - 5y = 4sin(3t)$
\eject
Example (cont):  Solve $y'' - 4y' - 5y = 4sin(3t)$
\eject
Example (cont):  Solve $y'' - 4y' - 5y = 4sin(3t)$
\eject

Thm:  Suppose $c_1 \phi_1(t) + c_2 \phi_2(t)$ is a general solution to

\centerline{$ay'' + by' + cy = 0$,}

If $\psi$ is a solution to

\centerline{$ay'' + by' + cy = g(t)$ [*],}
Then $\psi + c_1 \phi_1(t) + c_2 \phi_2(t)$ is also a solution to [*].

Moreover if $\gamma$ is also a solution to [*], then there exist constants $c_1, c_2$ such that 

\centerline{$\gamma = \psi + c_1 \phi_1(t) + c_2 \phi_2(t)$}

Or in other words, $\psi + c_1 \phi_1(t) + c_2 \phi_2(t)$ is a general solution to [*].


Proof:  

Define   $L(f) = af'' + bf' + cf$.

Recall $L$ is a linear function.

 Let $h = c_1 \phi_1(t) + c_2 \phi_2(t)$.  Since $h$ is a solution to the differential 
equation, $ay'' + by' + cy = 0$,

\vskip 0.5in

Since $\psi$ is a solution to 
 $ay'' + by' + cy = g(t)$,

\vskip 0.7in

We will now show that $\psi + c_1 \phi_1(t) + c_2 \phi_2(t) = \psi + h$ is also a solution to 
[*].

\vskip 0.7in



Since $\gamma$ a solution to  $ay'' + by' + cy = g(t)$,

\vskip 0.5in

We will first show that $\gamma - \psi$ is a solution to the differential equation
 $ay'' + by' + cy = 0$.

\vskip 0.7in
\vfill

Since $\gamma - \psi$ is a solution to $ay'' + by' + cy = 0$ and 

$c_1 \phi_1(t) + c_2 \phi_2(t)$ is a general solution to

\centerline{$ay'' + by' + cy = 0$,}

there exist constants $c_1, c_2$ such that

\vskip 10pt
\centerline{$\gamma - \psi =\underline{\hskip 2in}$}


Thus
$\gamma = \psi + c_1 \phi_1(t) + c_2 \phi_2(t)$.


\end 

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

