\magnification 2200
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\hsize 7truein
\vsize 9.2truein
\def\u{\vskip -10pt}
\def\v{\vskip -6pt}


{\bf Solving first order differential equation:}

Method 1 (sect. 2.2):  Separate variables.

Method 2 (sect. 2.1):  If linear [$y'(t) + p(t)y(t) = g(t)$], multiply equation by 
an integrating factor \hfil \break $u(t) = 
e^{\int p(t)dt}$.  
\vskip 10pt
\centerline{$y' + py = g$}
\centerline{$y'u + upy = ug$}

\centerline{$ (uy)' = ug$}

\centerline{$ \int(uy)' = \int ug$}

\centerline{$ uy = \int ug$}

\centerline{etc...}

Method 3 (sect. 2.4):  Solve Bernoulli's equation, 
$$y' + p(t)y = g(t)y^n,$$ 

when $n > 1$ by changing it
to a linear equation by
substituting $v = y^{1-n}$
\vskip 5pt

\hrule

If $v = {dx \over dt}$, can use the following to simplify (especially if there are 3 
variables).
\vskip 3pt
\centerline{${dv \over dt}= {dv \over dx}{dx \over dt} = v{dv \over dx}$}


\eject
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integration techniques:  $u$-substitution, integration by parts, partial fractions.

\vskip 5pt
\hrule


direction field = slope field = graph of ${dv \over dt}$ in $t, v$-plane.

*** can use slope field to determine behavior of $v$ including as $t 
\rightarrow \infty$.

\vskip 5pt


Equilibrium Solution = constant solution

stable, unstable, semi-stable.

%%\eject

\vskip 5pt
\hrule


{\bf Solving second order differential equation:}

p. 133: $y'' = f(t, y')$, $y'' = f(y, y')$, 


Transform to first order:  Let $v = y'$.  

If needed, note $v' = {dv \over dt} = 
{dv \over dt}{dy \over dy} = {dv \over dy}{dy \over dt} =  {dv \over dy}v$.

Note this trick sometimes helpful for first order equations.


\eject

Ch 3:  linear

$ay'' + by' + cy = 0$, ~~$y = e^{rt}$, then 
\vskip -8pt
$ar^2e^{rt} + bre^{rt} + ce^{rt} = 0$ implies $ar^2 + br + c = 0$,
\v
Suppose $r = r_1, r_2$ are solutions to $ar^2 + br + c = 0$

\centerline{$r_1, r_2 = {-b \pm \sqrt{b^2 - 4ac} \over 2a}$}


If $r_1 \not= r_2$, then $b^2 - 4ac \not= 0$.  Hence a general solution is
{$y = c_1e^{r_1t} + c_2e^{r_2t}$}
\vskip 5pt
\hrule

If $b^2 - 4ac > 0$, general solution is $y = c_1e^{r_1t} + c_2e^{r_2t}$.

\vskip 5pt
\hrule

If $b^2 - 4ac < 0$, change format to linear combination of real-valued functions instead of 
complex valued functions by using Euler's formula.

general solution is $y = c_1 e^{dt} cos (nt) + c_2 e^{dt} sin (nt)$ where $r = d \pm in$

\vskip 5pt
\hrule

If $b^2 - 4ac = 0$, $r_1 = r_2$, so need 2nd (independent) solution:  $te^{r_1t}$

Hence general solution is $y = c_1e^{r_1t} + c_2te^{r_1t}$.


\eject
To solve $ay'' + by' + cy = g_1(t) + g_2(t) + ... g_n(t)$ [**]

1.)  Find the general solution to $ay'' + by' + cy = 0$:  

\centerline{$c_1\phi_1 + c_2\phi_2$}

2.)  For each $g_i$, find a solution to $ay'' + by' + cy = g_i$:  

\centerline{$\psi_i$}

This includes plugging guessed solution into \break
$ay'' + by' + cy = g_i$ to 
find constant(s).
\vskip 10pt

The general solution to [**] is

\vskip 10pt
\centerline{$c_1\phi_1 + c_2\phi_2 + \psi_1 + \psi_2 + ... \psi_n$}

\vskip 10pt

3.)  If initial value problem:

{Once general solution is known, can solve initial value 
problem (i.e., use initial conditions to find $c_1, c_2$).}



\eject
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Thm:  Suppose that  
$f_1$ is a a solution to
$ay'' + by' + cy = g_1(t)$
and $f_2$ is a a solution to
$ay'' + by' + cy = g_2(t)$, then $f_1 + f_2$ is a solution to
$ay'' + by' + cy = g_1(t) + g_2(t)$


Proof:

Define $L(f) = af'' + bf' + cf$.  Note that $L$ is a linear function.

Since $f_1$ is a solution to 
 $ay'' + by' + cy = g_1(t)$,  $L(f_1) = af_1'' + bf_1' + cf_1 = g_1(t) $.   

Since $f_2$ is a solution to 
 $ay'' + by' + cy = g_2(t)$,  $L(f_2) = af_2'' + bf_2' + cf_2 = g_2(t)$.   

We will now show that $f_1 + f_2$ is a solution to $ay'' + by' + cy = g_1(t) + g_2(t)$.

$L(f_1 + f_2) = L(f_1) + L(f_2) = g_1(t) + g_2(t)$.  Thus $f_1 + f_2$ is a solution to $ay'' + by' + cy = g_1(t) + g_2(t)$.
 

Sidenote:  The proofs above work even if $a, b, c$ are functions of $t$ instead of constants.

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\eject

Existence and Uniqueness


{\bf 1st order LINEAR differential equation:}


Thm 2.4.1:  If $p:(a, b) \rightarrow R$ and $g:(a, b) \rightarrow R$ are continuous and $a < 
t_0 < b$, then 
there exists a unique function $y = \phi(t)$, $\phi:(a, b) \rightarrow R$ that satisfies the 
initial value problem

\centerline{$y' + p(t) y = g(t)$,} 

\centerline{$y(t_0) = y_0$}



{\bf 2nd order LINEAR differential equation:}


Thm 3.2.1:  If $p:(a, b) \rightarrow R$, $q:(a, b) \rightarrow R$, and $g:(a, b) \rightarrow R$ 
are 
continuous and $a < t_0 < b$, then
there exists a unique function $y = \phi(t)$, $\phi:(a, b) \rightarrow R$ that satisfies the
initial value problem

\centerline{$y'' + p(t) y' + q(t)y = g(t)$,}

\centerline{$y(t_0) = y_0$,}

\centerline{$y'(t_0) = y_0'$}


Definition:  The Wronskian of two differential functions, $f$ and $g$ is

\centerline{$W(f, g) = fg' - f'g = \big|\matrix{f & g \cr f' & g'}\big|$}

\vskip -7pt
sections 3.2, 3.3

\end