\magnification 2000
\parindent 0pt
\parskip 10pt
\pageno=1
\hsize 7.5truein
\vsize 10.8truein
\hoffset -0.5truein
\voffset -0.87truein
\def\u{\vskip -10pt}
\def\v{\vskip -6pt}
\def\s{\vskip 4pt}

\def\t{\vskip -4pt}

\def\hb{\hfil \break}
\def\b{\hfill \break}
\def\hr{\vskip 7pt \hrule \vskip -3pt}
\input graphicx


\def\s{\vskip -1pt}




\def\u{\vskip -10pt}
\def\v{\vskip -6pt}
\def\s{\vskip -1pt}
\def\t{\vskip -4pt}
\def\b{\hfill \break}
\def\hr{\vskip 7pt \hrule \vskip -3pt}
\input epsf

To solve 1rst order linear DE: 
%\centerline
{ $1y' + p(t)y = g(t)$ }

Create product rule by using integrating factor:  $u(t) = e^{\int p(t)dt}$


Ex 1:  $ty' + 9y = {e^{2t} \over t^5}$  ~~~~~$\Rightarrow$~~~~~~
$1y' + {9 \over t}  y = {e^{2t} \over t^6}$

Create product rule by using integrating factor: 

 $u(t) = e^{\int p(t)dt}  = e^{\int  {9 \over t}  dt}
 = e^{9 ln(t)  dt} = e^{ln(t)^9} = t^9$

$t^9(y' + {9 \over t}  y) = t^9({e^{2t} \over t^6})$

$t^9y' + {9t^8 }  y = t^3e^{2t}$

$(t^9y)' = t^3e^{2t}$
~~~~~$\Rightarrow$~~~~~~
$\int (t^9y)' = \int  t^3e^{2t}dt$

$t^9y = \int  t^3e^{2t}dt = t^3e^{2t} - \int 3t^2{e^{2t} \over 2}dt$

$t^9y =   t^3{e^{2t} \over 2} - {3t^2e^{2t} \over 4} + {6te^{2t} \over 8} 
- {6e^{2t} \over 16} + C $

General solution:
$y =   {e^{2t} \over 2t^6} - {3e^{2t} \over 4t^7} + {3e^{2t} \over 4t^8} 
- {3e^{2t} \over 8t^9} + {C \over t^9} $

Use integration by parts on RHS

$u = t^3 ~~~~~~~~~~dv = e^{2t}$

$du = 3t^2~~~~~~~~~~v = {e^{2t} \over 2}$

$d^2u = 6t~~~~~~~~~~\int v = {e^{2t} \over 4}$

$d^3u = 6~~~~~~~~~~\int\int v = {e^{2t} \over 8}$

$d^3u = 0~~~~~~~~~~\int\int\int v = {e^{2t} \over 16}$


{ Ex 2:  See ex 1 in https://homepage.math.uiowa.edu/~idarcy/\hfil \break
COURSES/100/SPRING21/2\_1notesPM.pdf}

Ex3:  $2y' - {4 \over t}y = {6t^2 \over (t-1)(t+2)}   $ 

$1y' - {2 \over t}y = {3t^2 \over (t-1)(t+2)}   $ 

Create product rule by using integrating factor: 

 $u(t) = e^{\int p(t)dt}  = e^{\int  {-2 \over t}  dt}
 = e^{-2 ln(t)  dt} = e^{ln(t)^{-2}} = t^{-2}$


$t^{-2}(y' - {2 \over t}y) = t^{-2}({3t^2 \over (t-1)(t+2)})   $ 

$t^{-2}y' - {2 \over t^{-3}}y = {3 \over (t-1)(t+2)} $

$(t^{-2}y)'  = {3 \over (t-1)(t+2)} $

$\int (t^{-2}y)'  = \int {3 \over (t-1)(t+2)} dt$

$t^{-2}y  = \int {3 \over (t-1)(t+2)} dt$


Integration by partial fractions:

 ${3 \over (t-1)(t+2)} = {A \over t-1} + {B \over t + 2}$

$3 = A(t+2) + B(t-1)$ 

Method 1 for finding $A$ and $B$:

Let $t = 1$:   ~~~~$3 = 3A$ and thus $A = 1$

Let $t = -2$:  ~~~~$3 = -3B$ and thus $B = -1$

Method 2 for finding $A$ and $B$:

$0t +  3 = (A+ B)t+2A - B$

Comparing like terms:  

$A + B = 0$ and \hfill \break
$2A - B = 3$

$3A = 3$ and thus $A = 1$

$B = -A = -1$


$t^{-2}y  = \int {3 \over (t-1)(t+2)}dt  =  \int ({1 \over t-1} - {1 \over t + 2})dt
= ln|t-1| - ln|t+2| + C$

General solutions:  $y = t^2ln|t-1| - t^2ln|t+2| + Ct^2$




\eject



{\bf Solve via separation of variables (since not linear):}   

\vskip 5pt

\centerline{$y' = {1 \over 2 - 2t -y + ty}$}

${dy \over dt}  = {1 \over 2(1 - t) -y(1 - t)}$

${dy \over dt} = {1 \over (1 - t)(2 - y)}$
\v
$\int(2 - y)dy = \int {dt \over 1 - t}$

$2y - {y^2 \over 2} = -ln|1 - t| + C$

$y^2 - 4y - 2ln|1 - t| + C = 0$

$y = {4 \pm \sqrt{16 + 4(2ln|1 - t| + C)} \over 2}$
$= 2  \pm \sqrt{4 + 2ln|1 - t| + C}$
\vskip 10pt
\centerline{General solution:  $y= 2  \pm \sqrt{2ln|1 - t| + C}$}

IVP:  $y(0) = 1$

$1= 2  \pm \sqrt{2ln|1 - 0| + C}$

$1 - 2=  \pm \sqrt{2ln|1| + C}$

$-1=  \pm \sqrt{C}$

Choose negative sign since both sides must be negative.

$-1=  - \sqrt{ C}$. ~~~~ Thus $C = 1$

\vskip 10pt
\centerline{IVP solution:  $y= 2  - \sqrt{2ln|1 - t| + 1}$}


{\bf Find domain:} 
\hb $2ln|1 - t| + 1 \geq 0$ and $t \not= 1$ and $y \not= 2$

{\bf NOTE:  the convention in this class to to choose largest possible connected domain where tangent line to solution is never vertical.}

$2ln|1 - t|  \geq -1$ and $t \not= 1$ and $y \not= 2$ implies

$ln|1 - t| > -{1 \over 2}$ 

 $|1 - t| > e^{-{1 \over 2}}$  since $e^x$ is an increasing function.


 $1 - t < -e^{-{1 \over 2}}$ or  $1 - t > e^{-{1 \over 2}}$


 $- t < -e^{-{1 \over 2}} - 1$ or  $-t > e^{-{1 \over 2}}- 1$

 Domain: $\cases{  t > e^{-{1 \over 2}} + 1 & if $t_0 > 1$ \cr
t < -e^{-{1 \over 2}} + 1 &  if $t_0 < 1$.}$

Since $t_0 = 0$, Domain is $t < -e^{-{1 \over 2}} + 1$


\end
Note:  Domain is much easier to determine when the ODE is linear.

\hr
 
{\bf Find C given $y(t_0) = y_0$:}  $y_0 =  2  \pm \sqrt{2ln|1 - t_0| + C}$

$\pm(y_0 -2) =    \sqrt{2ln|1 - t_0| + C}$

$(y_0 -2)^2 - 2ln|1 - t_0| = C$

$y = 2  \pm \sqrt{2ln|1 - t| + C}$


$y = 2  \pm \sqrt{2ln|1 - t| + (y_0 -2)^2 - 2ln|1 - t_0|}$
 
$y = 2  \pm \sqrt{ (y_0 -2)^2 + ln{(1 - t)^2 \over (1-t_0)^2}}$


 Domain: $\cases{  t > e^{-{C \over 2}} + 1 & if $t_0 > 1$ \cr
t < -e^{-{C \over 2}} + 1 &  if $t_0 < 1$.}$

$e^{-{C \over 2}} = e^{-{(y_0 -2)^2 - 2ln|1 - t_0| \over 2}}
= |1 - t_0|{e^{-{(y_0 -2)^2  \over 2}} }$

 Domain: $\cases{  t > 1 + |1 - t_0|{e^{-{(y_0 -2)^2  \over 2}} } & if $t_0 > 1$ \cr
t < 1 - |1 - t_0|{e^{-{(y_0 -2)^2  \over 2}} } &  if $t_0 < 1$.}$

%\includegraphics[width=20ex]{grapht.eps}

%\eject
\end
\t
Ex 3 (IVP): $ {dy \over dx} = y (1 + {1  \over x})$, $y(0) = 1$ 
\t
%%$ {dy \over y} =  (1 + {1  \over x})dx$

$ \int {dy \over y} =  \int (1 + {1  \over x})dx$
~~~implies~~~
$ln|y| = x + ln|x| + C$
\t
$ |y| = e^{x + ln|x|  + C} = e^x e^{ ln|x|} e^C = C|x|e^x  = Cxe^x$
\t
$y = \pm Cxe^x$ 
implies
$y = Cxe^x$
\t
 $y(0) = 1$:  ~~~$1 = C(0)e^0 = 0$ implies 
\vskip 6pt
\centerline{IVP   $ {dy \over dx} = y (1 + {1  \over x})$, 
$y(0) = 1$ has no solution.}


\hr
\end

2.4 \#27b.  Solve Bernoulli's equation, 
\vskip 5pt
\centerline{$y' + p(t)y = g(t)y^n,$}
\t
when $n \not= 0, 1$ by changing it 
\vskip 5pt
\centerline{$y^{-n}y' + p(t)y^{1-n} = g(t)$}
\t
when $n \not = 0, 1$ by changing it 
to a linear equation by 
substituting $v = y^{1-n}$

\vskip 5pt
\hrule
\t
Example: Solve $ty' + 2t^{-2}y = 2t^{-2}y^5$

\eject
Section 2.5:  Autonomous equations: $y' = f(y)$
\v\hr
Example:  Exponential Growth/Decay 

\centerline{Example:  population growth/radioactive decay}
\vskip 5pt
\centerline{$y' = ry$, $y(0) = y_0$ implies $y = y_0 e^{rt}$}

$r > 0$ \hskip 1.2in $r < 0$
\vskip 0.5in
\hr
Example:  Logistic growth:  $y' = h(y)y$

\vskip 5pt
\centerline{Example:  $y' = r(1 - {y \over K})y$}

$y$ vs $f(y)$ \hskip 1in slope field:

\vskip 0.6in
\vfill

Equilibrium solutions:


As $t \rightarrow \infty$, if $y > 0$, $y \rightarrow $


Solution:  $y = {y_0K \over y_0 + (K - y_0)e^{-rt}}$

\eject


Section 2.5  Autonomous equations: $y' = f(y)$


If given either differential equation $y' = f(y)$ 

 \rightline{ OR direction field:}
\vskip -5pt
~~~~~~Find  equilibrium solutions and determine if 

\centerline{stable, unstable, semi-stable.}

~~~~~~Understand what the above means.

\hr
Asymptotically stable:
\vfill\vfill
\hr
Asymptotically unstable:
\vfill\vfill
\hr
Asymptotically semi-stable:

\vfill


\end




\end