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CH 8:  Solve ${dy \over dt} = f(t, y)$
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***8.1:  Direction Fields ****************************
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Ch 5:  Solve ${dy \over dt} = f(t)$

Suppose $f$ is continuous
\u
 $$dy = f(t)dt$$
\u
 $$\int dy = \int f(t)dt$$
\u
$y = F(t) + C$ where $F$ is any anti-derivative of $F$.

Initial Value Problem (IVP):  $y(t_0) = y_0$
\vskip 5pt
\centerline{$y_0 = F(t_0) + C$  implies $C = y_0 - F(t_0)$}
\v

Hence unique solution (if domain connected) to IVP: 

\centerline{$y = F(t) + y_0 - F(t_0)$}

\hr


Section 8.3:  Solve ${dy \over dt} = f(y)$

If given either differential equation $y' = f(y)$  OR direction field:
\u
~~~~~~Find  equilibrium solutions and determine if stable, 
\u
~~~~~~unstable, semi-stable.

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

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8.2  Linear vs Non-linear
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nonlinear:  $y' = y^2$

linear:  $a_0(t) y^{(n)} + ... + a_n(t)y = g(t)$
\v
Ex:  $t y'' - t^3y' - 3y = sin(t)$
\v
Ex:  $ 2y'' - 3y' - 3y^2 = 0$



First order linear eqn:  $y' + p(t)y = g(t)$
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Ex 1:  $y' = ay + b$
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Ex 2:  $y' + 3t^2y = t^2$, $y(0) = 0$

Note:  could use section 8.4 method, separation of variables to solve ex 1 and 2.

Ex 3:  $t^2y' + 2ty = sin(t)$

\hr
FYI:  Solve $t^2y' + 2ty = sin(t)$ \hfil \break
(note, cannot use separation of variables).

$t^2y' + 2ty = sin(t)$

$ (t^2y)' = sin(t)$

$ \int (t^2y)' dt = \int sin(t) dt$

$ (t^2y) = -cos(t) + C$
implies
$ y = -t^{-2}cos(t) + Ct^{-2}$

\eject
FYI:  Solve $y' + p(x)y = g(x)$

Let $F(x)$ be an anti-derivative of $p(x)$

$e^{F(x)} y' + [p(x) e^{ F(x)} ] y = g(x)e^{F(x)}$

$e^{F(x)} y' + [F'(x) e^{ F(x)} ] y = g(x)e^{F(x)}$

$[e^{F(x)} y]' = g(x)e^{F(x)}$


$e^{F(x)} y = \int g(x)e^{F(x)} dx$

$ y = e^{-F(x)} \int g(x)e^{F(x)} dx$



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8.4:  Solve ${dy \over dt} = ay + b$ by separating variables:
${dy \over ay + b} = dt$
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$\int {dy \over ay + b} = \int dt$

${ln|ay + b| \over a} = t + C$


${ln|ay + b|} = at + C$


$e^{ln|ay + b|} = e^{at + C}$


$|ay + b| = e^C e^{at}$


$ay + b = \pm(e^C e^{at})$

$ay  = C e^{at} - b$
implies
$y  = C e^{at} - {b \over a}$


\eject
Ex 2:  $y' + 3t^2y = t^2$, $y(0) = 0$

 ${dy \over dt} + 3t^2y = t^2$

${dy \over dt} = t^2 - 3t^2y$

${dy} = (t^2 - 3t^2y)dt$

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

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

$-{1 \over 3} ln|1 - 3y| = {t^3 \over 3} + C$ 

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

$ |1 - 3y| = e^{-t^3  + C}$ \hfill  
$|1 - 3y|^{-{1 \over 3}} =  e^{-{t^3 \over 3}  + C}$

$ |1 - 3y| = e^Ce^{-t^3} $ \hfill  
$|1 - 3y|^{-{1 \over 3}} =   e^Ce^{{t^3 \over 3}}$

$ 1 - 3y = \pm e^Ce^{-t^3} $ \hfill  
$(|1 - 3y|^{-{1 \over 3}})^{-3} =   (e^Ce^{t^3 \over 3})^{-3}$

$ 1 - 3y = Ce^{-t^3} $  \hfill  
$|1 - 3y|=   (e^{-3C}e^{-t^3})$

$ - 3y = Ce^{-t^3}  - 1$ \hfill
$ 1 - 3y = Ce^{-t^3} $ 

$ y = {1 - Ce^{-t^3}  \over 3} = {1 \over 3} - Ce^{-t^3}$


**************Existence of  solution**************

Ch 5)  $y' = f(t)$, solution exists if $f$ is continuous \hfil \break
$y = F(t) + C$ where $F$ is any anti-derivative of $f$.

8.2)  linear:  $y' + p(x)y = g(x)$, solution exists if $p$ and $g$ are continuous

Ch 8):  $y' = f(t, y)$, solution may or may not exist.


Ex:  $y' = y' + 1$
\v
Ex:  $(y')^2 = -1$

IVP ex: $ {dy \over dx} = y (1 + {1  \over x})$, $y(0) = 1$ 

%%$ {dy \over y} =  (1 + {1  \over x})dx$

$ \int {dy \over y} =  \int (1 + {1  \over x})dx$

$ ln|y| = x + ln|x| + C$

$ |y| = e^{x + ln|x|  + C} = e^x e^{ ln|x|} e^C = C|x|e^x  = Cxe^x$

$y = \pm Cxe^x$ 
implies
$y = Cxe^x$

 $y(0) = 1$:  ~~~$1 = C(0)e^0 = 0$ implies 
\vskip 10 pt
\centerline{IVP   $ {dy \over dx} = y (1 + {1  \over x})$, 
$y(0) = 1$ has no solution.}

See direction field created using \hfil \break www.math.rutgers.edu/$\sim$ sontag/JODE/JOdeApplet.html

%%\rightline{\includegraphics[width = 2.2in]{slopefieldIVPnosoln}}  %% export as pdf %%file 

%%\epsfbox{slopefieldIVPnosoln.eps}
\eject
***************Uniqueness of solution***************

Given an initial value problem,

Ch 5)  $y' = f(t)$, $y(t_0) = y_0$:   if $f$ continuous, then on appropriate domain, unique solution $y = F(t) + y_0 - F(t_0)$.


8.2)  linear:  $y' + p(x)y = g(x)$,  then on appropriate domain, unique solution if $p$ and $g$ are continuous .

Ch 8):  $y' = f(t, y)$, solution may or may not be unique.

Ex: $ y' = y^{1 \over 3}$  

Note $y = 0$ is a solution to $ y' = y^{1 \over 3}$  since $y' = 0 = 0^{1 \over 3} =  y^{1 \over 3}$  


Suppose $y \not= 0$.  Then ${dy \over dx} = y^{1 \over 3}$ 
  implies
$y^{-{1 \over 3}}dy = dx$ 

$\int y^{-{1 \over 3}}dy = \int dx$ 
implies ${3 \over 2} y^{{2 \over 3}} = x  + C$ 

$y^{{2 \over 3}} = {2 \over 3}x  + C$ implies
$y = \pm  \sqrt{ ({2 \over 3}x  + C)^3}$ 

Suppose $y(3) = 0$.  Then $0 =  \sqrt{(2 +  C)^3}$ implies $C = -2$.

Thus initial value problem, $ y' = y^{1 \over 3}$,   $y(3) = 0$, has 3 
sol'ns:
\vskip 5pt
\centerline{$y = 0$,  ~~$y =  \sqrt{ ({2 \over 3}x - 2)^3}$, ~~$y = - 
\sqrt{ ({2 \over 3}x - 2)^3}$}  



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