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Quiz 5 ~~~SHOW ALL WORK\hfil \break
Nov 30, 2018

[13]~~ 1.)  Find all equilibrium solutions and classify them (stable, 
asymptotically stable, semi-stable, unstable and if system of DEs, 
node, saddle, spiral, center).  For the non-linear system of DEs, state all possibilities for type of equilibrium solution.  

1a.)  $y' = (y - 3)(y - 5)^8$
~~~~~
$y = 3$ is unstable, $y = 5$ is semi-stable.
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1b.)  $x' = y - 1$, ~$y' = (x-3)y$
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If $y -1= 0$, then $y =  1$

If $y = 1$, then $(x-1)y =  x-3 = 0$.  Thus $x = 3$.

Jacobian matrix: $\left[\matrix{0  & 1 \cr  y & x-3 }\right] $

For $(x, y) =  (1, 3)$, Jacobian matrix is $\left[\matrix{0  & 1 \cr  1 & 0 }\right] $

$\left|\matrix{-r & 1 \cr  1 & -r }\right| = r^2 - 1 = (r -1)(r+1) = 0$.

Thus $r = -1. 1$, i.e, one positive and one negative eigenvalue.


~~~Thus $(x(t), y(t)) = (1, 3)$ is an unstable saddle.
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[7]~~ 2.)   The slope field for a first order differential equation is shown below.
In addition to determining and classifying all equilibrium solutions (if any), also draw the trajectories satisfying the initial values $y(0) = 3$ and $y(1) = 0$.   
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~~~~~~~~{\includegraphics[width=21ex]{quiz6b.PNG}}
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\hskip 2in No equilibrium solution.
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2a.)  % {1, (x-1)(x+2)}/(100*sqrt(1 + [(x-1)(x+2)]^2))
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~~~~~~~~~{\includegraphics[width=20ex]{quiz6a.PNG}}
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\hskip 2in Unstable quilibrium solution:  $y = 2$.
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2b.)   % {1, (x-1)(x+2)}/(100*sqrt(1 + [(x-1)(x+2)]^2))
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[8]~ 3.)  The stream plot in the $x_1$-$x_2$ plane for a system of two first order differential equations is shown below.
In addition to determining and classifying the {\bf 4} equilibrium solutions, also draw the trajectory satisfying the initial value $(x_1(0), x_2(0) )=(0, -2)$.   Also describe the basins of attraction for each asymptotically stable equilibrium solutions.

~~~~~~~~~~{\includegraphics[width=25ex]{quiz6c.PNG}}

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\hskip 2.2in $ (x_1(t), x_2(t) )= (0, 0)$ is an unstable saddle. 

\hskip 2.2in $ (x_1(t), x_2(t) )= (2, 0)$ is an unstable saddle. 

\hskip 2.2in $ (x_1(t), x_2(t) )= (1, 1)$ is an asymptotically stable spiral. 

\hskip 3in basin of attraction:  $x_1 < 2$ and $x_2 > 0$.

\hskip 2.2in $ (x_1(t), x_2(t) )= (-2, -2)$ is an asymptotically stable spiral.

\hskip 3in basin of attraction:  $x_1 < 2$ and $x_2  < 0$.

[5]~ 4.)  Use  Picard's iteration method to find a degree 3  polynomial approximation for the solution to
 the initial value problem,  $y'= y + 6t^2$, $y(0) = 0$.  Start with $\phi_0(t) = 0$.


$\phi_1(t) =$ $ \int_0^t f(s, \phi_0(s))ds = \int_0^t(\phi_0(s) + 6s^2)ds = 
\int_0^t( 6s^2)ds = \int_0^t 2s^3|_0^2 = 2t^3$
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\centerline{Approximation: $\underline{~~y = 2t^3~~}$}


[7]~ 5.)  Using power series to find a degree 3  polynomial approximation for the general solution to
$y' - y= 6x^2$ for $x$ near $0$

\def\So{\Sigma_{n = 0}^\infty}
\def\S{\Sigma_{n = 1}^\infty}
\def\St{\Sigma_{n = 2}^\infty}

$y = \So a_n x ^n$, $y' = \S a_n nx^{n-1}$,

$ \S a_{n} (n)x^{n-1} - \So a_n x ^n = 6x^2$

$ \So a_{n+1} (n+1)x^{n} - \So a_n x ^n = 6x^2$

$ \So [a_{n+1} (n+1) - a_n] x ^n = 6x^2$

$a_{n+1} (n+1) - a_n = 0$ for $n = 0, 1$ and $n > 2$

$a_{n+1} = { a_n\over n+1}$ for $n = 0, 1$ and $n > 2$

For $n = 0$: ~  $a_1 = { a_0 \over 1}$ 

For $n = 1$: ~  $a_2 = { a_1\over 2} = { a_0\over 2}$

For $n = 2$: ~  $a_{3} (3) - a_2] x ^2 = 6x^2$ implies  $3a_{3}  - a_2 = 6$ and thus 
 $a_{3}  = {a_2 +6 \over 3} = {a_2 \over 3} + 2 ={a_0 \over 6} + 2  $

\v\v
\centerline{Answer: $\underline{~~y = a_0 + a_0x +{ a_0\over 2}x^2 + ({a_0 \over 6} + 2)x^3 ~~}$}

Note for this particular example, the approximation for the IVP solutions are the same for both 4 and 5.  This is NOT usually the case.  Different methods to approximate solutions will generaly give different approximations, but in the limit they should be the same.


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