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\usepackage{graphicx}
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%\AppendGraphicsExtensions{.gif}
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\def\hb{\hfil \break}
\def\So{\sum_{n=0}^\infty}
\def\S{\sum_{n=1}^\infty}

\def\Sn{\sum_{i=1}^n}
\def\Sno{\sum_{i=0}^n}

\def\v{\vskip 10pt}
\def\e{\eject}

%15 + 24 + 20 + 20 + 21  = 15 + 85

%1) 14
%2) 14
%3) 18
%4)  6

%32 + 20 = 52

%3 + 4 + 1 + 5 + 3 + 4  = 20

%6) 10 + 4 + 4
%7) 10

%28


%80

%8)  10

%36


%80 + 18   + 2

\begin{document}




MATH:3600:0002 Final Exam \hfil \break
Dec. 12, 2016 \hfill SHOW ALL STEPS \hfill


[6] ~1.)  The  Wronskian of $e^{-t}cos(t)$,  $e^{-t}sin(t)$ is 
~$W(e^{-t}cos(t),  e^{-t}sin(t)) = \underline{~~e^{-2t}~~}$



$W(e^{-t}cos(t),  e^{-t}sin(t)) = \left|\matrix{e^{-t}cos(t) &  e^{-t}sin(t) \cr 
 - e^{-t}sin(t)  - e^{-t}cos(t) &  e^{-t}cos(t)  -e^{-t}sin(t)}\right|$
\v
\hskip 1.3in $= e^{-t}cos(t)[e^{-t}cos(t)  -e^{-t}sin(t)]
- e^{-t}sin(t) [ -e^{-t}sin(t)  - e^{-t}cos(t) ]$
\v
\hskip 1.3in $= e^{-2t}cos^2(t)  -e^{-2t}cos(t)sin(t)
+  e^{-2t}sin^2(t)  + e^{-2t}cos(t) sin(t)$
\v
\hskip 1.3in $= e^{-2t}[cos^2(t) 
+  sin^2(t)] = e^{-2t}$



\vfill

[14]~  2.)  Solve $t \frac{dy}{dt} + y = 8 t^2$

$t y' + y = 8 t^2$ is a first order linear DE.

Short method:   


$[ty]' = t y' + y = 8 t^2$ 

$[ty]' = 8 t^2$ 


$\int [ty]' = \int 8 t^2$

$ty = \frac{8}{3}t^3 + C$

Thus $y = \frac{8}{3}t^2 + Ct^{-1}$

Longer method:

$ y' + \frac{1}{t}y = 8 t$

Integrating factor:  $u(t) = e^{\int \frac{dt}{t}} = e^{ln|t|} = |t|$

Multiply both sides  by $t$:  ~~ $t y' + y = 8 t^2$ 
and continue as above.


\vfill
\centerline{Answer: \underbar{~~ $y = \frac{8}{3}t^2 + Ct^{-1}$~~}}

\e

[14]~ 3.) Solve  $(3x^4 + 2y)dx + (2x +4y^3)dy= 0$



$M_y = 2$  $N_x = 2$

If $\psi_x = 3x^4 + 2y$, then $\psi = \int(3x^4 + 2y)dx = \frac{3}{5}x^5 + 2xy + h(y)$.

$\psi_y = 2x + h'(y) =2x +4y^3 $.  Thus $h'(y) = 4y^3$ and $h(y) = y^4$


Answer:  $\frac{3}{5}x^5 + 2xy + y^4 = C$

Check:   $\frac{d}{dx}(\frac{3}{5}x^5 + 2xy + y^4) = \frac{d}{dx}(C)$

   ${3}x^4 + 2y + 2x\frac{dy}{dx} + 4y^3\frac{dy}{dx} = 0$

\vfill
\centerline{Answer: \underbar{\hskip 5in}}
\e

[18]~ 4.)  Solve $y''' - 4y' = 30e^{3t} + 8t$, $y(0) = 0$, $y'(0) = 6$, $y''(0) = 16$.

Solve homogeneous: $y''' - 4y' = 0$

$y = e^{rt}$:   $r^3 - 4r = r(r^2 - 4) = r(r - 2)(r + 2) = 0$.  Thus $r = 0, 2, -2$

General homogeneous solution:  $y = c_1 e^{0t} + c_2 e^{2t} + c_3 e^{-2t}$

Guess:  $y = Ae^{3t}$:  $y''' - 4y' = 27Ae^{3t} - 12e^{3t} = 15Ae{3t} = 30e^{3t}$.  Thus $A = 2$.


Guess $y = At^2$.  Thus $y' = 2At$ and $y''' = 0$.
$0 - 4(2At) = 8t$.  Thus $A = -1$ 

Answer:  $y = y(t) = c_1 e^{2 t} + c_2 e^{-2 t} + c_3 - t^2 + 2 e^{3 t}$


$0 =  c_1  + c_2  + c_3 + 2$

$6 =  2c_1  - 2c_2  + 6$

$16 =  4c_1  + 4c_2  -2 + 18$


Answer:  $y = y(t) =   -2 - t^2 + 2 e^{3 t}$
\vfill
\centerline{Answer: \underbar{\hskip 5in}}





\e






[10]~ 5a.)  Solve  ${\bf x}' = \left(\matrix{ 4 & 5 \cr 4 & 3}\right){\bf x}$




Find eigenvalues:

$\left|\matrix{ 4 - r & 5 \cr 4 & 3 -r}\right| = (4-r)(3-r) - 20
= r^2 - 7r + 12 - 20 = r^2 -7r - 8 = (r-8)(r+1) = 0$

$r = 8, -1$.

Find eigenvectors:

$r = 8$:  $\left(\matrix{ 4 - 8 & 5 \cr 4 & 3 -8}\right)$

$\left(\matrix{ -4 & 5 \cr 4 & -5}\right)
\left(\matrix{  5 \cr 4 }\right) = \left(\matrix{ 0 \cr 0}\right)$

$r = -1$:  $\left(\matrix{ 4 +1 & 5 \cr 4 & 3 +1}\right)$

$\left(\matrix{ 5 & 5 \cr 4 & 5}\right)
\left(\matrix{  ~~1 \cr -1 }\right) = \left(\matrix{ 0 \cr 0}\right)$



\vfill
\centerline{Answer:  $\underline{~~y = c_1e^{8t}\left(\matrix{  5 \cr 4 }\right)  + c_2e^{-t}\left(\matrix{  ~~1 \cr -1 }\right) ~~}$}

\v
[4]~  5b.)  Find all equilibrium solutions and determine whether the critical point is asymptotically stable, stable, or unstable.  Also classify it as to type (nodal source, nodal sink, saddle point, spiral source, spiral sink, center).

\centerline{Equilibrium solution: $\underline{\left(\matrix{  x_1 \cr x_2 }\right) = \left(\matrix{  0 \cr 0 }\right) }$}
\v\v

\centerline{~~~Stability : $\underline{~~unstable~~}$ \hfill Type: $\underline{~~saddle~~}$~~~}

\v
[4]~  5c.)   Plot a phase portrait for this system
\v

\centerline{\includegraphics[width = 2.7in]{grapht.eps}}

\e

[10]~  6.)  Given the recursive relation  $a_n = { 4a_{n-1} \over  (n+1)n }$ where $a_0 = 1$,  

use induction to prove that $a_n = { 4^n  \over  (n+1)!n!}$. 


Proof outline (thus this answer is very incomplete):

${ 4^0  \over  (1)!0!} = 1 = a_0$



$a_n = { 4a_{n-1} \over  (n+1)n }$
$ = { 4( 4^{n-1})  \over  n!(n-1)!(n+1)n }$
$ = { 4^{n}  \over  (n+1)!n! }$


\e


7.) Given the differential equation: $xy'' + 4y = 0$





[3]~~ 7a.)  Show that $x = 0$ is a regular singular point of the above differential equation.

\vskip 40pt


[4]~~ 7b.)  The indicial equation is \underbar{\hskip 2in}

[1]~~ 7c.)  The roots are the indicial equation are \underbar{\hskip 2in}

[5]~~ 7d.)  After plugging in the larger root, the recurrence relation is \underbar{\hskip 2in}

\def\h{\hskip 0.7in}

[3]~~ 7e.)  For the larger root, if $a_0 = 1$, then $a_1 = $ \underbar{\h}, ~~
$a_2 = $ \underbar{\h}, ~~$a_3 = $ \underbar{\h}.

[4]~~ 7f.)  Use your answer to 5e to give an approximation of one solution to the above differential equation. 

Answer for 7f:  \underbar{\hskip 5in}


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


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


Let $y = \So a_n x^{n+r}$

$xy'' + 4y = x\So a_n (n+r)(n+r-1)x^{n+r-2} + 4\So a_n x^{n+r}$

~~~~~~~~~~~~$ = \So a_n (n+r)(n+r-1)x^{n+r-1} + \So 4 a_n x^{n+r}$

~~~~~~~~~~~~$ = \So a_n (n+r)(n+r-1)x^{n+r-1} + \S 4a_{n-1} x^{n+r-1}$

~~~~~~~~~~~~$ = a_0 r(r-1)x^{r-1} + \S a_n (n+r)(n+r-1)x^{n+r-1} + \S 4a_{n-1} x^{n+r-1}$

~~~~~~~~~~~~$ = a_0 r(r-1)x^{r-1} + \S [a_n (n+r)(n+r-1) +  4a_{n-1}] x^{n+r-1}$

$r(r-1) = 0$ implies $r = 0, 1$

$a_n (n+r)(n+r-1) +  4a_{n-1} = 0$  implies 
$a_n = { -4a_{n-1} \over  (n+r)(n+r-1) }$

Sidenote:  When $r = 0$, $n = 1$: 
$a_1 (0) +  4a_{0} = 0$ implies $a_0 = 0$, a contradiction.


When $r = 1$,  $a_n = { 4a_{n-1} \over  (n+1)n }$

$n = 1$:~~ $a_1 = { -4a_{0} \over  2 } = -2a_0$


$n = 2$:~~ $a_2 = { -4a_{1} \over  (3)2 } =   { 4^2a_{0} \over  3!2 } = {4a_0 \over 3} $

$n = 3$:~~ $a_3 = { -4a_{2} \over  (4)3 } = {- 4^3a_{0}  \over  (4)3(2)3! } = { - 4^3a_{0}  \over  4!3! }  = {-  4^2a_{0}  \over  3!3! }  
= { - 4a_{0}  \over  9 } $


$n = 4$:~~ $a_4 = { -4a_{3} \over  (5)4 } = { 4^4a_{0}  \over  (5)4 (4!3! ) } = { 4^4a_{0}  \over  5!4!}  $


\end{document}
[10]~ 8.) Choose one of the following three problems:

8A.)  Finish problem

8B.)  9.3

8C.)  Prove.





















\end{document}

[15]~  1.)  Solve $y'' - 4y' + 13y = 2 + 9e^{2t}$,  ~~$y(0) = \frac{15}{13}$, $y'(0) = 8$ 

%$r^2 $

%$4 \pm \sqrt{16 - 4(

%4(4 - 13
\vfill
\centerline{Answer:  $\underline{\hskip 5in}$}

\eject

[12]~2a.)  Solve:  $t(y' - 4y) = te^{2t}$

\vfill
\centerline{Answer:  $\underline{\hskip 5in}$}

2b.)    Does this  differential equation  have any equilibrium solutions?  If so, state what they are and whether they are stable, semi-stable or unstable.

\vskip 0.4in

[12]~ 3a.)  Solve $\frac{y'}{(y^2 - 9)} = 1$
\vfill
\centerline{Answer:  $\underline{\hskip 5in}$}

3b.)    Does this  differential equation  have any equilibrium solutions?  If so, state what they are and whether they are stable, semi-stable or unstable.

\vskip 0.5in

% 2.)   Circle T for True and F for false.
%\rightline{T \hskip 0.8in F}

\eject

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




4.) Given the differential equation $ 3xy'' - (2 + x)y = 0$,

[5]~  i.)  Determine if $x = 0$ is an ordinary point, regular singular point or irregular singular point.


%$lim_{x \rightarrow 0} \frac{-(1 + x)x}{x} =  lim_{x \rightarrow 0} -(1 + x)$
Answer: $ y'' - \left( \frac{2 + x}{3x}\right)y = 0$ implies $x = 0$ is a singular point since setting denominators = 0 implies $3x = 0$ implies $x = 0$.

$lim_{x \rightarrow 0} 0 =  
0$


$lim_{x \rightarrow 0} -\frac{x^2(1 + x)}{3x} = lim_{x \rightarrow 0} -\frac{x(1 + x)}{3} = 0$

    Since both limits are finite, $x = 0$ is a regular singular point.  



\vskip 70pt


[15]~ ii.) Determine the indicial equation, the roots of the indicial equation, and the recurrence relation.


Let $y = \So a_n x^{n + r}$ where $a_0 \not= 0$.

Then $y' = \So (n+r) a_n x^{n+r-1}$, and $y'' = \So (n+ r)(n+ r-1) a_n x^{n+r-2}$,

 $ 3xy'' - (2 + x)y = 3x \So (n+ r)(n+ r-1) a_n x^{n+r-2} - (2+x) \So a_n x^{n + r}$ 
 
 $\hskip 1.2in =  \So 3(n+ r)(n+ r-1) a_n x^{n+r-1} -  \So 2a_n x^{n + r} - 
 \So a_n x^{n + r+1}$ 
 
  $\hskip 1.2in =  {\sum_{n=-1}^\infty} 3(n+1+ r)(n+ r) a_{n+1} x^{n+r} -  \So 2a_n x^{n + r} - 
 \S a_{n-1} x^{n + r}$ 
 
 
   $ =   3(r)( r-1) a_{0} x^{r-1}  + 3(1+ r)( r) a_{1} x^{r} 
    -  2a_0 x^{r}$ 
  \vskip 5pt 
\rightline{$    +  \S 3(n+1+ r)(n+ r) a_{n+1} x^{n+r} -  \S 2a_n x^{n + r} - 
 \S a_{n-1} x^{n + r}$ }
 
 
    $ =   3r( r-1) a_{0} x^{r-1}  + [3r(1+ r) a_{1} - 2a_0]x^{r}$ 
 {$    +  \S[3(n+1+ r)(n+ r) a_{n+1} -   2a_n - 
  a_{n-1}] x^{n + r}$ }

 
$  3r(1+ r) a_{1} - 2a_0 = 0$ implies $a_1 = \frac{2a_0}{3r(1 + r)}$
\vfill




indicial equation: $\underline{\hskip 2.5in}$ \hfill 
 roots of the indicial equation: $\underline{\hskip 1in}$
\v\v\v\v
recurrence relation: $\underline{\hskip 5in}$
\eject

[10]~ 5a.)  Solve  ${\bf x}' = \left(\matrix{1 & -2 \cr 4 & -5}\right){\bf x}$

%$ \left(\matrix{1 - \lambda & -2 \cr 4 & -5 - \lambda}\right)$

%$(1 - \lambda)(-5 - \lambda) + 8 = \lambda^2 + 4\lambda - 5 + 8= \lambda^2 + 4\lambda + 3 $$= (\lambda + 1)(\lambda + 3)= 0$


% $ \left(\matrix{1 +1 & -2 \cr 4 & -5 +1}\right)$ =$\left(\matrix{2 & -2 \cr 4 & -4}\right)$
 
%$ \left(\matrix{1 +3 & -2 \cr 4 & -5 +3}\right)$ =$  \left(\matrix{4 & -2 \cr 4 & -2}\right)$


\vfill
\centerline{Answer:  $\underline{\hskip 5in}$}

[5]~  5b.)  Find all equilibrium solutions and determine whether the critical point is asymptotically stable, stable, or unstable.  Also classify it as to type (nodal source, nodal sink, saddle point, spiral source, spiral sink, center).

\centerline{Equilibrium solution: $\underline{\hskip 3in}$}
\v\v

\centerline{~~~Stability : $\underline{\hskip 3in}$ \hfill Type: $\underline{\hskip 3in}$~~~}

[5]~  5c.)   Plot a phase portrait for this system

\centerline{\includegraphics[width = 3.2in]{100/grapht.eps}}

\eject

[5]~  6a.) Show that $L({\bf x}) = {\bf x}' - A{\bf x}$ is a linear function.
\vfill\vfill
[8]~ 6b.)  Suppose ${\bf x} = {\bf \phi_1}(t)$ and ${\bf x} = {\bf \phi_2}(t)$ are solutions to the homogeneous linear equation 

${\bf x}' - A{\bf x} = 0$.  
Then $L(\phi_i) = \underline{\hskip .5in}$


Show that ${\bf x} = c_1\phi_1 + c_2\phi_2$ is also a solution to ${\bf x}' - A{\bf x} = 0$ where $c_1, c_2$ are real numbers. 

\vfill\vfill
[8]~  6c.)  Suppose ${\bf x} = \psi(t)$ is a solution to ${\bf x}' - A{\bf x} = {\bf b}(t)$.  

Then $L(\psi) = \underline{\hskip .5in}$

Show that ${\bf x} = c_1\phi_1(t) + c_2\phi_2(t) +  \psi(t)$ is also a solution to ${\bf x}' - A{\bf x} = {\bf b}(t)$.

\vfill\vfill

\end{document}

15 + 20 + 20 + 21 =   76
24

1-3           Ch 1:  1-3
4 - 10       CH 2:  1 - 7
11 - 16      Ch 3 1 -6
17            4.1
18 - 23      7.1 - 6
24 - 26       9/1 - 3


\eject
--------------------------------------------


[13]~ 1.)   Solve:  $ty' + 6t^2y - t^2 = 0$, $y(1) = 3$


$y' + 6ty  = t$, $\int 6tdt = 3t^2$


$e^{3t^2}y' + 6te^{3t^2} = $

$[e^{3t^2}y]' = te^{3t^2}$
\vfill

\centerline{Answer: \underline{\hskip 3in}}

\eject

[13]~ 2.)  Solve:  $(2x^3y^2 - 3x)dx + (x^4y - y^{-1}) dy = 0$


Let $\psi_x = 2x^3y^2 - 3x$.  Then $\psi = \frac{x^4y^2 - 3x^2}{2} + h(y)$

$psi_y = x^4y + h'(y) = x^4y - y^{-1}$

$\int h'(y)dy = \int -y^{-1}dy = -ln|y|$ 

$ \frac{x^4y^2 - 3x^2}{2} -ln|y| = C$


\vfill

\centerline{Answer: \underline{\hskip 3in}}
\eject
[14]~ 3.)  Use the infinite series solution method (ch 5) about the point $x_0 = 0$ to solve the differential equation, $(x+1) y' - 2y = 0$.

Note:  You must use the infinite series method.  However, this is still a relatively short problem as most terms will cancel out (so no need for induction proof/ratio test).


Suppose $y = \So a_nx^n$.  Then $y' = \S na_n x^{n-1}$


 $(x+1)  \S na_n x^{n-1} - 2\So a_nx^n = 0$.


 $  \S (x+1)na_n x^{n-1} - 2\So a_nx^n = 0$.

 $\S na_n x^{n} + \S na_n x^{n-1} - 2\So a_nx^n = 0$.

$\S na_n x^{n} + \So (n+1)a_{n+1} x^{n} - 2\So a_nx^n = 0$.

$\S na_n x^{n} + \S (n+1)a_{n+1} x^{n} - 2\S a_nx^n  + a_1 - 2a_0= 0$.

$\S [na_n + (n+1)a_{n+1}  - 2a_n]x^n  + a_1 - 2a_0= 0$.

hence $a_1 - 2a_0= 0$ and $na_n + (n+1)a_{n+1}  - 2a_n = 0$ for all $n \geq 1$

hence $a_1 = 2a_0$ and $a_{n+1} = \frac{(2-n)a_n}{n+1}$ for all $n \geq 1$

Thus $a_2 = \frac{(2-1)a_1}{2} =  \frac{(a_1}{2} =  \frac{(2a_0}{2} = a_0 $.

Thus $y = a_0x^2 + 2a_0x + a_0$

Check $y' = 2a_0x + 2a_0$

 $(x+1) ( 2a_0x + 2a_0) - 2(a_0x^2 + 2a_0x + a_0)$.

 $(x+1) ( a_0x + a_0) - (a_0x^2 + 2a_0x + a_0)$.
 
  $a_0x^2 + a_0x + a_0x + a_0  - a_0x^2 - 2a_0x - a_0 = 0$.
  
 .
 
\vfill

\centerline{Answer: \underline{\hskip 3in}}
\eject

[20]~ 4.)  Solve:  $ y'' - 3y' - 4y = 8t$, $y(0) = 5$, $y'(0) = 3$


$r^2 - 3r - 4 = 0$  

$(r - 4)(r + 1) = 0$.  Thus $r = 4, -1$.

general homogeneous solution:  $y = c_1e^{4t} + c_2e^{-t}$


Let $y = At + B$.  Then $y' = A$ and $y'' = 0$

$-3A - 4(At + B) = 8t$

$A = -2$, $6 - 4B = 0$.  Thus $B = \frac{3}{2}$


Thus $\psi = -2t + \frac{3}{2}$ is a non-homogeneous solution.

General solution to non-homogeneous DE:   $y = c_1e^{4t} + c_2e^{-t} - 2t + \frac{3}{2}$

$y' =  4c_1e^{4t} - c_2e^{-t} - 2$

$5 = c_1 + c_2 + \frac{3}{2}$

$3 = 4c_1 - c_2 - 2$





\vfill

\centerline{Answer: \underline{\hskip 3in}}
\eject

[20]~ 5.)  Suppose $\frac{dx}{dt} = (x+3)(y+2)$ and $\frac{dy}{dt} = 2x - y $. 

(a.)  State all critical points of this system of differential equations  \underline{\hskip 3in} 

(b.)  State two solutions to this system of differential equations.
\\
\\
\\
  
(c.)  For the critical point with positive $x$ coordinate, find the corresponding linear system.\hb
\vfill
\vfill
(d.)  Based on the linear translated approximation found in part b, what conclusions regarding stability can you draw about the nonlinear system near the critical point with positive $x$ coordinate?  
\vfill


\eject


[20]~  6.)  Choose 2 from the following three problems (6A, 6B, 6C).  Clearly indicate your TWO choices. You may do more than 2
problems in which case I may substitute your unchosen problem for one of your two choices
(with a penalty) if it improves your grade.

Your two choices:   \underline{\hskip 3in}



6A.)  Suppose $y = \phi_1(t)$, $y = \phi_2(t)$, and $y = \phi_3(t)$ are all solutions to the homogeneous \break
differential equation $y''' + p(t)y'' + q(t)y' + r(t)y = 0$ and suppose that $y = \psi(t)$ is a solution to  $y''' + p(t)y'' + q(t)y' + r(t)y = g(t)$.  Prove that $y = c_1\phi_1(t) + c_2\phi_2(t) +  c_3\phi_3(t) + \psi(t)$ is a solution to  $y''' + p(t)y'' + q(t)y' + r(t)y = g(t)$.
\eject

6B.)  Suppose $a_n = \frac{-2a_{n-1}}{n+1}$ for all $n \geq 1$.  Use induction to prove that $a_n = 
\frac{(-2)^n a_0}{(n+1)!}$ for all $n \geq 0$
\vfill
\vfill
\vfill
6C.)  Transform the given initial value problem into an equivalent problem with initial point at the origin:
\vskip -5pt
\centerline{$\frac{dy}{dt} = 2t - 3(y+1)^2$, $y(8) = 10$}
\vfill
\vfill
\eject

[5 ~ Extra Credit]  Prove the following theorem for up to 5 points extra credit.


Suppose for each $i = 1, ..., n$, $y =  \phi_i(t) $ are solutions to the homogeneous linear differential equation $\Sno p_i(t) y^{(i)}(t) = 0$.  Suppose also that $\{\phi_1, ..., \phi_n\}$ are linearly independent.  Show there is a unique choice of constants $c_i$ such that $\Sn c_i \phi_i$ is a solution to the initial value  problem $\Sno p_i(t) y^{(i)} = 0$, $  y^{(i)}(t_0) = y_i$, $i = 0, ..., n-1$.  

Note you may earn partial credit by proving special cases (eg $n = 1$ or $n = 2$) or just for starting the proof, stating definitions, etc..

\eject
~
\eject
~


\end{document}

Suppose $M(x, y) + N(x, y)y' = 0$ is an exact differential equation.  Prove that this equation has a solution if $M, N, M_y, N_x$ are continuous.
