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



\begin{document}




Math 3600 Differential Equations Exam \#2 \hfil \break
Oct 26, 2018 \hfill SHOW ALL STEPS \hfill

 1.)  Circle T for true and F for false.  

[4]~ 1a.) Suppose $f(x) = \Sigma a_n (x - 3)^n$ has a radius of convergence = $r$ about  $
3$.
Then
we can define the domain of $f$ to be $(3- r, 3 + r)$.
\u
\rightline{T~~~~~~~~~~~F}  
\v
[4]~ 1b.) If $b^2 - 4ac < 0$, then the solution to the initial value problem 
$ay'' +
by' + cy = 0$, $y(0) = 2$, $y'(0) = 1$ is a complex valued function.
\u
\rightline{T~~~~~~~~~~~F}  
\v

[4]~ 1c.)  If $b^2 - 4ac < 0$, then the solution to the characteristic equation $ar^2 + br + c = 0$ is complex valued.
\u
\rightline{T~~~~~~~~~~~F}  
\v

[4]~ 1d.) $D(f) = f'$ is a linear function.
\vskip -10pt
\rightline{T~~~~~~~~~~~F}  
\v

[4]~ 1e.)  There is a unique solution to the differential equation %\hfil \break
 $ay'' + by' + cy = g(t), y(0) = 1, y(1) = 0$

\rightline{T~~~~~~~~~~~F}  
\v\v
[7]~ 2.)  The  eigenvalues of $ \left(\matrix{ 3 & -2 \cr 1 & 5}\right)$ are  $\underline{\hskip 4in}$


%\vfill

\vskip 2.5in



%\centerline{Answer: $\underline{~~\hskip 5in~~}$}


[7] 3.)  Suppose
$A   \left[\matrix{4 \cr 12}\right] = \left[\matrix{-3 \cr 11}\right]$,
$A   \left[\matrix{1 \cr 7}\right] = \left[\matrix{3 \cr 21}\right]$,
$A   \left[\matrix{-2 \cr 2}\right] = \left[\matrix{9 \cr 31}\right]$,
$A   \left[\matrix{3 \cr 5}\right] = \left[\matrix{-6 \cr -10}\right]$
\v
State the 2 eigenvalues of $A$:

\vskip 30pt
State 5 eigenvectors of $A$:


\vskip 30pt


\eject
[20]~ 4.)  Using power series, find a degree 5  polynomial approximation for the solution to
$y'' - y = 4x$ for $x$ near $0$

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

\centerline{Approximation: $\underline{\hskip 6in}$}


\eject

[22]~ 5.)  Solve $y'' - y = e^t + 2$, ~~$y(0) = 1$, $y'(0) = 2$

%$y = At + B, $y ' = A, y'' = 0$.  


%$r^2 -1 - 0,$ 


%$y = Ate^t$, 

\vfill

\centerline{Solution: $\underline{\hskip 6in}$}


\eject
[24]~  6.)  Solve {\bf two} of the following (from this page and the next page).  If you solve all 4, I will grade your best 2 and will give 1 (or 2) points extra credit for 3 (or 4) correct problems):

6a.)  If $y = \psi(t)$ is a solution to $py'' + qy' + ry = g(t)$, show that $y = 2\psi(t)$ is a solution to $py'' + qy' + ry = 2g(t)$.
Hint use linearity OR plug in.
\vfill\vfill
6b.)  Use your work in problem 5 to solve  $y'' - y = 3e^t + 10$ for the general solution.

\vfill

6c.)  Given $a_0$, $a_1$ and $a_{n+2} = 2a_{n+1} - a_n$,  determine $a_n$ in terms of $a_0$ and $a_1$.

%$a_2 =  2a_1 - a_0$

%$a_3 =  2a_2 - a_1=  2( 2a_1 - a_0) - a_1 = 3a_1 - 2a_0$

%$a_4 =  2a_3 - a_2=  2(3a_1 - 2a_0) - (2a_1 - a_0) = 4a_1 -3a_0$

%$a_5 =  2a_4 - a_3=  2(4a_1 -3a_0) - (3a_1 - 2a_0) = 5a_1 -4a_0$

%$a_{n+2} = 2a_{n+1} - a_n = a_{n+2} = 2(2a_{n} - a_{n-1}) - ( 2a_{n-1} - a_{n-2})$

\vfill\vfill\vfill\vfill
\eject
6d.)  Use the ratio test to determine the radius of convergence for the power series
$\So {3^n \over 2n -1}x^n$.  For what values of $x$ does this series converge?
\vfill\vfill

%$ {3^{n+1} \over 2(n+1) -1}x^{n+1}}  {(2n -1)x^n  \over3^n }= {3(2n-1)x^n \over  2(n+1) -1}$


\end{document}


$y = \So a_n x ^n$, $y' = \S a_n nx^{n-1}$,  $y'' = \St a_n n(n-1)x^{n-2}$.

$ \St a_n n(n-1)x^{n-2} -  \So a_n x ^n = 4x$

$ \St a_n n(n-1)x^{n-2} - \So a_n x ^{n} = 4x$

$ \So a_{n+2} (n+2)(n+1)x^{n} - \So a_n x ^{n} = 4x$

$ \So [a_{n+2} (n+2)(n+1) - a_n] x ^{n} = 4x$


Thus $a_{n+2} (n+2)(n+1) - a_n = 0$ for $n \not= 1$. Thus $a_{n+2} = \frac{a_n}{(n+2)(n+1)}$

For $n = 1$:  $a_3(3)(2) - a_1 = 4$.  Thus $a_3 = \frac{a_1 + 4}{6}$
 

$a_2 = {a_0 \over 2}$
$a_4= {a_2 \over 12} = $
 

Approximation:  $y = a_0 + a_1x + {a_0 \over 2}x^2 +  \frac{a_1 + 4}{6}x^3 + $
------------------------------------------

%$a_{n+2} = {-a_n \over 3(n+1)}= $a_{n+2} = {(-1)^n a_{n-2} \over 3^2(n+1)(n-1)}$

%$a_2 = {a_1 - a_0 \over 3}$

%$a_3 = {a_2 - a_1 \over 3(2)} = {({a_1 - a_0 \over 3}) - a_1 \over 3(2)}
= {(a_1 - a_0) - 3a_1 \over 3^2(2)} 
= {- 2a_1 - a_0  \over 3^2(2)}   $

%$a_4 = {a_3 - a_2 \over 3(3)} =  {({(- 2a_1 - a_0  } ) - 6({a_1 - a_0 }) \over 3(3)3^2(2)}
=  {({- 2a_1 - a_0 } ) - 6({a_1 - a_0 }) \over 3(3)3^2(2)}
= {- 8a_1 + 5a_0   \over 3(3)3^2(2)}$

%$a_5 = {a_4 - a_3 \over 3(4)} = 
{({- 8a_1 + 5a_0   \over 3(3)3^2(2)}) -( {- 2a_1 - a_0  \over 3^2(2)}) \over 3(4)}
= {({- 8a_1 + 5a_0}) -9( {- 2a_1 - a_0  }) \over 3(4) 3(3)3^2(2)}$



1.)  The solution to $y'' + 16y =  36cos(2t) $ ~is~ $y = c_1cos(4t) +  c_2sin(4t) + 3cos(2t)$
\hfil \break
Use this fact to answer the following two questions.



[5]~ 1a.)  Guess a possible non-homog soln for the following differential equation 
(do
\hfill\break
 not solve):  ~~ $y'' + 16y = 3sin(4t) -  e^{4t} $
\vfill
\v

[3]~ 1b.)  The general solution to  ~$y'' + 16y = cos(2t) +  32$ ~is
\vfill
\v

Proof







\e






\e


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

[6]~~ 3a.)  Show that $x = 0$ is a regular singular point of the above differential equation.

\vskip 80pt


[6]~~ 3b.)  The indicial equation is \underbar{\hskip 2in}

[2]~~ 3c.)  The roots are the indicial equation are \underbar{\hskip 2in}

[6]~~ 3d.)  After plugging in the larger root, the recurrence relation is \underbar{\hskip 2in}

\def\h{\hskip 0.7in}

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

[6]~~ 3f.)  Use your answer to 3e to give an approximation of one solution to the above differential equation. 

Answer for 3f:  \underbar{\hskip 5in}


\e



\end{document}



29

recur  20

ch 7:  30



