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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\l{\lambda}
\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~~~~~~~~~~~}  
\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{~~~~~~~~~~~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~~~~~~~~~~~}  
\v

[4]~ 1d.) $D(f) = f'$ is a linear function.
\vskip -10pt
\rightline{T~~~~~~~~~~~}  
\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{~~~~~~~~~~~F}  
\v\v
[7]~ 2.)  The  eigenvalues of $ \left(\matrix{ 3 & -2 \cr 1 & 5}\right)$ are  $\underline{~~4 \pm i~~}$


$ \left|\matrix{ 3 - \l & -2 \cr 1 & 5 - \l}\right| = (3 - \l)(5 - \l) + 2 = 15 - 8\l + \l^2 + 2 = \l^2 - 8\l + 17$ 

$\l = \frac{8 \pm \sqrt{8^2 - 4(17)}}{2} = \frac{8 \pm 2\sqrt{2(8) - 17}}{2}
 = 4 \pm \sqrt{- 1} = 4 \pm i$

%\vfill



\vskip .1in



%\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
$A   \left[\matrix{1 \cr 7}\right] = \left[\matrix{3 \cr 21}\right] = 3\left[\matrix{1 \cr 7}\right]$,

$A   \left[\matrix{3 \cr 5}\right] = \left[\matrix{-6 \cr -10}\right] = -2 \left[\matrix{3 \cr 5}\right] $

State the 2 eigenvalues of $A$:

3, -2

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

$ \left[\matrix{1 \cr 7}\right], \left[\matrix{2 \cr 14}\right] ,  \left[\matrix{-1 \cr -7}\right], \left[\matrix{-3 \cr -21}\right],  \left[\matrix{3 \cr 5}\right]$, etc. 



\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}

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

$ \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$


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

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

 For $n = 0$:   $a_{2} = \frac{a_0}{(2)(1)}$

For $n = 2$:   $a_{4} = \frac{a_2}{(4)(3)} =\frac{a_0}{(4)(3)(2)(1)} $

For $n = 3$:   $a_{5} = \frac{a_3}{(5)(4)} = \frac{a_1 + 4}{6(5)(4)} $

\vskip 15pt
 

\centerline{Approximation:  $y = a_0 + a_1x + {a_0 \over 2}x^2 +  \frac{a_1 + 4}{6}x^3 +\frac{a_0}{4!}x^4 
+\frac{a_1 + 4}{120} x^5$}



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


Solve homogeneous:  Guess $y = e^{rt}$ and plug into $y'' - y =  0$:  $r^2e^{rt} - e^{rt} = 0$. 

 ~~~Thus $r^2 - 1 = (r+1)(r-1)= 0$.  Thus $r = 1, -1$.

~~~Homogeneous solution:  $c_1e^t + c_2e^{-t}$


Solve $y'' - y = e^t$


~~~$y = e^t$ is a homogeneous solution, so guess $y = Ate^t$.  Then $y' = Ae^t + Ate^t$ and 
$y'' = Ae^t +  Ae^t +Ate^t =  2Ae^t +Ate^t $.

~~~Plug into $y'' - y = e^t$:

~~~$2Ae^t +Ate^t - Ate^t = e^t$ implies $2Ae^t = e^t$.  Thus $2A = 1$ and $A = \frac{1}{2}$.

~~Thus $y = \frac{1}{2}te^t$ is one solution to $y'' - y = e^t$


Solve $y'' - y = 2$ 

~~~Guess $y = B$, then $y ' = 0, y'' = 0$.  

~~~Plug in:  $0 - B = 2$.  Thus $B = -2$.

~~Thus $y = -2$ is one solution to $y'' - y = 2$


Hence general solution to $y'' - y = e^t + 2$ is $y = c_1e^t + c_2e^{-t} +  \frac{1}{2}te^t - 2$



%Solve IVP:  $y(0) = 1$, $y'(0) = 2$. % (see next page)

%\eject




Solve IVP:  $y(0) = 1$, $y'(0) = 2$.

$y = c_1e^t + c_2e^{-t} +  \frac{1}{2}te^t - 2$

$y' = c_1e^t - c_2e^{-t} +  \frac{1}{2}te^t +  \frac{1}{2}e^t$

$y(0) = 1$:  $1 = c_1 + c_2 - 2$ implies  $3 = c_1 + c_2$

$y'(0) = 2$:  $2 =  c_1 - c_2 +  \frac{1}{2}$ implies $ \frac{3}{2} =  c_1 - c_2$

Add equations: $ \frac{9}{2} = 2c_1$.  Thus $c_1 =\frac{9}{4} $ 

Subtract equations:  $ \frac{3}{2} = 2c_2$.  Thus $c_2 =\frac{3}{4} $ 


\vskip 10pt

\centerline{Solution: $\underline{~~ y = \frac{9}{4}e^t + \frac{3}{4}e^{-t} +  \frac{1}{2}te^t - 2~~}$}


\vfill
\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.

{\bf Using linearity:}   Recall that $L(y) = py'' + qy' + ry $ is a linear function.  

Since  $y = \psi(t)$ is a solution to $py'' + qy' + ry = g(t)$, ~$L(\psi(t)) = g(t)$.  Since $L$ is a linear function, $L(2\psi(t)) =2L(\psi(t)) =  2g(t)$.  Thus $y = 2\psi(t)$ is a solution to $py'' + qy' + ry = 2g(t)$.


{\bf Plugging in:}  Since  $y = \psi(t)$ is a solution to $py'' + qy' + ry = g(t)$,  
~~  $p\psi''(t) + q\psi'(t) + r\psi(t) = g(t)$.   

Thus  $p[2\psi''(t)] + q[2\psi'(t)] + r[2\psi(t)] =  2[p\psi''(t) + q\psi'(t) + r\psi(t)] =2g(t)$.

 Thus $y = 2\psi(t)$ is a solution to $py'' + qy' + ry = 2g(t)$.


\vfill\vfill
6b.)  Use your work in problem 5 to solve  $y'' - y = 3e^t + 10$ for the general solution.


Homogeneous solution:  $c_1e^t + c_2e^{-t}$


Since  $y = \frac{1}{2}te^t$ is one solution to $y'' - y = e^t$, 
 ~$y = \frac{3}{2}te^t$ is one solution to $y'' - y = 3e^t$


Since $y = -2$ is one solution to $y'' - y = 2$, ~$y = -10$ is one solution to $y'' - y = 10$

Thus general solution to 
 $y'' - y = 3e^t + 10$~ is ~~$y = c_1e^t + c_2e^{-t} +  \frac{3}{2}te^t - 10$

\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 $a_1$.

$a_2 =  2a_1 - a_0$

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

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

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

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


Answer:  ~~$a_n = na_1 - (n-1)a_0$

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?

\u\vskip -5pt
$$\lim_{n \rightarrow \infty} \left| \left({3^{n+1}x^{n+1} \over 2(n+1) -1}\right)  \left({2n -1  \over3^n x^n}\right)\right| =\lim_{n \rightarrow \infty} \left|{3(2n-1)x \over  2(n+1) -1}\right| =\lim_{n \rightarrow \infty}\left|{3(2n-1)x \over  2n+1}\right| =
|3x|   \lim_{n \rightarrow \infty}  {2n-1\over  2n+1} = |3x| < 1$$
\vskip -5pt

Thus $|x| < \frac{1}{3}$.  Thus radius of convergence is  $\frac{1}{3}$ and the series converges for all $x \in ( -\frac{1}{3},  \frac{1}{3})$ and the series diverges if $|x| > \frac{1}{3}$ 

If $x = \frac{1}{3}$:   $\So {3^n \over 2n -1}x^n$ = $\So {3^n \over 2n -1}( \frac{1}{3})^n$
= $\So {1 \over 2n -1} > \So {1 \over 2n} =  {1 \over n} \So {1 \over n}$

Since $\So {1 \over n}$ diverges,  $\So {3^n \over 2n -1}( \frac{1}{3})^n$ diverges.


If $ x= -\frac{1}{3}$:    $\So {3^n \over 2n -1}x^n$ = $\So {3^n \over 2n -1}( -\frac{1}{3})^n$
= $\So {(-1)^n \over 2n -1}. $  
Since $ \lim\limits_{n \rightarrow \infty}{1 \over 2n -1} = 0$ and  $ {1 \over 2n -1}$ is a decreasing sequence, $\So {(-1)^n \over 2n -1}$ 
converges by the alternating series test.
%= -1 -1 + {1 \over 3} - {1 \over 5} + {1 \over 7} - {1 \over 9} + ...$

%Recall $\S {(-1)^n  \over n} = -1 + {1 \over 2} - {1 \over 3} + {1 \over 4} - {1 \over 5} + ...$ converges.


Thus the series $\So {3^n \over 2n -1}x^n$ converges for all $x \in [ -\frac{1}{3},  \frac{1}{3})$.

\end{document}



------------------------------------------

%$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)}   $

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

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







