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\pagestyle{fancy} \lhead{{\sf Homework 4 $|$ Due February 11 (Friday) }}
\rhead{\thepage} \cfoot{{\sf Math 4340/6340 $|$ Advanced Engineering
    Mathematics $|$ Spring 2022 $|$ Dr.~Macauley}}


\begin{document}

$\;$ \vspace{-6mm}

\begin{center}
  \textsc{Topics: Cauchy-Euler equations and power series solutions to ODEs}
\end{center}


%\noindent Read the following:
%\begin{itemize}
%\item Chapters 1 \& 2 of \emph{Visual Group Theory} (VGT). \vspace{-2mm}
%\item VGT Exercises 1.1--1.4, 1.8--1.12, 2.13--2.17. \vspace{-2mm}
%\item The article \emph{Group Think} by Steven Strogatz, which
%  appeared in the NY Times in 2010. \vspace{-2mm}
%\item Francis Su's essay on good mathematical writing.
%\end{itemize} 

\begin{enumerate}


\item For each of the \emph{Cauchy-Euler equations} below, look for a
  solution of the form $y(x)=x^r$, and plug this back in and find
  $r$. Find a basis of the solution space consisting of two
  real-valued functions, and use this to write the general solution.
  \begin{enumerate}
  \item $x^2y''-xy'-3y=0$
  \item $x^2y''-xy'+5y=0$
  \item $x^2y''-3xy'+4y=0$
  \end{enumerate}

\item Write each of the following as a single series of the form
  $\displaystyle\sum f(n) x^n$. That is, $f(n)$ is the coefficient
  of $x^n$. You may need to additionally ``pull out'' the first
  term(s) from one of the sums.
  \begin{multicols}{2}
    \begin{enumerate}
    \item $\displaystyle \sum_{n=0}^5 x^{n-1}$
    \item $\displaystyle \sum_{n=0}^5 x^{n+1}$
    \item $\displaystyle\sum_{n=0}^\infty na_nx^{n-1}+\sum_{n=0}^\infty
      a_n x^n$
    \item $\displaystyle \sum_{n=0}^\infty a_nx^n+\sum_{n=0}^\infty
      b_nx^{n-1}+\sum_{n=0}^\infty c_nx^{n+1}$
    \item $\displaystyle 5\sum_{n=0}^\infty n(n-1)x^{n-2}
      +\sum_{n=0}^\infty nx^{n-1}-\sum_{n=0}^\infty x^n$.
    \end{enumerate}
  \end{multicols}
  


\item Consider the ODE $y''-2xy'+10y=0$. Note that unlike the equation
  in the first problem, there will not longer be a simple solution of
  the form $x^r$. However, we know that the solution space is
  $2$-dimensional, and most ``nice'' functions can be written as a
  power series. Therefore, we'll look for a solution of the form
  $\displaystyle y(x)=\sum_{n=0}^\infty a_n x^n$.
  \begin{enumerate}
  \item Plug $y(x)$ back into the ODE and find a recurrence relation
    for $a_{n+2}$ in terms of $a_n$ and $a_{n+1}$.
  \item Explicitly write out the coefficients $a_n$ for $n\leq 9$, in
    terms of $a_0$ and $a_1$. Write down formulas for $a_{2n}$ and
    $a_{2n+1}$ in terms of $a_0$ and $a_1$.
  \item Since the solution space to this ODE is $2$-dimensional, the
    general solution you found in Part~(a) can be written as
    $y(x)=C_0y_0(x)+C_1y_1(x)$.  Find such a \emph{basis},
    $\{y_0(x),y_1(x)\}$.
  \item Find a non-zero \emph{polynomial} solution. [\textit{Hint}:
    Make a good choice for $a_0$ and $a_1$.]
  \item Are there any other polynomial solutions, excluding scalar
    multiples of the one you found in (d)? Why or why not?
  \item Consider the initial value problem
    \[
    y''-2xy'+10y=0\,,\qquad y(0)=x_0\,,\quad y'(0)=v_0\,.
    \]
    What are $x_0$ and $v_0$ in terms of the coefficients $a_n$?
  \end{enumerate}


\item The differential equation
  $(1-x^2)y''-2xy'+\nu(\nu+1)y=0$, where $\nu$ is a
  constant, is known as \textit{Legendre's equation}. It is used for
  modeling sperically symmetric potentials in the theory of Newtonian
  gravitation and in electricity and magnetism.
  \begin{enumerate}
  \item Assume that the general solution has the form $\displaystyle
    y(x)=\sum_{n=0}^\infty a_n x^n$, and find the recursion formula
    for $a_{n+2}$ in terms of $a_n$ and $a_{n+1}$.
  \item Use the recursion formula to determine $a_n$ in terms of
    $a_0$ and $a_1$, for $2\leq n\leq 9$.
  \item For each $\nu\in\N$, there will be a single (up to scalar
    multiples) nonzero polynomial solution $P_\nu(x)$, called the
    \emph{Legendre polynomial} of degree $\nu$. Find the Legendre
    polynomial of degree $\nu=3$.
  \item Find a basis for the solution space to
    $(1-x^2)y''-2xy'+12y=0$. \bigskip
  \end{enumerate}
  
  
\item The differential equation $y''-xy=0$ is called \textit{Airy's equation},
  and is used in physics to model the refraction of light.
  \begin{enumerate}
  \item Assume a power series solution, and find the recurrence
    relation of the coefficients. [\textit{Hint}: When shifting the
      indices, one way is to let $m=n-3$, then factor out $x^{n+1}$
      and find $a_{n+3}$ in terms of $a_n$. Alternatively, you can
      find $a_{n+2}$ in terms of $a_{n-1}$.]
  \item Show that $a_2=0$. [\textit{Hint}: the two series for
    $y''$ and $xy$ don't ``start'' at the same power of $x$, but for any
    solution, each term must be zero. (Why?)]
  \item Find the particular solution when $y(0)=1$, $y'(0)=0$,
    as well as the particular solution when $y(0)=0$, $y'(0)=1$. \bigskip
  \end{enumerate}



\item Consider the following initial value problem:
  \[
  y'''-y=0\,,\qquad y(0)=1\,,\qquad y'(0)=0\,,\qquad y''(0)=0\,.
  \]
  \begin{enumerate}
  \item Assume there is a solution of the form $y(x)=e^{rx}$, and
    plug this back in and solve for $r$. Use this to write the general
    solution. [\emph{Hint}: The equation $r^3=1$ has 3 distinct
      solutions over $\C$, called the \emph{3rd roots of
        unity}: $r_1=e^{0\pi i/3}=1$, $r_2=e^{2\pi i/3}$, and
      $r_2=e^{4\pi i/3}$.]
  \item From here, you could solve the initial value problem by
    plugging in the initial conditions, but you'll get a system of
    three equations with three unknowns, and involving complex
    numbers. Here's another way: look for a power series solution,
    $\displaystyle y(x)=\sum_{n=0}^\infty a_nx^n$. Plug this back into
    the ODE and find the recurrence relation for the coefficients.
  \item Compute $a_n$ for $n\leq 10$, and once you see the pattern,
    write down the general solution to the ODE as a power
    series. [\emph{Hint}: It should look familiar!] 
  \item Write down a \emph{basis} for the solution space. 
  \item Plug in the initial conditions and find the particular
    solution to the IVP.
  \item Solve a similar initial value problem:
    \[
    y'''-y=0\,,\qquad y(0)=1\,,\qquad y'(0)=1\,,\qquad y''(0)=1\,.
    \]
\end{enumerate}



\end{enumerate}

\end{document}
