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\pagestyle{fancy} \lhead{{\sf Homework 4 $|$ Due February 14
    (Tuesday) }} \rhead{\thepage} \cfoot{{\sf MthSc 434 $|$
    Advanced Engineering Mathematics $|$ Spring 2012 $|$ M.~Macauley}}


\begin{document}

$\;$

\begin{enumerate}

\item For each of the Cauchy-Euler equations, look for a solution of
  the form $y(t)=t^r$, and plug this back in to determine $r$. Find a
  basis of the solution space consisting of two real-valued functions,
  and use this to write the general solution.
  \begin{enumerate}
  \item $t^2y''-ty'-3y=0$
  \item $t^2y''-ty'+5y=0$
  \item $t^2y''-3ty'+4y=0$
  \end{enumerate}
  
  
\item Solve the following differential equations. You may freely use
  your answers to the previous problem without having to re-derive
  them.
  \begin{enumerate}
  \item $t^2y''-ty'-3y=6$
  \item $(t+2)^2y''-(t+2)y'-3y=0$
  \item $(t+2)^2y''-(t+2)y'-3y=6$.
  \end{enumerate}
  
  
\item For each of the Cauchy-Euler equations, make the substitution
  $x=\ln t$ and write the resulting differential equation in $y(x)$
  instead of $y(t)$. [No need to solve them.]
  \begin{enumerate}
  \item $t^2y''-ty'-3y=0$
  \item $t^2y''-ty'+5y=0$
  \item $t^2y''-3ty'+4y=0$
  \end{enumerate}
  
  
\item Write each of the following as a single series of the form
  $\displaystyle\sum f(n) t^n$. That is, $f(n)$ is the coefficient
  of $t^n$. You may need to additionally ``pull out'' the first
  term(s) from one of the sums.
  \begin{enumerate}
  \item $\displaystyle \sum_{n=0}^5 t^{n-1}$
  \item $\displaystyle \sum_{n=0}^5 t^{n+1}$
  \item $\displaystyle\sum_{n=0}^\infty na_nt^{n-1}+\sum_{n=0}^\infty
    a_n t^n$
  \item $\displaystyle \sum_{n=0}^\infty a_nt^n+\sum_{n=0}^\infty
    b_nt^{n-1}+\sum_{n=0}^\infty c_nt^{n+1}$
  \item $\displaystyle 5\sum_{n=0}^\infty n(n-1)t^{n-2}
    +3\sum_{n=0}^\infty nt^{n-1}-2\sum_{n=0}^\infty t^n$.
  \end{enumerate}
  

\item Consider the ODE $y''-2ty'+10y=0$. Note that unlike the equation
  in the first problem, there will not longer be a simple solution of
  the form $t^r$. 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
  $\displaystyle y(t)=\sum_{n=0}^\infty a_n t^n$.
  \begin{enumerate}
  \item Plug $y(t)$ 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
    $y(t)=C_0y_0(t)+C_1y_1(t)$.  Find such a \emph{basis},
    $\{y_0(t),y_1(t)\}$.
  \item Find a non-zero \emph{polynomial} solution for this
    ODE. [\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
    \[
    y''-2ty'+10y=0\,,\qquad y(0)=x_0\,,\quad y'(0)=v_0\,.
    \]
    What are $x_0$ and $v_0$ in terms of the coefficients $a_n$?
  \end{enumerate}


\item The differential equation
  $(1-t^2)y''-2ty'+\lambda(\lambda+1)y=0$, where $\lambda$ 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 $\displaystyle
    y(t)=\sum_{n=0}^\infty a_n t^n$, 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 $\lambda\in\N$, there will be a single (up to scalar
    multiples) nonzero polynomial solution $P_\lambda(t)$, called the
    \emph{Legendre polynomial} of degree $\lambda$. Find the Legendre
    polynomial of degree $\lambda=3$.
  \item Find a basis for the solution space to $(1-t^2)y''-2ty'+12y=0$.
  \end{enumerate}
  
  
\item The differential equation $y''-ty=0$ is called \textit{Airy's equation},
  and is used in physics to model the refraction of light.
  \begin{enumerate}
  \item Assume the solution is a power series, 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 $t^{n+1}$
      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 $ty$ don't ``start'' at the same power of $t$, 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 $y(0)=0$, $y'(0)=1$.
  \end{enumerate}

  
\end{enumerate}

\end{document}
