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 \centerline{\large Math.\ 401, Sec.\ 500\hfill    Spring, 2005}
                                                           
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\centerline{\bfseries\large Homework 6, due March 4}
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\begin{enumerate}

 \item {\it[Bush, p.~156, (i) and (ii)]}\quad
 Find the (lowest-order) composite expansion for the solution of
\begin{enumerate}
 \item\quad $\displaystyle \epsilon y'' + y' +\frac y{x+1} 
=2$.
 \item\quad$\displaystyle \epsilon y'' - y' +\frac y{x+1} 
=2$.
\end{enumerate}
\noindent In both cases [(a) and (b)] consider 
 $$0<\epsilon\ll 1, \qquad 0<x<1, \qquad f(0)=0, \quad f(1)=3. 
$$


 \item  {\it[Bush, p.~156, (iii)--(v)]}\quad
 Find the  composite expansion for the solution of
\begin{enumerate}
 \item \quad$\displaystyle\epsilon y'' + xy' +y =0$,
$$0<\epsilon\ll 1, \qquad 2<x<4, \qquad f(2)=0, \quad f(4)=1. $$
 \item \quad $\displaystyle\epsilon y'' - xy' +y =0$,
$$0<\epsilon\ll 1, \qquad 2<x<4, \qquad f(2)=0, \quad f(4)=1. $$
\item \quad$\displaystyle \epsilon y''+xy'+y=0$,
$$0<\epsilon\ll 1, \qquad -4<x<-2, \qquad f(-4)=1, \quad f(-2)=0. 
$$
\end{enumerate}
\noindent{\bf Do not} work out (c) from the beginning;
 knowing the answers to (a) and (b), you can solve (c) in two or 
three lines by a change of variable.

 \item {\it [Bush, p.~172, (i)]}\quad
$\displaystyle\epsilon y'' -y' +\frac1y=0$,
$$0<\epsilon\ll 1, \qquad 0<x<1, \qquad f(0)=2, \quad f(1)=1. $$

  \item {\it [Bush, p.~172, (ii)]}\quad
$\displaystyle \epsilon y'' +y' +e^y=0$,
$$0<\epsilon\ll 1, \qquad 0<x<1, \qquad f(0)=1, \quad f(1)=-\ln 
2. $$


\item {\it [Logan, p.\ 69,  Ex.\ 3.2(a)]\/} 
\quad Find the lowest-order composite expansion for each sign of 
$\epsilon$.  Compare the result with the exact solution, either 
numerically or in terms of power series.
 $$\epsilon y'' + 2y' +y=0, \qquad y(0)=0, \quad y(1) =1.$$





\item {\it [Logan, p.\ 69,  Ex.\ 3.4]\/}
\quad Find the lowest-order composite expansion for each sign of 
$\epsilon$.  Compare the result with the exact solution, either 
numerically or in terms of power series.
$$\epsilon y'' - y' = 2t, 
\qquad y(0) = y(1) = 1.$$

\item {\it [Logan, p.\ 69,  Ex.\ 3.9(a)]\/}
\quad $\epsilon y'' +(t^2 +1) y' -t^3 y =0$,
$$0<\epsilon\ll 1, \qquad y(0) = y(1) = 1.$$

 
  \item {\it [Logan, p.\ 69, Ex.\ 3.9(b)]\/}
\quad $\epsilon y'' +(\cosh t)y' -y=0$,
$$0<\epsilon\ll 1, \qquad
y(0) = y(1) = 1.$$
  {\sl Hint:\/} You may find 
some help with the integral in a handbook under ``Gudermannian
function'' or ``Lobachevsky's angle of 
parallelism''\negthinspace.

\end{enumerate}

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