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\begin{document}

\begin{center}
{\Large Math 2233 \\[0pt]
Solutions to Homework Set 9}
\end{center}

1. Combine each of the following power series expressions into a single
power series.

\begin{enumerate}
\item[(a)] $\displaystyle{\sum_{n=1}^{\infty
}(n+1)(x-1)^{n-1}+\sum_{n=0}^{\infty }n(x-1)^{n}}$

\begin{itemize}
\item First we shift the summation index of the first power series up by 1
(remembering to shift its starting point down by 1); 
\begin{equation*}
{\sum_{n=1}^{\infty }(n+1)(x-1)^{n-1}+\sum_{n=0}^{\infty
}n(x-1)^{n}=\sum_{n=0}^{\infty }(n+2)(x-1)^{n}+\sum_{n=0}^{\infty }n(x-1)^{n}%
}
\end{equation*}
Now both power series start off at the same value of $n$ and involve the
same powers of $\left( x-1\right) $; so we can combine them into a single
power series by simply adding the coefficients of like powers of $(x-1)$.
Hence, 
\begin{eqnarray*}
{\sum_{n=1}^{\infty }(n+1)(x-1)^{n-1}+\sum_{n=0}^{\infty }nx^{n}} &=&{%
\sum_{n=0}^{\infty }}\left( (n+2)+n\right) {(x-1)^{n}} \\
&=&\sum_{n=0}^{\infty }(2n+2)(x-1)^{n}
\end{eqnarray*}
\end{itemize}

\item[(b)] $\displaystyle{\sum_{n=0}^{\infty
}(n+1)a_{n+2}x^{n+1}+\sum_{n=0}^{\infty }na_{n}x^{n-1}}$

\begin{itemize}
\item This problem can be handled several different ways. We could shift the
summation index of the first series down by 2, or we could shift the
summation index of the second series up by 2, or we could shift the first up
by 1 and the second down by 1. Let's do it the first way 
\begin{eqnarray*}
\sum_{n=0}^{\infty }(n+1)a_{n+2}x^{n+1}+\sum_{n=0}^{\infty }na_{n}x^{n-1}
&=&\sum_{n=2}^{\infty }(n-1)a_{n}x^{n-1}+\sum_{n=0}^{\infty }na_{n}x^{n-1} \\
&=&\sum_{n=2}^{\infty
}(n-1)a_{n}x^{n-1}+0a_{0}x^{-1}+1a_{1}x^{0}+\sum_{n=2}^{\infty }na_{n}x^{n-1}
\\
&=&a_{1}+\sum_{n=2}^{\infty }\left( (n-1)a_{n}+na_{n}\right) x^{n-1} \\
&=&a_{1}+\sum_{n=2}^{\infty }(2n-1)a_{n}x^{n-1}
\end{eqnarray*}
In the last step we shifted the summation index again just so that it would
be easy to identify the total coefficient of the $n^{th}$ power of $x$ in
terms of $n$.
\end{itemize}

\item[(c)] $\displaystyle{(x-1)\sum_{n=0}^{\infty
}na_{n}x^{n-1}+\sum_{n=0}^{\infty }a_{n}x^{n}}$

\begin{itemize}
\item 
\begin{eqnarray*}
{(x-}1{)\sum_{n=0}^{\infty }{}_{n}}na_{n}{x^{n-1}+\sum_{n=0}^{\infty }}a_{n}{%
x^{n}} &=&{x\sum_{n=0}^{\infty }}na{_{n}x^{n-1}-}\sum_{n=0}^{\infty
}na_{n}x^{n-1}{+\sum_{n=0}^{\infty }}a{_{n}}x{^{n}} \\
&=&{\sum_{n=0}^{\infty }}na{_{n}x^{n}-}\sum_{n=0}^{\infty }na_{n}x^{n-1}{%
+\sum_{n=0}^{\infty }}a{_{n}}x{^{n}} \\
&=&{\sum_{n=0}^{\infty }}na{_{n}x^{n}-}\sum_{n=-1}^{\infty }(n+1)a_{n+1}x^{n}%
{+\sum_{n=0}^{\infty }}a{_{n}}x{^{n}} \\
&=&{\sum_{n=0}^{\infty }}na{_{n}x^{n}-}\left(
(-1+1)a_{-1+1}x^{-1}+\sum_{n=0}^{\infty }(n+1)a_{n+1}x^{n}\right) \\
&&{+\sum_{n=0}^{\infty }}a{_{n}}x{^{n}} \\
&=&{\sum_{n=0}^{\infty }}na{_{n}x^{n}-}\sum_{n=0}^{\infty }(n+1)a_{n+1}x^{n}{%
+\sum_{n=0}^{\infty }}a{_{n}}x{^{n}} \\
&=&\sum_{n=0}^{\infty }\left( na_{n}-(n+1)a_{n+1}+a_{n}\right) x^{n} \\
&=&\sum_{n=0}^{\infty }\left( n+1\right) (a_{n}-a_{n+1})x^{n}
\end{eqnarray*}
\end{itemize}

\item[(d)] $\displaystyle{x\sum_{n=0}^{\infty
}na_{n}(x-1)^{n-1}+\sum_{n=0}^{\infty }a_{n}(x-1)^{n}}$

\begin{itemize}
\item 
\begin{eqnarray*}
x\sum_{n=0}^{\infty }na_{n}(x-1)^{n-1}+\sum_{n=0}^{\infty }a_{n}(x-1)^{n}
&=&x\sum_{n=0}^{\infty }na_{n}(x-1)^{n-1}+\sum_{n=0}^{\infty }a_{n}(x-1)^{n}
\\
&=&\left[ (x-1)+1\right] \sum_{n=0}^{\infty
}na_{n}(x-1)^{n-1}+\sum_{n=0}^{\infty }a_{n}(x-1)^{n} \\
&=&\sum_{n=0}^{\infty }na_{n}(x-1)^{n}+\sum_{n=0}^{\infty
}na_{n}(x-1)^{n-1}+\sum_{n=0}^{\infty }a_{n}(x-1)^{n} \\
&=&\sum_{n=0}^{\infty }na_{n}(x-1)^{n}+\sum_{n=-1}^{\infty
}(n+1)a_{n+1}(x-1)^{n}+\sum_{n=0}^{\infty }a_{n}(x-1)^{n} \\
&=&\sum_{n=0}^{\infty }na_{n}(x-1)^{n}+\left(
(0)a_{0}(x-1)^{-1}+\sum_{n=0}^{\infty }(n+1)a_{n+1}(x-1)^{n}\right) \\
&&+\sum_{n=0}^{\infty }a_{n}(x-1)^{n} \\
&=&\sum_{n=0}^{\infty }na_{n}(x-1)^{n}+\sum_{n=0}^{\infty
}(n+1)a_{n+1}(x-1)^{n}+\sum_{n=0}^{\infty }a_{n}(x-1)^{n} \\
&=&\sum_{n=0}^{\infty }\left( na_{n}+(n+1)a_{n+1}+a_{n}\right) (x-1)^{n} \\
&=&\sum_{n=0}^{\infty }\left( (n+1)a_{n+1}+(n+1)a_{n}\right) (x-1)^{n}
\end{eqnarray*}

\qed
\end{itemize}

\item[(e)] $\displaystyle{x^{2}\sum_{n=0}^{\infty }n(n-1)a}_{n}{(x-1)^{n-2}} 
$
\end{enumerate}

\begin{itemize}
\item We shall begin by calculating the Taylor expansion of $x^{2}$ about $%
x_{o}=1.$%
\begin{eqnarray*}
x^{2} &=&f(x)=f(1)+f^{\prime }(1)(x-1)+\frac{f^{\prime \prime }(1)}{2!}%
(x-1)^{2}+\frac{f^{\prime \prime \prime }(1)}{3!}(x-1)^{3}+\cdots \\
&=&1+2(x-1)+\frac{2}{2!}(x-1)^{2}+\frac{0}{3!}(x-1)^{3}+\frac{0}{4!}%
(x-1)^{4}+\cdots \\
&=&1+2(x-1)+(x-1)^{2}
\end{eqnarray*}
Therefore 
\begin{eqnarray*}
x^{2}\sum_{n=0}^{\infty }n(n-1)a_{n}(x-1)^{n-2} &=&\left(
1+2(x-1)+(x-1)^{2}\right) \sum_{n=0}^{\infty }n(n-1)a_{n}(x-1)^{n-2} \\
&=&\sum_{n=0}^{\infty }n(n-1)a_{n}(x-1)^{n-2}+2(x-1)\sum_{n=0}^{\infty
}n(n-1)a_{n}(x-1)^{n-2} \\
&&+(x-1)^{2}\sum_{n=0}^{\infty }n(n-1)a_{n}(x-1)^{n-2} \\
&=&\sum_{n=0}^{\infty }n(n-1)a_{n}(x-1)^{n-2}+\sum_{n=0}^{\infty
}2n(n-1)a_{n}(x-1)^{n-1} \\
&&+\sum_{n=0}^{\infty }n(n-1)a_{n}(x-1)^{n} \\
&=&\sum_{n=-2}^{\infty }(n+2)(n+1)a_{n+2}(x-1)^{n}+\sum_{n=-1}^{\infty
}2(n+1)(n)a_{n+1}(x-1)^{n} \\
&&+\sum_{n=0}^{\infty }n(n-1)a_{n}(x-1)^{n} \\
&=&\left( 0+0+\sum_{n=0}^{\infty }(n+2)(n+1)a_{n+2}(x-1)^{n}\right) \\
&&+\left( 0+\sum_{n=0}^{\infty }2(n+1)(n)a_{n+1}(x-1)^{n}\right) \\
&&+\sum_{n=0}^{\infty }n(n-1)a_{n}(x-1)^{n} \\
&=&\sum_{n=0}^{\infty }\left[ (n+2)(n+1)a_{n+2}+2n(n+1)a_{n+1}+n(n-1)a_{n}%
\right] (x-1)^{n}
\end{eqnarray*}

\item \qed
\end{itemize}

2. Find the recursion relations for the power series solutions $%
y(x)=\sum_{n=0}^\infty a_n(x-x_o)^n$ of the following differential equations

\begin{enumerate}
\item[(a)] $y^{\prime \prime }-xy^{\prime }-y$, $x_{o}=0$

\begin{itemize}
\item Setting 
\begin{equation*}
y(x)=\sum_{n=0}^{\infty }a_{n}\left( x-0\right) ^{n}=\sum_{n=0}^{\infty
}a_{n}x^{n}
\end{equation*}
into the differential equation (\ref{p-5.2.1a}), we get 
\begin{eqnarray*}
0 &=&\sum_{n=0}^{\infty }n(n-1)a_{n}x^{n-2}-x\sum_{n=0}^{\infty
}na_{n}x^{n-1}-\sum_{n=0}^{\infty }a_{n}x^{n}\quad \\
&=&\sum_{n=0}^{\infty }n(n-1)a_{n}x^{n-2}-\sum_{n=0}^{\infty
}na_{n}x^{n}-\sum_{n=0}^{\infty }a_{n}x^{n}\quad .
\end{eqnarray*}
We now shift the first series by $2$ by setting 
\begin{eqnarray*}
k &=&n-2 \\
n &=&k+2
\end{eqnarray*}
and simply relabel the counting index of the last two series by $k$. 
\begin{eqnarray*}
0 &=&\sum_{k=-2}^{\infty }(k+2)(k+1)a_{k+2}x^{k}-\sum_{k=0}^{\infty
}ka_{k}x^{k}-\sum_{k=0}^{\infty }a_{k}x^{k} \\
&=&(-2+2)(-2+1)a_{0}x^{-2}+(-1+2)(-1+1)a_{1}x^{-1}+\sum_{k=0}^{\infty
}(k+2)(k+1)a_{k+2}x^{k} \\
&&\quad \quad -\sum_{k=0}^{\infty }ka_{k}x^{k}-\sum_{k=0}^{\infty }a_{k}x^{k}
\\
&& \\
&=&0+0+\sum_{k=0}^{\infty }(k+2)(k+1)a_{k+2}x^{k}-\sum_{k=0}^{\infty
}ka_{k}x^{k}-\sum_{k=0}^{\infty }a_{k}x^{k} \\
&=&\sum_{k=0}^{\infty }\left( (k+2)(k+1)a_{k+2}-ka_{k-}a_{k}\right) x^{k} \\
&&
\end{eqnarray*}
In order to ensure that the right hand side vanish for all $x$ we now demand
the total coefficient of $x^{k}$ vanish for all $k$. Thus 
\begin{equation*}
(k+2)(k+1)a_{k+2}-(k+1)a_{k}=0\quad \quad ,\quad \quad \text{for all $k$}%
\quad ,
\end{equation*}
or, after solving for $a_{k+2}$ and then replacing $k$ by $n$ 
\begin{equation*}
a_{n+2}=\frac{a_{n}}{(n+2)}\quad .
\end{equation*}
The above equation is the recursion relation for the coefficients $a_{k}$.%
\qed
\end{itemize}

\item[(b)] $y^{\prime \prime }-xy^{\prime }-y=0$, $x_{o}=1$

\begin{itemize}
\item We set 
\begin{equation*}
y(x)=\sum_{n=0}^{\infty }a_{n}(x-1)^{n}
\end{equation*}
and plug into the differential equation: 
\begin{equation*}
0=\sum_{n=2}^{\infty }n(n-1)a_{n}(x-1)^{n-2}-x\sum_{n=1}^{\infty
}na_{n}(x-1)^{n-1}-\sum_{n=0}^{\infty }a_{n}(x-1)^{n}\quad .
\end{equation*}
The hardest part about this problem will be to combine the three power
series appearing on the right hand side of the equation above into a single
power series. We have 
\begin{eqnarray*}
0 &=&\sum_{n=2}^{\infty }n(n-1)a_{n}(x-1)^{n-2}-x\sum_{n=1}^{\infty
}na_{n}(x-1)^{n-1}-\sum_{n=0}^{\infty }a_{n}(x-1)^{n} \\
&=&\sum_{n=2}^{\infty }n(n-1)a_{n}(x-1)^{n-2}-(x-1+1)\sum_{n=2}^{\infty
}na_{n}(x-1)^{n-1}-\sum_{n=0}^{\infty }a_{n}(x-1)^{n} \\
&=&\sum_{n=1}^{\infty }n(n-1)a_{n}(x-1)^{n-2}-(x-1)\sum_{n=1}^{\infty
}na_{n}(x-1)^{n-1}-\sum_{n=1}na_{n}(x-1)^{n-1} \\
&&\quad \quad -\sum_{n=0}^{\infty }a_{n}(x-1)^{n} \\
&=&\sum_{n=2}^{\infty }n(n-1)a_{n}(x-1)^{n-2}-\sum_{n=1}^{\infty
}na_{n}(x-1)^{n}-\sum_{n=1}^{\infty }na_{n}(x-1)^{n-1} \\
&&\quad \quad -\sum_{n=0}^{\infty }a_{n}(x-x_{o})^{n} \\
&=&\sum_{m=0}^{\infty }(m+2)(m+1)a_{m+2}(x-1)^{m}-\sum_{n=0}^{\infty
}na_{n}(x-1)^{n} \\
&&\quad \quad -\sum_{k=0}^{\infty }(k+1)a_{k+1}(x-1)^{k}-\sum_{n=0}^{\infty
}a_{n}(x-1)^{n} \\
&=&\sum_{n=0}^{\infty }(n+2)(n+1)a_{n+2}(x-1)^{n}-\sum_{n=0}na_{n}(x-1)^{n}
\\
&&\quad \quad -\sum_{n=0}^{\infty }(n+1)a_{n+1}(x-1)^{n}-\sum_{n=0}^{\infty
}a_{n}(x-1)^{n} \\
&=&\sum_{n=0}^{\infty }\left(
(n+2)(n+1)a_{n+2}-na_{n}-(n+1)a_{n+1}+a_{n}\right) (x-1)^{n}
\end{eqnarray*}
We must therefore have 
\begin{equation*}
0=(n+2)(n+1)a_{n+2}-na_{n}-(n+1)a_{n+1}-a_{n}
\end{equation*}
or 
\begin{equation*}
a_{n+2}=\frac{a_{n}+a_{n+1}}{(n+2)}\quad .
\end{equation*}
\qed
\end{itemize}

\item[(c)] $(1-x)y^{\prime \prime }+y=0$, $x_{o}=0$

\begin{itemize}
\item 
\begin{equation*}
(1-x)y^{\prime \prime }+y=0\quad \quad ,\quad \quad x_{o}=0\quad
\end{equation*}
Since $x_{o}=0$, we set 
\begin{equation*}
y(x)=\sum_{n=0}^{\infty }a_{n}x^{n}
\end{equation*}
and plug this expression for $y(x)$ into the differential equation. This
yields 
\begin{eqnarray*}
0 &=&(1-x)\sum_{n=0}^{\infty }n(n-1)a_{n}x^{n-2}+\sum_{n=0}^{\infty
}a_{n}x^{n} \\
&=&\sum_{n=0}^{\infty }n(n-1)a_{n}x^{n-2}-\sum_{n=0}^{\infty
}n(n-1)a_{n}x^{n-1}+\sum_{n=0}^{\infty }a_{n}x^{n}
\end{eqnarray*}
We now perform the usual shifts of summation indices (in order to put each
series in standard form) and then combine the above sum of three series into
a single series. 
\begin{eqnarray*}
0 &=&\sum_{k=-2}^{\infty }(k+2)(k+1)a_{k+2}x^{k}-\sum_{k=-1}^{\infty
}(k+1)(k)a_{k+1}x^{k}+\sum_{k=0}^{\infty }a_{k}x^{k} \\
&=&0+0+\sum_{k=0}^{\infty }(k+2)(k+1)a_{k+2}x^{k}-0-\sum_{k=0}^{\infty
}(k+1)(k)a_{k+1}x^{k}+\sum_{k=0}^{\infty }a_{k}x^{k} \\
&=&\sum_{k=0}^{\infty }\left( (k+2)(k+1)a_{k+2}-k(k+1)a_{k+1}+a_{k}\right)
x^{k}
\end{eqnarray*}
We can now read off the recursion relation: 
\begin{equation*}
(k+2)(k+1)a_{k+2}-k(k+1)a_{k+1}+a_{k}=0
\end{equation*}
or, after solving for $a_{k+2}$ and then replacing $k$ by $n$%
\begin{equation*}
a_{n+2}=\frac{n(n+1)a_{n+1}-a_{n}}{(n+2)(n+1)}
\end{equation*}
\qed
\end{itemize}

\item[(d)] $y^{\prime \prime }+xy^{\prime }+2y=0$, $x_{o}=0$

\begin{itemize}
\item Since $x_{o}=0$, we set 
\begin{equation*}
y(x)=\sum_{n=0}^{\infty }a_{n}x^{n}
\end{equation*}%
and plug this expression for $y(x)$ into the differential equation. This
yields 
\begin{eqnarray*}
0 &=&\sum_{n=0}^{\infty }n(n-1)a_{n}x^{n-2}+x\sum_{n=0}^{\infty
}na_{n}x^{n-1}+2\sum_{n=0}^{\infty }a_{n}x^{n} \\
&=&\sum_{n=0}^{\infty }n(n-1)a_{n}x^{n-2}+\sum_{n=0}^{\infty
}na_{n}x^{n}+\sum_{n=0}^{\infty }2a_{n}x^{n}
\end{eqnarray*}%
We now perform the usual shifts of summation indices (in order to put each
series in standard form) and then combine the above sum of three series into
a single series. 
\begin{eqnarray*}
0 &=&\sum_{k=-2}^{\infty }(k+2)(k+1)a_{k+2}x^{k}+\sum_{k=0}^{\infty
}ka_{k}x^{k}+\sum_{k=0}^{\infty }2a_{k}x^{k} \\
&=&0+0+\sum_{k=0}^{\infty }(k+2)(k+1)a_{k+2}x^{k}+\sum_{k=0}^{\infty
}ka_{k}x^{k}+\sum_{k=0}^{\infty }2a_{k}x^{k} \\
&=&\sum_{k=0}^{\infty }\left( (k+2)(k+1)a_{k+2}+(k+2)a_{k}\right) x^{k}
\end{eqnarray*}%
Our recursion relations are thus, 
\begin{equation*}
(k+2)(k+1)a_{k+2}+(k+2)a_{k}=0
\end{equation*}%
or, after solving for $a_{k+2}$ and then replacing $k$ by $n$ 
\begin{equation*}
a_{n+2}=-\frac{a_{n}}{n+1}\quad .
\end{equation*}%
\qed
\end{itemize}

\item[(e)] $(1+x^{2})y^{\prime \prime }-4xy^{\prime }+6y=0$, $x_{o}=0$

\begin{itemize}
\item Since $x_{o}=0$, we set 
\begin{equation*}
y(x)=\sum_{n=0}^{\infty }a_{n}x^{n}
\end{equation*}
and plug this expression for $y(x)$ into the differential equation. This
yields 
\begin{eqnarray*}
0 &=&(1+x^{2})\sum_{n=0}^{\infty }n(n-1)a_{n}x^{n-2}-4x\sum_{n=0}^{\infty
}na_{n}x^{n}+6\sum_{n=0}^{\infty }a_{n}x^{n} \\
&=&\sum_{n=0}^{\infty }n(n-1)a_{n}x^{n-2}+\sum_{n=0}^{\infty
}n(n-1)a_{n}x^{n}-\sum_{n=0}^{\infty }4na_{n}x^{n}+\sum_{n=0}^{\infty
}6a_{n}x^{n}
\end{eqnarray*}
We now perform the usual shifts of summation indices (in order to put each
series in standard form) and then combine the above sum of three series into
a single series. 
\begin{eqnarray*}
0 &=&\sum_{k=-2}^{\infty }(k+2)(k+1)a_{k+2}x^{k}+\sum_{k=0}^{\infty
}k(k-1)a_{k}x^{k}-\sum_{k=0}^{\infty }4ka_{k}x^{k}+\sum_{k=0}^{\infty
}6a_{k}x^{k} \\
&=&0+0+\sum_{k=0}^{\infty }(k+2)(k+1)a_{k+2}x^{k}+\sum_{k=0}^{\infty
}k(k-1)a_{k}x^{k}-\sum_{k=0}^{\infty }4ka_{k}x^{k}+\sum_{k=0}^{\infty
}6a_{k}x^{k} \\
&=&\sum_{k=0}^{\infty }\left( (k+2)(k+1)a_{k+2}+\left( k^{2}-5k+6\right)
a_{k}\right) x^{k} \\
&=&\sum_{k=0}^{\infty }\left( (k+2)(k+1)a_{k+2}+(k-2)(k-3)a_{k}\right) x^{k}
\end{eqnarray*}
We thus arrive at the following recursion relation $%
(k+2)(k+1)a_{k+2}+(k-2)(k-3)a_{k}=0$. After solving for $a_{k+2}$ and then
replacing $k$ by $n$ 
\begin{equation*}
a_{n+2}=\frac{-(n-2)(n-3)a_{n}}{(n+2)(n+1)}\quad .
\end{equation*}
\qed
\end{itemize}
\end{enumerate}

3. Find power series expressions for the general solutions of the following
differential equations. (You may utilize recursion relations found in
Problem 3.)

\begin{enumerate}
\item[(a)] $y^{\prime \prime }-xy^{\prime }-y=0$, $x_{o}=0$

\begin{itemize}
\item In Problem 3a we found that the recursion relations for a power series
solution about $x_{o}=0$ for this differential equation are 
\begin{equation*}
a_{n+2}=\frac{a_{n}}{n+2}\quad
\end{equation*}
We will now apply these recursion relations for $n=0,1,2,3,\ldots $ to get
expressions for the coefficients $a_{2},a_{3},a_{4},a_{5},\ldots $ in terms
of the first two coefficients $a_{0}$ and $a_{1}$. 
\begin{equation*}
\begin{array}{lll}
n=0 & \Rightarrow & a_{2}=a_{0+2}=\frac{a_{0}}{0+2}=\frac{a_{0}}{2} \\ 
n=1 & \Rightarrow & a_{3}=a_{1+2}=\frac{a_{1}}{1+2}=\frac{a_{1}}{3} \\ 
n=2 & \Rightarrow & a_{4}=a_{2+2}=\frac{a_{2}}{2+2}=\frac{a_{2}}{4}=\frac{%
a_{0}}{2\cdot 4} \\ 
n=3 & \Rightarrow & a_{5}=a_{3+2}=\frac{a_{3}}{3+2}=\frac{a_{3}}{5}=\frac{%
a_{1}}{3\cdot 5} \\ 
n=4 & \Rightarrow & a_{6}=a_{4+2}=\frac{a_{4}}{4+2}=\frac{a_{4}}{6}=\frac{%
a_{0}}{2\cdot 4\cdot 6}%
\end{array}%
\end{equation*}
Hence 
\begin{eqnarray*}
y(x) &=&\sum_{n=0}^{\infty }a_{n}x^{n} \\
&=&a_{0}+a_{1}x+a_{2}x^{2}+a_{3}x^{3}+a_{4}x^{4}+a_{5}x^{5}+a_{6}x^{6}+\cdots
\\
&=&a_{0}+a_{1}x+\frac{a_{0}}{2}x^{2}+\frac{a_{1}}{3}x^{3}+\frac{a_{0}}{%
2\cdot 4}x^{4}+\frac{a_{1}}{3\cdot 5}x^{5}+\frac{a_{0}}{2\cdot 4\cdot 6}%
x^{6}+\cdots \\
&=&a_{0}\left( 1+\frac{1}{2}x^{2}+\frac{1}{2\cdot 4}x^{4}+\frac{1}{2\cdot
4\cdot 6}x^{6}+\cdots \right) \\
&&+a_{1}\left( x+\frac{1}{3}x^{3}+\frac{1}{3\cdot 5}x^{5}+\cdots \right)
\end{eqnarray*}
\qed
\end{itemize}

\item[(b)] $y^{\prime \prime }-xy-y=0$, $x_{o}=1$
\end{enumerate}

\begin{itemize}
\item n Problem 3b we found that the recursion relations for a power series
solution about $x_{o}=0$ for this differential equation are 
\begin{equation*}
a_{n+2}=\frac{a_{n}+a_{n+1}}{(n+2)}\quad
\end{equation*}
We will now apply these recursion relations for $n=0,1,2,\ldots $ to get
expressions for the coefficients $a_{2},a_{3},a_{4},\ldots $ in terms of the
first two coefficients $a_{0}$ and $a_{1}$. 
\begin{equation*}
\begin{array}{lll}
n=0 & \Rightarrow & a_{2}=a_{0+2}=\frac{a_{0}+a_{1}}{0+2}=\frac{a_{0}}{2}+%
\frac{a_{1}}{2} \\ 
n=1 & \Rightarrow & a_{3}=a_{1+2}=\frac{a_{1}+a_{2}}{1+2}=\frac{a_{1}}{3}+%
\frac{1}{3}\left( \frac{a_{0}}{2}+\frac{a_{1}}{2}\right) =\frac{a_{0}}{6}+%
\frac{a_{1}}{2} \\ 
n=2 & \Rightarrow & a_{4}=a_{2+2}=\frac{a_{2}+a_{3}}{2+2}=\frac{1}{4}\left( 
\frac{a_{0}}{2}+\frac{a_{1}}{2}\right) +\frac{1}{4}\left( \frac{a_{0}}{6}+%
\frac{a_{1}}{2}\right) =\frac{a_{0}}{6}+\frac{a_{1}}{4} \\ 
&  &  \\ 
&  & 
\end{array}%
\end{equation*}
Therefore 
\begin{eqnarray*}
y(x) &=&\sum_{n=0}^{\infty }a_{n}(x-1)^{n} \\
&=&a_{0}+a_{1}(x-1)+a_{2}(x-1)^{2}+a_{3}(x-1)^{3}+a_{4}(x-1)^{4}+\cdots \\
&=&a_{0}+a_{1}(x-1)+\left( \frac{a_{0}}{2}+\frac{a_{1}}{2}\right)
(x-1)^{2}+\left( \frac{a_{0}}{6}+\frac{a_{1}}{2}\right) (x-1)^{3}+\left( 
\frac{a_{0}}{6}+\frac{a_{1}}{4}\right) (x-1)^{4}+\cdots \\
&=&a_{0}\left( 1+\frac{1}{2}(x-1)^{2}+\frac{1}{6}(x-1)^{3}+\frac{1}{6}%
(x-1)^{4}+\cdots \right) + \\
&&a_{1}\left( (x-1)+\frac{1}{2}(x-1)^{2}+\frac{1}{2}(x-1)^{3}+\frac{1}{4}%
(x-1)^{4}+\cdots \right)
\end{eqnarray*}
\qed
\end{itemize}

4. Find power series expressions for the solutions to the following initial
value problems. (You may utilize recursion relations found in Problem 3.)

\begin{enumerate}
\item[(a)] $(1-x)y^{\prime \prime }+y=0$, $y(0)=2$, $y^{\prime }(0)=1$

\begin{itemize}
\item In Problem 3c we obtained the following recursion relations for power
series solutions of this differential equation (expanded about $x_{o}=0$). 
\begin{equation*}
a_{n+2}=\frac{n(n+1)a_{n+1}-a_{n}}{(n+2)(n+1)}
\end{equation*}
For such solutions 
\begin{equation*}
y(x)=\sum_{n=0}^{\infty }a_{n}x^{n}
\end{equation*}
the initial conditions imply 
\begin{eqnarray*}
2 &=&y(0)=a_{0} \\
1 &=&y^{\prime }(0)=a_{1}
\end{eqnarray*}
We now can apply the recursion relations to get numerical values for $%
a_{2},a_{3},a_{4},\ldots $%
\begin{equation*}
\begin{array}{lll}
n=0 & \Rightarrow & a_{2}=\frac{0(0+1)a-+a_{0}}{2\cdot 1}=-\frac{a_{0}}{2}=-1
\\ 
n=1 & \Rightarrow & a_{3}=\frac{1(1+1)a_{2}-a_{1}}{3\cdot 2}=\frac{%
2a_{2}-a_{1}}{6}=\frac{-2-1}{6}=-\frac{1}{2} \\ 
n=2 & \Rightarrow & a_{4}=\frac{2(2+1)a_{3}-a_{2}}{4\cdot 3}=\frac{6a-a_{2}}{%
12}=\frac{-3+1}{12}=-\frac{1}{6} \\ 
n=3 & \Rightarrow & a_{5}=\frac{3(3+1)a_{4}-a_{3}}{5\cdot 4}=\frac{%
12a_{4}-a_{3}}{20}=\frac{-2+\frac{1}{2}}{20}=-\frac{3}{40}%
\end{array}%
\end{equation*}
so 
\begin{eqnarray*}
y(x) &=&\sum_{n=0}^{\infty }a_{n}x^{n} \\
&=&a_{0}+a_{1}x+a_{2}x^{2}+a_{3}x^{3}+a_{4}x^{4}+a_{5}x^{5} \\
&=&2+x-x^{2}-\frac{1}{2}x^{3}-\frac{1}{6}x^{4}-\frac{3}{40}x^{5}+\cdots
\end{eqnarray*}
\qed
\end{itemize}

\item[(b)] $y^{\prime \prime }-xy^{\prime }-y=0$, $y(1)=1,$ $y^{\prime }(1)=2
$
\end{enumerate}

\begin{itemize}
\item In Problem 3b we found that the recursion relations for a power series
solution about $x_{o}=1$ for this differential equation are 
\begin{equation*}
a_{n+2}=\frac{a_{n}+a_{n+1}}{(n+2)}\quad 
\end{equation*}%
Also, also in Problem 4b we found that the power series expansion about $%
x_{0}=1$ of the general solution to this differential equation is 
\begin{equation*}
y(x)=a_{0}+a_{1}(x-1)+\left( \frac{a_{0}}{2}+\frac{a_{1}}{2}\right)
(x-1)^{2}+\left( \frac{a_{0}}{6}+\frac{a_{1}}{2}\right) (x-1)^{3}+\left( 
\frac{a_{0}}{6}+\frac{a_{1}}{4}\right) (x-1)^{4}+\cdots 
\end{equation*}%
All that remains to be done is to use the initial conditions to fix the
values of the first two coefficients. 
\begin{equation*}
\begin{array}{lll}
y(1)=1 & \Rightarrow  & a_{0}=1 \\ 
y^{\prime }(1)=2 & \Rightarrow  & a_{1}=2%
\end{array}%
\end{equation*}%
Therefore, the solution to the initial value problem is 
\begin{eqnarray*}
y(x) &=&1+2(x-1)+\left( \frac{1}{2}+\frac{2}{2}\right) (x-1)^{2}+\left( 
\frac{1}{6}+\frac{2}{2}\right) (x-1)^{3}+\left( \frac{1}{6}+\frac{2}{4}%
\right) (x-1)^{4}+\cdots  \\
&=&1+2(x-1)+\frac{3}{2}(x-1)^{2}+\frac{7}{6}(x-1)^{3}+\frac{2}{3}%
(x-1)^{4}+\cdots 
\end{eqnarray*}%
\qed
\end{itemize}

\end{document}
