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\title {Note on Singular Solutions}

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\noindent  Looking at prob. 21 on pg. 54,

\begin{equation} \label{DE}
\frac{dy}{dx}= x\sqrt{1-y^2}
\end{equation}

\noindent we can solve it via separation of variables to get the
family of solutions

\begin{equation} \label{fam}
y = \sin(\frac{x^2}{2} + c)
\end{equation}

\noindent Now we can see that $y = 1$ and $y = -1$ are both constant
solutions, since when we plug them into the original DE (\ref{DE})
we get $\frac{dy}{dx} = 0$.  Now we want to check to see if they are
included in the family of solutions (\ref{fam}).  If this isn't true
for one of them, we have found a singular solution.  So let's take
$y = 1$, giving us an initial value $y(0) = 1$, and try to solve for
$c$:

\begin{equation*}
y(0) = \sin(\frac{0^2}{2} + c) = \sin c = 1
\end{equation*}

\noindent Note that it is possible to solve for $c$ here, taking for
example $c = \pi/2$.  However, this does \textit{not} automatically
mean that $y=1$ is not a singular solution.  Plugging $c = \pi/2$
back into our family of solutions (\ref{fam}), we get

\begin{equation*}
y = \sin(\frac{x^2}{2} + \frac{\pi}{2})
\end{equation*}

\noindent which is certainly not the same thing as $y = 0$.  So even
though we were able to solve for $c$, we still couldn't come up with
the desired solution. Thus $y = 1$ is a singular solution.  We can
do the same thing for $y = -1$ and find that it is also a singular
solution.

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