Example 7.3.1.
Consider the simple first order equation
\begin{equation*}
2 x y' - y = 0 .
\end{equation*}
Note that \(x=0\) is a singular point. Setting \(x=0\) in the equation, we find that any solution defined near zero satisfies \(y(0)=0\text{,}\) but it is even worse. If we try to plug in
\begin{equation*}
y = \sum_{k=0}^\infty a_k x^k ,
\end{equation*}
we obtain
\begin{equation*}
\begin{split}
0 = 2 xy'-y &=
2x \, \Biggl( \sum_{k=1}^\infty k a_k x^{k-1} \Biggr)
-
\Biggl( \sum_{k=0}^\infty a_k x^k \Biggr)
\\
& =
a_0 +
\sum_{k=1}^\infty (2 k a_k - a_k) \, x^{k} .
\end{split}
\end{equation*}
First, \(a_0 = 0\text{.}\) Next, the only way to solve \(0 = 2 k a_k - a_k = (2k-1) \, a_k\) for \(k = 1,2,3,\dots\) is for \(a_k = 0\) for all \(k\text{.}\) Therefore, in this manner we only get the trivial solution \(y=0\text{.}\) We need a nonzero solution to get the general solution to the equation.
Let us try \(y=x^r\) for some real number \(r\text{.}\) Consequently our solution—if we can find one—may only make sense for positive \(x\text{.}\) Then \(y' = r x^{r-1}\text{.}\) So
\begin{equation*}
0 = 2 x y' - y = 2 x r x^{r-1} - x^r = (2r-1) x^r .
\end{equation*}
Thus \(r= \nicefrac{1}{2}\) and so \(y = x^{1/2}\text{.}\) As the equation is linear, the general solution for positive \(x\) is
\begin{equation*}
y = C x^{1/2} .
\end{equation*}
If \(C \not= 0\text{,}\) then the derivative of the solution “blows up” at \(x=0\) (the singular point). There is only one solution that is differentiable at \(x=0\) and that’s the trivial solution \(y=0\text{.}\)
Not every problem with a singular point has a solution of the form \(y=x^r\text{,}\) of course. But perhaps we can combine the methods. What we will do is to try a solution of the form
\begin{equation*}
y = x^r f(x)
\end{equation*}
for positive \(x\text{,}\) where \(f(x)\) is an analytic function (a power series).
Example 7.3.2.
Consider the equation
\begin{equation*}
4 x^2 y'' - 4 x^2 y' + (1-2x)y = 0,
\end{equation*}
and again note that \(x=0\) is a singular point.
Let us try
\begin{equation*}
y = x^r \sum_{k=0}^\infty a_k x^k
= \sum_{k=0}^\infty a_k x^{k+r} ,
\end{equation*}
where \(r\) is a real number, not necessarily an integer. Again if such a solution exists, it may only exist for positive \(x\text{.}\) First we find the derivatives
\begin{equation*}
\begin{aligned}
y' & = \sum_{k=0}^\infty (k+r)\, a_k x^{k+r-1} , \\
y'' & = \sum_{k=0}^\infty (k+r)\,(k+r-1)\, a_k x^{k+r-2} .
\end{aligned}
\end{equation*}
We plug those into our equation:
\begin{equation*}
\begin{split}
0 & = 4x^2y''-4x^2y'+(1-2x)y
\\
&=
4x^2 \, \Biggl( \sum_{k=0}^\infty (k+r)\,(k+r-1) \, a_k x^{k+r-2} \Biggr)
-
4x^2 \, \Biggl( \sum_{k=0}^\infty (k+r) \, a_k x^{k+r-1} \Biggr)
+
(1-2x)
\Biggl( \sum_{k=0}^\infty a_k x^{k+r} \Biggr)
\\
&=
\Biggl( \sum_{k=0}^\infty 4 (k+r)\,(k+r-1) \, a_k x^{k+r} \Biggr)
\\
& \phantom{mmm}
-
\Biggl( \sum_{k=0}^\infty 4 (k+r) \, a_k x^{k+r+1} \Biggr)
+
\Biggl( \sum_{k=0}^\infty a_k x^{k+r} \Biggr)
-
\Biggl( \sum_{k=0}^\infty 2a_k x^{k+r+1} \Biggr)
\\
&=
\Biggl( \sum_{k=0}^\infty 4 (k+r)\,(k+r-1) \, a_k x^{k+r} \Biggr)
\\
& \phantom{mmm}
-
\Biggl( \sum_{k=1}^\infty 4 (k+r-1) \, a_{k-1} x^{k+r} \Biggr)
+
\Biggl( \sum_{k=0}^\infty a_k x^{k+r} \Biggr)
-
\Biggl( \sum_{k=1}^\infty 2a_{k-1} x^{k+r} \Biggr)
\\
&=
4r(r-1) \, a_0 x^r + a_0 x^r +
\sum_{k=1}^\infty
\Bigl( 4 (k+r)\,(k+r-1) \, a_k
-
4 (k+r-1) \, a_{k-1}
+
a_k
-
2a_{k-1} \Bigr) \, x^{k+r}
\\
&=
\bigl( 4r(r-1) + 1 \bigr) \, a_0 x^r +
\sum_{k=1}^\infty
\Bigl( \bigl( 4 (k+r)\,(k+r-1) + 1 \bigr) \, a_k
-
\bigl( 4 (k+r-1) + 2 \bigr) \, a_{k-1} \Bigr) \, x^{k+r} .
\end{split}
\end{equation*}
First, to have a solution we must have \(\bigl( 4r(r-1) + 1 \bigr) \, a_0 = 0\text{.}\) Supposing \(a_0 \not= 0\text{,}\)
\begin{equation*}
4r(r-1) + 1 = 0 .
\end{equation*}
This equation is called the indicial equation. This particular indicial equation has a double root at \(r = \nicefrac{1}{2}\text{.}\)
OK, so we know what \(r\) has to be. That knowledge we obtained simply by looking at the coefficient of \(x^r\text{.}\) All other coefficients of \(x^{k+r}\) also have to be zero so
\begin{equation*}
\bigl( 4 (k+r)\,(k+r-1) + 1 \bigr) \, a_k
-
\bigl( 4 (k+r-1) + 2 \bigr) \, a_{k-1} = 0 .
\end{equation*}
If we plug in \(r=\nicefrac{1}{2}\) and solve for \(a_k\text{,}\) we get
\begin{equation*}
a_k
=
\frac{4 (k+\nicefrac{1}{2}-1) + 2}{4 (k+\nicefrac{1}{2})\,(k+\nicefrac{1}{2}-1) + 1} \, a_{k-1}
=
\frac{1}{k} \, a_{k-1} .
\end{equation*}
Let us set \(a_0 = 1\text{.}\) Then
\begin{equation*}
a_1 = \frac{1}{1} a_0 = 1 ,
\qquad
a_2 = \frac{1}{2} a_1 = \frac{1}{2} ,
\qquad
a_3 = \frac{1}{3} a_2 = \frac{1}{3 \cdot 2} ,
\qquad
a_4 = \frac{1}{4} a_3 = \frac{1}{4 \cdot 3 \cdot 2} ,
\qquad \cdots
\end{equation*}
Extrapolating, we notice that
\begin{equation*}
a_k = \frac{1}{k(k-1)(k-2) \cdots 3 \cdot 2} = \frac{1}{k!} .
\end{equation*}
In other words,
\begin{equation*}
y =
\sum_{k=0}^\infty a_k x^{k+r}
=
\sum_{k=0}^\infty \frac{1}{k!} x^{k+1/2}
=
x^{1/2}
\sum_{k=0}^\infty \frac{1}{k!} x^{k}
=
x^{1/2}
e^x .
\end{equation*}
That was lucky! In general, we will not be able to write the series in terms of elementary functions.
We have one solution, let us call it \(y_1 = x^{1/2} e^x\text{.}\) But what about a second solution? If we want a general solution, we need two linearly independent solutions. Picking \(a_0\) to be a different constant only gets us a constant multiple of \(y_1\text{,}\) and we do not have any other \(r\) to try; we only have one solution to the indicial equation. Well, there are powers of \(x\) floating around and we are taking derivatives, perhaps the logarithm (the antiderivative of \(x^{-1}\)) is around as well. It turns out we want to try for another solution of the form
\begin{equation*}
y_2 = \sum_{k=0}^\infty b_k x^{k+r} + (\ln x) y_1 ,
\end{equation*}
which in our case is
\begin{equation*}
y_2 = \sum_{k=0}^\infty b_k x^{k+1/2} + (\ln x) x^{1/2} e^x .
\end{equation*}
We now differentiate this equation, substitute into the differential equation and solve for \(b_k\text{.}\) A long computation ensues and we obtain some recursion relation for \(b_k\text{.}\) The reader can (and should) try this to obtain for example the first three terms
\begin{equation*}
b_1 = b_0 -1 , \qquad b_2 = \frac{2b_1-1}{4} , \qquad b_3 =
\frac{6b_2-1}{18} , \qquad \ldots
\end{equation*}
We then fix \(b_0\) and obtain a solution \(y_2\text{.}\) Then we write the general solution as \(y = A y_1 + B y_2\text{.}\)
