2.3.2.0 Example 2 \(4x^{2}y^{\prime \prime }+4xy^{\prime }+\left ( 4x^{2}-1\right ) y=0\)

\[ 4x^{2}y^{\prime \prime }+4xy^{\prime }+\left ( 4x^{2}-1\right ) y=0 \]

Comparing the ode to

\[ y^{\prime \prime }+p\left ( x\right ) y^{\prime }+q\left ( x\right ) y=0 \]

Hence \(p\left ( x\right ) =\frac {1}{x},q\left ( x\right ) =\frac {4x^{2}-1}{4x^{2}}\). Therefore \(p_{0}=\lim _{x\rightarrow 0}xp\left ( x\right ) =\lim _{x\rightarrow 0}1=1\) and \(q_{0}=\lim _{x\rightarrow 0}x^{2}q\left ( x\right ) =\lim _{x\rightarrow 0}\frac {4x^{2}-1}{4}=-\frac {1}{4}\). Hence the indicial equation is

\begin{align*} r\left ( r-1\right ) +p_{0}r+q_{0} & =0\\ r\left ( r-1\right ) +r-\frac {1}{4} & =0\\ r^{2}-\frac {1}{4} & =0\\ r & =-\frac {1}{2},\frac {1}{2}\end{align*}

Therefore \(r_{1}=\frac {1}{2},r_{2}=-\frac {1}{2}\).  They differ by an integer \(N=1\). Therefore two linearly independent solutions can be constructed using

\begin{align*} y_{1} & =\sum _{n=0}^{\infty }a_{n}x^{n+r_{1}}\\ y_{2} & =Cy_{1}\ln \left ( x\right ) +\sum _{n=0}^{\infty }b_{n}x^{n+r_{2}}\end{align*}

We start by expanding

\begin{align*} y & =\sum _{n=0}^{\infty }a_{n}x^{n+r}\\ y^{\prime } & =\sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r-1}\\ y^{\prime \prime } & =\sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r-2}\end{align*}

The ode becomes

\begin{align} 4x^{2}\sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r-2}+4x\sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r-1}+4x^{2}\sum _{n=0}^{\infty }a_{n}x^{n+r}-\sum _{n=0}^{\infty }a_{n}x^{n+r} & =0\nonumber \\ \sum _{n=0}^{\infty }4\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r}+\sum _{n=0}^{\infty }4\left ( n+r\right ) a_{n}x^{n+r}+\sum _{n=0}^{\infty }4a_{n}x^{n+r+2}-\sum _{n=0}^{\infty }a_{n}x^{n+r} & =0\nonumber \\ \sum _{n=0}^{\infty }\left ( 4\left ( n+r\right ) \left ( n+r-1\right ) +4\left ( n+r\right ) -1\right ) a_{n}x^{n+r}+\sum _{n=0}^{\infty }4a_{n}x^{n+r+2} & =0\nonumber \\ \sum _{n=0}^{\infty }\left ( 4n^{2}+8nr+4r^{2}-1\right ) a_{n}x^{n+r}+\sum _{n=0}^{\infty }4a_{n}x^{n+r+2} & =0\nonumber \\ \sum _{n=0}^{\infty }\left ( 4\left ( n+r\right ) ^{2}-1\right ) a_{n}x^{n+r}+\sum _{n=0}^{\infty }4a_{n}x^{n+r+2} & =0 \tag {1}\end{align}

Re indexing to lowest powers on \(x\) gives

\begin{equation} \sum _{n=0}^{\infty }\left ( 4\left ( n+r\right ) ^{2}-1\right ) a_{n}x^{n+r}+\sum _{n=2}^{\infty }4a_{n-2}x^{n+r}=0 \tag {2}\end{equation}

\(n=0\) is not used since was used to find roots. We let \(a_{0}=1\). For \(n=1\) EQ. (2) gives

\begin{align} \left ( 4\left ( 1+r\right ) ^{2}-1\right ) a_{1} & =0\tag {3}\\ \left ( 4\left ( 1+\left ( \frac {1}{2}\right ) \right ) ^{2}-1\right ) a_{1} & =0\nonumber \\ 8a_{1} & =0\nonumber \\ a_{1} & =0\nonumber \end{align}

The recurrence relation is when \(n\geq 2\) from (2) is given by

\begin{align} \left ( 4\left ( n+r\right ) ^{2}-1\right ) a_{n}+4a_{n-2} & =0\nonumber \\ a_{n} & =\frac {-4}{4\left ( n+r\right ) ^{2}-1}a_{n-2} \tag {4}\end{align}

For \(n=2\) and from (4)

\begin{align*} a_{2} & =\frac {-4}{4\left ( 2+r\right ) ^{2}-1}a_{0}\\ & =\frac {-4}{4\left ( 2+r\right ) ^{2}-1}\\ & =\frac {-4}{4\left ( 2+\frac {1}{2}\right ) ^{2}-1}\\ & =-\frac {1}{6}\end{align*}

For \(n=3\) Eq (5) gives (and since \(a_{1}=0\))

\begin{align*} a_{3} & =\frac {-4}{4\left ( 3+r\right ) ^{2}-1}a_{1}\\ & =0 \end{align*}

And so on. Hence

\begin{align*} y_{1} & =\sum _{n=0}^{\infty }a_{n}x^{n+\frac {1}{2}}\\ & =x^{\frac {1}{2}}\left ( a_{0}+a_{1}x+a_{2}x^{2}+a_{3}x^{3}+a_{4}x^{4}+\cdots \right ) \\ & =\sqrt {x}\left ( 1-\frac {1}{6}x^{2}+\frac {1}{120}x^{4}+\cdots \right ) \end{align*}

Now we need to find \(y_{2}\). We first check if \(y_{2}\) can be found using standard method as was done above for \(y_{1}\). To find if \(C=0\) or not, We need to evaluate

\begin{align*} \lim _{r\rightarrow r_{2}}a_{N}\left ( r\right ) & =\lim _{r\rightarrow -\frac {1}{2}}a_{1}\left ( r\right ) \\ & =\lim _{r\rightarrow -\frac {1}{2}}0\\ & =0 \end{align*}

Since limit exist then \(C=0\) and we do not need the log term.

\[ y_{2}=\sum _{n=0}^{\infty }b_{n}x^{n+r_{2}}\]

Using same recursive relation for \(a_{n}\) above, but using \(b_{n}\) instead and use \(r=\frac {-1}{2}\) instead of \(r=\frac {1}{2}\) we now can find all \(b_{n}\).

The recurive relation becomes

\begin{align} b_{n} & =\frac {-4}{4\left ( n-\frac {1}{2}\right ) ^{2}-1}b_{n-2}\nonumber \\ & =-\frac {1}{n\left ( n-1\right ) }b_{n-2} \tag {6}\end{align}

For \(n=2\) Eq (6) gives (and using \(b_{0}=1\))

\begin{align*} b_{2} & =-\frac {1}{2\left ( 2-1\right ) }b_{0}\\ & =-\frac {1}{2}\end{align*}

For \(n=3\) Eq (6) gives

\begin{align*} b_{3} & =-\frac {1}{3\left ( 3-1\right ) }b_{1}\\ & =0 \end{align*}

For \(n=4\) Eq (6) gives

\begin{align*} b_{4} & =-\frac {1}{4\left ( 4-1\right ) }b_{2}\\ & =-\frac {1}{12}\left ( -\frac {1}{2}\right ) \\ & =\frac {1}{24}\end{align*}

And so on. Hence

\begin{align*} y_{2} & =\sum _{n=0}^{\infty }b_{n}x^{n-\frac {1}{2}}\\ & =\frac {1}{\sqrt {x}}\left ( b_{0}+b_{1}x+b_{2}x^{2}+\cdots \right ) \\ & =\frac {1}{\sqrt {x}}\left ( 1-\frac {1}{2}x^{2}+\frac {1}{24}x^{4}+\cdots \right ) \end{align*}

Therefore the final solution is

\begin{align*} y & =c_{1}y_{1}+c_{2}y_{2}\\ & =c_{1}\sqrt {x}\left ( 1-\frac {1}{6}x^{2}+\frac {1}{120}x^{4}+\cdots \right ) +c_{2}\frac {1}{\sqrt {x}}\left ( 1-\frac {1}{2}x^{2}+\frac {1}{24}x^{4}+\cdots \right ) \end{align*}