2.5.2.6 Example 6 \(2xy^{\prime \prime }+y^{\prime }+y=0\)

\[ 2xy^{\prime \prime }+y^{\prime }+y=0 \]

Comparing the ode to

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

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

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

Therefore \(r_{1}=0,r_{2}=\frac {1}{2}\). Expansion around \(x=x_{0}=0\). This is regular singular point. Hence Frobenius is needed.  First we find \(y_{h}\). Let \(y=\sum _{n=0}^{\infty }a_{n}x^{n+r}\), hence

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

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

\[ \sum _{n=0}^{\infty }2\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r-1}+\sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r-1}+\sum _{n=1}^{\infty }a_{n-1}x^{n+r-1}=0 \]

When \(n=0\)

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

Since \(a_{0}\neq 0\) then \(r\left ( 2r-1\right ) =0\), hence \(r=0,r=\frac {1}{2}\) as was found above. Therefore the homogenous ode satisfies

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

Where the RHS will be zero when \(r=1,r=\frac {1}{2}\).  For \(n\geq 1\) the recurrence relation is

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

For \(r=0\) the above becomes

\[ a_{n}=\frac {-a_{n-1}}{n\left ( 2n-1\right ) }\]

For \(n=1\) and using \(a_{0}=1\)

\[ a_{1}=\frac {-a_{0}}{n\left ( 2n-1\right ) }=-1 \]

For \(n=2\)

\[ a_{2}=\frac {-a_{1}}{2\left ( 3\right ) }=\frac {1}{6}\]

For \(n=3\)

\[ a_{3}=\frac {-a_{2}}{3\left ( 5\right ) }=\frac {-\frac {1}{6}}{15}=-\frac {1}{90}\]

And so on. Hence

\begin{align*} y_{1} & =\sum _{n=0}^{\infty }a_{n}x^{n+r_{1}}\\ & =a_{0}+a_{1}x+a_{2}x^{2}+\cdots \\ & =1-x+\frac {1}{6}x^{2}-\frac {1}{90}x^{3}+\cdots \end{align*}

To find \(y_{2}\), using (1) but replacing \(a\) by \(b\) and using \(r=\frac {1}{2}\) and letting \(b_{0}=1\) and following the above process gives

\[ b_{n}=\frac {-b_{n-1}}{2n^{2}+4n\left ( \frac {1}{2}\right ) -n+2\left ( \frac {1}{2}\right ) ^{2}-\frac {1}{2}}=-\frac {b_{n-1}}{2n^{2}+n}\]

For \(n=1\)

\[ b_{1}=-\frac {b_{0}}{3}=-\frac {1}{3}\]

For \(n=2\)

\[ b_{2}=-\frac {b_{1}}{8+2}=-\frac {b_{1}}{10}=-\frac {-\frac {1}{3}}{10}=\frac {1}{30}\]

And so on. Hence we obtain

\begin{align*} y_{2} & =\sqrt {x}\sum _{n=0}^{\infty }b_{n}x^{n}\\ & =\sqrt {x}\left ( b_{0}+b_{1}x+b_{2}x^{2}+\cdots \right ) \\ & =\sqrt {x}\left ( 1-\frac {1}{3}x+\frac {1}{30}x^{2}+\cdots \right ) \end{align*}

Therefore the solution is

\begin{align*} y & =c_{1}y_{1}+c_{2}y_{1}\\ & =c_{1}\left ( 1-x+\frac {1}{6}x^{2}-\frac {1}{90}x^{3}+\cdots \right ) +c_{2}\left ( \sqrt {x}\left ( 1-\frac {1}{3}x+\frac {1}{30}x^{2}+\cdots \right ) \right ) \end{align*}