1.2.2.2.8 Example 8 \(xy^{\prime \prime }+y^{\prime }-xy=0\)

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

Comparing the ode to

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

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

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

Therefore \(r_{1}=0,r_{2}=0\).  Expansion around \(x=0\). This is regular singular point. Hence Frobenius is needed. Let

\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*} x\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}-x\sum _{n=0}^{\infty }a_{n}x^{n+r} & =0\\ \sum _{n=0}^{\infty }\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+1} & =0 \end{align*}

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

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

The indicial equation is obtained from \(n=0\). The above reduces to

\[ r^{2}a_{0}x^{n+r-1}=0 \]

Since \(a_{0}\neq 0\) then

\[ r^{2}=0 \]

Hence \(r_{1}=0,r_{2}=0\) as found earlier. Since the roots are repeated then two linearly independent solutions can be constructed using

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

\(n=1\) gives

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

Hence \(a_{1}=0\). The recurrence relation is obtained for \(n\geq 2\). From (1)

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

Since we need to differentiate \(y_{1}\) to obtain \(y_{2}\) and the differentiation is w.r.t \(r\), we will carry the calculations with \(r\) in place and at the end replace \(r\) by its value (which happened to be \(zero\) in this example). We do this only in the case of repeated roots.

For \(n=2\)

\[ a_{2}=\frac {a_{0}}{\left ( 2+r\right ) ^{2}}=\frac {1}{\left ( 2+r\right ) ^{2}}\]

For \(n=3\)

\[ a_{3}=\frac {a_{1}}{\left ( 3+r\right ) ^{2}}=0 \]

For \(n=4\)

\[ a_{4}=\frac {a_{2}}{\left ( 4+r\right ) ^{2}}=\frac {\frac {1}{\left ( 2+r\right ) ^{2}}}{\left ( 4+r\right ) ^{2}}=\frac {1}{\left ( 2+r\right ) ^{2}\left ( 4+r\right ) ^{2}}\]

For \(n=5\), we will find \(a_{5}=0\) (for all odd \(n\) this is the case). For \(n=6\)

\[ a_{6}=\frac {a_{4}}{\left ( 6+r\right ) ^{2}}=\frac {1}{\left ( 2+r\right ) ^{2}\left ( 4+r\right ) ^{2}\left ( 6+r\right ) ^{2}}\]

And so on. We see that \(n^{th}\) term is \(a_{n}=\Pi _{j=1}^{k}\frac {1}{\left ( 2j+r\right ) ^{2}}\). Now we can substitute the \(r=0\) value into the above to obtain

\begin{align*} a_{2} & =\frac {1}{4}\\ a_{4} & =\frac {1}{64}\\ a_{6} & =\frac {1}{2304}\end{align*}

Hence

\begin{align*} y_{1} & =\sum a_{n}x^{n}\\ & =a_{0}+a_{1}x+a_{2}x^{2}+\cdots \\ & =1+\frac {1}{4}x^{2}+\frac {1}{64}x^{4}+\frac {1}{2304}x^{6}+\cdots \end{align*}

To find \(y_{2}\) we use \(b_{n}=\frac {d}{dr}a_{n}\) and evaluate this at \(r=r_{2}\) which in this case is zero. Hence

\begin{align*} b_{2} & =\frac {d}{dr}a_{2}=\frac {d}{dr}\left ( \frac {1}{\left ( 2+r\right ) ^{2}}\right ) =\left ( -\frac {2}{\left ( r+2\right ) ^{3}}\right ) _{r=0}=-\frac {2}{8}=-\frac {1}{4}\\ b_{4} & =\frac {d}{dr}a_{4}=\frac {d}{dr}\left ( \frac {1}{\left ( 2+r\right ) ^{2}\left ( 4+r\right ) ^{2}}\right ) =\left ( -4\frac {r+3}{\left ( r^{2}+6r+8\right ) ^{3}}\right ) _{r=0}=\left ( -4\frac {3}{\left ( 8\right ) ^{3}}\right ) =-\frac {3}{128}\\ b_{6} & =\frac {d}{dr}a_{6}\\ & =\frac {d}{dr}\left ( \frac {1}{\left ( 2+r\right ) ^{2}\left ( 4+r\right ) ^{2}\left ( 6+r\right ) ^{2}}\right ) \\ & =\left ( -2\frac {3r^{2}+24r+44}{\left ( r^{3}+12r^{2}+44r+48\right ) ^{3}}\right ) _{r=0}\\ & =-2\frac {44}{\left ( 48\right ) ^{3}}\\ & =-\frac {11}{13\,824}\end{align*}

And so on. Hence

\begin{align*} y_{1} & =y_{1}\ln \left ( x\right ) +\sum _{n=1}^{\infty }b_{n}x^{n+r_{2}}\\ & =y_{1}\ln \left ( x\right ) +\sum _{n=1}^{\infty }b_{n}x^{n}\\ & =y_{1}\ln \left ( x\right ) +\left ( b_{2}x^{2}+b_{4}x^{4}+b_{6}x^{6}+\cdots \right ) \\ & =y_{1}\ln \left ( x\right ) +\left ( -\frac {1}{4}x^{2}-\frac {3}{128}x^{4}+-\frac {11}{13\,824}x^{6}+\cdots \right ) \end{align*}

Therefore the complete solution is

\begin{align*} y & =c_{1}y_{1}+c_{2}y_{2}\\ & =c_{1}\left ( 1+\frac {1}{4}x^{2}+\frac {1}{64}x^{4}+\frac {1}{2304}x^{6}+\cdots \right ) \\ & +c_{2}\left ( \ln \left ( x\right ) \left ( 1+\frac {1}{4}x^{2}+\frac {1}{64}x^{4}+\frac {1}{2304}x^{6}+\cdots \right ) +\left ( -\frac {1}{4}x^{2}-\frac {3}{128}x^{4}+-\frac {11}{13\,824}x^{6}+\cdots \right ) \right ) \end{align*}