2.2.2.0 Example 4 \(x^{2}y^{\prime \prime }+xy^{\prime }+xy=0\)

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

Comparing the ode to

\[ y^{\prime \prime }+py^{\prime }+qy=0 \]

Hence \(p=\frac {1}{x},q=\frac {1}{x}\). Therefore \(p_{0}=\lim _{x\rightarrow 0}xp=\lim _{x\rightarrow 0}1=1\) and \(q_{0}=\lim _{x\rightarrow 0}x^{2}q=\lim _{x\rightarrow 0}x=0\). 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^{2}\sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r-2}+x\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}+\sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r}+\sum _{n=0}^{\infty }a_{n}x^{n+r+1} & =0 \end{align*}

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

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

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

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

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

\[ r^{2}=0 \]

Hence \(r_{1}=0,r_{2}=0\). 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*}

For \(n\geq 1\) the recurrence relation is

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

Starting with \(y_{1}\).  From (1) with \(r=0\) gives

\[ a_{n}=-\frac {a_{n-1}}{n^{2}}\]

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

\[ a_{1}=-1 \]

For \(n=2\)

\[ a_{2}=-\frac {a_{1}}{4}=\frac {1}{4}\]

And so on. Hence

\begin{align*} y_{1} & =\sum _{n=0}^{\infty }a_{n}x^{n}\\ & =a_{0}+a_{1}x+a_{2}x^{2}+a_{3}x^{3}+\cdots \\ & =1-x+\frac {1}{4}x^{2}-\frac {1}{36}x^{3}+\cdots \end{align*}

In the case of duplicate roots, \(b_{n}\) is found using \(b_{n}=\frac {d}{dr}a_{n}\left ( r\right ) \). And this is evaluated at \(r=r_{0}=0\) in this case since \(r_{0}=0\) here. So we need to find \(a_{n}\left ( r\right ) \). This is done from (1). For \(n=1\)

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

Evaluated at \(r=0\) gives

\[ b_{1}=2 \]

For \(n=2\,\ \) then (2) becomes

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

At \(r=0\) the above becomes

\begin{align*} b_{2} & =-2\frac {3}{\left ( 2\right ) ^{3}}\\ & =-\frac {3}{4}\end{align*}

And so on. Just remember when replacing the \(a_{n}\) in the above, is to use the original \(a_{n}\left ( r\right ) \) as function of \(r\) and not the actual \(a_{n}\) values from above. It has to be function of \(r\) first before taking derivatives, Hence

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

Therefore the general solution is

\begin{align*} y & =c_{1}y_{1}+c_{2}y_{2}\\ & =c_{1}\left ( 1-x+\frac {1}{4}x^{2}-\frac {1}{36}x^{3}+\cdots \right ) +c_{2}\left ( y_{1}\ln \left ( x\right ) +\left ( 2x-\frac {3}{4}x^{2}+\cdots \right ) \right ) \end{align*}