Example 5 2x2y′′ − xy′ + ( 1 − x2) y = x2

Hence \(p\left ( x\right ) =\frac {-x}{2x^{2}}=-\frac {1}{2x},q\left ( x\right ) =\frac {\left ( 1-x^{2}\right ) }{2x^{2}}\). Therefore \(p_{0}=\lim _{x\rightarrow 0}xp\left ( x\right ) =\lim _{x\rightarrow 0}\frac {-1}{2}=\frac {-1}{2}\) and \(q_{0}=\lim _{x\rightarrow 0}x^{2}q\left ( x\right ) =\lim _{x\rightarrow 0}\frac {\left ( 1-x^{2}\right ) }{2}=\frac {1}{2}\). 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+\frac {1}{2} & =0\\ r^{2}-\frac {3}{2}r+\frac {1}{2} & =0\\ r & =1,\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*}

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

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

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

Since \(a_{0}\neq 0\) then \(2r^{2}-3r+1=0\), hence \(r=1,r=\frac {1}{2}\) as was found above. Therefore the homogenous ode satisﬁes

\[ 2x^{2}y^{\prime \prime }-xy^{\prime }+\left ( 1-x^{2}\right ) y=\left ( 2r^{2}-3r+1\right ) a_{0}x^{r}\]

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

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

\[ a_{4}=\frac {a_{2}}{4\left ( 8+1\right ) }=\frac {\frac {1}{10}}{4\left ( 8+1\right ) }=\frac {1}{360}\]

\begin{align*} y_{1} & =\sum _{n=0}^{\infty }a_{n}x^{n+r}=x\sum _{n=0}^{\infty }a_{n}x^{n}\\ & =x\left ( a_{0}+a_{1}x+a_{2}x^{2}+\cdots \right ) \\ & =x\left ( 1+\frac {x^{2}}{10}+\frac {x^{4}}{360}+\cdots \right ) \end{align*}

And for \(r=\frac {1}{2}\) the recurrence relation (1) becomes, and using \(b\) instead of \(a\)

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

Notice also that \(b_{1}=0\) just like \(a_{1}=0\) from above. Now, for \(n=2\) and using \(b_{0}=1\)

\[ b_{n}=\frac {b_{2}}{4\left ( 8-1\right ) }=\frac {\frac {1}{6}}{4\left ( 8-1\right ) }=\frac {1}{168}\]

\begin{align*} y_{2} & =\sum _{n=0}^{\infty }b_{n}x^{n+r_{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}{6}x^{2}+\frac {1}{168}x^{4}+\cdots \right ) \end{align*}

\begin{align*} y_{h} & =c_{1}y_{1}+c_{2}y_{2}\\ & =c_{1}\left ( x\left ( 1+\frac {x^{2}}{10}+\frac {x^{4}}{360}+\cdots \right ) \right ) +c_{2}\sqrt {x}\left ( 1+\frac {1}{6}x^{2}+\frac {1}{168}x^{4}+\cdots \right ) \\ & =c_{1}\left ( x+\frac {x^{3}}{10}+\frac {x^{5}}{360}+\cdots \right ) +c_{2}\sqrt {x}\left ( 1+\frac {1}{6}x^{2}+\frac {1}{168}x^{4}+\cdots \right ) \end{align*}

\[ 2x^{2}y^{\prime \prime }-xy^{\prime }+\left ( 1-x^{2}\right ) y=\left ( 2r^{2}-3r+1\right ) a_{0}x^{r}\]

To ﬁnd \(y_{p}\), and relabeling \(r\) as \(m\) and \(a\) as \(c\) so not to confuse terms used for \(y_{h}\). Then the above becomes

\[ 2x^{2}y^{\prime \prime }-xy^{\prime }+\left ( 1-x^{2}\right ) y=\left ( 2m^{2}-3m+1\right ) c_{0}x^{m}\]

Therefore \(\left ( 2m^{2}-3m+1\right ) c_{0}=1\), or \(\left ( 8-6+1\right ) c_{0}=1\) or

The recurrence relation (1) from above now becomes (using \(m\) for \(r\) and \(c_{0}\) for \(a_{0}\))

\begin{align*} c_{n} & =\frac {c_{n-2}}{2\left ( n+2\right ) \left ( n+1\right ) -\left ( n+2\right ) +1}\\ & =\frac {c_{n-2}}{2n^{2}+5n+3}\end{align*}

For \(n=1\) we use \(c_{1}=0\) the same as was found for \(a_{1},b_{1}\). For \(n\geq 2\) the above is used. Hence for \(n=2\)

\begin{align*} y_{p} & =\sum _{n=0}^{\infty }c_{n}x^{n+m}=x^{2}\sum _{n=0}^{\infty }c_{n}x^{n}\\ & =x^{2}\left ( c_{0}+c_{1}x+c_{2}x^{2}+\cdots \right ) \\ & =x^{2}\left ( \frac {1}{3}+\frac {1}{63}x^{2}+\frac {1}{3465}x^{4}+\cdots \right ) \\ & =\frac {1}{3}x^{2}+\frac {1}{63}x^{4}+\frac {1}{3465}x^{6}+\cdots \end{align*}

\begin{align*} y & =y_{h}+y_{p}\\ & =c_{1}\left ( x+\frac {x^{3}}{10}+\frac {x^{5}}{360}+\cdots \right ) +c_{2}\sqrt {x}\left ( 1+\frac {1}{6}x^{2}+\frac {1}{168}x^{4}+\cdots \right ) +\left ( \frac {1}{3}x^{2}+\frac {1}{63}x^{4}+\frac {1}{3465}x^{6}+\cdots \right ) \end{align*}

Alternative way to ﬁnd \(y_{p}\) is the the following. Let \(y_{p}=c_{0}+c_{1}x+c_{2}x^{2}+c_{3}x^{3}+\cdots \) then \(y_{p}^{\prime }=c_{1}+2c_{2}x+3c_{3}x^{2}+\cdots \) and \(y_{p}^{\prime \prime }=2c_{2}+6c_{3}x+\cdots \). Hence the ode becomes

\begin{align*} 2x^{2}\left ( 2c_{2}+6c_{3}x+\cdots \right ) -x\left ( c_{1}+2c_{2}x+3c_{3}x^{2}+\cdots \right ) +\left ( 1-x^{2}\right ) \left ( c_{0}+c_{1}x+c_{2}x^{2}+c_{3}x^{3}+\cdots \right ) & =x^{2}\\ c_{0}+x\left ( -c_{1}+c_{1}\right ) +x^{2}\left ( 4c_{2}-2c_{2}+c_{2}-c_{0}\right ) +x^{3}\left ( \cdots \right ) & =x^{2}\end{align*}

Hence \(c_{0}=0,4c_{2}-2c_{2}+c_{2}-c_{0}=1\) or \(3c_{2}-c_{0}=1\) or \(c_{2}=\frac {1}{3}\). We need to keep adding more equations and solving them simultaneously. This method is not as easy to use as the method used above, which uses the balance equation to ﬁnd to \(y_{p}\). Also this method could fail, since in practice we should not use undetermined coeﬃcients method (which is what this does) on an ode with variable coeﬃcients. So I will not use this any more.

2.5.2.5 Example 5 \(2x^{2}y^{\prime \prime }-xy^{\prime }+\left ( 1-x^{2}\right ) y=x^{2}\)