2.5.2.1 Example 1 \(2x^{2}y^{\prime \prime }+3xy^{\prime }-xy=x^{2}+2x\)

\[ 2x^{2}y^{\prime \prime }+3xy^{\prime }-xy=x^{2}+2x \]

Comparing the ode to

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

Hence \(p\left ( x\right ) =\frac {3}{2x},q\left ( x\right ) =\frac {-1}{2x}\). There is one singular point at \(x=0\). Therefore \(p_{0}=\lim _{x\rightarrow 0}xp\left ( x\right ) =\lim _{x\rightarrow 0}\frac {3}{2}=\frac {3}{2}\) 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 {3}{2}r+0 & =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*} 2x^{2}\sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r-2}+3x\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 }2\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r}+\sum _{n=0}^{\infty }3\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

\[ \sum _{n=0}^{\infty }2\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r}+\sum _{n=0}^{\infty }3\left ( n+r\right ) a_{n}x^{n+r}-\sum _{n=1}^{\infty }a_{n-1}x^{n+r}=0 \]

When \(n=0\)

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

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

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

Where the RHS will be zero when \(r=0\) or \(r=-\frac {1}{2}\). For \(n\geq 1\) the recurrence relation is

\begin{align} 2\left ( n+r\right ) \left ( n+r-1\right ) a_{n}+3\left ( n+r\right ) a_{n}-a_{n-1} & =0\nonumber \\ a_{n} & =\frac {a_{n-1}}{2\left ( n+r\right ) \left ( n+r-1\right ) +3\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

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

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

\[ a_{1}=\frac {1}{3}\]

For \(n=2\)

\[ a_{2}=\frac {a_{1}}{8+2}=\frac {a_{1}}{10}=\frac {1}{30}\]

For \(n=3\)

\[ a_{3}=\frac {a_{2}}{18+3}=\frac {a_{2}}{21}=\frac {1}{21\left ( 30\right ) }=\frac {1}{630}\]

And so on. Hence

\begin{align*} y_{1} & =\sum _{n=0}^{\infty }a_{n}x^{n+r}=\sum _{n=0}^{\infty }a_{n}x^{n}\\ & =a_{0}+a_{1}x+a_{2}x^{2}+\cdots \\ & =1+\frac {1}{3}x+\frac {1}{30}x^{2}+\frac {1}{630}x^{3}+\cdots \end{align*}

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

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

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

\[ b_{1}=-\frac {b_{0}}{1-2}=1 \]

For \(n=2\)

\[ b_{2}=-\frac {b_{1}}{2-8}=-\frac {1}{2-8}=\frac {1}{6}\]

For \(n=3\)

\[ b_{3}=-\frac {b_{2}}{3-18}=-\frac {\frac {1}{6}}{3-18}=\frac {1}{90}\]

And so on. Hence

\begin{align*} y_{2} & =\sum _{n=0}^{\infty }b_{n}x^{n+r_{2}}\\ & =\frac {1}{\sqrt {x}}\sum _{n=0}^{\infty }b_{n}x^{n}\\ & =\frac {1}{\sqrt {x}}\left ( b_{0}+b_{1}x+b_{2}x^{2}+\cdots \right ) \\ & =\frac {1}{\sqrt {x}}\left ( 1+x+\frac {1}{6}x^{2}+\frac {1}{90}x^{3}+\cdots \right ) \end{align*}

Hence

\begin{align*} y_{h} & =c_{1}y_{1}+c_{2}y_{2}\\ & =c_{1}\left ( 1+\frac {1}{3}x+\frac {1}{30}x^{2}+\frac {1}{630}x^{3}+\cdots \right ) +c_{2}\frac {1}{\sqrt {x}}\left ( 1+x+\frac {1}{6}x^{2}+\frac {1}{90}x^{3}+\cdots \right ) \end{align*}

Now we find \(y_{p}\). Since ode satisfies

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

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

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

The RHS is \(x^{2}+2x\). We balance each term at a time, this finds a particular solution for each term on the RHS, then these particular solutions are added at the end. For the input \(2x\) the balance equation is

\[ \left ( m\left ( 2m+1\right ) \right ) c_{0}x^{m}=2x \]

This implies that

\[ m=1 \]

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

\[ c_{0}=\frac {2}{3}\]

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

\[ c_{n}=\frac {c_{n-1}}{2n^{2}+4nm+n+2m^{2}+m}\]

For \(m=1\) the above becomes

\[ c_{n}=\frac {c_{n-1}}{2n^{2}+5n+3}\]

For \(n=1\) and using \(c_{0}=\frac {2}{3}\)

\[ c_{1}=\frac {\frac {2}{3}}{2+5+3}=\frac {1}{15}\]

For \(n=2\)

\[ c_{2}=\frac {c_{1}}{8+10+3}=\frac {\frac {1}{15}}{8+10+3}=\frac {1}{315}\]

For \(n=3\)

\[ c_{3}=\frac {c_{2}}{18+15+3}=\frac {\frac {1}{315}}{18+15+3}=\frac {1}{11\,340}\]

And so on. Hence

\begin{align*} y_{p_{1}} & =\sum _{n=0}^{\infty }c_{n}x^{n+m}=x\sum _{n=0}^{\infty }c_{n}x^{n}\\ & =x\left ( c_{0}+c_{1}x+c_{2}x^{2}+\cdots \right ) \\ & =x\left ( \frac {2}{3}+\frac {1}{15}x+\frac {1}{315}x^{2}+\frac {1}{11\,340}x^{3}+\cdots \right ) \\ & =\left ( \frac {2}{3}x+\frac {1}{15}x^{2}+\frac {1}{315}x^{3}+\frac {1}{11\,340}x^{4}+\cdots \right ) \end{align*}

The second term \(x^{2}\) is now balanced \(x^{2}\). The balance equation is

\[ \left ( m\left ( 2m+1\right ) \right ) c_{0}x^{m}=x^{2}\]

Therefore \(m=2\) and \(\left ( m\left ( 2m+1\right ) \right ) c_{0}=1\). Hence

\begin{align*} \left ( 2\left ( 4+1\right ) \right ) c_{0} & =1\\ c_{0} & =\frac {1}{10}\end{align*}

The recurrence relation becomes for \(m=2\)

\[ c_{n}=\frac {c_{n-1}}{2n^{2}+4nm+n+2m^{2}+m}\]

For \(m=2\) the above becomes

\[ c_{n}=\frac {c_{n-1}}{2n^{2}+9n+10}\]

For \(n=1\) and using \(c_{0}=\frac {1}{10}\)

\[ c_{1}=\frac {\frac {1}{10}}{2+9+10}=\frac {1}{210}\]

For \(n=2\)

\[ c_{2}=\frac {c_{1}}{8+18+10}=\frac {\frac {1}{210}}{8+18+10}=\frac {1}{7560}\]

For \(n=3\)

\[ c_{3}=\frac {c_{2}}{18+27+10}=\frac {\frac {1}{7560}}{18+27+10}=\frac {1}{415\,800}\]

And so on. Hence

\begin{align*} y_{p_{2}} & =\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}{10}+\frac {1}{210}x+\frac {1}{7560}x^{2}+\frac {1}{415\,800}x^{3}+\cdots \right ) \\ & =\left ( \frac {1}{10}x^{2}+\frac {1}{210}x^{3}+\frac {1}{7560}x^{4}+\frac {1}{415\,800}x^{5}+\cdots \right ) \end{align*}

The particular solution is the sum of all the particular solutions found above, which is

\begin{align*} y_{p} & =y_{p_{1}}+y_{p_{2}}\\ & =\left ( \frac {2}{3}x+\frac {1}{15}x^{2}+\frac {1}{315}x^{3}+\frac {1}{11\,340}x^{4}+\cdots \right ) +\left ( \frac {1}{10}x^{2}+\frac {1}{210}x^{3}+\frac {1}{7560}x^{4}+\frac {1}{415\,800}x^{5}+\cdots \right ) \\ & =\frac {2}{3}x+\left ( \frac {1}{15}+\frac {1}{10}\right ) x^{2}+\left ( \frac {1}{315}+\frac {1}{210}\right ) x^{3}+\left ( \frac {1}{11\,340}+\frac {1}{7560}\right ) x^{4}+\cdots \\ & =\frac {2}{3}x+\frac {1}{6}x^{2}+\frac {1}{126}x^{3}+\frac {1}{4536}x^{4}+\cdots \end{align*}

Hence the complete solution is

\begin{align*} y & =y_{h}+y_{p}\\ & =c_{1}\left ( 1+\frac {1}{3}x+\frac {1}{30}x^{2}+\frac {1}{630}x^{3}+\cdots \right ) +c_{2}\frac {1}{\sqrt {x}}\left ( 1+x+\frac {1}{6}x^{2}+\frac {1}{90}x^{3}+\cdots \right ) +\frac {2}{3}x+\frac {1}{6}x^{2}+\frac {1}{126}x^{3}+\frac {1}{4536}x^{4}+\cdots \end{align*}