5.3.2.1 Example 1

Let

\[ y^{\prime \prime }=\frac {4x^{6}-8x^{5}+12x^{4}+4x^{3}+7x^{2}-20x+4}{4x^{4}}y \]

Therefore

\begin{align} r & =\frac {s}{t}\nonumber \\ & =\frac {4x^{6}-8x^{5}+12x^{4}+4x^{3}+7x^{2}-20x+4}{4x^{4}}\nonumber \\ & =x^{2}-2x+3+\frac {1}{x}+\frac {7}{x^{2}}-\frac {5}{x^{3}}+\frac {1}{x^{4}} \tag {1}\end{align}

Step 1 In this we find all \(\left [ \sqrt {r}\right ] _{c}\) and associated \(\alpha _{c}^{\pm }\) for each pole. There is only one pole at \(x=0\) of order 4. Hence \(2v=4\). And \(v=2\). This is step (C3) now in the paper (1).

We need now to find Laurent series of \(\sqrt {r}\) expanded around \(x=c=0\). This is given by (using series command on the computer)

\begin{equation} \frac {1}{x^{2}}-\frac {5}{2}\frac {1}{x}-\frac {9}{4}-\frac {41}{8}x-\frac {443}{32}x^{2}+\cdots \tag {2}\end{equation}

We need to add all terms in the Laurent series expansion of \(\sqrt {r}\) from \(v=2\) down to \(2\). Hence

\begin{equation} \left [ \sqrt {r}\right ] _{c}=\frac {1}{\left ( x-0\right ) ^{2}} \tag {3}\end{equation}

Is only term from 2. Comparing the above to \(\frac {a}{\left ( x-0\right ) ^{2}}\) shows that

\begin{equation} a=1 \tag {4}\end{equation}

Hence

\begin{align} \left [ \sqrt {r}\right ] _{c} & =\frac {1}{x^{2}}\tag {5}\\ \alpha _{c}^{+} & =\frac {1}{2}\left ( \frac {b}{a}+v\right ) \nonumber \\ \alpha _{c}^{-} & =\frac {1}{2}\left ( -\frac {b}{a}+v\right ) \tag {6}\end{align}

Where \(v=2\) and \(a=1\). We still need to find \(b\). But \(b\) is the coefficient of the term \(\frac {1}{\left ( x-0\right ) ^{v+1}}\) in \(r\) minus the coefficient of \(\frac {1}{\left ( x-0\right ) ^{v+1}}\) in \(\left [ \sqrt {r}\right ] _{c}\) which we just found above. Looking at \(r\) from Eq (1) we see that the term \(\frac {1}{\left ( x-0\right ) ^{v+1}}=\frac {1}{\left ( x-0\right ) ^{3}}\) has coefficient \(-5\). And looking at Eq (3) we see that there is no term \(\frac {1}{\left ( x-0\right ) ^{3}}\) in it. Hence

\begin{align} b & =-5-0\nonumber \\ & =-5 \tag {7}\end{align}

Now we found \(a,b\), then (5,6) becomes (since \(v=2\))

\begin{align} \alpha _{c}^{+} & =\frac {1}{2}\left ( -5+2\right ) =\frac {-3}{2}\tag {8}\\ \alpha _{c}^{-} & =\frac {1}{2}\left ( 5+2\right ) =\frac {7}{2} \tag {9}\end{align}

We are done with all the poles.

Now we consider \(O\left ( \infty \right ) \) which is \(\deg \left ( t\right ) -\deg \left ( s\right ) =4-6=-2\). Since this is even order and negative then \(-2v=-2\) or

\[ v=1 \]

We need the Laurent series of \(\sqrt {r}\) around \(\infty \). Using the computer this is

\[ \left [ \sqrt {r}\right ] _{\infty }=x-1+\frac {1}{x}+\frac {3}{2}\frac {1}{x^{2}}+\frac {15}{8x^{3}}+\cdots \]

Now we only want the terms \(\,x^{i}\) where \(0\leq i\leq v\). This implies the above is reduced to

\[ \left [ \sqrt {r}\right ] _{\infty }=x-1 \]

The \(a\) is the coefficient of \(x^{v}=x\) which is

\[ a=1 \]

Now we need to find \(\alpha _{\infty }^{\pm }\) associated with \(\left [ \sqrt {r}\right ] _{\infty }\). For this we need to first find \(b\). Recall from above that \(b\) is the coefficient of \(x^{v-1}\) or \(x^{0}\) in \(r\) minus the coefficient of \(x^{v-1}=x^{0}\) in \(\left ( \left [ \sqrt {r}\right ] _{\infty }\right ) ^{2}\). Since \(v=1\) then we want the coefficient of \(x^{0}\) in \(r\) and subtract from it the coefficient of \(x^{0}\) in \(\left ( \left [ \sqrt {r}\right ] _{\infty }\right ) ^{2}\). But

\begin{align*} \left ( \left [ \sqrt {r}\right ] _{\infty }\right ) ^{2} & =\left ( x-1\right ) ^{2}\\ & =x^{2}+1-2x \end{align*}

Hence the coefficient of \(x^{0}\) in \(\left ( \left [ \sqrt {r}\right ] _{\infty }\right ) ^{2}\) is \(1\).  To find the coefficient of \(x^{0}\) in \(r\) long division is done

\begin{align*} r & =\frac {s}{t}\\ & =\frac {4x^{6}-8x^{5}+12x^{4}+4x^{3}+7x^{2}-20x+4}{4x^{4}}\\ & =Q+\frac {R}{4x^{2}}\end{align*}

Where \(Q\) is the quotient and \(R\) is the remainder. This gives

\[ r=\left ( x^{2}-2x+3\right ) +\frac {4x^{3}+7x^{2}-20x+4}{4x^{2}}\]

For the case of \(v\neq 0\) then the coefficient is read from \(Q\) above. Which is \(3\). Hence

\begin{align*} b & =3-1\\ & =2 \end{align*}

For the other case of \(v=0\) then the coefficient of \(x^{-1}\) in \(r\) is found using \(\frac {lcoeff\left ( R\right ) }{lcoeff\left ( t\right ) }\) which will give \(1\) in this case. (More examples below).

Now that we found \(a,b\), then from the above section describing the algorithm, we see in this case that

\begin{align*} \alpha _{\infty }^{+} & =\frac {1}{2}\left ( \frac {b}{a}-v\right ) =\frac {1}{2}\left ( \frac {2}{1}-1\right ) =\frac {1}{2}\\ \alpha _{\infty }^{-} & =\frac {1}{2}\left ( -\frac {b}{a}-v\right ) =\frac {1}{2}\left ( -\frac {2}{1}-1\right ) =-\frac {3}{2}\end{align*}

This completes step 1 of the solution. We have found \(\left [ \sqrt {r}\right ] _{c}\) and its associated \(\alpha _{c}^{\pm }\) and found \(\left [ \sqrt {r}\right ] _{\infty }\) and its associated \(\alpha _{\infty }^{\pm }\). Now we go to step 2 which is to find the \(d^{\prime }s\).

step 2 Since we have a pole at zero, and we have one \(O\left ( \infty \right ) \), each with \(\pm \) signs, then we set up this table to make it easier to work with. This implements

\[ d=\alpha _{\infty }^{\pm }-\sum _{i=1}^{1}\alpha _{c_{i}}^{\pm }\]

Therefore we obtain 4 possible \(d\) values.

pole \(c\)	\(\alpha _{c}\) value	\(O\left ( \infty \right ) \) value	\(d\)	\(d\) value
\(x=0\)	\(\alpha _{c}^{+}=\frac {-3}{2}\)	\(\alpha _{\infty }^{+}=\frac {1}{2}\)	\(\alpha _{\infty }^{+}-\left ( \alpha _{c}^{+}\right ) =\frac {1}{2}-\left ( \frac {-3}{2}\right ) \)	\(2\)
\(x=0\)	\(\alpha _{c}^{+}=\frac {-3}{2}\)	\(\alpha _{\infty }^{-}=-\frac {3}{2}\)	\(\alpha _{\infty }^{-}-\left ( \alpha _{c}^{+}\right ) =-\frac {3}{2}-\left ( \frac {-3}{2}\right ) \)	\(0\)
\(x=0\)	\(\alpha _{c}^{-}=\frac {7}{2}\)	\(\alpha _{\infty }^{+}=\frac {1}{2}\)	\(\alpha _{\infty }^{+}-\left ( \alpha _{c}^{-}\right ) =\frac {1}{2}-\left ( \frac {7}{2}\right ) \)	\(-3\)
\(x=0\)	\(\alpha _{c}^{-}=\frac {7}{2}\)	\(\alpha _{\infty }^{-}=-\frac {3}{2}\)	\(\alpha _{\infty }^{-}-\left ( \alpha _{c}^{-}\right ) =-\frac {3}{2}-\left ( \frac {7}{2}\right ) \)	\(-5\)

We see from the above that we took each pole in this problem (there is only one pole here at \(x=0\)) and its associated \(\alpha _{c}^{\pm }\) with each \(\alpha _{\infty }^{\pm }\) and generated all possible \(d\) values from all the combinations. Hence we obtain 4 possible \(d\) values in this case. If we had 2 poles, then we would have 8 possible \(d\) values. Hence the maximum possible \(d\) values we can get is \(2^{p+1}\) where \(p\) is number of poles. Now we remove all negative \(d\) values. Hence the trial \(d\) values remaining is

\[ d=\left \{ 0,2\right \} \]

Now for each \(d\) value, we generate \(\omega \) using

\[ \omega =\left ( \sum _{c}s\left ( c\right ) \left [ \sqrt {r}\right ] _{c}+\frac {\alpha _{c}}{x-c}\right ) +s\left ( \infty \right ) \left [ \sqrt {r}\right ] _{\infty }\]

To apply the above, we update the table above, now using only \(d=0,d=2\) values, but also add columns for \(\left [ \sqrt {r}\right ] _{c},\left [ \sqrt {r}\right ] _{\infty }\) to make the computation easier. Here is the new table

pole \(c\)	\(\alpha _{c}\) value	\(s\left ( c\right ) \)	\(\left [ \sqrt {r}\right ] _{c}\)	\(O\left ( \infty \right ) \) value	\(s\left ( \infty \right ) \)	\(\left [ \sqrt {r}\right ] _{\infty }\)	\(d\) value	\(\omega \) value \(\left ( \sum _{c}s\left ( c\right ) \left [ \sqrt {r}\right ] _{c}+\frac {\alpha _{c}}{x-c}\right ) +s\left ( \infty \right ) \left [ \sqrt {r}\right ] _{\infty }\)
\(x=0\)	\(\alpha _{c}^{+}=\frac {-3}{2}\)	\(+\)	\(\frac {1}{x^{2}}\)	\(\alpha _{\infty }^{-}=-\frac {3}{2}\)	\(-\)	\(x-1\)	\(0\)	\(\left ( +\left ( \frac {1}{x^{2}}\right ) +\frac {\frac {-3}{2}}{x-0}\right ) +\left ( -\right ) \left ( x-1\right ) =\frac {1}{x^{2}}-\frac {3}{2x}-x+1\)
\(x=0\)	\(\alpha _{c}^{+}=\frac {-3}{2}\)	\(+\)	\(\frac {1}{x^{2}}\)	\(\alpha _{\infty }^{+}=\frac {1}{2}\)	\(+\)	\(x-1\)	\(2\)	\(\left ( +\left ( \frac {1}{x^{2}}\right ) +\frac {\frac {-3}{2}}{x-0}\right ) +\left ( +\right ) \left ( x-1\right ) =x-\frac {3}{2x}+\frac {1}{x^{2}}-1\)

The above are the two candidate \(\omega \) values. For each \(\omega \) we need to find polynomial \(P\) by solving

\begin{equation} P^{\prime \prime }+2\omega P^{\prime }+\left ( \omega ^{\prime }+\omega ^{2}-r\right ) P=0 \tag {8}\end{equation}

If we are able to find \(P\), then we stop and the ode \(y^{\prime \prime }=ry\) is solved. If we try all candidate \(\omega \) and can not find \(P\) then this case is not successful and we go to the next case.

step 3 Now for each candidate \(\omega \) we solve the above Eq (8). Starting with \(\omega =\frac {1}{x^{2}}-\frac {3}{2x}-x+1\) associated with \(d=0\) in the table, then (8) becomes

\begin{align*} P^{\prime \prime }+2\left ( \frac {1}{x^{2}}-\frac {3}{2x}-x+1\right ) P^{\prime }+\left ( \left ( \frac {3}{2x^{2}}-\frac {2}{x^{3}}-1\right ) +\left ( \frac {1}{x^{2}}-\frac {3}{2x}-x+1\right ) ^{2}-\left ( \frac {4x^{6}-8x^{5}+12x^{4}+4x^{3}+7x^{2}-20x+4}{4x^{4}}\right ) \right ) P & =0\\ P^{\prime \prime }+\left ( \frac {2}{x^{2}}-\frac {3}{2x}-2x+2\right ) P^{\prime }+\left ( \frac {4}{x^{2}}-\frac {6}{x}\right ) P & =0 \end{align*}

Since this the case for \(d=0\), then \(P\) has zero degree, Hence \(P\) is constant. Therefore the above simplifies to

\[ \left ( \frac {4}{x^{2}}-\frac {6}{x}\right ) P=0 \]

Which means

\[ \frac {4}{x^{2}}-\frac {6}{x}=0 \]

Which is not possible for all \(x\). Hence \(d=0\) do not generate valid \(P\) polynomial. We now try the case of \(d=2\). Since \(d=2\), it means the polynomial \(d\) is of degree two. Let

\[ P=x^{2}+ax+b \]

Substituting this in (8) using \(\omega =x-\frac {3}{2x}+\frac {1}{x^{2}}-1\). This gives

\begin{align*} P^{\prime \prime }+2\left ( x-\frac {3}{2x}+\frac {1}{x^{2}}-1\right ) P^{\prime }+\left ( \left ( \frac {3}{2x^{2}}-\frac {2}{x^{3}}+1\right ) +\left ( x-\frac {3}{2x}+\frac {1}{x^{2}}-1\right ) ^{2}-\left ( \frac {4x^{6}-8x^{5}+12x^{4}+4x^{3}+7x^{2}-20x+4}{4x^{4}}\right ) \right ) P & =0\\ P^{\prime \prime }+2\left ( x-\frac {3}{2x}+\frac {1}{x^{2}}-1\right ) P^{\prime }+\left ( \frac {4}{x}-4\right ) P & =0 \end{align*}

Using \(P=x^{2}+ax+b\) the above becomes

\begin{align*} 2+2\left ( x-\frac {3}{2x}+\frac {1}{x^{2}}-1\right ) \left ( 2x+a\right ) +\left ( \frac {4}{x}-4\right ) \left ( x^{2}+ax+b\right ) & =0\\ 2a-4b-3\frac {a}{x}+2\frac {a}{x^{2}}+4\frac {b}{x}-2ax+\frac {4}{x}-4 & =0\\ \left ( 2a-4b-4\right ) +\frac {1}{x}\left ( -\frac {3}{4}a+4b+4\right ) +\frac {1}{x^{2}}\left ( 2a\right ) -2ax & =0 \end{align*}

Therefore

\begin{align*} 2a-4b-4 & =0\\ -\frac {3}{4}a+4b+4 & =0\\ 2a & =0\\ 2a & =0 \end{align*}

hence \(a=0\) from last equation. Using first or second equation gives \(b=-1\). Therefore a solution is found. Hence

\[ p\left ( x\right ) =x^{2}-1 \]

Therefore the solution to \(y^{\prime \prime }=ry\) is

\begin{align*} y & =p\left ( x\right ) e^{\int \omega dx}\\ & =\left ( x^{2}-1\right ) e^{\int x-\frac {3}{2x}+\frac {1}{x^{2}}-1\ dx}\\ & =\left ( x^{2}-1\right ) e^{-\frac {1}{x}-\frac {3}{2}\ln x+\frac {x^{2}}{2}-x}\\ & =\left ( x^{2}-1\right ) x^{\frac {-3}{2}}e^{-\frac {1}{x}+\frac {x^{2}}{2}-x}\end{align*}

The second solution can be found by reduction of order.