3.2.2 Example 2

Let

\[ y^{\prime \prime }=\left ( \frac {2}{x^{2}}-1\right ) y \]

Therefore

\begin{align} r & =\frac {s}{t}\nonumber \\ & =\frac {2-x^{2}}{x^{2}} \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 one pole at \(x=0\) of order 2. In this case, from the description of the algorithm earlier, we write

\begin{align*} \left [ \sqrt {r}\right ] _{c} & =0\\ \alpha _{c}^{+} & =\frac {1}{2}+\frac {1}{2}\sqrt {1+4b}\\ \alpha _{c}^{-} & =\frac {1}{2}-\frac {1}{2}\sqrt {1+4b}\end{align*}

Where \(b\) is the coefficient of \(\frac {1}{\left ( x-0\right ) ^{2}}\) in the partial fraction decomposition of \(r\) which is \(\frac {2}{x^{2}}-1\). Hence \(b=2\). Therefore

\begin{align*} \left [ \sqrt {r}\right ] _{c} & =0\\ \alpha _{c}^{+} & =\frac {1}{2}+\frac {1}{2}\sqrt {1+8}=\frac {1}{2}+\frac {3}{2}=2\\ \alpha _{c}^{-} & =\frac {1}{2}-\frac {1}{2}\sqrt {1+8}=\frac {1}{2}-\frac {3}{2}=-1 \end{align*}

We are done with all the poles. Now we consider \(O\left ( \infty \right ) \) which is \(\deg \left ( t\right ) -\deg \left ( s\right ) =2-2=0\). This falls in the case \(-2v\leq 0\). Hence

\[ v=0 \]

We need the Laurent series of \(\sqrt {r}\) around \(\infty \). Using the computer this is \(i-\frac {i}{x^{2}}-\frac {1}{2x^{4}}i+\cdots \). Hence we need the coefficient of \(x^{0}\) in this series (\(0\) because that is value of \(v\)).

\[ \left [ \sqrt {r}\right ] _{\infty }=ix^{0}\]

Recall that \(\left [ \sqrt {r}\right ] _{\infty }\) is the sum of terms of \(x^{j}\) for \(0\leq j\leq v\) or for \(j=0\) since \(v=0\). Looking at the series above, we see that

\[ a=i \]

Which is the coefficient of the term \(x^{0}\). Now we need to find \(\alpha _{\infty }^{\pm }\) associated with \(\left [ \sqrt {r}\right ] _{\infty }\). For this we need to first find \(b\) which is the coefficient of \(x^{v-1}=\frac {1}{x}\) in \(r\) minus the coefficient of \(x^{v-1}=\frac {1}{x}\) in \(\left ( \left [ \sqrt {r}\right ] _{\infty }\right ) ^{2}\). But

\[ \left ( \left [ \sqrt {r}\right ] _{\infty }\right ) ^{2}=i^{2}=-1 \]

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

\begin{align*} r & =\frac {s}{t}\\ & =\frac {2-x^{2}}{x^{2}}\\ & =Q+\frac {R}{x^{2}}\end{align*}

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

\[ r=-1+\frac {2}{x^{2}}\]

For the other case of \(v=0\) then the coefficient of \(x^{-1}\) is found by looking up the coefficient in \(R\) of \(x\) to the degree of of \(t\) then subtracting one and dividing result by \(lcoeff\left ( t\right ) \). But degree of \(t\) is \(2\). Therefore we want the coefficient of \(x^{2-1}\) or \(x\) in \(R\) which is zero. Hence \(b=0-0=0\).

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 ( 0-0\right ) =0\\ \alpha _{\infty }^{-} & =\frac {1}{2}\left ( -\frac {b}{a}-v\right ) =\frac {1}{2}\left ( -0-0\right ) =0 \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
\(x=0\)	\(\alpha _{c}^{+}=2\)	\(\alpha _{\infty }^{+}=0\)	\(\alpha _{\infty }^{+}-\left ( \alpha _{c}^{+}\right ) =0-\left ( 2\right ) \)	\(-2\)
\(x=0\)	\(\alpha _{c}^{+}=2\)	\(\alpha _{\infty }^{-}=0\)	\(\alpha _{\infty }^{-}-\left ( \alpha _{c}^{+}\right ) =0-\left ( 2\right ) \)	\(-2\)
\(x=0\)	\(\alpha _{c}^{-}=-1\)	\(\alpha _{\infty }^{+}=0\)	\(\alpha _{\infty }^{+}-\left ( \alpha _{c}^{-}\right ) =0-\left ( -1\right ) \)	\(1\)
\(x=0\)	\(\alpha _{c}^{-}=-1\)	\(\alpha _{\infty }^{-}=0\)	\(\alpha _{\infty }^{-}-\left ( \alpha _{c}^{-}\right ) =0-\left ( -1\right ) \)	\(1\)

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

\[ d=\left \{ 1\right \} \]

There is one \(d\) value to try. We can pick any one of the two values of \(d\) generated since there are both \(d=1\). Both will give same solution. 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 the first \(d=1\) value in the above table. (selecting the second \(d=1\) row, will not change the final solution). but we 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 \(\left ( \sum _{c}s\left ( c\right ) \left [ \sqrt {r}\right ] _{c}+\frac {\alpha _{c}}{x-c}\right ) +s\left ( \infty \right ) \left [ \sqrt {r}\right ] _{\infty }\)
\(x=0\)	\(\alpha _{c}^{-}=-1\)	\(-\)	\(0\)	\(\alpha _{\infty }^{+}=0\)	\(+\)	\(i\)	\(1\)	\(\left ( -\left ( 0\right ) +\frac {-1}{x-0}\right ) +\left ( +\right ) \left ( i\right ) =\frac {-1}{x}+i\)

The above gives one candidate \(\omega \) value to try. For this \(\omega \) 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 \(\omega =\frac {-1}{x}+i\) associated with first \(d=1\) in the table, then (8) becomes

\begin{align*} P^{\prime \prime }+2\left ( \frac {-1}{x}+i\right ) P^{\prime }+\left ( \left ( \frac {-1}{x}+i\right ) ^{\prime }+\left ( \frac {-1}{x}+i\right ) ^{2}-\left ( \frac {2}{x^{2}}-1\right ) \right ) P & =0\\ P^{\prime \prime }+2\left ( \frac {-1}{x}+i\right ) P^{\prime }+\left ( \frac {-2i}{x}\right ) P & =0 \end{align*}

This needs to be solved for \(P\). Since degree of \(p\left ( x\right ) \) is \(d=1\). Let \(p=x+a\). The above becomes

\begin{align*} 2\left ( \frac {-1}{x}+i\right ) +\left ( \frac {-2i}{x}\right ) \left ( x+a\right ) & =0\\ \frac {-2}{x}+2i-2i-\frac {2ia}{x} & =0\\ \frac {-2}{x}-\frac {2ia}{x} & =0 \end{align*}

Which means

\[ a=i \]

Hence we found the polynomial

\[ p\left ( x\right ) =x+i \]

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

\begin{align*} y & =pe^{\int \omega dx}\\ & =\left ( x+i\right ) e^{\int \frac {-1}{x}+i\ dx}\\ & =\left ( x+i\right ) \frac {1}{x}e^{ix}\\ & =\frac {x+i}{x}\left ( \cos x+i\sin x\right ) \end{align*}

The second solution can be found by reduction of order. The full general solution to \(y^{\prime \prime }=\left ( \frac {2}{x^{2}}-1\right ) y\) is

\[ y\left ( x\right ) =\frac {c_{1}}{x}\left ( x\cos x-\sin x\right ) +\frac {c_{2}}{x}\left ( \cos x+x\sin x\right ) \]