3.2.4 Example 4

Let

\[ y^{\prime \prime }=\frac {2}{x^{2}}y \]

Therefore

\begin{align} r & =\frac {s}{t}\nonumber \\ & =\frac {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 \(\frac {2}{x^{2}}\). 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 \(\deg \left ( t\right ) -\deg \left ( s\right ) =2-0=2\). Since \(O\left ( \infty \right ) =2\) then from the algorithm above

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

Now we calculate \(b\) for this case. This is given by the leading coefficient of \(s\) divided by the leading coefficient of \(t\) when \(\gcd \left ( s,t\right ) =1\). In this case \(r=\frac {2}{x^{2}}\) , hence \(b=\frac {2}{1}=2\). Therefore

\begin{align*} \left [ \sqrt {r}\right ] _{\infty } & =0\\ \alpha _{\infty }^{+} & =\frac {1}{2}+\frac {1}{2}\sqrt {1+4b}=\frac {1}{2}+\frac {1}{2}\sqrt {1+8}=2\\ \alpha _{\infty }^{-} & =\frac {1}{2}-\frac {1}{2}\sqrt {1+4b}=\frac {1}{2}-\frac {1}{2}\sqrt {1+8}=-1 \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
\(x=0\)	\(\alpha _{c}^{+}=2\)	\(\alpha _{\infty }^{+}=2\)	\(\alpha _{\infty }^{+}-\left ( \alpha _{c}^{+}\right ) =2-\left ( 2\right ) \)	\(0\)
\(x=0\)	\(\alpha _{c}^{+}=2\)	\(\alpha _{\infty }^{-}=-1\)	\(\alpha _{\infty }^{-}-\left ( \alpha _{c}^{+}\right ) =-1-\left ( 2\right ) \)	\(-3\)
\(x=0\)	\(\alpha _{c}^{-}=-1\)	\(\alpha _{\infty }^{+}=2\)	\(\alpha _{\infty }^{+}-\left ( \alpha _{c}^{-}\right ) =2-\left ( -1\right ) \)	\(3\)
\(x=0\)	\(\alpha _{c}^{-}=-1\)	\(\alpha _{\infty }^{-}=-1\)	\(\alpha _{\infty }^{-}-\left ( \alpha _{c}^{-}\right ) =-1-\left ( -1\right ) \)	\(0\)

Hence the trial \(d\) values which are not negative are

\[ d=\left \{ 0,3\right \} \]

For \(d=0\,\), since it shows in two rows, we take the first row. Now we generate \(\omega \) for each \(d\) 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, 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 \(\left ( \sum _{c}s\left ( c\right ) \left [ \sqrt {r}\right ] _{c}+\frac {\alpha _{c}}{x-c}\right ) +s\left ( \infty \right ) \left [ \sqrt {r}\right ] _{\infty }\)
\(x=0\)	\(\alpha _{c}^{+}=2\)	\(+\)	\(0\)	\(\alpha _{\infty }^{+}=2\)	\(+\)	\(0\)	\(0\)	\(\left ( +\left ( 0\right ) +\frac {2}{x-0}\right ) +\left ( +\right ) \left ( 0\right ) =\frac {2}{x}\)
\(x=0\)	\(\alpha _{c}^{-}=-1\)	\(-\)	\(0\)	\(\alpha _{\infty }^{+}=2\)	\(+\)	\(0\)	\(3\)	\(\left ( -\left ( 0\right ) +\frac {-1}{x-0}\right ) +\left ( +\right ) \left ( 0\right ) =\frac {-1}{x}\)

The above gives two candidate \(\omega =\left \{ \frac {2}{x},\frac {-1}{x}\right \} \) 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 \(\omega =\frac {2}{x}\) associated with first \(d=0\) in the table, then (8) becomes

\begin{align*} P^{\prime \prime }+2\left ( \frac {2}{x}\right ) P^{\prime }+\left ( \left ( \frac {2}{x}\right ) ^{\prime }+\left ( \frac {2}{x}\right ) ^{2}-\left ( \frac {2}{x^{2}}\right ) \right ) P & =0\\ P^{\prime \prime }+\frac {4}{x}P^{\prime }+\left ( -\frac {2}{x^{2}}+\frac {4}{x^{2}}-\frac {2}{x^{2}}\right ) P & =0\\ P^{\prime \prime }+\frac {4}{x}P^{\prime } & =0 \end{align*}

This needs to be solved for \(P\). Since degree of \(p\left ( x\right ) \) is \(d=0\). Let \(p=a\). The above becomes

\[ 0=0 \]

No unique solution. Hence \(d=0\) did not work. Now we try the second \(\omega =\frac {-1}{x}\) associated with \(d=3\). Substituting in 8 gives

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

Since \(d=3\), let

\[ p\left ( x\right ) =x^{3}+ax^{2}+bx+c \]

Then \(P^{\prime \prime }-\frac {2}{x}P^{\prime }=0\) becomes

\begin{align*} \left ( 6x+2a\right ) -\frac {2}{x}\left ( 3x^{2}+2ax+b\right ) & =0\\ -2a-2\frac {b}{x} & =0 \end{align*}

Hence \(a=0,b=0\) is solution. \(c\) is arbitrary. Taking \(c=0\) then

\[ p\left ( x\right ) =x^{3}\]

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

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

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

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