3.2.6 Example 6

Let

\[ y^{\prime \prime }=y \]

Therefore

\[ r=\frac {s}{t}=1 \]

The necessary conditions for case 1 are met since zero order pole and \(O\left ( \infty \right ) =0\).

Step 1 In this we find all \(\left [ \sqrt {r}\right ] _{c}\) and associated \(\alpha _{c}^{\pm }\) for each pole. There are no poles. In this case\(\ \left [ \sqrt {r}\right ] _{c}=0\) and \(\alpha _{c}^{\pm }=0\). Since \(O\left ( \infty \right ) =0\), we are in case \(2v\leq 0\). Hence \(v=0\). Then now \(\left [ \sqrt {r}\right ] _{\infty }\) is the sum of all terms \(x^{i}\) for \(0\leq i\leq v\) in the Laurent series expansion of \(\sqrt {r}\) at \(\infty \).

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

Hence \(a=1\).  And \(b\) is the coefficient of \(x^{v-1}=x^{-1}\) in \(r\) minus the coefficient of \(x^{v-1}=x^{-1}\) in \(\left ( \left [ \sqrt {r}\right ] _{\infty }\right ) ^{2}\). Hence \(b=0\). Then

\begin{align*} \alpha _{\infty }^{+} & =\frac {1}{2}\left ( \frac {b}{a}-v\right ) =0\\ \alpha _{\infty }^{-} & =\frac {1}{2}\left ( -\frac {b}{a}-v\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 pole at \(x=1\), 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

\[ d=\alpha _{\infty }^{\pm }-\sum _{i=1}^{0}\alpha _{c_{i}}^{\pm }\]

Therefore we obtain 2 possible \(d\) values.

pole \(c\)	\(\alpha _{c}\) value	\(O\left ( \infty \right ) \) value	\(d\)	\(d\) value
\(x=NA\)	\(\alpha _{c}=0\)	\(\alpha _{\infty }^{+}=0\)	\(\alpha _{\infty }^{+}=0\)	\(0\)
\(x=NA\)	\(\alpha _{c}=0\)	\(\alpha _{\infty }^{-}=0\)	\(\alpha _{\infty }^{+}=0\)	\(0\)

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

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

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 \(\left ( \sum _{c}s\left ( c\right ) \left [ \sqrt {r}\right ] _{c}+\frac {\alpha _{c}}{x-c}\right ) +s\left ( \infty \right ) \left [ \sqrt {r}\right ] _{\infty }\)
\(x=NA\)	\(\alpha _{c}=0\)	\(+\)	\(0\)	\(\alpha _{\infty }^{+}=0\)	\(+\)	\(0\)	\(0\)	\(\left ( +\left ( 0\right ) +0\right ) +\left ( +\right ) \left ( 1\right ) =1\)

The above gives candidate \(\omega =1\) 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.

step 3 Now for each candidate \(\omega \) we solve the above Eq (8). Starting with \(\omega =1\) associated with \(d=0\) in the table. Let \(p\left ( x\right ) =1\) since degree is zero, then (8) becomes

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

Hence \(p\left ( x\right ) =1\) is valid solution. Therefore the solution to \(y^{\prime \prime }=y\) is

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

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

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