3.2.3 Example 3

Let

\[ y^{\prime \prime }=\left ( x^{2}+3\right ) y \]

Therefore

\begin{align} r & =\frac {s}{t}\nonumber \\ & =\frac {x^{2}+3}{1} \tag {1}\end{align}

Step 1 In this step we find all \(\left [ \sqrt {r}\right ] _{c}\) and associated \(\alpha _{c}^{\pm }\) for each pole. There are no poles. In this case\(\ \alpha _{c}^{\pm }=0\) (paper was not explicit in saying this, but from example 3 in paper this can be inferred). Hence the value of \(d\) is decided by \(\alpha _{\infty }^{\pm }\) only in this case.

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

\[ v=1 \]

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

\[ x+\frac {3}{2x}-\frac {9}{8x^{3}}+\cdots \]

Hence we need the coefficient of \(x^{1}\) in this series (\(1\) because that is value of \(v\)). Recall that \(\left [ \sqrt {r}\right ] _{\infty }\) is the sum of terms of \(x^{j}\) for \(0\leq j\leq v\) or for \(j=0,1\) since \(v=1\). Looking at the series above, we see that

\[ a=1 \]

Which is the coefficient of the term \(x\). There is no term \(x^{0}\). Hence

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

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

\begin{align*} r & =\frac {s}{t}\\ & =\frac {x^{2}+3}{1}\\ & =Q+\frac {R}{1}\end{align*}

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

\[ r=x^{2}+3+\frac {0}{1}\]

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

\begin{align*} b & =3-0\\ & =3 \end{align*}

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

This completes step 1 of the solution. We have found \(\left [ \sqrt {r}\right ] _{c}\) and its associated \(\alpha _{c}^{\pm }\) (these are zero, in this example, since there are no poles) 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 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 \(c\)	\(\alpha _{c}\) values (all zero)	\(O\left ( \infty \right ) \) value	\(d\)	\(d\) value
\(x=\ \)N/A	\(\alpha _{c}=0\)	\(\alpha _{\infty }^{+}=1\)	\(\alpha _{\infty }^{+}=1\)	\(1\)
\(x=\ \)N/A	\(\alpha _{c}=0\)	\(\alpha _{\infty }^{-}=-2\)	\(\alpha _{\infty }^{-}=-2\)	\(-2\)

Picking the positive \(d\) integers, this gives

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

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

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 \(\omega \) (there is only one in this example) we solve the above Eq (8). Starting with \(\omega =x\) associated with first \(d=1\) in the table, then (8) becomes

\begin{align*} P^{\prime \prime }+2\left ( x\right ) P^{\prime }+\left ( \left ( x\right ) ^{\prime }+\left ( x\right ) ^{2}-\left ( x^{2}+3\right ) \right ) P & =0\\ P^{\prime \prime }+2xP^{\prime }+\left ( 1+x^{2}-x^{2}-3\right ) P & =0\\ P^{\prime \prime }+2xP^{\prime }-2P & =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*} 2x-2\left ( x+a\right ) & =0\\ 2x-2x-2a & =0\\ 2a & =0 \end{align*}

Which means

\[ a=0 \]

Hence we found the polynomial

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

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

\begin{align*} y & =pe^{\int \omega dx}\\ & =xe^{\int x\ dx}\\ & =xe^{\frac {x^{2}}{2}}\end{align*}

The second solution can be found by reduction of order. The full general solution to \(y^{\prime \prime }=\left ( x^{2}+3\right ) y\) is

\[ y\left ( x\right ) =c_{1}xe^{\frac {x^{2}}{2}}+c_{2}\left ( xe^{\frac {x^{2}}{2}}\sqrt {\pi }\operatorname {erf}\left ( x\right ) +e^{\frac {-x^{2}}{2}}\right ) \]