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