5.3.2.1 Example 1
Let
\[ y^{\prime \prime }=\frac {4x^{6}-8x^{5}+12x^{4}+4x^{3}+7x^{2}-20x+4}{4x^{4}}y \]
Therefore
\begin{align} r & =\frac {s}{t}\nonumber \\ & =\frac {4x^{6}-8x^{5}+12x^{4}+4x^{3}+7x^{2}-20x+4}{4x^{4}}\nonumber \\ & =x^{2}-2x+3+\frac {1}{x}+\frac {7}{x^{2}}-\frac {5}{x^{3}}+\frac {1}{x^{4}} \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 only one pole at \(x=0\) of order
4. Hence \(2v=4\). And \(v=2\). This is step (C3) now in the paper (1).
We need now to find Laurent series of \(\sqrt {r}\) expanded around \(x=c=0\). This is given by (using series
command on the computer)
\begin{equation} \frac {1}{x^{2}}-\frac {5}{2}\frac {1}{x}-\frac {9}{4}-\frac {41}{8}x-\frac {443}{32}x^{2}+\cdots \tag {2}\end{equation}
We need to add all terms in the Laurent series expansion of
\(\sqrt {r}\)
from
\(v=2\) down to
\(2\). Hence
\begin{equation} \left [ \sqrt {r}\right ] _{c}=\frac {1}{\left ( x-0\right ) ^{2}} \tag {3}\end{equation}
Is only term from 2. Comparing the above to
\(\frac {a}{\left ( x-0\right ) ^{2}}\) shows that
\begin{equation} a=1 \tag {4}\end{equation}
Hence
\begin{align} \left [ \sqrt {r}\right ] _{c} & =\frac {1}{x^{2}}\tag {5}\\ \alpha _{c}^{+} & =\frac {1}{2}\left ( \frac {b}{a}+v\right ) \nonumber \\ \alpha _{c}^{-} & =\frac {1}{2}\left ( -\frac {b}{a}+v\right ) \tag {6}\end{align}
Where \(v=2\) and \(a=1\). We still need to find \(b\). But \(b\) is the coefficient of the term \(\frac {1}{\left ( x-0\right ) ^{v+1}}\) in \(r\) minus the
coefficient of \(\frac {1}{\left ( x-0\right ) ^{v+1}}\) in \(\left [ \sqrt {r}\right ] _{c}\) which we just found above. Looking at \(r\) from Eq (1) we see that the term
\(\frac {1}{\left ( x-0\right ) ^{v+1}}=\frac {1}{\left ( x-0\right ) ^{3}}\) has coefficient \(-5\). And looking at Eq (3) we see that there is no term \(\frac {1}{\left ( x-0\right ) ^{3}}\) in it. Hence
\begin{align} b & =-5-0\nonumber \\ & =-5 \tag {7}\end{align}
Now we found \(a,b\), then (5,6) becomes (since \(v=2\))
\begin{align} \alpha _{c}^{+} & =\frac {1}{2}\left ( -5+2\right ) =\frac {-3}{2}\tag {8}\\ \alpha _{c}^{-} & =\frac {1}{2}\left ( 5+2\right ) =\frac {7}{2} \tag {9}\end{align}
We are done with all the poles.
Now we consider \(O\left ( \infty \right ) \) which is \(\deg \left ( t\right ) -\deg \left ( s\right ) =4-6=-2\). Since this is even order and negative then \(-2v=-2\) or
\[ v=1 \]
We need the
Laurent series of
\(\sqrt {r}\) around
\(\infty \). Using the computer this is
\[ \left [ \sqrt {r}\right ] _{\infty }=x-1+\frac {1}{x}+\frac {3}{2}\frac {1}{x^{2}}+\frac {15}{8x^{3}}+\cdots \]
Now we only want the terms
\(\,x^{i}\)
where
\(0\leq i\leq v\). This implies the above is reduced to
\[ \left [ \sqrt {r}\right ] _{\infty }=x-1 \]
The
\(a\) is the coefficient of
\(x^{v}=x\) which is
\[ a=1 \]
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
\(\left ( \left [ \sqrt {r}\right ] _{\infty }\right ) ^{2}\). Since
\(v=1\)
then we want the coefficient of
\(x^{0}\) in
\(r\) and subtract from it the coefficient of
\(x^{0}\) in
\(\left ( \left [ \sqrt {r}\right ] _{\infty }\right ) ^{2}\).
But
\begin{align*} \left ( \left [ \sqrt {r}\right ] _{\infty }\right ) ^{2} & =\left ( x-1\right ) ^{2}\\ & =x^{2}+1-2x \end{align*}
Hence the coefficient of \(x^{0}\) in \(\left ( \left [ \sqrt {r}\right ] _{\infty }\right ) ^{2}\) is \(1\). To find the coefficient of \(x^{0}\) in \(r\) long division is
done
\begin{align*} r & =\frac {s}{t}\\ & =\frac {4x^{6}-8x^{5}+12x^{4}+4x^{3}+7x^{2}-20x+4}{4x^{4}}\\ & =Q+\frac {R}{4x^{2}}\end{align*}
Where \(Q\) is the quotient and \(R\) is the remainder. This gives
\[ r=\left ( x^{2}-2x+3\right ) +\frac {4x^{3}+7x^{2}-20x+4}{4x^{2}}\]
For the case of
\(v\neq 0\) then the
coefficient is read from
\(Q\) above. Which is
\(3\). Hence
\begin{align*} b & =3-1\\ & =2 \end{align*}
For the other case of \(v=0\) then the coefficient of \(x^{-1}\) in \(r\) is found using \(\frac {lcoeff\left ( R\right ) }{lcoeff\left ( t\right ) }\) which will give \(1\) in this
case. (More examples below).
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 ( \frac {2}{1}-1\right ) =\frac {1}{2}\\ \alpha _{\infty }^{-} & =\frac {1}{2}\left ( -\frac {b}{a}-v\right ) =\frac {1}{2}\left ( -\frac {2}{1}-1\right ) =-\frac {3}{2}\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 |
| | | | |
| \(x=0\) |
\(\alpha _{c}^{+}=\frac {-3}{2}\) |
\(\alpha _{\infty }^{+}=\frac {1}{2}\) |
\(\alpha _{\infty }^{+}-\left ( \alpha _{c}^{+}\right ) =\frac {1}{2}-\left ( \frac {-3}{2}\right ) \) |
\(2\) |
| | | | |
| \(x=0\) |
\(\alpha _{c}^{+}=\frac {-3}{2}\) | \(\alpha _{\infty }^{-}=-\frac {3}{2}\) | \(\alpha _{\infty }^{-}-\left ( \alpha _{c}^{+}\right ) =-\frac {3}{2}-\left ( \frac {-3}{2}\right ) \) | \(0\) |
| | | | |
| \(x=0\) | \(\alpha _{c}^{-}=\frac {7}{2}\) | \(\alpha _{\infty }^{+}=\frac {1}{2}\) | \(\alpha _{\infty }^{+}-\left ( \alpha _{c}^{-}\right ) =\frac {1}{2}-\left ( \frac {7}{2}\right ) \) | \(-3\) |
| | | | |
| \(x=0\) |
\(\alpha _{c}^{-}=\frac {7}{2}\) |
\(\alpha _{\infty }^{-}=-\frac {3}{2}\) |
\(\alpha _{\infty }^{-}-\left ( \alpha _{c}^{-}\right ) =-\frac {3}{2}-\left ( \frac {7}{2}\right ) \) |
\(-5\) |
| | | | |
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
\[ d=\left \{ 0,2\right \} \]
Now for each
\(d\) value, 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
\(d=0,d=2\)
values, but 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 | | \(\left ( \sum _{c}s\left ( c\right ) \left [ \sqrt {r}\right ] _{c}+\frac {\alpha _{c}}{x-c}\right ) +s\left ( \infty \right ) \left [ \sqrt {r}\right ] _{\infty }\) |
|
| | | | | | | | |
| \(x=0\) | \(\alpha _{c}^{+}=\frac {-3}{2}\) | \(+\) | \(\frac {1}{x^{2}}\) | \(\alpha _{\infty }^{-}=-\frac {3}{2}\) | \(-\) | \(x-1\) | \(0\) | \(\left ( +\left ( \frac {1}{x^{2}}\right ) +\frac {\frac {-3}{2}}{x-0}\right ) +\left ( -\right ) \left ( x-1\right ) =\frac {1}{x^{2}}-\frac {3}{2x}-x+1\) |
| | | | | | | | |
| \(x=0\) | \(\alpha _{c}^{+}=\frac {-3}{2}\) | \(+\) | \(\frac {1}{x^{2}}\) | \(\alpha _{\infty }^{+}=\frac {1}{2}\) | \(+\) | \(x-1\) | \(2\) | \(\left ( +\left ( \frac {1}{x^{2}}\right ) +\frac {\frac {-3}{2}}{x-0}\right ) +\left ( +\right ) \left ( x-1\right ) =x-\frac {3}{2x}+\frac {1}{x^{2}}-1\) |
| | | | | | | | |
The above are the two candidate \(\omega \) values. For each \(\omega \) 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 \(\omega =\frac {1}{x^{2}}-\frac {3}{2x}-x+1\) associated with \(d=0\)
in the table, then (8) becomes
\begin{align*} P^{\prime \prime }+2\left ( \frac {1}{x^{2}}-\frac {3}{2x}-x+1\right ) P^{\prime }+\left ( \left ( \frac {3}{2x^{2}}-\frac {2}{x^{3}}-1\right ) +\left ( \frac {1}{x^{2}}-\frac {3}{2x}-x+1\right ) ^{2}-\left ( \frac {4x^{6}-8x^{5}+12x^{4}+4x^{3}+7x^{2}-20x+4}{4x^{4}}\right ) \right ) P & =0\\ P^{\prime \prime }+\left ( \frac {2}{x^{2}}-\frac {3}{2x}-2x+2\right ) P^{\prime }+\left ( \frac {4}{x^{2}}-\frac {6}{x}\right ) P & =0 \end{align*}
Since this the case for \(d=0\), then \(P\) has zero degree, Hence \(P\) is constant. Therefore the above
simplifies to
\[ \left ( \frac {4}{x^{2}}-\frac {6}{x}\right ) P=0 \]
Which means
\[ \frac {4}{x^{2}}-\frac {6}{x}=0 \]
Which is not possible for all
\(x\). Hence
\(d=0\) do not generate valid
\(P\)
polynomial. We now try the case of
\(d=2\). Since
\(d=2\), it means the polynomial
\(d\) is of degree two. Let
\[ P=x^{2}+ax+b \]
Substituting this in (8) using
\(\omega =x-\frac {3}{2x}+\frac {1}{x^{2}}-1\). This gives
\begin{align*} P^{\prime \prime }+2\left ( x-\frac {3}{2x}+\frac {1}{x^{2}}-1\right ) P^{\prime }+\left ( \left ( \frac {3}{2x^{2}}-\frac {2}{x^{3}}+1\right ) +\left ( x-\frac {3}{2x}+\frac {1}{x^{2}}-1\right ) ^{2}-\left ( \frac {4x^{6}-8x^{5}+12x^{4}+4x^{3}+7x^{2}-20x+4}{4x^{4}}\right ) \right ) P & =0\\ P^{\prime \prime }+2\left ( x-\frac {3}{2x}+\frac {1}{x^{2}}-1\right ) P^{\prime }+\left ( \frac {4}{x}-4\right ) P & =0 \end{align*}
Using \(P=x^{2}+ax+b\) the above becomes
\begin{align*} 2+2\left ( x-\frac {3}{2x}+\frac {1}{x^{2}}-1\right ) \left ( 2x+a\right ) +\left ( \frac {4}{x}-4\right ) \left ( x^{2}+ax+b\right ) & =0\\ 2a-4b-3\frac {a}{x}+2\frac {a}{x^{2}}+4\frac {b}{x}-2ax+\frac {4}{x}-4 & =0\\ \left ( 2a-4b-4\right ) +\frac {1}{x}\left ( -\frac {3}{4}a+4b+4\right ) +\frac {1}{x^{2}}\left ( 2a\right ) -2ax & =0 \end{align*}
Therefore
\begin{align*} 2a-4b-4 & =0\\ -\frac {3}{4}a+4b+4 & =0\\ 2a & =0\\ 2a & =0 \end{align*}
hence \(a=0\) from last equation. Using first or second equation gives \(b=-1\). Therefore a solution is
found. Hence
\[ p\left ( x\right ) =x^{2}-1 \]
Therefore the solution to
\(y^{\prime \prime }=ry\) is
\begin{align*} y & =p\left ( x\right ) e^{\int \omega dx}\\ & =\left ( x^{2}-1\right ) e^{\int x-\frac {3}{2x}+\frac {1}{x^{2}}-1\ dx}\\ & =\left ( x^{2}-1\right ) e^{-\frac {1}{x}-\frac {3}{2}\ln x+\frac {x^{2}}{2}-x}\\ & =\left ( x^{2}-1\right ) x^{\frac {-3}{2}}e^{-\frac {1}{x}+\frac {x^{2}}{2}-x}\end{align*}
The second solution can be found by reduction of order.