5.3.2.4 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
\[ y\left ( x\right ) =c_{1}\frac {1}{x}+c_{2}x^{2}\]