6.3.3.2 Example 2
The ode is
\begin{align*} x^2 (1+x) y''+x (2 x +1) y''-(4+6 x) y&=0 \end{align*}
Converting it \(y''+ a y' + b y=0\) gives
\begin{align*} y''+\frac {2 x +1}{x (1+x)} y'-\frac {4+6 x}{x^2 (1+x)} y=0 \end{align*}
Where \(a=\frac {2 x +1}{x (1+x)}, b=-\frac {4+6 x}{x^2 (1+x)}\). Applying the transformation \(z=y e^{\frac {1}{2} \int {a\,dx}}\) results in \(z''=r z\) where \(r = \frac {1}{4} a^2 + \frac {1}{2} a' - b\) where
\begin{align*} r &= \frac {s}{t} \\ &= \frac {24 x^2+40 x +15}{4 \left ( x(x+1) \right )^2} \end{align*}
There is a pole at \(x=0\) of order \(2\) and a pole at \(x=-1\) of order \(2\). Since there is no odd order pole larger
than \(2\) and the order at \(\infty \) is \(2\) then the necessary conditions for case one are satisfied.
Since there is a pole of order \(2\) then necessary conditions for case two are also
satisfied. Since pole order is not larger than \(2\) and the order at \(\infty \) is \(2\) then the necessary
conditions for case three are also satisfied. This is now solved as case three for
illustration.
Starting with \(n=4\), and for the pole of order \(2\) at \(x=-1\)
\begin{alignat*}{2} E_{-1} &=\left \{ 6 + \frac {12 k}{n} \sqrt {1+ 4 b}\right \} \qquad \text {for} \quad k&&=-\frac {n}{2}\cdots \frac {n}{2} \end{alignat*}
This simplifies to
\begin{alignat*}{2} E_{-1} &=\left \{ 6 + 3 k \sqrt {1+ 4 b}\right \} \qquad \text {for} \quad k&&=-2\cdots 2 \tag {1} \end{alignat*}
\(b\) is the coefficient of \(\frac {1}{ \left (1+x \right )^{2}}\) in the partial fractions decomposition of \(r\) given by
\begin{align*} r = -\frac {1}{4 \left (1+x \right )^{2}}+\frac {15}{4 x^{2}}-\frac {5}{2 \left (1+x \right )}+\frac {5}{2 x} \end{align*}
The above shows that \(b=-\frac {1}{4}\). Equation (1) becomes
\begin{alignat*}{2} E_{-1} &=\left \{ 6 + 3 k \sqrt {1- 4 \left (\frac {1}{4}\right ) }\right \} \qquad \text {for} \quad k&=-2\cdots 2\\ &= \{6\} \end{alignat*}
For the pole at \(x=0\) of order \(2\), \(b\) is the coefficient of \(\frac {1}{ x^{2}}\) in the above partial fractions decomposition
of \(r\). This shows that \(b={\frac {15}{4}}\). Hence
\begin{alignat*}{2} E_{0} &=\left \{ 6 + 3 k \sqrt {1+ 4 \left ( \frac {15}{4}\right ) }\right \} \qquad \text {for} \quad k&=-2\cdots 2 \\ &= \{-18, -6, 6, 18, 30\} \end{alignat*}
\(E_\infty \) is found using equation (1) but with different \(b\). In this case \(b\) is given by \(b=\frac {\operatorname {lcoef}(s)}{\text {lcoeff}(t)}\) where \(r=\frac {s}{t}\). \(\operatorname {lcoef}(s)\) is the
leading coefficient of \(s\) and \(\operatorname {lcoef}(t)\) is the leading coefficient of \(t\). Since \(r=\frac {24 x^2+40 x +15}{4 x^4+8x^3+4 x^2}\) then \(b =6\). Equation (1) becomes
\begin{alignat*}{2} E_\infty &=\left \{ 6 + 3 k \sqrt {1+ 4 \left (6\right ) } \right \} \qquad \text {for} \quad k&&=-2\cdots 2 \\ &= \{-24, -9, 6, 21, 36\} \end{alignat*}
The following table summarizes step \(1\) results using \(n=4\).
The next step is to determine a non negative integer \(d\) using
\begin{align*} d &= \frac {n}{12} \left ( e_\infty - \sum _{c \in \Gamma } e_c \right ) \end{align*}
Where in the above \(e_c\) is a distinct element from each corresponding \(E_c\). This means all
possible tuples \(\{e_{c_1},e_{c_2},\dots ,e_{c_n}\}\) are tried in the sum above, where \(e_{c_i}\) is one element of each \(E_c\) found
earlier.
This results in the following values for \(d\) using \(n=4\).
\begin{alignat*}{2} d &= \frac {1}{3} \left ( -24 - (6 -18) \right ) &= -4\\ &= \frac {1}{3} \left ( -24 - (6 +6) \right ) &=-12\\ &= \frac {1}{3} \left ( -24 - (6 -6) \right ) &= -8\\ &=\frac {1}{3} \left ( -24 - (6 +18) \right ) &= -16\\ &= \frac {1}{3} \left ( -24 - (6 +30) \right ) &= -20\\ &= \frac {1}{3} \left ( -9 - (6 -18) \right ) &=1\\ &= \frac {1}{3} \left ( -9 - (6 +6) \right ) &=-7\\ &= \frac {1}{3} \left ( -9 - (6 -6) \right ) &= -3\\ &= \frac {1}{3} \left ( -9 - (6 +18) \right ) &= -11\\ &= \frac {1}{3} \left ( -9 - (6 +30) \right ) &= -15\\ &= \frac {1}{3} \left ( 6 - (6 -18) \right ) &= 6\\ &= \frac {1}{3} \left ( 6 - (6 +6) \right ) &= -2\\ &= \frac {1}{3} \left ( 6 - (6 -6) \right ) &=2 \\ &= \frac {1}{3}\left ( 6 - (6 +18) \right ) &= -6\\ &= \frac {1}{3} \left ( 6 - (6 +30) \right ) &=-10 \\ &= \frac {1}{3} \left ( 21 - (6 -18) \right ) &= 11\\ &= \frac {1}{3} \left ( 21 - (6 +6) \right ) &= 3\\ &= \frac {1}{3} \left ( 21 - (6 -6) \right ) &= 7\\ &= \frac {1}{3} \left ( 21 - (6 +18) \right ) &= -1\\ &= \frac {1}{3} \left ( 21 - (6 +30) \right ) &= -5\\ &= \frac {1}{3} \left ( 36 - (6 -18) \right ) &= 16\\ &= \frac {1}{3} \left ( 36 - (6 +6) \right ) &=8\\ &= \frac {1}{3} \left ( 36 - (6 -6) \right ) &=12\\ &=\frac {1}{3} \left ( 36 - (6 +18) \right ) &= 4\\ &= \frac {1}{3} \left ( 36 - (6 +30) \right ) &= 0 \end{alignat*}
From the above, the following families all produce non-negative \(d\)
| \(e_\infty =-9\) |
\(e_{-1}=6\) |
\(e_{0}=-18\) |
\(d=1\) |
| \(e_\infty =6\) |
\(e_{-1}=6\) |
\(e_{0}=-18\) |
\(d=6\) |
| \(e_\infty =6\) |
\(e_{-1}=6\) |
\(e_{0}=-6\) |
\(d=2\) |
| \(e_\infty =21\) |
\(e_{-1}=6\) |
\(e_{0}=-18\) |
\(d=11\) |
| \(e_\infty =21\) |
\(e_{-1}=6\) |
\(e_{0}=6\) |
\(d=3\) |
| \(e_\infty =21\) |
\(e_{-1}=6\) | \(e_{0}=-6\) | \(d=7\) |
| \(e_\infty =36\) | \(e_{-1}=6\) | \(e_{0}=-18\) | \(d=16\) |
| \(e_\infty =36\) |
\(e_{-1}=6\) |
\(e_{0}=6\) |
\(d=8\) |
| \(e_\infty =36\) |
\(e_{-1}=6\) |
\(e_{0}=-6\) |
\(d=12\) |
| \(e_\infty =36\) |
\(e_{-1}=6\) |
\(e_{0}=18\) |
\(d=4\) |
| \(e_\infty =36\) |
\(e_{-1}=6\) |
\(e_{0}=30\) |
\(d=0\) |
Starting with the smallest \(d=0\) as that is the least computationally expensive with its
corresponding \(e_{-1}=6,e_0=30,e_\infty =36\), the following rational function is now formed
\begin{align*} \theta &= \frac {n}{12} \sum _{c \in \Gamma } \frac {e_c}{x-c} \\ &= \frac {4}{12} \left (\frac {6}{x+1}+\frac {30}{x}\right ) \\ &= \frac {12 x +10}{x \left (1+x \right )} \end{align*}
And
\begin{align*} S &= \prod _{c\in \Gamma } (x-c) \\ &= (x+1)x \end{align*}
This completes the step 2 of the algorithm.
Polynomial \(p(x)\) is now determined. Since the degree of the polynomial is \(d=0\), then
\begin{align*} p(x) = 1 \end{align*}
The \(P_i(x)\) polynomials are generated using
\begin{align*} P_n &= - p(x) \\ P_{i-1} &= - S p'_{i} + ((n-i)S' -S\theta ) P_i - (n-1)(i+1) S^2 r P_{i+1} \qquad i=n,n-1,\dots , 0 \end{align*}
The above results in the following set
\begin{align*} P_{4} &= -1\\ P_{3} &= 12 x +10\\ P_{2} &= -3 \left (6 x +5\right )^{2}\\ P_{1} &= 3 \left (6 x +5\right )^{3}\\ P_{0} &= -\frac {3 \left (6 x +5\right )^{4}}{2}\\ P_{-1} &= 0 \end{align*}
There is coefficient for \(p(x)\) to solve for from the last equation \(P_{-1} = 0\) as \(p(x)=1\) is already known because
the degree \(d\) is zero. \(\omega \) is now determined as the solution to the following equation, using \(n=4\)
\begin{align*} \sum _{i=0}^{n} S^i \frac {P_i}{(n-i)!} \omega ^i &= 0 \\ \frac {P_0}{4!} + \frac {S P_1}{3!} \omega +\frac {S^2 P_2}{2!} \omega ^2 +\frac {S^3 P_3}{1!} \omega ^3 + +\frac {S^4 P_4}{0!} \omega ^4 &=0\\ -\frac {1}{16} \left (2 \omega \,x^{2}+2 x \omega -6 x -5\right )^{4} &=0 \end{align*}
Solving the above and using any one of the roots gives
\begin{align*} \omega =\frac {x-2}{2 x \left (x -1\right )} \end{align*}
This \(\omega \) is used to find a solution to \(z''=r z\) from
\begin{align*} z &= e^{ \int \omega \,dx}\\ &= {\mathrm e}^{\int \frac {6 x +5}{2 x \left (1+x \right )}d x}\\ &= x^{\frac {5}{2}} \sqrt {1+x} \end{align*}
The first solution to the original ode in \(y\) is found from
\begin{align*} y &= z e^{ \int -\frac {1}{2} a \,dx}\\ &= z e^{ -\int \frac {1}{2} \frac {2 x^{2}+x}{x^{2} \left (1+x \right )} \,dx}\\ &= z e^{-\frac {\ln \left (x \left (1+x \right )\right )}{2}}\\ &=\left ( x^{\frac {5}{2}} \sqrt {1+x}\right ) \left (\frac {1}{\sqrt {x \left (1+x \right )}}\right ) \end{align*}
Which simplifies to
\begin{align*} y &= x^{2} \end{align*}
The second solution to the original ode is found using reduction of order.