6.3.1.2 Example 2
Given the ode
\begin{align*} x^2 \left (x^2-2 x +1\right ) y''-x (3+x) y'+(4+x) y=0 \end{align*}
Converting it \(y''+ a y' + b y=0\) gives
\begin{align*} y''- \frac {x(3+x)}{x^2 \left (x^2-2 x +1\right )} y'+ \frac {4+x}{x^2 \left (x^2-2 x +1\right )} y=0 \end{align*}
Where \(a=- \frac {x (3+x) }{x^2 \left (x^2-2 x +1\right )}, b=\frac {4+x}{x^2 \left (x^2-2 x +1\right )}\). Applying the transformation \(z=y e^{\frac {1}{2} \int {a\,dx}}\) gives \(z''=r z\) where \(r = \frac {1}{4} a^2 + \frac {1}{2} a' - b\). This results in
\begin{align*} r &= \frac {s}{t} \\ &= \frac {7 x^{2}+10 x -1}{4 x^{2} \left (x -1\right )^{4}} \end{align*}
There is one pole at \(x=0\) of order \(2\) and pole at \(x=1\) of order \(4\), hence \(\Gamma =\{0,1\}\), and \(\mathcal {O}(\infty )=\deg (t)-\deg (s)=6-2=4\). Table 6.1 shows that
the necessary conditions for case one and two are both satisfied. For the pole at \(0\) and since
its order is \(2\) then
\begin{align*}\tag {1} \left [ \sqrt {r}\right ]_{0} & =0\\ \alpha _{0}^{+} & =\frac {1}{2}+\frac {1}{2}\sqrt {1+4b}\\ \alpha _{0}^{-} & =\frac {1}{2}-\frac {1}{2}\sqrt {1+4b} \end{align*}
Where \(b\) is the coefficient of \(\frac {1}{(x-0)^{2}}=\frac {1}{x^{2}}\) in the partial fraction decomposition of \(r\) which is
\begin{align*} r &= \frac {4}{\left (x -1\right )^{4}}-\frac {2}{\left (x -1\right )^{3}}-\frac {1}{4 x^{2}}-\frac {3}{2 \left (x -1\right )}+\frac {3}{2 x}+\frac {7}{4 \left (x -1\right )^{2}} \tag {2} \end{align*}
The above shows that \(b=-\frac {1}{4}\). Equation (1) becomes becomes
\begin{align*} \left [ \sqrt {r}\right ]_{0} & =0\\ \alpha _{0}^{+} & =\frac {1}{2}+\frac {1}{2}\sqrt {1-4\frac {1}{4}} = \frac {1}{2}\\ \alpha _{0}^{-} & =\frac {1}{2}-\frac {1}{2}\sqrt {1-4\frac {1}{4}} = \frac {1}{2} \end{align*}
For the second pole at \(x=1\), since its order is \(4\), then \(2v=4\) or \(v=2\). Therefore the corresponding \(\left [ \sqrt {r}\right ]_{1}\) is the
sum of all terms involving \(\frac {1}{ (x-1)^i}\) for \(2\leq i \leq v\) or \(2\leq i \leq 2\) in the Laurent series expansion of \(\sqrt {r}\) (not \(r\)) around this
pole. This results in
\begin{align*} [\sqrt {r}]_{1} &= \sum _{i=2}^{2} \frac {a_i}{(x-1)^i}\\ &= \frac {a_2}{(x-1)^2} \tag {3} \end{align*}
\(a_2\) is found using
\begin{align*} a_2 &= \lim _{x\rightarrow 1} (x-1)^2 \sqrt {r}\\ &= \lim _{x\rightarrow 1} (x-1)^2 \sqrt { \frac {7 x^{2}+10 x -1}{4 x^{2} \left (x -1\right )^{4}} }\\ &= 2 \end{align*}
Therefore
\begin{alignat*}{3} [\sqrt r]_{1} &= \frac {2}{ (x-1)^2} \\ \alpha _{1}^{+} &= \frac {1}{2} \left ( \frac {b}{a} + v \right ) &&= \frac {1}{2} \left ( \frac {b}{2} + 2 \right ) \\ \tag {4} \alpha _{1}^{-} &= \frac {1}{2} \left (- \frac {b}{a} + v \right ) &&= \frac {1}{2} \left (- \frac {b}{2} + 2 \right ) \end{alignat*}
What remains is to determine \(b\). This is the coefficient of \(\frac {1}{ (x-c)^{v+1}}=\frac {1}{ (x-1)^{3}}\) in the partial fraction
decomposition of \(r\) which from (2) is \(-2\) minus the coefficient of same term in \([\sqrt r]_{1}\) which from (3)
is zero. Therefore \(b=-2-0=-2\). (4) now becomes
\begin{alignat*}{3} [\sqrt r]_{1} &= \frac {2}{ (x-1)^2} \\ \alpha _{1}^{+} &= \frac {1}{2} \left ( \frac {b}{a} + v \right ) &&= \frac {1}{2} \left ( \frac {-2}{2} + 2 \right ) &&=\frac {1}{2} \\ \tag {5} \alpha _{1}^{-} &= \frac {1}{2} \left (- \frac {b}{a} + v \right ) &&= \frac {1}{2} \left (- \frac {-2}{2} + 2 \right ) &&=\frac {3}{2} \end{alignat*}
The above completes finding \([\sqrt r]_{c},\alpha _{c}^{+},\alpha _{c}^{+}\) for all poles in the set \(\Gamma \).
Since the order of \(r\) at \(\infty \) is \(4 > 2\) then
\begin{alignat*}{2} [\sqrt r]_\infty &= 0 \\ \alpha _{\infty }^{+} &= 0 \\ \alpha _{\infty }^{-} &= 1 \end{alignat*}
This completes the first step of the solution. The following tables summarizes the findings
so far
This completes step 1 for case one. Step 2 searches for non-negative integer \(d\) using
\begin{align*} d&=\alpha _{\infty }^{\pm }-\sum _{c\in \Gamma }\alpha _{c}^{\pm } \end{align*}
Where the above is carried over all possible combinations resulting in the following 8
possibilities (in this example, there are two poles, hence the sum contains two term) and
the number of possible combinations is therefore \(2^3=8\)
\begin{alignat*}{3} d &= \alpha _{\infty }^{+} - \left ( \alpha _{0}^{+} + \alpha _{1}^{+}\right ) &&= 0 -\left ( \frac {1}{2} + \frac {1}{2} \right ) &&= -1 \\ d &= \alpha _{\infty }^{+} - \left ( \alpha _{0}^{+} + \alpha _{1}^{-}\right ) &&= 0 -\left ( \frac {1}{2} + \frac {3}{2} \right ) &&= -2\\ d &= \alpha _{\infty }^{+} - \left ( \alpha _{0}^{-} + \alpha _{1}^{+}\right ) &&= 0 -\left ( \frac {1}{2} + \frac {1}{2} \right ) &&= -1\\ d &= \alpha _{\infty }^{+} - \left ( \alpha _{0}^{-} + \alpha _{1}^{-}\right ) &&= 0 -\left ( \frac {1}{2} + \frac {3}{2} \right ) &&= -2\\ d &= \alpha _{\infty }^{-} - \left ( \alpha _{0}^{+} + \alpha _{1}^{+}\right ) &&= 1 -\left ( \frac {1}{2} + \frac {1}{2} \right ) &&= 0 \\ d &= \alpha _{\infty }^{-} - \left ( \alpha _{0}^{+} + \alpha _{1}^{-}\right ) &&= 1 -\left ( \frac {1}{2} + \frac {3}{2} \right ) &&= -1 \\ d &= \alpha _{\infty }^{-} - \left ( \alpha _{0}^{-} + \alpha _{1}^{+}\right ) &&= 1 -\left ( \frac {1}{2} + \frac {1}{2} \right ) &&= 0 \\ d &= \alpha _{\infty }^{-} - \left ( \alpha _{0}^{-} + \alpha _{1}^{-}\right ) &&= 1 -\left ( \frac {1}{2} + \frac {3}{2} \right ) &&= -1 \\ \end{alignat*}
There is only one possible \(d=0\) values to use. Candidate \(\omega \) are now found using
\begin{align*} \omega &=\sum _{c}\left ( (\pm ) \left [ \sqrt {r}\right ]_{c}+\frac {\alpha _{c}^{\pm }}{x-c}\right ) + (\pm ) \left [\sqrt {r}\right ]_{\infty } \end{align*}
Which gives
\begin{align*} \omega _1&= \left ( (+) \left [ \sqrt {r}\right ]_{0}+\frac {\alpha _{0}^{+}}{x}\right ) + \left ( (+) \left [ \sqrt {r}\right ]_{1}+\frac {\alpha _{1}^{+}}{x-1}\right ) + (+) \left [\sqrt {r}\right ]_{\infty }\\ \omega _2 &= \left ( (+) \left [ \sqrt {r}\right ]_{0}+\frac {\alpha _{0}^{+}}{x}\right ) + \left ( (-) \left [ \sqrt {r}\right ]_{1}+\frac {\alpha _{1}^{-}}{x-1}\right ) + (+) \left [\sqrt {r}\right ]_{\infty }\\ \omega _3 &= \left ( (-) \left [ \sqrt {r}\right ]_{0}+\frac {\alpha _{0}^{-}}{x}\right ) + \left ( (+) \left [ \sqrt {r}\right ]_{1}+\frac {\alpha _{1}^{+}}{x-1}\right ) + (+) \left [\sqrt {r}\right ]_{\infty }\\ \omega _4 &= \left ( (-) \left [ \sqrt {r}\right ]_{0}+\frac {\alpha _{0}^{-}}{x}\right ) + \left ( (-) \left [ \sqrt {r}\right ]_{1}+\frac {\alpha _{1}^{-}}{x-1}\right ) + (+) \left [\sqrt {r}\right ]_{\infty }\\ \omega _5&= \left ( (+) \left [ \sqrt {r}\right ]_{0}+\frac {\alpha _{0}^{+}}{x}\right ) + \left ( (+) \left [ \sqrt {r}\right ]_{1}+\frac {\alpha _{1}^{+}}{x-1}\right ) + (-) \left [\sqrt {r}\right ]_{\infty }\\ \omega _6 &= \left ( (+) \left [ \sqrt {r}\right ]_{0}+\frac {\alpha _{0}^{+}}{x}\right ) + \left ( (-) \left [ \sqrt {r}\right ]_{1}+\frac {\alpha _{1}^{-}}{x-1}\right ) + (-) \left [\sqrt {r}\right ]_{\infty }\\ \omega _7 &= \left ( (-) \left [ \sqrt {r}\right ]_{0}+\frac {\alpha _{0}^{-}}{x}\right ) + \left ( (+) \left [ \sqrt {r}\right ]_{1}+\frac {\alpha _{1}^{+}}{x-1}\right ) + (-) \left [\sqrt {r}\right ]_{\infty }\\ \omega _8 &= \left ( (-) \left [ \sqrt {r}\right ]_{0}+\frac {\alpha _{0}^{-}}{x}\right ) + \left ( (-) \left [ \sqrt {r}\right ]_{1}+\frac {\alpha _{1}^{-}}{x-1}\right ) + (-) \left [\sqrt {r}\right ]_{\infty } \end{align*}
Substituting values found in step 1 into the above gives
\begin{alignat*}{3} \omega _1&= \left ( \frac {\frac {1}{2}}{x}\right ) + \left ( (+) \frac {2}{ (x-1)^2} + \frac {\frac {1}{2} }{x-1}\right ) &&= \frac {2 x^{2}+x +1}{2 x \left (x -1\right )^{2}}\\ \omega _2 &= \left ( \frac {\frac {1}{2}}{x} \right ) + \left ( (-) \frac {2}{ (x-1)^2} +\frac {\frac {3}{2} }{x-1}\right ) &&=\frac {4 x^{2}-9 x +1}{2 x \left (x -1\right )^{2}}\\ \omega _3 &= \left ( \frac {\frac {1}{2}}{x}\right ) + \left ( (+) \frac {2}{ (x-1)^2} +\frac {\frac {1}{2}}{x-1}\right ) &&=\frac {2 x^{2}+x +1}{2 x \left (x -1\right )^{2}} \\ \omega _4 &= \left ( \frac {\frac {1}{2}}{x}\right ) + \left ( (-) \frac {2}{ (x-1)^2} +\frac {\frac {3}{2}}{x-1}\right ) &&= \frac {4 x^{2}-9 x +1}{2 x \left (x -1\right )^{2}}\\ \omega _5&= \left ( \frac {\frac {1}{2}}{x} \right ) + \left ( (+) \frac {2}{ (x-1)^2} +\frac {\frac {1}{2}}{x-1}\right ) &&=\frac {2 x^{2}+x +1}{2 x \left (x -1\right )^{2}}\\ \omega _6 &= \left ( \frac {\frac {1}{2}}{x} \right ) + \left ( (-) \frac {2}{ (x-1)^2} +\frac {\frac {3}{2}}{x-1}\right ) &&= \frac {4 x^{2}-9 x +1}{2 x \left (x -1\right )^{2}}\\ \omega _7 &= \left ( \frac {\frac {1}{2}}{x}\right ) + \left ( (+) \frac {2}{ (x-1)^2} +\frac {\frac {1}{2}}{x-1}\right ) &&= \frac {2 x^{2}+x +1}{2 x \left (x -1\right )^{2}}\\ \omega _8 &= \left ( \frac {\frac {1}{2}}{x}\right ) + \left ( (-) \frac {2}{ (x-1)^2} +\frac {\frac {3}{2}}{x-1}\right ) &&=\frac {4 x^{2}-9 x +1}{2 x \left (x -1\right )^{2}} \end{alignat*}
Which shows there are only two different \(\omega \) to try, these are \(\omega _1,\omega _2\). This complete step 2. For each
trial, step 3 is now invoked. Starting with \(d=0\) and \(\omega =\omega _1=\frac {2 x^{2}+x +1}{2 x \left (x -1\right )^{2}}\). Since the degree \(d=0\) then \(p(x)=1\). This polynomial
needs to satisfy
\begin{align*} p''+2 \omega p' + (\omega '+\omega ^2-r) p &= 0 \tag {6}\\ (\omega '+\omega ^2-r) p &= 0\\ \frac {d}{dx}\left ( \frac {2 x^{2}+x +1}{2 x \left (x -1\right )^{2}} \right ) + \left (\frac {2 x^{2}+x +1}{2 x \left (x -1\right )^{2}}\right )^2- \frac {7 x^{2}+10 x -1}{4 x^{2} \left (x -1\right )^{4}} &=0\\ 0 &= 0 \end{align*}
Because the equation is satisfied, the polynomial \(p(x)=1\) can be used. The solution to \(z''=rz\) is now
found from
\begin{align*} z&=p e^{\int \omega \, dx}\\ &=e^{\int \frac {2 x^{2}+x +1}{2 x \left (x -1\right )^{2}} \, dx}\\ &=e^{\frac {\ln \left (x -1\right )}{2}-\frac {2}{x -1}+\frac {\ln \left (x \right )}{2}}\\ &=\sqrt {x -1}\, \sqrt {x}\, {\mathrm e}^{-\frac {2}{x -1}} \end{align*}
Given this solution for \(z(x)\), the first basis solution of the original ode in \(y\) is found using the
inverse of the original transformation used to generate the \(z\) ode which is \(z = y e^{\frac {1}{2}\int a\,dx}\), therefore
\begin{align*} y_1 &= z e^{-\frac {1}{2}\int a\,dx}\\ &= \sqrt {x -1} \sqrt {x} {\mathrm e}^{-\frac {2}{x -1}} {\mathrm e}^{-\frac {1}{2}\int - \frac {x (3+x) }{x^2 \left (x^2-2 x +1\right )} \,dx} \end{align*}
Which simplifies to
\begin{align*} y_1 &= \frac {x^2}{x-1} {\mathrm e}^{-\frac {4}{x -1}} \end{align*}
The second solution \(y_2\) to the original ode is found using reduction of order as was done in
the first example.