6.3.2.1 Example 1

Given the ode

\begin{align*} 2 x^2 y''-x y'+(1+x) y&=0 \end{align*}

Converting it \(y''+ a y' + b y=0\) gives

\begin{align*} y''-\frac {1}{2 x} y'+\frac {1+x}{2 x^2} y&=0 \end{align*}

Where \(a=-\frac {1}{2 x}, b=\frac {1+x}{2 x^2}\). Applying the transformation \(z=y e^{\frac {1}{2} \int {a\,dx}}\) gives \(z''=r z\) and \(r = \frac {1}{4} a^2 + \frac {1}{2} a' - b\). Therefore

\begin{align*} r &=\frac {s}{t} \\ &=\frac {-3-8 x}{16 x^2} \end{align*}

There is one pole at \(x=0\) of order \(2\) and \(\mathcal {O}(\infty )=\deg (t)-\deg (s)=2-1=1\). Table 6.1 shows that necessary conditions for only case two are satisfied. Since pole is order \(2\), the set \(E_0=\{2, 2+ 2\sqrt {1+ 4b}, 2-2 \sqrt {1+ 4b}\}\) where \(b\) is the coefficient of \(\frac {1}{(x)^2}\) in the partial fraction decomposition of \(r\) given by

\begin{align*} r &= -\frac {3}{16 x^{2}}-\frac {1}{2 x} \end{align*}

Therefore \(b=-{\frac {3}{16}}\). Hence

\begin{align*} E_0 &=\left \{ 2, 2+ 2\sqrt {1- 4 \frac {3}{16}}, 2-2 \sqrt {1- 4 \frac {3}{16}}\right \} \\ &=\left \{ 2,3, 1 \right \} \end{align*}

Since \(\mathcal {O}(\infty )=1<2\) then \(E_{\infty }=\mathcal {O}(\infty )=\{1\}\). This completes step \(1\) of case two. Step \(2\) is used to determine a non-negative integer \(d\). Using \(e_\infty =1\) gives

\begin{align*} d&=\frac {1}{2} \left ( e_{\infty } - \sum _{c\in \Gamma } e_c \right ) \end{align*}

There is only one pole, so the sum contains only one term. There are 3 possible combinations to try, using either \(e_0=2\), \(e_0=3\) or \(e_0=1\). Therefore

\begin{alignat*}{3} d&=\frac {1}{2} \left ( e_\infty - ( e_0 ) \right ) &&=\frac {1}{2} \left ( 1- 2 \right ) &&= - \frac {1}{2}\\ &=\frac {1}{2} \left ( e_\infty - ( e_0 ) \right ) &&=\frac {1}{2} \left ( 1- 3 \right ) &&= -1\\ &=\frac {1}{2} \left ( e_\infty - ( e_0 ) \right ) &&=\frac {1}{2} \left ( 1- 1 \right ) &&= 0 \end{alignat*}

The above shows that only the family \(\{e_\infty =1,e_0=1\}\) generated non-negative \(d=0\). \(\theta \) is now found. In the following sum, only \(e_c\) retained from the above are used. In this example, this will be \(e_0=1\) since it is the member of \(E_0\) which generated non-negative integer. If there were more than one \(e_i\) found, then each would be tried at time.

\begin{align*} \theta &= \frac {1}{2} \sum _{c\in \Gamma } \frac {e_c}{x-c}\\ &= \frac {1}{2} \left ( \frac {e_0}{x-0} \right )\\ &= \frac {1}{2 x} \end{align*}

This completes step \(2\). Step \(3\) finds polynomial \(p(x)=a_0+a_1 x+ a_2 x^2 + \dots + x^d\) of degree \(d\). Since \(d=0\) then \(p(x)=1\). This polynomial has to satisfy the following

\begin{align*} p'''+3 \theta p'' \left (3 \theta ^2 + 3 \theta ' - 4 r \right ) p' + \left (\theta ''+ 3 \theta \theta ' + \theta ^3 - 4 r \theta -2 r' \right )p=0 \end{align*}

Substituting \(p=1, \theta =\frac {1}{2 x}\) into the above and simplifying gives

\begin{align*} 0 &= 0 \end{align*}

Since \(p(x)=1\) is verified, then

\begin{align*} \phi = \theta + \frac {p'}{p} \\ = \frac {1}{2 x} \end{align*}

Next, \(\omega \) solution is found using

\begin{align*} \omega ^2 - \phi \omega + \left (\frac {1}{2} \phi ' +\frac {1}{2} \phi ^2 -r\right ) &=0 \end{align*}

Substituting the values for \(\phi \) and \(r\) into the above gives

\begin{align*} w^{2}-\frac {w}{2 x}+\frac {1+8 x}{16 x^{2}} = 0 \end{align*}

Solving for \(\omega \) gives two roots, either one can be used. Using

\begin{align*} \omega &= \frac {1+2 \sqrt {2}\, \sqrt {-x}}{4 x} \end{align*}

Therefore the first solution to the ode \(z'' = r z\) is

\begin{align*} z(x) &= e^{ \int \omega \,dx} \\ &= {\mathrm e}^{\int \frac {1+2 \sqrt {2}\, \sqrt {-x}}{4 x}d x}\\ &= x^{\frac {1}{4}} {\mathrm e}^{\sqrt {2}\, \sqrt {-x}} \end{align*}

The first solution to the original ode in \(y\) is now found from

\begin{align*} y_1 &= z e^{ \int -\frac {1}{2} a \,dx}\\ &= z e^{ -\int \frac {1}{2} \frac {-x}{2 x^{2}} \,dx}\\ &= z e^{\frac {\ln \left (x \right )}{4}}\\ &= \left (x^{\frac {1}{4}} {\mathrm e}^{\sqrt {2}\, \sqrt {-x}}\right ) x^{\frac {1}{4}}\\ &= \sqrt {x}\, {\mathrm e}^{\sqrt {2}\, \sqrt {-x}} \end{align*}

The second solution \(y_2\) to the original ode can be found using reduction of order.