6.3.1.1 Example 1

Given the ode

\begin{align*} (2 x + 1)y''-2 y'- (2x +3)y=0 \end{align*}

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

\begin{align*} y''-\frac {2}{2 x + 1} y'-\frac {2x +3}{2 x + 1} y=0 \end{align*}

Where \(a=-\frac {2}{2 x + 1}, b=-\frac {2x +3}{2 x + 1}\). 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 {4x^2+8x+6}{(2x+1)^2}=\frac {s}{t} \end{align*}

There is one pole at \(x=-\frac {1}{2}\), hence \(\Gamma =\{-\frac {1}{2}\}\). The order is \(2\) and \(\mathcal {O}(\infty )=\deg (t)-\deg (s)=0\). Table 6.1 shows that the necessary conditions for case one and two are both satisfied. This is solved first using case one. Since the order of the pole is \(2\), then

\begin{align*}\tag {1} \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 \(\frac {1}{(x-c)^{2}}=\frac {1}{(x+\frac {1}{2})^{2}}\) in the partial fraction decomposition of \(r\) which is \(r=1+\frac {3}{4}\frac {1}{(x+\frac {1}{2})^2}+\frac {1}{x+\frac {1}{2}}\). Therefore \(b=\frac {3}{4}\) and the above becomes

\begin{align*} \left [ \sqrt {r}\right ]_{c} & =0\\ \alpha _{c}^{+} & =\frac {1}{2}+\frac {1}{2}\sqrt {1+4\left (\frac {3}{4}\right )} = \frac {3}{2}\\ \alpha _{c}^{-} & =\frac {1}{2}-\frac {1}{2}\sqrt {1+4\left (\frac {3}{4}\right )} = -\frac {1}{2} \end{align*}

Since \(\mathcal {O}(\infty )= 0\) then \(v =0\) and \(\left [ \sqrt {r}\right ] _{\infty }\) is the sum of all terms \(x^i\) for \(0\leq i\leq v\) in the Laurent series expansion of \(\sqrt {r}\) at \(\infty \) which is found as follows. Since \(\sqrt {r} =\sqrt {\frac {4x^2+8x+6}{(2x+1)^2}}\) then setting \(x=\frac {1}{y}\) gives \(\sqrt {r(y)} =\sqrt {\frac {4 \left (\frac {1}{y}\right )^2+8\frac {1}{y}+6}{\left (2\frac {1}{y}+1\right )^2}}\) and since \(v=0\) then the constant term is \(\lim _{y\rightarrow 0} \sqrt {r(y)}= 1\). Therefore \(\left [ \sqrt {r(x)}\right ] _{\infty }=1\). Hence \(a=1\).

\(b\) is the coefficient of \(x^{v-1}=\frac {1}{x}\) in \(r\) minus the coefficient of \(\frac {1}{x}\) in \(\left (\left [ \sqrt {r}\right ] _{\infty }\right )^2=1\) which is zero since there is no term \(\frac {1}{x}\). Because \(v=0\), long division is used is used find the coefficient of \(\frac {1}{x}\) in \(r\).

\begin{align*} r &= \frac {s}{t}\\ &= \frac {4 x^2 + 8 x + 6}{4 x^2 + 4 x +1}\\ &= Q + \frac {R}{t}\\ &= 1 + \frac {4 x + 5}{4 x^2+4 x+1} \end{align*}

The coefficient of \(\frac {1}{x}\) in \(r\) is the leading coefficient in \(R\) minus the leading coefficient in \(t\) which gives \(\frac {4}{4}=1\). Therefore \(b=1-0=1\). This results in

\begin{alignat*}{3} \left [ \sqrt {r}\right ] _{\infty }&=1\\ \alpha _{\infty }^{+} & =\frac {1}{2}\left ( \frac {b}{a_v}-v\right ) &&= \frac {1}{2}\left ( \frac {1}{1}-0\right ) &&= \frac {1}{2}\\ \alpha _{\infty }^{-} & =\frac {1}{2}\left ( -\frac {b}{a_v}-v\right ) &&= \frac {1}{2}\left ( -\frac {1}{1}-0\right ) &&= -\frac {1}{2} \end{alignat*}

The above completes step 1 for case one. Step \(2\) searches for a non-negative integer \(d\) using

\begin{align*} d&=\alpha _{\infty }^{\pm }-\sum _{c\in \Gamma }\alpha _{c}^{\pm } \tag {2} \end{align*}

Where the above is carried over all possible combinations resulting in the following 4 possibilities (in this example, there is only one pole, hence the sum contains only one term) and the number of possible combinations is therefore \(2^2=4\)

\begin{alignat*}{3} d &= \alpha _{\infty }^{+} - \alpha _{c}^{+} &&= \frac {1}{2} - \left ( \frac {3}{2}\right ) &&= -1\\ d &= \alpha _{\infty }^{+} - \alpha _{c}^{-} &&= \frac {1}{2} - \left (- \frac {1}{2}\right ) &&= 1 \\ d &= \alpha _{\infty }^{-} - \alpha _{c}^{+} &&= -\frac {1}{2} - \left ( \frac {3}{2}\right ) &&= -2 \\ d &= \alpha _{\infty }^{-} - \alpha _{c}^{-} &&= -\frac {1}{2} - \left (- \frac {1}{2}\right ) &&= 0 \end{alignat*}

The above shows there are two possible \(d\) values to use. \(d=1\) or \(d=0\). Each is tried until one produces a solution or all fail to do so. For each valid \(d\) found an \(\omega \) is 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*}

But \(\left [ \sqrt {r}\right ]_{c}=0\) in this example, hence the above simplifies to

\begin{align*} \omega &=\sum _{c}\left ( \frac {\alpha _{c}^{\pm }}{x-c}\right ) + (\pm ) \left [\sqrt {r}\right ]_{\infty } \end{align*}

Since there is one pole, then the candidate \(\omega \) to try are the following

\begin{align*} \omega &= \frac {\alpha _{c}^{+}}{x-c} + (+1) \left [\sqrt {r}\right ]_{\infty }\\ \omega &= \frac {\alpha _{c}^{+}}{x-c} + (-1) \left [\sqrt {r}\right ]_{\infty }\\ \omega &= \frac {\alpha _{c}^{-}}{x-c} + (+1) \left [\sqrt {r}\right ]_{\infty }\\ \omega &= \frac {\alpha _{c}^{-}}{x-c} + (-1) \left [\sqrt {r}\right ]_{\infty } \end{align*}

Substituting the known values found in step 1 into the above gives

\begin{align*} \omega &= \frac { \frac {3}{2} } {x+\frac {1}{2}} + (+1) (1) = \frac {6+2 x}{2 x +3} \\ \omega &= \frac { \frac {3}{2} } {x+\frac {1}{2}} + (-1) (1) = -\frac {2 x}{2 x +3}\\ \omega &= \frac { -\frac {1}{2} } {x+\frac {1}{2}} + (+1) (1) = \frac {2 x}{2 x +1}\\ \omega &= \frac { -\frac {1}{2} } {x+\frac {1}{2}} + (-1) (1) = -\frac {2 \left (1+x \right )}{2 x +1} \end{align*}

So there are two possible \(d\) values to try, and for each, there are \(4\) possible \(w(x)\), which gives \(8\) possible tries. This completes step 2. For each trial, step 3 is now invoked.

Starting with \(d=0\) and using \(\omega =\frac {2 x}{2 x +1}\), and since the degree is \(d=0\) then \(p(x)=1\). This polynomial is now checked to see if it satisfies

\begin{align*} p''+2 \omega p' + (\omega '+\omega ^2-r) p &= 0 \tag {3}\\ (\omega '+\omega ^2-r) p &= 0 \\ \frac {d}{dx} \left (\frac {2 x}{2 x +1}\right ) +\left (\frac {2 x}{2 x +1}\right )^2- \frac {4x^2+8x+6}{(2x+1)^2} &=0\\ -\frac {4}{2 x +1} &=0 \end{align*}

Since the left side is not identically zero, then this candidate \(\omega \) has failed. Carrying out this process for the other 3 possible \(\omega \) values shows that non are satisfied as well. \(d=1\) is now tried. This implies the polynomial is \(p(x)=a_0 + x\). The coefficient \(a_0\) needs to be determined such that \(p''+2 \omega p' + (\omega '+\omega ^2-r) p= 0\) is satisfied. Starting with \(\omega = \frac {2 x}{2 x +1}\) gives

\begin{align*} p''+2 \omega p' + (\omega '+\omega ^2-r) p &= 0 \\ 2 \omega p' + (\omega '+\omega ^2-r) p &= 0 \end{align*}

Substituting \(p=a_0+x\) and \(\omega = \frac {2 x}{2 x +1}\) and \(r= \frac {4x^2+8x+6}{(2x+1)^2}\) into the above and simplifying gives \(-\frac {4 a_0}{2 x +1}=0\). This implies that (3) can be satisfied for \(a_0=0\). Therefore the polynomial of degree one is found and given by

\begin{align*} p(x)&=x \end{align*}

Therefore the solution to \(z''=rz\) is

\begin{align*} z&=p e^{\int \omega \, dx}\\ &=x e^{\int \frac {2 x}{2 x +1} \, dx}\\ &=x e^{x -\frac {\ln \left (2 x +1\right )}{2}}\\ &=\frac {x e^x}{\sqrt {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}\\ &= \frac {x e^x}{\sqrt {2 x + 1}} e^{-\frac {1}{2}\int -\frac {2}{2 x + 1} \,dx}\\ &= \frac {x e^x}{\sqrt {2 x + 1}} e^{\frac {\ln (2x+1}{2}}\\ &= \frac {x e^x}{\sqrt {2 x + 1}} \sqrt {2 x + 1}\\ &= x e^x \end{align*}

The second basis solution is found using reduction of order

\begin{align*} y_2 &= y_1 \int { \frac { e^{\int {-a \,dx}}\,dx } {y_{1}^{2}} \, dx } \\ &= x e^x \int { \frac { e^{\int {- \frac {-2}{2 x + 1} \,dx}}\,dx } {(x e^x)^{2}} \, dx } \\ &= x e^x \int { \frac { e^{\ln (2 x + 1)}} {(x e^x)^{2}} \, dx } \\ &= -e^{-x} \end{align*}

Therefore the general solution to the original ode \((2 x + 1)y''-2 y'- (2x +3)y=0\) is

\begin{align*} y(x) &= c_1 y_1 + x_2 y_2 \\ y(x) &= c_1 xe^x - c_2 e^{-x} \end{align*}

This completes the solution.