Internal problem ID [6442]
Internal file name [OUTPUT/5690_Sunday_June_05_2022_03_47_07_PM_58129404/index.tex]
Book: Differential Equations: Theory, Technique, and Practice by George Simmons, Steven
Krantz. McGraw-Hill NY. 2007. 1st Edition.
Section: Chapter 4. Power Series Solutions and Special Functions. Section 4.4. REGULAR
SINGULAR POINTS. Page 175
Problem number: 1(b).
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second order series method. Regular singular point. Difference not integer"
Maple gives the following as the ode type
[[_2nd_order, _with_linear_symmetries]]
\[ \boxed {x^{2} \left (x^{2}-1\right ) y^{\prime \prime }-x \left (1-x \right ) y^{\prime }+2 y=0} \] With the expansion point for the power series method at \(x = 0\).
The type of the expansion point is first determined. This is done on the homogeneous part of the ODE. \[ \left (x^{4}-x^{2}\right ) y^{\prime \prime }+\left (x^{2}-x \right ) y^{\prime }+2 y = 0 \] The following is summary of singularities for the above ode. Writing the ode as \begin {align*} y^{\prime \prime }+p(x) y^{\prime } + q(x) y &=0 \end {align*}
Where \begin {align*} p(x) &= \frac {1}{x \left (x +1\right )}\\ q(x) &= \frac {2}{x^{2} \left (x^{2}-1\right )}\\ \end {align*}
| \(p(x)=\frac {1}{x \left (x +1\right )}\) | |
| singularity | type |
| \(x = -1\) | \(\text {``regular''}\) |
| \(x = 0\) | \(\text {``regular''}\) |
| \(q(x)=\frac {2}{x^{2} \left (x^{2}-1\right )}\) | |
| singularity | type |
| \(x = -1\) | \(\text {``regular''}\) |
| \(x = 0\) | \(\text {``regular''}\) |
| \(x = 1\) | \(\text {``regular''}\) |
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \([-1, 0, 1, \infty ]\)
Irregular singular points : \([]\)
Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ x^{2} \left (x^{2}-1\right ) y^{\prime \prime }+\left (x^{2}-x \right ) y^{\prime }+2 y = 0 \] Let the solution be represented as Frobenius power series of the form \[ y = \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r} \] Then \begin{align*} y^{\prime } &= \moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1} \\ y^{\prime \prime } &= \moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} x^{n +r -2} \\ \end{align*} Substituting the above back into the ode gives \begin{equation} \tag{1} x^{2} \left (x^{2}-1\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} x^{n +r -2}\right )+\left (x^{2}-x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+2 \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\right ) = 0 \end{equation} Which simplifies to \begin{equation} \tag{2A} \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r +2} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{1+n +r} a_{n} \left (n +r \right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}2 a_{n} x^{n +r}\right ) = 0 \end{equation} The next step is to make all powers of \(x\) be \(n +r\) in each summation term. Going over each summation term above with power of \(x\) in it which is not already \(x^{n +r}\) and adjusting the power and the corresponding index gives \begin{align*} \moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r +2} a_{n} \left (n +r \right ) \left (n +r -1\right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}a_{n -2} \left (n +r -2\right ) \left (n -3+r \right ) x^{n +r} \\ \moverset {\infty }{\munderset {n =0}{\sum }}x^{1+n +r} a_{n} \left (n +r \right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}a_{n -1} \left (n +r -1\right ) x^{n +r} \\ \end{align*} Substituting all the above in Eq (2A) gives the following equation where now all powers of \(x\) are the same and equal to \(n +r\). \begin{equation} \tag{2B} \left (\moverset {\infty }{\munderset {n =2}{\sum }}a_{n -2} \left (n +r -2\right ) \left (n -3+r \right ) x^{n +r}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}a_{n -1} \left (n +r -1\right ) x^{n +r}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}2 a_{n} x^{n +r}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ -x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )-x^{n +r} a_{n} \left (n +r \right )+2 a_{n} x^{n +r} = 0 \] When \(n = 0\) the above becomes \[ -x^{r} a_{0} r \left (-1+r \right )-x^{r} a_{0} r +2 a_{0} x^{r} = 0 \] Or \[ \left (-x^{r} r \left (-1+r \right )-x^{r} r +2 x^{r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ -\left (r^{2}-2\right ) x^{r} = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ -r^{2}+2 = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= -\sqrt {2}\\ r_2 &= \sqrt {2} \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ -\left (r^{2}-2\right ) x^{r} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \(\left [-\sqrt {2}, \sqrt {2}\right ]\).
Since \(r_1 - r_2 = -2 \sqrt {2}\) is not an integer, then we can construct two linearly independent solutions \begin {align*} y_{1}\left (x \right ) &= x^{r_{1}} \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n}\right )\\ y_{2}\left (x \right ) &= x^{r_{2}} \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n}\right ) \end {align*}
Or \begin {align*} y_{1}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n -\sqrt {2}}\\ y_{2}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +\sqrt {2}} \end {align*}
We start by finding \(y_{1}\left (x \right )\). Eq (2B) derived above is now used to find all \(a_{n}\) coefficients. The case \(n = 0\) is skipped since it was used to find the roots of the indicial equation. \(a_{0}\) is arbitrary and taken as \(a_{0} = 1\). Substituting \(n = 1\) in Eq. (2B) gives \[ a_{1} = \frac {r}{r^{2}+2 r -1} \] For \(2\le n\) the recursive equation is \begin{equation} \tag{3} a_{n -2} \left (n +r -2\right ) \left (n -3+r \right )-a_{n} \left (n +r \right ) \left (n +r -1\right )+a_{n -1} \left (n +r -1\right )-a_{n} \left (n +r \right )+2 a_{n} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = \frac {n^{2} a_{n -2}+2 n r a_{n -2}+r^{2} a_{n -2}-5 n a_{n -2}+n a_{n -1}-5 r a_{n -2}+r a_{n -1}+6 a_{n -2}-a_{n -1}}{n^{2}+2 n r +r^{2}-2}\tag {4} \] Which for the root \(r = -\sqrt {2}\) becomes \[ a_{n} = \frac {\left (-2 n a_{n -2}+5 a_{n -2}-a_{n -1}\right ) \sqrt {2}+n^{2} a_{n -2}+\left (-5 a_{n -2}+a_{n -1}\right ) n +8 a_{n -2}-a_{n -1}}{n \left (-2 \sqrt {2}+n \right )}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = -\sqrt {2}\) and after as more terms are found using the above recursive equation.
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {r}{r^{2}+2 r -1}\) | \(\frac {\sqrt {2}}{-1+2 \sqrt {2}}\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {r \left (r^{3}+r^{2}-2 r +2\right )}{\left (r^{2}+2 r -1\right ) \left (r^{2}+4 r +2\right )} \] Which for the root \(r = -\sqrt {2}\) becomes \[ a_{2}=\frac {\sqrt {2}}{-5+3 \sqrt {2}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {r}{r^{2}+2 r -1}\) | \(\frac {\sqrt {2}}{-1+2 \sqrt {2}}\) |
| \(a_{2}\) | \(\frac {r \left (r^{3}+r^{2}-2 r +2\right )}{\left (r^{2}+2 r -1\right ) \left (r^{2}+4 r +2\right )}\) | \(\frac {\sqrt {2}}{-5+3 \sqrt {2}}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=\frac {2 r \left (r^{4}+4 r^{3}+3 r^{2}+2\right )}{\left (r^{2}+2 r -1\right ) \left (r^{2}+4 r +2\right ) \left (r^{2}+6 r +7\right )} \] Which for the root \(r = -\sqrt {2}\) becomes \[ a_{3}=-\frac {2 \sqrt {2}}{15-9 \sqrt {2}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {r}{r^{2}+2 r -1}\) | \(\frac {\sqrt {2}}{-1+2 \sqrt {2}}\) |
| \(a_{2}\) | \(\frac {r \left (r^{3}+r^{2}-2 r +2\right )}{\left (r^{2}+2 r -1\right ) \left (r^{2}+4 r +2\right )}\) | \(\frac {\sqrt {2}}{-5+3 \sqrt {2}}\) |
| \(a_{3}\) | \(\frac {2 r \left (r^{4}+4 r^{3}+3 r^{2}+2\right )}{\left (r^{2}+2 r -1\right ) \left (r^{2}+4 r +2\right ) \left (r^{2}+6 r +7\right )}\) | \(-\frac {2 \sqrt {2}}{15-9 \sqrt {2}}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {r \left (r^{7}+10 r^{6}+36 r^{5}+58 r^{4}+41 r^{3}+20 r^{2}+42 r +40\right )}{\left (r^{2}+2 r -1\right ) \left (r^{2}+4 r +2\right ) \left (r^{2}+6 r +7\right ) \left (r^{2}+8 r +14\right )} \] Which for the root \(r = -\sqrt {2}\) becomes \[ a_{4}=\frac {-69+49 \sqrt {2}}{-1104+780 \sqrt {2}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {r}{r^{2}+2 r -1}\) | \(\frac {\sqrt {2}}{-1+2 \sqrt {2}}\) |
| \(a_{2}\) | \(\frac {r \left (r^{3}+r^{2}-2 r +2\right )}{\left (r^{2}+2 r -1\right ) \left (r^{2}+4 r +2\right )}\) | \(\frac {\sqrt {2}}{-5+3 \sqrt {2}}\) |
| \(a_{3}\) | \(\frac {2 r \left (r^{4}+4 r^{3}+3 r^{2}+2\right )}{\left (r^{2}+2 r -1\right ) \left (r^{2}+4 r +2\right ) \left (r^{2}+6 r +7\right )}\) | \(-\frac {2 \sqrt {2}}{15-9 \sqrt {2}}\) |
| \(a_{4}\) | \(\frac {r \left (r^{7}+10 r^{6}+36 r^{5}+58 r^{4}+41 r^{3}+20 r^{2}+42 r +40\right )}{\left (r^{2}+2 r -1\right ) \left (r^{2}+4 r +2\right ) \left (r^{2}+6 r +7\right ) \left (r^{2}+8 r +14\right )}\) | \(\frac {-69+49 \sqrt {2}}{-1104+780 \sqrt {2}}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=\frac {r \left (3 r^{8}+48 r^{7}+306 r^{6}+996 r^{5}+1749 r^{4}+1616 r^{3}+866 r^{2}+680 r +496\right )}{\left (r^{2}+2 r -1\right ) \left (r^{2}+4 r +2\right ) \left (r^{2}+6 r +7\right ) \left (r^{2}+8 r +14\right ) \left (r^{2}+10 r +23\right )} \] Which for the root \(r = -\sqrt {2}\) becomes \[ a_{5}=\frac {-414+293 \sqrt {2}}{6108 \sqrt {2}-8640} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {r}{r^{2}+2 r -1}\) | \(\frac {\sqrt {2}}{-1+2 \sqrt {2}}\) |
| \(a_{2}\) | \(\frac {r \left (r^{3}+r^{2}-2 r +2\right )}{\left (r^{2}+2 r -1\right ) \left (r^{2}+4 r +2\right )}\) | \(\frac {\sqrt {2}}{-5+3 \sqrt {2}}\) |
| \(a_{3}\) | \(\frac {2 r \left (r^{4}+4 r^{3}+3 r^{2}+2\right )}{\left (r^{2}+2 r -1\right ) \left (r^{2}+4 r +2\right ) \left (r^{2}+6 r +7\right )}\) | \(-\frac {2 \sqrt {2}}{15-9 \sqrt {2}}\) |
| \(a_{4}\) | \(\frac {r \left (r^{7}+10 r^{6}+36 r^{5}+58 r^{4}+41 r^{3}+20 r^{2}+42 r +40\right )}{\left (r^{2}+2 r -1\right ) \left (r^{2}+4 r +2\right ) \left (r^{2}+6 r +7\right ) \left (r^{2}+8 r +14\right )}\) | \(\frac {-69+49 \sqrt {2}}{-1104+780 \sqrt {2}}\) |
| \(a_{5}\) | \(\frac {r \left (3 r^{8}+48 r^{7}+306 r^{6}+996 r^{5}+1749 r^{4}+1616 r^{3}+866 r^{2}+680 r +496\right )}{\left (r^{2}+2 r -1\right ) \left (r^{2}+4 r +2\right ) \left (r^{2}+6 r +7\right ) \left (r^{2}+8 r +14\right ) \left (r^{2}+10 r +23\right )}\) | \(\frac {-414+293 \sqrt {2}}{6108 \sqrt {2}-8640}\) |
For \(n = 6\), using the above recursive equation gives \[ a_{6}=\frac {r \left (r^{11}+27 r^{10}+314 r^{9}+2064 r^{8}+8439 r^{7}+22209 r^{6}+37650 r^{5}+40744 r^{4}+30972 r^{3}+26532 r^{2}+26728 r +13520\right )}{\left (r^{2}+2 r -1\right ) \left (r^{2}+4 r +2\right ) \left (r^{2}+6 r +7\right ) \left (r^{2}+8 r +14\right ) \left (r^{2}+10 r +23\right ) \left (r^{2}+12 r +34\right )} \] Which for the root \(r = -\sqrt {2}\) becomes \[ a_{6}=\frac {3898-2757 \sqrt {2}}{114408-80892 \sqrt {2}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {r}{r^{2}+2 r -1}\) | \(\frac {\sqrt {2}}{-1+2 \sqrt {2}}\) |
| \(a_{2}\) | \(\frac {r \left (r^{3}+r^{2}-2 r +2\right )}{\left (r^{2}+2 r -1\right ) \left (r^{2}+4 r +2\right )}\) | \(\frac {\sqrt {2}}{-5+3 \sqrt {2}}\) |
| \(a_{3}\) | \(\frac {2 r \left (r^{4}+4 r^{3}+3 r^{2}+2\right )}{\left (r^{2}+2 r -1\right ) \left (r^{2}+4 r +2\right ) \left (r^{2}+6 r +7\right )}\) | \(-\frac {2 \sqrt {2}}{15-9 \sqrt {2}}\) |
| \(a_{4}\) | \(\frac {r \left (r^{7}+10 r^{6}+36 r^{5}+58 r^{4}+41 r^{3}+20 r^{2}+42 r +40\right )}{\left (r^{2}+2 r -1\right ) \left (r^{2}+4 r +2\right ) \left (r^{2}+6 r +7\right ) \left (r^{2}+8 r +14\right )}\) | \(\frac {-69+49 \sqrt {2}}{-1104+780 \sqrt {2}}\) |
| \(a_{5}\) | \(\frac {r \left (3 r^{8}+48 r^{7}+306 r^{6}+996 r^{5}+1749 r^{4}+1616 r^{3}+866 r^{2}+680 r +496\right )}{\left (r^{2}+2 r -1\right ) \left (r^{2}+4 r +2\right ) \left (r^{2}+6 r +7\right ) \left (r^{2}+8 r +14\right ) \left (r^{2}+10 r +23\right )}\) | \(\frac {-414+293 \sqrt {2}}{6108 \sqrt {2}-8640}\) |
| \(a_{6}\) | \(\frac {r \left (r^{11}+27 r^{10}+314 r^{9}+2064 r^{8}+8439 r^{7}+22209 r^{6}+37650 r^{5}+40744 r^{4}+30972 r^{3}+26532 r^{2}+26728 r +13520\right )}{\left (r^{2}+2 r -1\right ) \left (r^{2}+4 r +2\right ) \left (r^{2}+6 r +7\right ) \left (r^{2}+8 r +14\right ) \left (r^{2}+10 r +23\right ) \left (r^{2}+12 r +34\right )}\) | \(\frac {3898-2757 \sqrt {2}}{114408-80892 \sqrt {2}}\) |
For \(n = 7\), using the above recursive equation gives \[ a_{7}=\frac {4 r \left (r^{12}+36 r^{11}+569 r^{10}+5196 r^{9}+30327 r^{8}+118064 r^{7}+310235 r^{6}+544884 r^{5}+625540 r^{4}+476164 r^{3}+306608 r^{2}+226776 r +104600\right )}{\left (r^{2}+2 r -1\right ) \left (r^{2}+4 r +2\right ) \left (r^{2}+6 r +7\right ) \left (r^{2}+8 r +14\right ) \left (r^{2}+10 r +23\right ) \left (r^{2}+12 r +34\right ) \left (r^{2}+14 r +47\right )} \] Which for the root \(r = -\sqrt {2}\) becomes \[ a_{7}=-\frac {\sqrt {2}\, \left (-77567+54843 \sqrt {2}\right )}{126 \left (-1+2 \sqrt {2}\right ) \left (\sqrt {2}-1\right ) \left (-3+2 \sqrt {2}\right ) \left (-2+\sqrt {2}\right ) \left (-5+2 \sqrt {2}\right ) \left (-3+\sqrt {2}\right ) \left (-7+2 \sqrt {2}\right )} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {r}{r^{2}+2 r -1}\) | \(\frac {\sqrt {2}}{-1+2 \sqrt {2}}\) |
| \(a_{2}\) | \(\frac {r \left (r^{3}+r^{2}-2 r +2\right )}{\left (r^{2}+2 r -1\right ) \left (r^{2}+4 r +2\right )}\) | \(\frac {\sqrt {2}}{-5+3 \sqrt {2}}\) |
| \(a_{3}\) | \(\frac {2 r \left (r^{4}+4 r^{3}+3 r^{2}+2\right )}{\left (r^{2}+2 r -1\right ) \left (r^{2}+4 r +2\right ) \left (r^{2}+6 r +7\right )}\) | \(-\frac {2 \sqrt {2}}{15-9 \sqrt {2}}\) |
| \(a_{4}\) | \(\frac {r \left (r^{7}+10 r^{6}+36 r^{5}+58 r^{4}+41 r^{3}+20 r^{2}+42 r +40\right )}{\left (r^{2}+2 r -1\right ) \left (r^{2}+4 r +2\right ) \left (r^{2}+6 r +7\right ) \left (r^{2}+8 r +14\right )}\) | \(\frac {-69+49 \sqrt {2}}{-1104+780 \sqrt {2}}\) |
| \(a_{5}\) | \(\frac {r \left (3 r^{8}+48 r^{7}+306 r^{6}+996 r^{5}+1749 r^{4}+1616 r^{3}+866 r^{2}+680 r +496\right )}{\left (r^{2}+2 r -1\right ) \left (r^{2}+4 r +2\right ) \left (r^{2}+6 r +7\right ) \left (r^{2}+8 r +14\right ) \left (r^{2}+10 r +23\right )}\) | \(\frac {-414+293 \sqrt {2}}{6108 \sqrt {2}-8640}\) |
| \(a_{6}\) | \(\frac {r \left (r^{11}+27 r^{10}+314 r^{9}+2064 r^{8}+8439 r^{7}+22209 r^{6}+37650 r^{5}+40744 r^{4}+30972 r^{3}+26532 r^{2}+26728 r +13520\right )}{\left (r^{2}+2 r -1\right ) \left (r^{2}+4 r +2\right ) \left (r^{2}+6 r +7\right ) \left (r^{2}+8 r +14\right ) \left (r^{2}+10 r +23\right ) \left (r^{2}+12 r +34\right )}\) | \(\frac {3898-2757 \sqrt {2}}{114408-80892 \sqrt {2}}\) |
| \(a_{7}\) | \(\frac {4 r \left (r^{12}+36 r^{11}+569 r^{10}+5196 r^{9}+30327 r^{8}+118064 r^{7}+310235 r^{6}+544884 r^{5}+625540 r^{4}+476164 r^{3}+306608 r^{2}+226776 r +104600\right )}{\left (r^{2}+2 r -1\right ) \left (r^{2}+4 r +2\right ) \left (r^{2}+6 r +7\right ) \left (r^{2}+8 r +14\right ) \left (r^{2}+10 r +23\right ) \left (r^{2}+12 r +34\right ) \left (r^{2}+14 r +47\right )}\) | \(-\frac {\sqrt {2}\, \left (-77567+54843 \sqrt {2}\right )}{126 \left (-1+2 \sqrt {2}\right ) \left (\sqrt {2}-1\right ) \left (-3+2 \sqrt {2}\right ) \left (-2+\sqrt {2}\right ) \left (-5+2 \sqrt {2}\right ) \left (-3+\sqrt {2}\right ) \left (-7+2 \sqrt {2}\right )}\) |
Using the above table, then the solution \(y_{1}\left (x \right )\) is \begin {align*} y_{1}\left (x \right )&= x^{-\sqrt {2}} \left (a_{0}+a_{1} x +a_{2} x^{2}+a_{3} x^{3}+a_{4} x^{4}+a_{5} x^{5}+a_{6} x^{6}+a_{7} x^{7}+a_{8} x^{8}\dots \right ) \\ &= x^{-\sqrt {2}} \left (1+\frac {\sqrt {2}\, x}{-1+2 \sqrt {2}}+\frac {\sqrt {2}\, x^{2}}{-5+3 \sqrt {2}}-\frac {2 \sqrt {2}\, x^{3}}{15-9 \sqrt {2}}+\frac {\left (-69+49 \sqrt {2}\right ) x^{4}}{-1104+780 \sqrt {2}}+\frac {\left (-414+293 \sqrt {2}\right ) x^{5}}{6108 \sqrt {2}-8640}+\frac {\left (3898-2757 \sqrt {2}\right ) x^{6}}{114408-80892 \sqrt {2}}-\frac {\sqrt {2}\, \left (-77567+54843 \sqrt {2}\right ) x^{7}}{126 \left (-1+2 \sqrt {2}\right ) \left (\sqrt {2}-1\right ) \left (-3+2 \sqrt {2}\right ) \left (-2+\sqrt {2}\right ) \left (-5+2 \sqrt {2}\right ) \left (-3+\sqrt {2}\right ) \left (-7+2 \sqrt {2}\right )}+O\left (x^{8}\right )\right ) \end {align*}
Now the second solution \(y_{2}\left (x \right )\) is found. Eq (2B) derived above is now used to find all \(b_{n}\) coefficients. The case \(n = 0\) is skipped since it was used to find the roots of the indicial equation. \(b_{0}\) is arbitrary and taken as \(b_{0} = 1\). Substituting \(n = 1\) in Eq. (2B) gives \[ b_{1} = \frac {r}{r^{2}+2 r -1} \] For \(2\le n\) the recursive equation is \begin{equation} \tag{3} b_{n -2} \left (n +r -2\right ) \left (n -3+r \right )-b_{n} \left (n +r \right ) \left (n +r -1\right )+b_{n -1} \left (n +r -1\right )-b_{n} \left (n +r \right )+2 b_{n} = 0 \end{equation} Solving for \(b_{n}\) from recursive equation (4) gives \[ b_{n} = \frac {n^{2} b_{n -2}+2 n r b_{n -2}+r^{2} b_{n -2}-5 n b_{n -2}+n b_{n -1}-5 r b_{n -2}+r b_{n -1}+6 b_{n -2}-b_{n -1}}{n^{2}+2 n r +r^{2}-2}\tag {4} \] Which for the root \(r = \sqrt {2}\) becomes \[ b_{n} = \frac {\left (2 n b_{n -2}-5 b_{n -2}+b_{n -1}\right ) \sqrt {2}+n^{2} b_{n -2}+\left (-5 b_{n -2}+b_{n -1}\right ) n +8 b_{n -2}-b_{n -1}}{n \left (2 \sqrt {2}+n \right )}\tag {5} \] At this point, it is a good idea to keep track of \(b_{n}\) in a table both before substituting \(r = \sqrt {2}\) and after as more terms are found using the above recursive equation.
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {r}{r^{2}+2 r -1}\) | \(\frac {\sqrt {2}}{1+2 \sqrt {2}}\) |
For \(n = 2\), using the above recursive equation gives \[ b_{2}=\frac {r \left (r^{3}+r^{2}-2 r +2\right )}{\left (r^{2}+2 r -1\right ) \left (r^{2}+4 r +2\right )} \] Which for the root \(r = \sqrt {2}\) becomes \[ b_{2}=\frac {\sqrt {2}}{5+3 \sqrt {2}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {r}{r^{2}+2 r -1}\) | \(\frac {\sqrt {2}}{1+2 \sqrt {2}}\) |
| \(b_{2}\) | \(\frac {r \left (r^{3}+r^{2}-2 r +2\right )}{\left (r^{2}+2 r -1\right ) \left (r^{2}+4 r +2\right )}\) | \(\frac {\sqrt {2}}{5+3 \sqrt {2}}\) |
For \(n = 3\), using the above recursive equation gives \[ b_{3}=\frac {2 r \left (r^{4}+4 r^{3}+3 r^{2}+2\right )}{\left (r^{2}+2 r -1\right ) \left (r^{2}+4 r +2\right ) \left (r^{2}+6 r +7\right )} \] Which for the root \(r = \sqrt {2}\) becomes \[ b_{3}=\frac {2 \sqrt {2}}{15+9 \sqrt {2}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {r}{r^{2}+2 r -1}\) | \(\frac {\sqrt {2}}{1+2 \sqrt {2}}\) |
| \(b_{2}\) | \(\frac {r \left (r^{3}+r^{2}-2 r +2\right )}{\left (r^{2}+2 r -1\right ) \left (r^{2}+4 r +2\right )}\) | \(\frac {\sqrt {2}}{5+3 \sqrt {2}}\) |
| \(b_{3}\) | \(\frac {2 r \left (r^{4}+4 r^{3}+3 r^{2}+2\right )}{\left (r^{2}+2 r -1\right ) \left (r^{2}+4 r +2\right ) \left (r^{2}+6 r +7\right )}\) | \(\frac {2 \sqrt {2}}{15+9 \sqrt {2}}\) |
For \(n = 4\), using the above recursive equation gives \[ b_{4}=\frac {r \left (r^{7}+10 r^{6}+36 r^{5}+58 r^{4}+41 r^{3}+20 r^{2}+42 r +40\right )}{\left (r^{2}+2 r -1\right ) \left (r^{2}+4 r +2\right ) \left (r^{2}+6 r +7\right ) \left (r^{2}+8 r +14\right )} \] Which for the root \(r = \sqrt {2}\) becomes \[ b_{4}=\frac {69+49 \sqrt {2}}{1104+780 \sqrt {2}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {r}{r^{2}+2 r -1}\) | \(\frac {\sqrt {2}}{1+2 \sqrt {2}}\) |
| \(b_{2}\) | \(\frac {r \left (r^{3}+r^{2}-2 r +2\right )}{\left (r^{2}+2 r -1\right ) \left (r^{2}+4 r +2\right )}\) | \(\frac {\sqrt {2}}{5+3 \sqrt {2}}\) |
| \(b_{3}\) | \(\frac {2 r \left (r^{4}+4 r^{3}+3 r^{2}+2\right )}{\left (r^{2}+2 r -1\right ) \left (r^{2}+4 r +2\right ) \left (r^{2}+6 r +7\right )}\) | \(\frac {2 \sqrt {2}}{15+9 \sqrt {2}}\) |
| \(b_{4}\) | \(\frac {r \left (r^{7}+10 r^{6}+36 r^{5}+58 r^{4}+41 r^{3}+20 r^{2}+42 r +40\right )}{\left (r^{2}+2 r -1\right ) \left (r^{2}+4 r +2\right ) \left (r^{2}+6 r +7\right ) \left (r^{2}+8 r +14\right )}\) | \(\frac {69+49 \sqrt {2}}{1104+780 \sqrt {2}}\) |
For \(n = 5\), using the above recursive equation gives \[ b_{5}=\frac {r \left (3 r^{8}+48 r^{7}+306 r^{6}+996 r^{5}+1749 r^{4}+1616 r^{3}+866 r^{2}+680 r +496\right )}{\left (r^{2}+2 r -1\right ) \left (r^{2}+4 r +2\right ) \left (r^{2}+6 r +7\right ) \left (r^{2}+8 r +14\right ) \left (r^{2}+10 r +23\right )} \] Which for the root \(r = \sqrt {2}\) becomes \[ b_{5}=\frac {414+293 \sqrt {2}}{6108 \sqrt {2}+8640} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {r}{r^{2}+2 r -1}\) | \(\frac {\sqrt {2}}{1+2 \sqrt {2}}\) |
| \(b_{2}\) | \(\frac {r \left (r^{3}+r^{2}-2 r +2\right )}{\left (r^{2}+2 r -1\right ) \left (r^{2}+4 r +2\right )}\) | \(\frac {\sqrt {2}}{5+3 \sqrt {2}}\) |
| \(b_{3}\) | \(\frac {2 r \left (r^{4}+4 r^{3}+3 r^{2}+2\right )}{\left (r^{2}+2 r -1\right ) \left (r^{2}+4 r +2\right ) \left (r^{2}+6 r +7\right )}\) | \(\frac {2 \sqrt {2}}{15+9 \sqrt {2}}\) |
| \(b_{4}\) | \(\frac {r \left (r^{7}+10 r^{6}+36 r^{5}+58 r^{4}+41 r^{3}+20 r^{2}+42 r +40\right )}{\left (r^{2}+2 r -1\right ) \left (r^{2}+4 r +2\right ) \left (r^{2}+6 r +7\right ) \left (r^{2}+8 r +14\right )}\) | \(\frac {69+49 \sqrt {2}}{1104+780 \sqrt {2}}\) |
| \(b_{5}\) | \(\frac {r \left (3 r^{8}+48 r^{7}+306 r^{6}+996 r^{5}+1749 r^{4}+1616 r^{3}+866 r^{2}+680 r +496\right )}{\left (r^{2}+2 r -1\right ) \left (r^{2}+4 r +2\right ) \left (r^{2}+6 r +7\right ) \left (r^{2}+8 r +14\right ) \left (r^{2}+10 r +23\right )}\) | \(\frac {414+293 \sqrt {2}}{6108 \sqrt {2}+8640}\) |
For \(n = 6\), using the above recursive equation gives \[ b_{6}=\frac {r \left (r^{11}+27 r^{10}+314 r^{9}+2064 r^{8}+8439 r^{7}+22209 r^{6}+37650 r^{5}+40744 r^{4}+30972 r^{3}+26532 r^{2}+26728 r +13520\right )}{\left (r^{2}+2 r -1\right ) \left (r^{2}+4 r +2\right ) \left (r^{2}+6 r +7\right ) \left (r^{2}+8 r +14\right ) \left (r^{2}+10 r +23\right ) \left (r^{2}+12 r +34\right )} \] Which for the root \(r = \sqrt {2}\) becomes \[ b_{6}=\frac {3898+2757 \sqrt {2}}{114408+80892 \sqrt {2}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {r}{r^{2}+2 r -1}\) | \(\frac {\sqrt {2}}{1+2 \sqrt {2}}\) |
| \(b_{2}\) | \(\frac {r \left (r^{3}+r^{2}-2 r +2\right )}{\left (r^{2}+2 r -1\right ) \left (r^{2}+4 r +2\right )}\) | \(\frac {\sqrt {2}}{5+3 \sqrt {2}}\) |
| \(b_{3}\) | \(\frac {2 r \left (r^{4}+4 r^{3}+3 r^{2}+2\right )}{\left (r^{2}+2 r -1\right ) \left (r^{2}+4 r +2\right ) \left (r^{2}+6 r +7\right )}\) | \(\frac {2 \sqrt {2}}{15+9 \sqrt {2}}\) |
| \(b_{4}\) | \(\frac {r \left (r^{7}+10 r^{6}+36 r^{5}+58 r^{4}+41 r^{3}+20 r^{2}+42 r +40\right )}{\left (r^{2}+2 r -1\right ) \left (r^{2}+4 r +2\right ) \left (r^{2}+6 r +7\right ) \left (r^{2}+8 r +14\right )}\) | \(\frac {69+49 \sqrt {2}}{1104+780 \sqrt {2}}\) |
| \(b_{5}\) | \(\frac {r \left (3 r^{8}+48 r^{7}+306 r^{6}+996 r^{5}+1749 r^{4}+1616 r^{3}+866 r^{2}+680 r +496\right )}{\left (r^{2}+2 r -1\right ) \left (r^{2}+4 r +2\right ) \left (r^{2}+6 r +7\right ) \left (r^{2}+8 r +14\right ) \left (r^{2}+10 r +23\right )}\) | \(\frac {414+293 \sqrt {2}}{6108 \sqrt {2}+8640}\) |
| \(b_{6}\) | \(\frac {r \left (r^{11}+27 r^{10}+314 r^{9}+2064 r^{8}+8439 r^{7}+22209 r^{6}+37650 r^{5}+40744 r^{4}+30972 r^{3}+26532 r^{2}+26728 r +13520\right )}{\left (r^{2}+2 r -1\right ) \left (r^{2}+4 r +2\right ) \left (r^{2}+6 r +7\right ) \left (r^{2}+8 r +14\right ) \left (r^{2}+10 r +23\right ) \left (r^{2}+12 r +34\right )}\) | \(\frac {3898+2757 \sqrt {2}}{114408+80892 \sqrt {2}}\) |
For \(n = 7\), using the above recursive equation gives \[ b_{7}=\frac {4 r \left (r^{12}+36 r^{11}+569 r^{10}+5196 r^{9}+30327 r^{8}+118064 r^{7}+310235 r^{6}+544884 r^{5}+625540 r^{4}+476164 r^{3}+306608 r^{2}+226776 r +104600\right )}{\left (r^{2}+2 r -1\right ) \left (r^{2}+4 r +2\right ) \left (r^{2}+6 r +7\right ) \left (r^{2}+8 r +14\right ) \left (r^{2}+10 r +23\right ) \left (r^{2}+12 r +34\right ) \left (r^{2}+14 r +47\right )} \] Which for the root \(r = \sqrt {2}\) becomes \[ b_{7}=\frac {\sqrt {2}\, \left (77567+54843 \sqrt {2}\right )}{126 \left (1+2 \sqrt {2}\right ) \left (1+\sqrt {2}\right ) \left (3+2 \sqrt {2}\right ) \left (2+\sqrt {2}\right ) \left (5+2 \sqrt {2}\right ) \left (3+\sqrt {2}\right ) \left (7+2 \sqrt {2}\right )} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {r}{r^{2}+2 r -1}\) | \(\frac {\sqrt {2}}{1+2 \sqrt {2}}\) |
| \(b_{2}\) | \(\frac {r \left (r^{3}+r^{2}-2 r +2\right )}{\left (r^{2}+2 r -1\right ) \left (r^{2}+4 r +2\right )}\) | \(\frac {\sqrt {2}}{5+3 \sqrt {2}}\) |
| \(b_{3}\) | \(\frac {2 r \left (r^{4}+4 r^{3}+3 r^{2}+2\right )}{\left (r^{2}+2 r -1\right ) \left (r^{2}+4 r +2\right ) \left (r^{2}+6 r +7\right )}\) | \(\frac {2 \sqrt {2}}{15+9 \sqrt {2}}\) |
| \(b_{4}\) | \(\frac {r \left (r^{7}+10 r^{6}+36 r^{5}+58 r^{4}+41 r^{3}+20 r^{2}+42 r +40\right )}{\left (r^{2}+2 r -1\right ) \left (r^{2}+4 r +2\right ) \left (r^{2}+6 r +7\right ) \left (r^{2}+8 r +14\right )}\) | \(\frac {69+49 \sqrt {2}}{1104+780 \sqrt {2}}\) |
| \(b_{5}\) | \(\frac {r \left (3 r^{8}+48 r^{7}+306 r^{6}+996 r^{5}+1749 r^{4}+1616 r^{3}+866 r^{2}+680 r +496\right )}{\left (r^{2}+2 r -1\right ) \left (r^{2}+4 r +2\right ) \left (r^{2}+6 r +7\right ) \left (r^{2}+8 r +14\right ) \left (r^{2}+10 r +23\right )}\) | \(\frac {414+293 \sqrt {2}}{6108 \sqrt {2}+8640}\) |
| \(b_{6}\) | \(\frac {r \left (r^{11}+27 r^{10}+314 r^{9}+2064 r^{8}+8439 r^{7}+22209 r^{6}+37650 r^{5}+40744 r^{4}+30972 r^{3}+26532 r^{2}+26728 r +13520\right )}{\left (r^{2}+2 r -1\right ) \left (r^{2}+4 r +2\right ) \left (r^{2}+6 r +7\right ) \left (r^{2}+8 r +14\right ) \left (r^{2}+10 r +23\right ) \left (r^{2}+12 r +34\right )}\) | \(\frac {3898+2757 \sqrt {2}}{114408+80892 \sqrt {2}}\) |
| \(b_{7}\) | \(\frac {4 r \left (r^{12}+36 r^{11}+569 r^{10}+5196 r^{9}+30327 r^{8}+118064 r^{7}+310235 r^{6}+544884 r^{5}+625540 r^{4}+476164 r^{3}+306608 r^{2}+226776 r +104600\right )}{\left (r^{2}+2 r -1\right ) \left (r^{2}+4 r +2\right ) \left (r^{2}+6 r +7\right ) \left (r^{2}+8 r +14\right ) \left (r^{2}+10 r +23\right ) \left (r^{2}+12 r +34\right ) \left (r^{2}+14 r +47\right )}\) | \(\frac {\sqrt {2}\, \left (77567+54843 \sqrt {2}\right )}{126 \left (1+2 \sqrt {2}\right ) \left (1+\sqrt {2}\right ) \left (3+2 \sqrt {2}\right ) \left (2+\sqrt {2}\right ) \left (5+2 \sqrt {2}\right ) \left (3+\sqrt {2}\right ) \left (7+2 \sqrt {2}\right )}\) |
Using the above table, then the solution \(y_{2}\left (x \right )\) is \begin {align*} y_{2}\left (x \right )&= x^{-\sqrt {2}} \left (b_{0}+b_{1} x +b_{2} x^{2}+b_{3} x^{3}+b_{4} x^{4}+b_{5} x^{5}+b_{6} x^{6}+b_{7} x^{7}+b_{8} x^{8}\dots \right ) \\ &= x^{\sqrt {2}} \left (1+\frac {\sqrt {2}\, x}{1+2 \sqrt {2}}+\frac {\sqrt {2}\, x^{2}}{5+3 \sqrt {2}}+\frac {2 \sqrt {2}\, x^{3}}{15+9 \sqrt {2}}+\frac {\left (69+49 \sqrt {2}\right ) x^{4}}{1104+780 \sqrt {2}}+\frac {\left (414+293 \sqrt {2}\right ) x^{5}}{6108 \sqrt {2}+8640}+\frac {\left (3898+2757 \sqrt {2}\right ) x^{6}}{114408+80892 \sqrt {2}}+\frac {\sqrt {2}\, \left (77567+54843 \sqrt {2}\right ) x^{7}}{126 \left (1+2 \sqrt {2}\right ) \left (1+\sqrt {2}\right ) \left (3+2 \sqrt {2}\right ) \left (2+\sqrt {2}\right ) \left (5+2 \sqrt {2}\right ) \left (3+\sqrt {2}\right ) \left (7+2 \sqrt {2}\right )}+O\left (x^{8}\right )\right ) \end {align*}
Therefore the homogeneous solution is \begin{align*} y_h(x) &= c_{1} y_{1}\left (x \right )+c_{2} y_{2}\left (x \right ) \\ &= c_{1} x^{-\sqrt {2}} \left (1+\frac {\sqrt {2}\, x}{-1+2 \sqrt {2}}+\frac {\sqrt {2}\, x^{2}}{-5+3 \sqrt {2}}-\frac {2 \sqrt {2}\, x^{3}}{15-9 \sqrt {2}}+\frac {\left (-69+49 \sqrt {2}\right ) x^{4}}{-1104+780 \sqrt {2}}+\frac {\left (-414+293 \sqrt {2}\right ) x^{5}}{6108 \sqrt {2}-8640}+\frac {\left (3898-2757 \sqrt {2}\right ) x^{6}}{114408-80892 \sqrt {2}}-\frac {\sqrt {2}\, \left (-77567+54843 \sqrt {2}\right ) x^{7}}{126 \left (-1+2 \sqrt {2}\right ) \left (\sqrt {2}-1\right ) \left (-3+2 \sqrt {2}\right ) \left (-2+\sqrt {2}\right ) \left (-5+2 \sqrt {2}\right ) \left (-3+\sqrt {2}\right ) \left (-7+2 \sqrt {2}\right )}+O\left (x^{8}\right )\right ) + c_{2} x^{\sqrt {2}} \left (1+\frac {\sqrt {2}\, x}{1+2 \sqrt {2}}+\frac {\sqrt {2}\, x^{2}}{5+3 \sqrt {2}}+\frac {2 \sqrt {2}\, x^{3}}{15+9 \sqrt {2}}+\frac {\left (69+49 \sqrt {2}\right ) x^{4}}{1104+780 \sqrt {2}}+\frac {\left (414+293 \sqrt {2}\right ) x^{5}}{6108 \sqrt {2}+8640}+\frac {\left (3898+2757 \sqrt {2}\right ) x^{6}}{114408+80892 \sqrt {2}}+\frac {\sqrt {2}\, \left (77567+54843 \sqrt {2}\right ) x^{7}}{126 \left (1+2 \sqrt {2}\right ) \left (1+\sqrt {2}\right ) \left (3+2 \sqrt {2}\right ) \left (2+\sqrt {2}\right ) \left (5+2 \sqrt {2}\right ) \left (3+\sqrt {2}\right ) \left (7+2 \sqrt {2}\right )}+O\left (x^{8}\right )\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} x^{-\sqrt {2}} \left (1+\frac {\sqrt {2}\, x}{-1+2 \sqrt {2}}+\frac {\sqrt {2}\, x^{2}}{-5+3 \sqrt {2}}-\frac {2 \sqrt {2}\, x^{3}}{15-9 \sqrt {2}}+\frac {\left (-69+49 \sqrt {2}\right ) x^{4}}{-1104+780 \sqrt {2}}+\frac {\left (-414+293 \sqrt {2}\right ) x^{5}}{6108 \sqrt {2}-8640}+\frac {\left (3898-2757 \sqrt {2}\right ) x^{6}}{114408-80892 \sqrt {2}}-\frac {\sqrt {2}\, \left (-77567+54843 \sqrt {2}\right ) x^{7}}{126 \left (-1+2 \sqrt {2}\right ) \left (\sqrt {2}-1\right ) \left (-3+2 \sqrt {2}\right ) \left (-2+\sqrt {2}\right ) \left (-5+2 \sqrt {2}\right ) \left (-3+\sqrt {2}\right ) \left (-7+2 \sqrt {2}\right )}+O\left (x^{8}\right )\right )+c_{2} x^{\sqrt {2}} \left (1+\frac {\sqrt {2}\, x}{1+2 \sqrt {2}}+\frac {\sqrt {2}\, x^{2}}{5+3 \sqrt {2}}+\frac {2 \sqrt {2}\, x^{3}}{15+9 \sqrt {2}}+\frac {\left (69+49 \sqrt {2}\right ) x^{4}}{1104+780 \sqrt {2}}+\frac {\left (414+293 \sqrt {2}\right ) x^{5}}{6108 \sqrt {2}+8640}+\frac {\left (3898+2757 \sqrt {2}\right ) x^{6}}{114408+80892 \sqrt {2}}+\frac {\sqrt {2}\, \left (77567+54843 \sqrt {2}\right ) x^{7}}{126 \left (1+2 \sqrt {2}\right ) \left (1+\sqrt {2}\right ) \left (3+2 \sqrt {2}\right ) \left (2+\sqrt {2}\right ) \left (5+2 \sqrt {2}\right ) \left (3+\sqrt {2}\right ) \left (7+2 \sqrt {2}\right )}+O\left (x^{8}\right )\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x^{-\sqrt {2}} \left (1+\frac {\sqrt {2}\, x}{-1+2 \sqrt {2}}+\frac {\sqrt {2}\, x^{2}}{-5+3 \sqrt {2}}-\frac {2 \sqrt {2}\, x^{3}}{15-9 \sqrt {2}}+\frac {\left (-69+49 \sqrt {2}\right ) x^{4}}{-1104+780 \sqrt {2}}+\frac {\left (-414+293 \sqrt {2}\right ) x^{5}}{6108 \sqrt {2}-8640}+\frac {\left (3898-2757 \sqrt {2}\right ) x^{6}}{114408-80892 \sqrt {2}}-\frac {\sqrt {2}\, \left (-77567+54843 \sqrt {2}\right ) x^{7}}{126 \left (-1+2 \sqrt {2}\right ) \left (\sqrt {2}-1\right ) \left (-3+2 \sqrt {2}\right ) \left (-2+\sqrt {2}\right ) \left (-5+2 \sqrt {2}\right ) \left (-3+\sqrt {2}\right ) \left (-7+2 \sqrt {2}\right )}+O\left (x^{8}\right )\right )+c_{2} x^{\sqrt {2}} \left (1+\frac {\sqrt {2}\, x}{1+2 \sqrt {2}}+\frac {\sqrt {2}\, x^{2}}{5+3 \sqrt {2}}+\frac {2 \sqrt {2}\, x^{3}}{15+9 \sqrt {2}}+\frac {\left (69+49 \sqrt {2}\right ) x^{4}}{1104+780 \sqrt {2}}+\frac {\left (414+293 \sqrt {2}\right ) x^{5}}{6108 \sqrt {2}+8640}+\frac {\left (3898+2757 \sqrt {2}\right ) x^{6}}{114408+80892 \sqrt {2}}+\frac {\sqrt {2}\, \left (77567+54843 \sqrt {2}\right ) x^{7}}{126 \left (1+2 \sqrt {2}\right ) \left (1+\sqrt {2}\right ) \left (3+2 \sqrt {2}\right ) \left (2+\sqrt {2}\right ) \left (5+2 \sqrt {2}\right ) \left (3+\sqrt {2}\right ) \left (7+2 \sqrt {2}\right )}+O\left (x^{8}\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} x^{-\sqrt {2}} \left (1+\frac {\sqrt {2}\, x}{-1+2 \sqrt {2}}+\frac {\sqrt {2}\, x^{2}}{-5+3 \sqrt {2}}-\frac {2 \sqrt {2}\, x^{3}}{15-9 \sqrt {2}}+\frac {\left (-69+49 \sqrt {2}\right ) x^{4}}{-1104+780 \sqrt {2}}+\frac {\left (-414+293 \sqrt {2}\right ) x^{5}}{6108 \sqrt {2}-8640}+\frac {\left (3898-2757 \sqrt {2}\right ) x^{6}}{114408-80892 \sqrt {2}}-\frac {\sqrt {2}\, \left (-77567+54843 \sqrt {2}\right ) x^{7}}{126 \left (-1+2 \sqrt {2}\right ) \left (\sqrt {2}-1\right ) \left (-3+2 \sqrt {2}\right ) \left (-2+\sqrt {2}\right ) \left (-5+2 \sqrt {2}\right ) \left (-3+\sqrt {2}\right ) \left (-7+2 \sqrt {2}\right )}+O\left (x^{8}\right )\right )+c_{2} x^{\sqrt {2}} \left (1+\frac {\sqrt {2}\, x}{1+2 \sqrt {2}}+\frac {\sqrt {2}\, x^{2}}{5+3 \sqrt {2}}+\frac {2 \sqrt {2}\, x^{3}}{15+9 \sqrt {2}}+\frac {\left (69+49 \sqrt {2}\right ) x^{4}}{1104+780 \sqrt {2}}+\frac {\left (414+293 \sqrt {2}\right ) x^{5}}{6108 \sqrt {2}+8640}+\frac {\left (3898+2757 \sqrt {2}\right ) x^{6}}{114408+80892 \sqrt {2}}+\frac {\sqrt {2}\, \left (77567+54843 \sqrt {2}\right ) x^{7}}{126 \left (1+2 \sqrt {2}\right ) \left (1+\sqrt {2}\right ) \left (3+2 \sqrt {2}\right ) \left (2+\sqrt {2}\right ) \left (5+2 \sqrt {2}\right ) \left (3+\sqrt {2}\right ) \left (7+2 \sqrt {2}\right )}+O\left (x^{8}\right )\right ) \] Verified OK.
Maple trace
`Methods for second order ODEs: --- Trying classification methods --- trying a quadrature checking if the LODE has constant coefficients checking if the LODE is of Euler type trying a symmetry of the form [xi=0, eta=F(x)] checking if the LODE is missing y -> Trying a Liouvillian solution using Kovacics algorithm <- No Liouvillian solutions exists -> Trying a solution in terms of special functions: -> Bessel -> elliptic -> Legendre -> Whittaker -> hyper3: Equivalence to 1F1 under a power @ Moebius -> hypergeometric -> heuristic approach -> hyper3: Equivalence to 2F1, 1F1 or 0F1 under a power @ Moebius -> Mathieu -> Equivalence to the rational form of Mathieu ODE under a power @ Moebius trying a solution in terms of MeijerG functions -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius <- Heun successful: received ODE is equivalent to the HeunG ODE, case a <> 0, e <> 0, g <> 0, c = 0 `
✓ Solution by Maple
Time used: 0.032 (sec). Leaf size: 654
Order:=8; dsolve(x^2*(x^2-1)*diff(y(x),x$2)-x*(1-x)*diff(y(x),x)+2*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = c_{1} x^{-\sqrt {2}} \left (1+\frac {\sqrt {2}}{-1+2 \sqrt {2}} x +\frac {\sqrt {2}}{\left (1-2 \sqrt {2}\right ) \left (\sqrt {2}-1\right )} x^{2}+\frac {6 \sqrt {2}-8}{57 \sqrt {2}-81} x^{3}+\frac {-49 \sqrt {2}+69}{1104-780 \sqrt {2}} x^{4}+\frac {293 \sqrt {2}-414}{6108 \sqrt {2}-8640} x^{5}+\frac {-2757 \sqrt {2}+3898}{114408-80892 \sqrt {2}} x^{6}+\frac {1}{126} \frac {77567 \sqrt {2}-109686}{\left (-1+2 \sqrt {2}\right ) \left (\sqrt {2}-1\right ) \left (2 \sqrt {2}-3\right ) \left (-2+\sqrt {2}\right ) \left (-5+2 \sqrt {2}\right ) \left (-3+\sqrt {2}\right ) \left (-7+2 \sqrt {2}\right )} x^{7}+\operatorname {O}\left (x^{8}\right )\right )+c_{2} x^{\sqrt {2}} \left (1+\frac {\sqrt {2}}{1+2 \sqrt {2}} x +\frac {\sqrt {2}}{5+3 \sqrt {2}} x^{2}+\frac {6 \sqrt {2}+8}{57 \sqrt {2}+81} x^{3}+\frac {49 \sqrt {2}+69}{1104+780 \sqrt {2}} x^{4}+\frac {293 \sqrt {2}+414}{6108 \sqrt {2}+8640} x^{5}+\frac {2757 \sqrt {2}+3898}{114408+80892 \sqrt {2}} x^{6}+\frac {1}{126} \frac {77567 \sqrt {2}+109686}{\left (1+2 \sqrt {2}\right ) \left (1+\sqrt {2}\right ) \left (3+2 \sqrt {2}\right ) \left (2+\sqrt {2}\right ) \left (5+2 \sqrt {2}\right ) \left (3+\sqrt {2}\right ) \left (7+2 \sqrt {2}\right )} x^{7}+\operatorname {O}\left (x^{8}\right )\right ) \]
✓ Solution by Mathematica
Time used: 0.007 (sec). Leaf size: 10352
AsymptoticDSolveValue[x^2*(x^2-1)*y''[x]-x*(1-x)*y'[x]+2*y[x]==0,y[x],{x,0,7}]
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