Internal problem ID [5561]
Internal file name [OUTPUT/4809_Sunday_June_05_2022_03_06_24_PM_93524911/index.tex]
Book: A FIRST COURSE IN DIFFERENTIAL EQUATIONS with Modeling Applications.
Dennis G. Zill. 9th edition. Brooks/Cole. CA, USA.
Section: Chapter 6. SERIES SOLUTIONS OF LINEAR EQUATIONS. Exercises. 6.2 page
239
Problem number: 6.
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]]
\[ \boxed {x^{2} \left (x -5\right )^{2} y^{\prime \prime }+4 y^{\prime } x +\left (x^{2}-25\right ) y=0} \] 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. \[ \left (x^{4}-10 x^{3}+25 x^{2}\right ) y^{\prime \prime }+4 y^{\prime } x +\left (x^{2}-25\right ) y = 0 \] 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 {4}{x \left (x -5\right )^{2}}\\ q(x) &= \frac {x +5}{\left (x -5\right ) x^{2}}\\ \end {align*}
| \(p(x)=\frac {4}{x \left (x -5\right )^{2}}\) | |
| singularity | type |
| \(x = 0\) | \(\text {``regular''}\) |
| \(x = 5\) | \(\text {``irregular''}\) |
| \(q(x)=\frac {x +5}{\left (x -5\right ) x^{2}}\) | |
| singularity | type |
| \(x = 0\) | \(\text {``regular''}\) |
| \(x = 5\) | \(\text {``regular''}\) |
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \([0, \infty ]\)
Irregular singular points : \([5]\)
Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ x^{2} \left (x^{2}-10 x +25\right ) y^{\prime \prime }+4 y^{\prime } x +\left (x^{2}-25\right ) y = 0 \] 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}-10 x +25\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} x^{n +r -2}\right )+4 \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right ) x +\left (x^{2}-25\right ) \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 (-10 x^{1+n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}25 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}4 x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r +2} a_{n}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-25 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 }}\left (-10 x^{1+n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-10 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) x^{n +r}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r +2} a_{n} &= \moverset {\infty }{\munderset {n =2}{\sum }}a_{n -2} 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 =1}{\sum }}\left (-10 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}25 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}4 x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}a_{n -2} x^{n +r}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-25 a_{n} x^{n +r}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ 25 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )+4 x^{n +r} a_{n} \left (n +r \right )-25 a_{n} x^{n +r} = 0 \] When \(n = 0\) the above becomes \[ 25 x^{r} a_{0} r \left (-1+r \right )+4 x^{r} a_{0} r -25 a_{0} x^{r} = 0 \] Or \[ \left (25 x^{r} r \left (-1+r \right )+4 x^{r} r -25 x^{r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ \left (25 r^{2}-21 r -25\right ) x^{r} = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ 25 r^{2}-21 r -25 = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= \frac {21}{50}+\frac {\sqrt {2941}}{50}\\ r_2 &= \frac {21}{50}-\frac {\sqrt {2941}}{50} \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ \left (25 r^{2}-21 r -25\right ) x^{r} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \(\left [\frac {21}{50}+\frac {\sqrt {2941}}{50}, \frac {21}{50}-\frac {\sqrt {2941}}{50}\right ]\).
Since \(r_1 - r_2 = \frac {\sqrt {2941}}{25}\) 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 +\frac {21}{50}+\frac {\sqrt {2941}}{50}}\\ y_{2}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +\frac {21}{50}-\frac {\sqrt {2941}}{50}} \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 \[ a_{1} = \frac {10 r \left (-1+r \right )}{25 r^{2}+29 r -21} \] 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 )-10 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right )+25 a_{n} \left (n +r \right ) \left (n +r -1\right )+4 a_{n} \left (n +r \right )+a_{n -2}-25 a_{n} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = -\frac {n^{2} a_{n -2}-10 n^{2} a_{n -1}+2 n r a_{n -2}-20 n r a_{n -1}+r^{2} a_{n -2}-10 r^{2} a_{n -1}-5 n a_{n -2}+30 n a_{n -1}-5 r a_{n -2}+30 r a_{n -1}+7 a_{n -2}-20 a_{n -1}}{25 n^{2}+50 n r +25 r^{2}-21 n -21 r -25}\tag {4} \] Which for the root \(r = \frac {21}{50}+\frac {\sqrt {2941}}{50}\) becomes \[ a_{n} = \frac {\left (\left (-25 a_{n -2}+250 a_{n -1}\right ) n +52 a_{n -2}-270 a_{n -1}\right ) \sqrt {2941}+\left (-625 a_{n -2}+6250 a_{n -1}\right ) n^{2}+\left (2600 a_{n -2}-13500 a_{n -1}\right ) n -3908 a_{n -2}+13080 a_{n -1}}{625 n \left (\sqrt {2941}+25 n \right )}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = \frac {21}{50}+\frac {\sqrt {2941}}{50}\) and after as more terms are found using the above recursive equation.
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {10 r \left (-1+r \right )}{25 r^{2}+29 r -21}\) | \(\frac {\left (21+\sqrt {2941}\right ) \left (-29+\sqrt {2941}\right )}{6250+250 \sqrt {2941}}\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {75 r^{4}-4 r^{3}-75 r^{2}-50 r +21}{\left (25 r^{2}+29 r -21\right ) \left (25 r^{2}+79 r +33\right )} \] Which for the root \(r = \frac {21}{50}+\frac {\sqrt {2941}}{50}\) becomes \[ a_{2}=\frac {\frac {717381}{15625}+\frac {7911 \sqrt {2941}}{15625}}{\left (25+\sqrt {2941}\right ) \left (50+\sqrt {2941}\right )} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {10 r \left (-1+r \right )}{25 r^{2}+29 r -21}\) | \(\frac {\left (21+\sqrt {2941}\right ) \left (-29+\sqrt {2941}\right )}{6250+250 \sqrt {2941}}\) |
| \(a_{2}\) | \(\frac {75 r^{4}-4 r^{3}-75 r^{2}-50 r +21}{\left (25 r^{2}+29 r -21\right ) \left (25 r^{2}+79 r +33\right )}\) | \(\frac {\frac {717381}{15625}+\frac {7911 \sqrt {2941}}{15625}}{\left (25+\sqrt {2941}\right ) \left (50+\sqrt {2941}\right )}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=\frac {500 r^{6}+1420 r^{5}+300 r^{4}-2580 r^{3}-2000 r^{2}-40 r +420}{\left (25 r^{2}+29 r -21\right ) \left (25 r^{2}+79 r +33\right ) \left (25 r^{2}+129 r +137\right )} \] Which for the root \(r = \frac {21}{50}+\frac {\sqrt {2941}}{50}\) becomes \[ a_{3}=\frac {\frac {906742764}{1953125}+\frac {15291084 \sqrt {2941}}{1953125}}{\left (25+\sqrt {2941}\right ) \left (50+\sqrt {2941}\right ) \left (75+\sqrt {2941}\right )} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {10 r \left (-1+r \right )}{25 r^{2}+29 r -21}\) | \(\frac {\left (21+\sqrt {2941}\right ) \left (-29+\sqrt {2941}\right )}{6250+250 \sqrt {2941}}\) |
| \(a_{2}\) | \(\frac {75 r^{4}-4 r^{3}-75 r^{2}-50 r +21}{\left (25 r^{2}+29 r -21\right ) \left (25 r^{2}+79 r +33\right )}\) | \(\frac {\frac {717381}{15625}+\frac {7911 \sqrt {2941}}{15625}}{\left (25+\sqrt {2941}\right ) \left (50+\sqrt {2941}\right )}\) |
| \(a_{3}\) | \(\frac {500 r^{6}+1420 r^{5}+300 r^{4}-2580 r^{3}-2000 r^{2}-40 r +420}{\left (25 r^{2}+29 r -21\right ) \left (25 r^{2}+79 r +33\right ) \left (25 r^{2}+129 r +137\right )}\) | \(\frac {\frac {906742764}{1953125}+\frac {15291084 \sqrt {2941}}{1953125}}{\left (25+\sqrt {2941}\right ) \left (50+\sqrt {2941}\right ) \left (75+\sqrt {2941}\right )}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {3125 r^{8}+24000 r^{7}+61766 r^{6}+33496 r^{5}-104033 r^{4}-168040 r^{3}-59654 r^{2}+22392 r +16569}{\left (25 r^{2}+29 r -21\right ) \left (25 r^{2}+79 r +33\right ) \left (25 r^{2}+129 r +137\right ) \left (25 r^{2}+179 r +291\right )} \] Which for the root \(r = \frac {21}{50}+\frac {\sqrt {2941}}{50}\) becomes \[ a_{4}=\frac {\frac {1473770634612}{244140625}+\frac {26407796172 \sqrt {2941}}{244140625}}{\left (25+\sqrt {2941}\right ) \left (50+\sqrt {2941}\right ) \left (75+\sqrt {2941}\right ) \left (100+\sqrt {2941}\right )} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {10 r \left (-1+r \right )}{25 r^{2}+29 r -21}\) | \(\frac {\left (21+\sqrt {2941}\right ) \left (-29+\sqrt {2941}\right )}{6250+250 \sqrt {2941}}\) |
| \(a_{2}\) | \(\frac {75 r^{4}-4 r^{3}-75 r^{2}-50 r +21}{\left (25 r^{2}+29 r -21\right ) \left (25 r^{2}+79 r +33\right )}\) | \(\frac {\frac {717381}{15625}+\frac {7911 \sqrt {2941}}{15625}}{\left (25+\sqrt {2941}\right ) \left (50+\sqrt {2941}\right )}\) |
| \(a_{3}\) | \(\frac {500 r^{6}+1420 r^{5}+300 r^{4}-2580 r^{3}-2000 r^{2}-40 r +420}{\left (25 r^{2}+29 r -21\right ) \left (25 r^{2}+79 r +33\right ) \left (25 r^{2}+129 r +137\right )}\) | \(\frac {\frac {906742764}{1953125}+\frac {15291084 \sqrt {2941}}{1953125}}{\left (25+\sqrt {2941}\right ) \left (50+\sqrt {2941}\right ) \left (75+\sqrt {2941}\right )}\) |
| \(a_{4}\) | \(\frac {3125 r^{8}+24000 r^{7}+61766 r^{6}+33496 r^{5}-104033 r^{4}-168040 r^{3}-59654 r^{2}+22392 r +16569}{\left (25 r^{2}+29 r -21\right ) \left (25 r^{2}+79 r +33\right ) \left (25 r^{2}+129 r +137\right ) \left (25 r^{2}+179 r +291\right )}\) | \(\frac {\frac {1473770634612}{244140625}+\frac {26407796172 \sqrt {2941}}{244140625}}{\left (25+\sqrt {2941}\right ) \left (50+\sqrt {2941}\right ) \left (75+\sqrt {2941}\right ) \left (100+\sqrt {2941}\right )}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=\frac {18750 r^{10}+271250 r^{9}+1552980 r^{8}+4225260 r^{7}+4278470 r^{6}-4527750 r^{5}-15744100 r^{4}-13518440 r^{3}-1814650 r^{2}+2790990 r +1132740}{\left (25 r^{2}+29 r -21\right ) \left (25 r^{2}+79 r +33\right ) \left (25 r^{2}+129 r +137\right ) \left (25 r^{2}+179 r +291\right ) \left (25 r^{2}+229 r +495\right )} \] Which for the root \(r = \frac {21}{50}+\frac {\sqrt {2941}}{50}\) becomes \[ a_{5}=\frac {\frac {10008934775328384}{152587890625}+\frac {181292058002304 \sqrt {2941}}{152587890625}}{\left (25+\sqrt {2941}\right ) \left (50+\sqrt {2941}\right ) \left (75+\sqrt {2941}\right ) \left (100+\sqrt {2941}\right ) \left (125+\sqrt {2941}\right )} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {10 r \left (-1+r \right )}{25 r^{2}+29 r -21}\) | \(\frac {\left (21+\sqrt {2941}\right ) \left (-29+\sqrt {2941}\right )}{6250+250 \sqrt {2941}}\) |
| \(a_{2}\) | \(\frac {75 r^{4}-4 r^{3}-75 r^{2}-50 r +21}{\left (25 r^{2}+29 r -21\right ) \left (25 r^{2}+79 r +33\right )}\) | \(\frac {\frac {717381}{15625}+\frac {7911 \sqrt {2941}}{15625}}{\left (25+\sqrt {2941}\right ) \left (50+\sqrt {2941}\right )}\) |
| \(a_{3}\) | \(\frac {500 r^{6}+1420 r^{5}+300 r^{4}-2580 r^{3}-2000 r^{2}-40 r +420}{\left (25 r^{2}+29 r -21\right ) \left (25 r^{2}+79 r +33\right ) \left (25 r^{2}+129 r +137\right )}\) | \(\frac {\frac {906742764}{1953125}+\frac {15291084 \sqrt {2941}}{1953125}}{\left (25+\sqrt {2941}\right ) \left (50+\sqrt {2941}\right ) \left (75+\sqrt {2941}\right )}\) |
| \(a_{4}\) | \(\frac {3125 r^{8}+24000 r^{7}+61766 r^{6}+33496 r^{5}-104033 r^{4}-168040 r^{3}-59654 r^{2}+22392 r +16569}{\left (25 r^{2}+29 r -21\right ) \left (25 r^{2}+79 r +33\right ) \left (25 r^{2}+129 r +137\right ) \left (25 r^{2}+179 r +291\right )}\) | \(\frac {\frac {1473770634612}{244140625}+\frac {26407796172 \sqrt {2941}}{244140625}}{\left (25+\sqrt {2941}\right ) \left (50+\sqrt {2941}\right ) \left (75+\sqrt {2941}\right ) \left (100+\sqrt {2941}\right )}\) |
| \(a_{5}\) | \(\frac {18750 r^{10}+271250 r^{9}+1552980 r^{8}+4225260 r^{7}+4278470 r^{6}-4527750 r^{5}-15744100 r^{4}-13518440 r^{3}-1814650 r^{2}+2790990 r +1132740}{\left (25 r^{2}+29 r -21\right ) \left (25 r^{2}+79 r +33\right ) \left (25 r^{2}+129 r +137\right ) \left (25 r^{2}+179 r +291\right ) \left (25 r^{2}+229 r +495\right )}\) | \(\frac {\frac {10008934775328384}{152587890625}+\frac {181292058002304 \sqrt {2941}}{152587890625}}{\left (25+\sqrt {2941}\right ) \left (50+\sqrt {2941}\right ) \left (75+\sqrt {2941}\right ) \left (100+\sqrt {2941}\right ) \left (125+\sqrt {2941}\right )}\) |
Using the above table, then the solution \(y_{1}\left (x \right )\) is \begin {align*} y_{1}\left (x \right )&= x^{\frac {21}{50}+\frac {\sqrt {2941}}{50}} \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}\dots \right ) \\ &= x^{\frac {21}{50}+\frac {\sqrt {2941}}{50}} \left (1+\frac {\left (21+\sqrt {2941}\right ) \left (-29+\sqrt {2941}\right ) x}{6250+250 \sqrt {2941}}+\frac {9 \left (79709+879 \sqrt {2941}\right ) x^{2}}{15625 \left (25+\sqrt {2941}\right ) \left (50+\sqrt {2941}\right )}+\frac {12 \left (75561897+1274257 \sqrt {2941}\right ) x^{3}}{1953125 \left (25+\sqrt {2941}\right ) \left (50+\sqrt {2941}\right ) \left (75+\sqrt {2941}\right )}+\frac {12 \left (122814219551+2200649681 \sqrt {2941}\right ) x^{4}}{244140625 \left (25+\sqrt {2941}\right ) \left (50+\sqrt {2941}\right ) \left (75+\sqrt {2941}\right ) \left (100+\sqrt {2941}\right )}+\frac {1152 \left (8688311436917+157371578127 \sqrt {2941}\right ) x^{5}}{152587890625 \left (25+\sqrt {2941}\right ) \left (50+\sqrt {2941}\right ) \left (75+\sqrt {2941}\right ) \left (100+\sqrt {2941}\right ) \left (125+\sqrt {2941}\right )}+O\left (x^{6}\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 \[ b_{1} = \frac {10 r \left (-1+r \right )}{25 r^{2}+29 r -21} \] 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 )-10 b_{n -1} \left (n +r -1\right ) \left (n +r -2\right )+25 b_{n} \left (n +r \right ) \left (n +r -1\right )+4 b_{n} \left (n +r \right )+b_{n -2}-25 b_{n} = 0 \end{equation} Solving for \(b_{n}\) from recursive equation (4) gives \[ b_{n} = -\frac {n^{2} b_{n -2}-10 n^{2} b_{n -1}+2 n r b_{n -2}-20 n r b_{n -1}+r^{2} b_{n -2}-10 r^{2} b_{n -1}-5 n b_{n -2}+30 n b_{n -1}-5 r b_{n -2}+30 r b_{n -1}+7 b_{n -2}-20 b_{n -1}}{25 n^{2}+50 n r +25 r^{2}-21 n -21 r -25}\tag {4} \] Which for the root \(r = \frac {21}{50}-\frac {\sqrt {2941}}{50}\) becomes \[ b_{n} = \frac {\left (\left (25 b_{n -2}-250 b_{n -1}\right ) n -52 b_{n -2}+270 b_{n -1}\right ) \sqrt {2941}+\left (-625 b_{n -2}+6250 b_{n -1}\right ) n^{2}+\left (2600 b_{n -2}-13500 b_{n -1}\right ) n -3908 b_{n -2}+13080 b_{n -1}}{625 n \left (-\sqrt {2941}+25 n \right )}\tag {5} \] At this point, it is a good idea to keep track of \(b_{n}\) in a table both before substituting \(r = \frac {21}{50}-\frac {\sqrt {2941}}{50}\) and after as more terms are found using the above recursive equation.
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {10 r \left (-1+r \right )}{25 r^{2}+29 r -21}\) | \(-\frac {\left (-21+\sqrt {2941}\right ) \left (29+\sqrt {2941}\right )}{-6250+250 \sqrt {2941}}\) |
For \(n = 2\), using the above recursive equation gives \[ b_{2}=\frac {75 r^{4}-4 r^{3}-75 r^{2}-50 r +21}{\left (25 r^{2}+29 r -21\right ) \left (25 r^{2}+79 r +33\right )} \] Which for the root \(r = \frac {21}{50}-\frac {\sqrt {2941}}{50}\) becomes \[ b_{2}=\frac {\frac {717381}{15625}-\frac {7911 \sqrt {2941}}{15625}}{\left (-25+\sqrt {2941}\right ) \left (-50+\sqrt {2941}\right )} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {10 r \left (-1+r \right )}{25 r^{2}+29 r -21}\) | \(-\frac {\left (-21+\sqrt {2941}\right ) \left (29+\sqrt {2941}\right )}{-6250+250 \sqrt {2941}}\) |
| \(b_{2}\) | \(\frac {75 r^{4}-4 r^{3}-75 r^{2}-50 r +21}{\left (25 r^{2}+29 r -21\right ) \left (25 r^{2}+79 r +33\right )}\) | \(\frac {\frac {717381}{15625}-\frac {7911 \sqrt {2941}}{15625}}{\left (-25+\sqrt {2941}\right ) \left (-50+\sqrt {2941}\right )}\) |
For \(n = 3\), using the above recursive equation gives \[ b_{3}=\frac {500 r^{6}+1420 r^{5}+300 r^{4}-2580 r^{3}-2000 r^{2}-40 r +420}{\left (25 r^{2}+29 r -21\right ) \left (25 r^{2}+79 r +33\right ) \left (25 r^{2}+129 r +137\right )} \] Which for the root \(r = \frac {21}{50}-\frac {\sqrt {2941}}{50}\) becomes \[ b_{3}=\frac {-\frac {906742764}{1953125}+\frac {15291084 \sqrt {2941}}{1953125}}{\left (-25+\sqrt {2941}\right ) \left (-50+\sqrt {2941}\right ) \left (-75+\sqrt {2941}\right )} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {10 r \left (-1+r \right )}{25 r^{2}+29 r -21}\) | \(-\frac {\left (-21+\sqrt {2941}\right ) \left (29+\sqrt {2941}\right )}{-6250+250 \sqrt {2941}}\) |
| \(b_{2}\) | \(\frac {75 r^{4}-4 r^{3}-75 r^{2}-50 r +21}{\left (25 r^{2}+29 r -21\right ) \left (25 r^{2}+79 r +33\right )}\) | \(\frac {\frac {717381}{15625}-\frac {7911 \sqrt {2941}}{15625}}{\left (-25+\sqrt {2941}\right ) \left (-50+\sqrt {2941}\right )}\) |
| \(b_{3}\) | \(\frac {500 r^{6}+1420 r^{5}+300 r^{4}-2580 r^{3}-2000 r^{2}-40 r +420}{\left (25 r^{2}+29 r -21\right ) \left (25 r^{2}+79 r +33\right ) \left (25 r^{2}+129 r +137\right )}\) | \(\frac {-\frac {906742764}{1953125}+\frac {15291084 \sqrt {2941}}{1953125}}{\left (-25+\sqrt {2941}\right ) \left (-50+\sqrt {2941}\right ) \left (-75+\sqrt {2941}\right )}\) |
For \(n = 4\), using the above recursive equation gives \[ b_{4}=\frac {3125 r^{8}+24000 r^{7}+61766 r^{6}+33496 r^{5}-104033 r^{4}-168040 r^{3}-59654 r^{2}+22392 r +16569}{\left (25 r^{2}+29 r -21\right ) \left (25 r^{2}+79 r +33\right ) \left (25 r^{2}+129 r +137\right ) \left (25 r^{2}+179 r +291\right )} \] Which for the root \(r = \frac {21}{50}-\frac {\sqrt {2941}}{50}\) becomes \[ b_{4}=\frac {\frac {1473770634612}{244140625}-\frac {26407796172 \sqrt {2941}}{244140625}}{\left (-25+\sqrt {2941}\right ) \left (-50+\sqrt {2941}\right ) \left (-75+\sqrt {2941}\right ) \left (-100+\sqrt {2941}\right )} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {10 r \left (-1+r \right )}{25 r^{2}+29 r -21}\) | \(-\frac {\left (-21+\sqrt {2941}\right ) \left (29+\sqrt {2941}\right )}{-6250+250 \sqrt {2941}}\) |
| \(b_{2}\) | \(\frac {75 r^{4}-4 r^{3}-75 r^{2}-50 r +21}{\left (25 r^{2}+29 r -21\right ) \left (25 r^{2}+79 r +33\right )}\) | \(\frac {\frac {717381}{15625}-\frac {7911 \sqrt {2941}}{15625}}{\left (-25+\sqrt {2941}\right ) \left (-50+\sqrt {2941}\right )}\) |
| \(b_{3}\) | \(\frac {500 r^{6}+1420 r^{5}+300 r^{4}-2580 r^{3}-2000 r^{2}-40 r +420}{\left (25 r^{2}+29 r -21\right ) \left (25 r^{2}+79 r +33\right ) \left (25 r^{2}+129 r +137\right )}\) | \(\frac {-\frac {906742764}{1953125}+\frac {15291084 \sqrt {2941}}{1953125}}{\left (-25+\sqrt {2941}\right ) \left (-50+\sqrt {2941}\right ) \left (-75+\sqrt {2941}\right )}\) |
| \(b_{4}\) | \(\frac {3125 r^{8}+24000 r^{7}+61766 r^{6}+33496 r^{5}-104033 r^{4}-168040 r^{3}-59654 r^{2}+22392 r +16569}{\left (25 r^{2}+29 r -21\right ) \left (25 r^{2}+79 r +33\right ) \left (25 r^{2}+129 r +137\right ) \left (25 r^{2}+179 r +291\right )}\) | \(\frac {\frac {1473770634612}{244140625}-\frac {26407796172 \sqrt {2941}}{244140625}}{\left (-25+\sqrt {2941}\right ) \left (-50+\sqrt {2941}\right ) \left (-75+\sqrt {2941}\right ) \left (-100+\sqrt {2941}\right )}\) |
For \(n = 5\), using the above recursive equation gives \[ b_{5}=\frac {18750 r^{10}+271250 r^{9}+1552980 r^{8}+4225260 r^{7}+4278470 r^{6}-4527750 r^{5}-15744100 r^{4}-13518440 r^{3}-1814650 r^{2}+2790990 r +1132740}{\left (25 r^{2}+29 r -21\right ) \left (25 r^{2}+79 r +33\right ) \left (25 r^{2}+129 r +137\right ) \left (25 r^{2}+179 r +291\right ) \left (25 r^{2}+229 r +495\right )} \] Which for the root \(r = \frac {21}{50}-\frac {\sqrt {2941}}{50}\) becomes \[ b_{5}=\frac {-\frac {10008934775328384}{152587890625}+\frac {181292058002304 \sqrt {2941}}{152587890625}}{\left (-25+\sqrt {2941}\right ) \left (-50+\sqrt {2941}\right ) \left (-75+\sqrt {2941}\right ) \left (-100+\sqrt {2941}\right ) \left (-125+\sqrt {2941}\right )} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {10 r \left (-1+r \right )}{25 r^{2}+29 r -21}\) | \(-\frac {\left (-21+\sqrt {2941}\right ) \left (29+\sqrt {2941}\right )}{-6250+250 \sqrt {2941}}\) |
| \(b_{2}\) | \(\frac {75 r^{4}-4 r^{3}-75 r^{2}-50 r +21}{\left (25 r^{2}+29 r -21\right ) \left (25 r^{2}+79 r +33\right )}\) | \(\frac {\frac {717381}{15625}-\frac {7911 \sqrt {2941}}{15625}}{\left (-25+\sqrt {2941}\right ) \left (-50+\sqrt {2941}\right )}\) |
| \(b_{3}\) | \(\frac {500 r^{6}+1420 r^{5}+300 r^{4}-2580 r^{3}-2000 r^{2}-40 r +420}{\left (25 r^{2}+29 r -21\right ) \left (25 r^{2}+79 r +33\right ) \left (25 r^{2}+129 r +137\right )}\) | \(\frac {-\frac {906742764}{1953125}+\frac {15291084 \sqrt {2941}}{1953125}}{\left (-25+\sqrt {2941}\right ) \left (-50+\sqrt {2941}\right ) \left (-75+\sqrt {2941}\right )}\) |
| \(b_{4}\) | \(\frac {3125 r^{8}+24000 r^{7}+61766 r^{6}+33496 r^{5}-104033 r^{4}-168040 r^{3}-59654 r^{2}+22392 r +16569}{\left (25 r^{2}+29 r -21\right ) \left (25 r^{2}+79 r +33\right ) \left (25 r^{2}+129 r +137\right ) \left (25 r^{2}+179 r +291\right )}\) | \(\frac {\frac {1473770634612}{244140625}-\frac {26407796172 \sqrt {2941}}{244140625}}{\left (-25+\sqrt {2941}\right ) \left (-50+\sqrt {2941}\right ) \left (-75+\sqrt {2941}\right ) \left (-100+\sqrt {2941}\right )}\) |
| \(b_{5}\) | \(\frac {18750 r^{10}+271250 r^{9}+1552980 r^{8}+4225260 r^{7}+4278470 r^{6}-4527750 r^{5}-15744100 r^{4}-13518440 r^{3}-1814650 r^{2}+2790990 r +1132740}{\left (25 r^{2}+29 r -21\right ) \left (25 r^{2}+79 r +33\right ) \left (25 r^{2}+129 r +137\right ) \left (25 r^{2}+179 r +291\right ) \left (25 r^{2}+229 r +495\right )}\) | \(\frac {-\frac {10008934775328384}{152587890625}+\frac {181292058002304 \sqrt {2941}}{152587890625}}{\left (-25+\sqrt {2941}\right ) \left (-50+\sqrt {2941}\right ) \left (-75+\sqrt {2941}\right ) \left (-100+\sqrt {2941}\right ) \left (-125+\sqrt {2941}\right )}\) |
Using the above table, then the solution \(y_{2}\left (x \right )\) is \begin {align*} y_{2}\left (x \right )&= x^{\frac {21}{50}+\frac {\sqrt {2941}}{50}} \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}\dots \right ) \\ &= x^{\frac {21}{50}-\frac {\sqrt {2941}}{50}} \left (1-\frac {\left (-21+\sqrt {2941}\right ) \left (29+\sqrt {2941}\right ) x}{-6250+250 \sqrt {2941}}+\frac {9 \left (79709-879 \sqrt {2941}\right ) x^{2}}{15625 \left (-25+\sqrt {2941}\right ) \left (-50+\sqrt {2941}\right )}+\frac {12 \left (-75561897+1274257 \sqrt {2941}\right ) x^{3}}{1953125 \left (-25+\sqrt {2941}\right ) \left (-50+\sqrt {2941}\right ) \left (-75+\sqrt {2941}\right )}+\frac {12 \left (122814219551-2200649681 \sqrt {2941}\right ) x^{4}}{244140625 \left (-25+\sqrt {2941}\right ) \left (-50+\sqrt {2941}\right ) \left (-75+\sqrt {2941}\right ) \left (-100+\sqrt {2941}\right )}+\frac {1152 \left (-8688311436917+157371578127 \sqrt {2941}\right ) x^{5}}{152587890625 \left (-25+\sqrt {2941}\right ) \left (-50+\sqrt {2941}\right ) \left (-75+\sqrt {2941}\right ) \left (-100+\sqrt {2941}\right ) \left (-125+\sqrt {2941}\right )}+O\left (x^{6}\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^{\frac {21}{50}+\frac {\sqrt {2941}}{50}} \left (1+\frac {\left (21+\sqrt {2941}\right ) \left (-29+\sqrt {2941}\right ) x}{6250+250 \sqrt {2941}}+\frac {9 \left (79709+879 \sqrt {2941}\right ) x^{2}}{15625 \left (25+\sqrt {2941}\right ) \left (50+\sqrt {2941}\right )}+\frac {12 \left (75561897+1274257 \sqrt {2941}\right ) x^{3}}{1953125 \left (25+\sqrt {2941}\right ) \left (50+\sqrt {2941}\right ) \left (75+\sqrt {2941}\right )}+\frac {12 \left (122814219551+2200649681 \sqrt {2941}\right ) x^{4}}{244140625 \left (25+\sqrt {2941}\right ) \left (50+\sqrt {2941}\right ) \left (75+\sqrt {2941}\right ) \left (100+\sqrt {2941}\right )}+\frac {1152 \left (8688311436917+157371578127 \sqrt {2941}\right ) x^{5}}{152587890625 \left (25+\sqrt {2941}\right ) \left (50+\sqrt {2941}\right ) \left (75+\sqrt {2941}\right ) \left (100+\sqrt {2941}\right ) \left (125+\sqrt {2941}\right )}+O\left (x^{6}\right )\right ) + c_{2} x^{\frac {21}{50}-\frac {\sqrt {2941}}{50}} \left (1-\frac {\left (-21+\sqrt {2941}\right ) \left (29+\sqrt {2941}\right ) x}{-6250+250 \sqrt {2941}}+\frac {9 \left (79709-879 \sqrt {2941}\right ) x^{2}}{15625 \left (-25+\sqrt {2941}\right ) \left (-50+\sqrt {2941}\right )}+\frac {12 \left (-75561897+1274257 \sqrt {2941}\right ) x^{3}}{1953125 \left (-25+\sqrt {2941}\right ) \left (-50+\sqrt {2941}\right ) \left (-75+\sqrt {2941}\right )}+\frac {12 \left (122814219551-2200649681 \sqrt {2941}\right ) x^{4}}{244140625 \left (-25+\sqrt {2941}\right ) \left (-50+\sqrt {2941}\right ) \left (-75+\sqrt {2941}\right ) \left (-100+\sqrt {2941}\right )}+\frac {1152 \left (-8688311436917+157371578127 \sqrt {2941}\right ) x^{5}}{152587890625 \left (-25+\sqrt {2941}\right ) \left (-50+\sqrt {2941}\right ) \left (-75+\sqrt {2941}\right ) \left (-100+\sqrt {2941}\right ) \left (-125+\sqrt {2941}\right )}+O\left (x^{6}\right )\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} x^{\frac {21}{50}+\frac {\sqrt {2941}}{50}} \left (1+\frac {\left (21+\sqrt {2941}\right ) \left (-29+\sqrt {2941}\right ) x}{6250+250 \sqrt {2941}}+\frac {9 \left (79709+879 \sqrt {2941}\right ) x^{2}}{15625 \left (25+\sqrt {2941}\right ) \left (50+\sqrt {2941}\right )}+\frac {12 \left (75561897+1274257 \sqrt {2941}\right ) x^{3}}{1953125 \left (25+\sqrt {2941}\right ) \left (50+\sqrt {2941}\right ) \left (75+\sqrt {2941}\right )}+\frac {12 \left (122814219551+2200649681 \sqrt {2941}\right ) x^{4}}{244140625 \left (25+\sqrt {2941}\right ) \left (50+\sqrt {2941}\right ) \left (75+\sqrt {2941}\right ) \left (100+\sqrt {2941}\right )}+\frac {1152 \left (8688311436917+157371578127 \sqrt {2941}\right ) x^{5}}{152587890625 \left (25+\sqrt {2941}\right ) \left (50+\sqrt {2941}\right ) \left (75+\sqrt {2941}\right ) \left (100+\sqrt {2941}\right ) \left (125+\sqrt {2941}\right )}+O\left (x^{6}\right )\right )+c_{2} x^{\frac {21}{50}-\frac {\sqrt {2941}}{50}} \left (1-\frac {\left (-21+\sqrt {2941}\right ) \left (29+\sqrt {2941}\right ) x}{-6250+250 \sqrt {2941}}+\frac {9 \left (79709-879 \sqrt {2941}\right ) x^{2}}{15625 \left (-25+\sqrt {2941}\right ) \left (-50+\sqrt {2941}\right )}+\frac {12 \left (-75561897+1274257 \sqrt {2941}\right ) x^{3}}{1953125 \left (-25+\sqrt {2941}\right ) \left (-50+\sqrt {2941}\right ) \left (-75+\sqrt {2941}\right )}+\frac {12 \left (122814219551-2200649681 \sqrt {2941}\right ) x^{4}}{244140625 \left (-25+\sqrt {2941}\right ) \left (-50+\sqrt {2941}\right ) \left (-75+\sqrt {2941}\right ) \left (-100+\sqrt {2941}\right )}+\frac {1152 \left (-8688311436917+157371578127 \sqrt {2941}\right ) x^{5}}{152587890625 \left (-25+\sqrt {2941}\right ) \left (-50+\sqrt {2941}\right ) \left (-75+\sqrt {2941}\right ) \left (-100+\sqrt {2941}\right ) \left (-125+\sqrt {2941}\right )}+O\left (x^{6}\right )\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x^{\frac {21}{50}+\frac {\sqrt {2941}}{50}} \left (1+\frac {\left (21+\sqrt {2941}\right ) \left (-29+\sqrt {2941}\right ) x}{6250+250 \sqrt {2941}}+\frac {9 \left (79709+879 \sqrt {2941}\right ) x^{2}}{15625 \left (25+\sqrt {2941}\right ) \left (50+\sqrt {2941}\right )}+\frac {12 \left (75561897+1274257 \sqrt {2941}\right ) x^{3}}{1953125 \left (25+\sqrt {2941}\right ) \left (50+\sqrt {2941}\right ) \left (75+\sqrt {2941}\right )}+\frac {12 \left (122814219551+2200649681 \sqrt {2941}\right ) x^{4}}{244140625 \left (25+\sqrt {2941}\right ) \left (50+\sqrt {2941}\right ) \left (75+\sqrt {2941}\right ) \left (100+\sqrt {2941}\right )}+\frac {1152 \left (8688311436917+157371578127 \sqrt {2941}\right ) x^{5}}{152587890625 \left (25+\sqrt {2941}\right ) \left (50+\sqrt {2941}\right ) \left (75+\sqrt {2941}\right ) \left (100+\sqrt {2941}\right ) \left (125+\sqrt {2941}\right )}+O\left (x^{6}\right )\right )+c_{2} x^{\frac {21}{50}-\frac {\sqrt {2941}}{50}} \left (1-\frac {\left (-21+\sqrt {2941}\right ) \left (29+\sqrt {2941}\right ) x}{-6250+250 \sqrt {2941}}+\frac {9 \left (79709-879 \sqrt {2941}\right ) x^{2}}{15625 \left (-25+\sqrt {2941}\right ) \left (-50+\sqrt {2941}\right )}+\frac {12 \left (-75561897+1274257 \sqrt {2941}\right ) x^{3}}{1953125 \left (-25+\sqrt {2941}\right ) \left (-50+\sqrt {2941}\right ) \left (-75+\sqrt {2941}\right )}+\frac {12 \left (122814219551-2200649681 \sqrt {2941}\right ) x^{4}}{244140625 \left (-25+\sqrt {2941}\right ) \left (-50+\sqrt {2941}\right ) \left (-75+\sqrt {2941}\right ) \left (-100+\sqrt {2941}\right )}+\frac {1152 \left (-8688311436917+157371578127 \sqrt {2941}\right ) x^{5}}{152587890625 \left (-25+\sqrt {2941}\right ) \left (-50+\sqrt {2941}\right ) \left (-75+\sqrt {2941}\right ) \left (-100+\sqrt {2941}\right ) \left (-125+\sqrt {2941}\right )}+O\left (x^{6}\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} x^{\frac {21}{50}+\frac {\sqrt {2941}}{50}} \left (1+\frac {\left (21+\sqrt {2941}\right ) \left (-29+\sqrt {2941}\right ) x}{6250+250 \sqrt {2941}}+\frac {9 \left (79709+879 \sqrt {2941}\right ) x^{2}}{15625 \left (25+\sqrt {2941}\right ) \left (50+\sqrt {2941}\right )}+\frac {12 \left (75561897+1274257 \sqrt {2941}\right ) x^{3}}{1953125 \left (25+\sqrt {2941}\right ) \left (50+\sqrt {2941}\right ) \left (75+\sqrt {2941}\right )}+\frac {12 \left (122814219551+2200649681 \sqrt {2941}\right ) x^{4}}{244140625 \left (25+\sqrt {2941}\right ) \left (50+\sqrt {2941}\right ) \left (75+\sqrt {2941}\right ) \left (100+\sqrt {2941}\right )}+\frac {1152 \left (8688311436917+157371578127 \sqrt {2941}\right ) x^{5}}{152587890625 \left (25+\sqrt {2941}\right ) \left (50+\sqrt {2941}\right ) \left (75+\sqrt {2941}\right ) \left (100+\sqrt {2941}\right ) \left (125+\sqrt {2941}\right )}+O\left (x^{6}\right )\right )+c_{2} x^{\frac {21}{50}-\frac {\sqrt {2941}}{50}} \left (1-\frac {\left (-21+\sqrt {2941}\right ) \left (29+\sqrt {2941}\right ) x}{-6250+250 \sqrt {2941}}+\frac {9 \left (79709-879 \sqrt {2941}\right ) x^{2}}{15625 \left (-25+\sqrt {2941}\right ) \left (-50+\sqrt {2941}\right )}+\frac {12 \left (-75561897+1274257 \sqrt {2941}\right ) x^{3}}{1953125 \left (-25+\sqrt {2941}\right ) \left (-50+\sqrt {2941}\right ) \left (-75+\sqrt {2941}\right )}+\frac {12 \left (122814219551-2200649681 \sqrt {2941}\right ) x^{4}}{244140625 \left (-25+\sqrt {2941}\right ) \left (-50+\sqrt {2941}\right ) \left (-75+\sqrt {2941}\right ) \left (-100+\sqrt {2941}\right )}+\frac {1152 \left (-8688311436917+157371578127 \sqrt {2941}\right ) x^{5}}{152587890625 \left (-25+\sqrt {2941}\right ) \left (-50+\sqrt {2941}\right ) \left (-75+\sqrt {2941}\right ) \left (-100+\sqrt {2941}\right ) \left (-125+\sqrt {2941}\right )}+O\left (x^{6}\right )\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{2} \left (x^{2}-10 x +25\right ) y^{\prime \prime }+4 y^{\prime } x +\left (x^{2}-25\right ) y=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & y^{\prime \prime } \\ \bullet & {} & \textrm {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & y^{\prime \prime }=-\frac {\left (x +5\right ) y}{\left (x -5\right ) x^{2}}-\frac {4 y^{\prime }}{x \left (x^{2}-10 x +25\right )} \\ \bullet & {} & \textrm {Group terms with}\hspace {3pt} y\hspace {3pt}\textrm {on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear}\hspace {3pt} \\ {} & {} & y^{\prime \prime }+\frac {4 y^{\prime }}{x \left (x^{2}-10 x +25\right )}+\frac {\left (x +5\right ) y}{\left (x -5\right ) x^{2}}=0 \\ \square & {} & \textrm {Check to see if}\hspace {3pt} x_{0}\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & \circ & \textrm {Define functions}\hspace {3pt} \\ {} & {} & \left [P_{2}\left (x \right )=\frac {4}{x \left (x^{2}-10 x +25\right )}, P_{3}\left (x \right )=\frac {x +5}{\left (x -5\right ) x^{2}}\right ] \\ {} & \circ & x \cdot P_{2}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =0 \\ {} & {} & \left (x \cdot P_{2}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}0}}}=\frac {4}{25} \\ {} & \circ & x^{2}\cdot P_{3}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =0 \\ {} & {} & \left (x^{2}\cdot P_{3}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}0}}}=-1 \\ {} & \circ & x =0\textrm {is a regular singular point}\hspace {3pt} \\ & {} & \textrm {Check to see if}\hspace {3pt} x_{0}\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & {} & x_{0}=0 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & y^{\prime \prime } x^{2} \left (x^{2}-10 x +25\right ) \left (x -5\right )+4 y^{\prime } \left (x -5\right ) x +\left (x +5\right ) \left (x^{2}-10 x +25\right ) y=0 \\ \bullet & {} & \textrm {Assume series solution for}\hspace {3pt} y \\ {} & {} & y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +r} \\ \square & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{m}\cdot y\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..3 \\ {} & {} & x^{m}\cdot y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +r +m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k -m \\ {} & {} & x^{m}\cdot y=\moverset {\infty }{\munderset {k =m}{\sum }}a_{k -m} x^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{m}\cdot y^{\prime }\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =1..2 \\ {} & {} & x^{m}\cdot y^{\prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) x^{k +r -1+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1-m \\ {} & {} & x^{m}\cdot y^{\prime }=\moverset {\infty }{\munderset {k =-1+m}{\sum }}a_{k +1-m} \left (k +1-m +r \right ) x^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{m}\cdot y^{\prime \prime }\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =2..5 \\ {} & {} & x^{m}\cdot y^{\prime \prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k +r -1\right ) x^{k +r -2+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2-m \\ {} & {} & x^{m}\cdot y^{\prime \prime }=\moverset {\infty }{\munderset {k =-2+m}{\sum }}a_{k +2-m} \left (k +2-m +r \right ) \left (k +1-m +r \right ) x^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & -5 a_{0} \left (25 r^{2}-21 r -25\right ) x^{r}+\left (-5 a_{1} \left (25 r^{2}+29 r -21\right )+a_{0} \left (75 r^{2}-71 r -25\right )\right ) x^{1+r}+\left (-5 a_{2} \left (25 r^{2}+79 r +33\right )+a_{1} \left (75 r^{2}+79 r -21\right )-5 a_{0} \left (3 r^{2}-3 r +1\right )\right ) x^{2+r}+\left (\moverset {\infty }{\munderset {k =3}{\sum }}\left (-5 a_{k} \left (25 k^{2}+50 k r +25 r^{2}-21 k -21 r -25\right )+a_{k -1} \left (75 \left (k -1\right )^{2}+150 \left (k -1\right ) r +75 r^{2}-71 k +46-71 r \right )-5 a_{k -2} \left (3 \left (k -2\right )^{2}+6 \left (k -2\right ) r +3 r^{2}-3 k +7-3 r \right )+a_{k -3} \left (\left (k -3\right )^{2}+2 \left (k -3\right ) r +r^{2}-k +4-r \right )\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & -125 r^{2}+105 r +125=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{\frac {21}{50}-\frac {\sqrt {2941}}{50}, \frac {21}{50}+\frac {\sqrt {2941}}{50}\right \} \\ \bullet & {} & \textrm {The coefficients of each power of}\hspace {3pt} x \hspace {3pt}\textrm {must be 0}\hspace {3pt} \\ {} & {} & \left [-5 a_{1} \left (25 r^{2}+29 r -21\right )+a_{0} \left (75 r^{2}-71 r -25\right )=0, -5 a_{2} \left (25 r^{2}+79 r +33\right )+a_{1} \left (75 r^{2}+79 r -21\right )-5 a_{0} \left (3 r^{2}-3 r +1\right )=0\right ] \\ \bullet & {} & \textrm {Solve for the dependent coefficient(s)}\hspace {3pt} \\ {} & {} & \left \{a_{1}=\frac {a_{0} \left (75 r^{2}-71 r -25\right )}{5 \left (25 r^{2}+29 r -21\right )}, a_{2}=\frac {6 a_{0} \left (625 r^{4}+50 r^{3}-989 r^{2}-464 r +175\right )}{25 \left (625 r^{4}+2700 r^{3}+2591 r^{2}-702 r -693\right )}\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & \left (-125 a_{k}+a_{k -3}-15 a_{k -2}+75 a_{k -1}\right ) k^{2}+\left (2 \left (-125 a_{k}+a_{k -3}-15 a_{k -2}+75 a_{k -1}\right ) r +105 a_{k}-7 a_{k -3}+75 a_{k -2}-221 a_{k -1}\right ) k +\left (-125 a_{k}+a_{k -3}-15 a_{k -2}+75 a_{k -1}\right ) r^{2}+\left (105 a_{k}-7 a_{k -3}+75 a_{k -2}-221 a_{k -1}\right ) r +125 a_{k}+13 a_{k -3}-95 a_{k -2}+121 a_{k -1}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +3 \\ {} & {} & \left (-125 a_{k +3}+a_{k}-15 a_{k +1}+75 a_{k +2}\right ) \left (k +3\right )^{2}+\left (2 \left (-125 a_{k +3}+a_{k}-15 a_{k +1}+75 a_{k +2}\right ) r +105 a_{k +3}-7 a_{k}+75 a_{k +1}-221 a_{k +2}\right ) \left (k +3\right )+\left (-125 a_{k +3}+a_{k}-15 a_{k +1}+75 a_{k +2}\right ) r^{2}+\left (105 a_{k +3}-7 a_{k}+75 a_{k +1}-221 a_{k +2}\right ) r +125 a_{k +3}+13 a_{k}-95 a_{k +1}+121 a_{k +2}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +3}=\frac {k^{2} a_{k}-15 k^{2} a_{k +1}+75 k^{2} a_{k +2}+2 k r a_{k}-30 k r a_{k +1}+150 k r a_{k +2}+r^{2} a_{k}-15 r^{2} a_{k +1}+75 r^{2} a_{k +2}-k a_{k}-15 k a_{k +1}+229 k a_{k +2}-r a_{k}-15 r a_{k +1}+229 r a_{k +2}+a_{k}-5 a_{k +1}+133 a_{k +2}}{5 \left (25 k^{2}+50 k r +25 r^{2}+129 k +129 r +137\right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {21}{50}-\frac {\sqrt {2941}}{50} \\ {} & {} & a_{k +3}=\frac {k^{2} a_{k}-15 k^{2} a_{k +1}+75 k^{2} a_{k +2}+2 k \left (\frac {21}{50}-\frac {\sqrt {2941}}{50}\right ) a_{k}-30 k \left (\frac {21}{50}-\frac {\sqrt {2941}}{50}\right ) a_{k +1}+150 k \left (\frac {21}{50}-\frac {\sqrt {2941}}{50}\right ) a_{k +2}+\left (\frac {21}{50}-\frac {\sqrt {2941}}{50}\right )^{2} a_{k}-15 \left (\frac {21}{50}-\frac {\sqrt {2941}}{50}\right )^{2} a_{k +1}+75 \left (\frac {21}{50}-\frac {\sqrt {2941}}{50}\right )^{2} a_{k +2}-k a_{k}-15 k a_{k +1}+229 k a_{k +2}-\left (\frac {21}{50}-\frac {\sqrt {2941}}{50}\right ) a_{k}-15 \left (\frac {21}{50}-\frac {\sqrt {2941}}{50}\right ) a_{k +1}+229 \left (\frac {21}{50}-\frac {\sqrt {2941}}{50}\right ) a_{k +2}+a_{k}-5 a_{k +1}+133 a_{k +2}}{5 \left (25 k^{2}+50 k \left (\frac {21}{50}-\frac {\sqrt {2941}}{50}\right )+25 \left (\frac {21}{50}-\frac {\sqrt {2941}}{50}\right )^{2}+129 k +\frac {9559}{50}-\frac {129 \sqrt {2941}}{50}\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\frac {21}{50}-\frac {\sqrt {2941}}{50} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +\frac {21}{50}-\frac {\sqrt {2941}}{50}}, a_{k +3}=\frac {k^{2} a_{k}-15 k^{2} a_{k +1}+75 k^{2} a_{k +2}+2 k \left (\frac {21}{50}-\frac {\sqrt {2941}}{50}\right ) a_{k}-30 k \left (\frac {21}{50}-\frac {\sqrt {2941}}{50}\right ) a_{k +1}+150 k \left (\frac {21}{50}-\frac {\sqrt {2941}}{50}\right ) a_{k +2}+\left (\frac {21}{50}-\frac {\sqrt {2941}}{50}\right )^{2} a_{k}-15 \left (\frac {21}{50}-\frac {\sqrt {2941}}{50}\right )^{2} a_{k +1}+75 \left (\frac {21}{50}-\frac {\sqrt {2941}}{50}\right )^{2} a_{k +2}-k a_{k}-15 k a_{k +1}+229 k a_{k +2}-\left (\frac {21}{50}-\frac {\sqrt {2941}}{50}\right ) a_{k}-15 \left (\frac {21}{50}-\frac {\sqrt {2941}}{50}\right ) a_{k +1}+229 \left (\frac {21}{50}-\frac {\sqrt {2941}}{50}\right ) a_{k +2}+a_{k}-5 a_{k +1}+133 a_{k +2}}{5 \left (25 k^{2}+50 k \left (\frac {21}{50}-\frac {\sqrt {2941}}{50}\right )+25 \left (\frac {21}{50}-\frac {\sqrt {2941}}{50}\right )^{2}+129 k +\frac {9559}{50}-\frac {129 \sqrt {2941}}{50}\right )}, a_{1}=\frac {a_{0} \left (75 \left (\frac {21}{50}-\frac {\sqrt {2941}}{50}\right )^{2}-\frac {2741}{50}+\frac {71 \sqrt {2941}}{50}\right )}{5 \left (25 \left (\frac {21}{50}-\frac {\sqrt {2941}}{50}\right )^{2}-\frac {441}{50}-\frac {29 \sqrt {2941}}{50}\right )}, a_{2}=\frac {6 a_{0} \left (625 \left (\frac {21}{50}-\frac {\sqrt {2941}}{50}\right )^{4}+50 \left (\frac {21}{50}-\frac {\sqrt {2941}}{50}\right )^{3}-989 \left (\frac {21}{50}-\frac {\sqrt {2941}}{50}\right )^{2}-\frac {497}{25}+\frac {232 \sqrt {2941}}{25}\right )}{25 \left (625 \left (\frac {21}{50}-\frac {\sqrt {2941}}{50}\right )^{4}+2700 \left (\frac {21}{50}-\frac {\sqrt {2941}}{50}\right )^{3}+2591 \left (\frac {21}{50}-\frac {\sqrt {2941}}{50}\right )^{2}-\frac {24696}{25}+\frac {351 \sqrt {2941}}{25}\right )}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {21}{50}+\frac {\sqrt {2941}}{50} \\ {} & {} & a_{k +3}=\frac {k^{2} a_{k}-15 k^{2} a_{k +1}+75 k^{2} a_{k +2}+2 k \left (\frac {21}{50}+\frac {\sqrt {2941}}{50}\right ) a_{k}-30 k \left (\frac {21}{50}+\frac {\sqrt {2941}}{50}\right ) a_{k +1}+150 k \left (\frac {21}{50}+\frac {\sqrt {2941}}{50}\right ) a_{k +2}+\left (\frac {21}{50}+\frac {\sqrt {2941}}{50}\right )^{2} a_{k}-15 \left (\frac {21}{50}+\frac {\sqrt {2941}}{50}\right )^{2} a_{k +1}+75 \left (\frac {21}{50}+\frac {\sqrt {2941}}{50}\right )^{2} a_{k +2}-k a_{k}-15 k a_{k +1}+229 k a_{k +2}-\left (\frac {21}{50}+\frac {\sqrt {2941}}{50}\right ) a_{k}-15 \left (\frac {21}{50}+\frac {\sqrt {2941}}{50}\right ) a_{k +1}+229 \left (\frac {21}{50}+\frac {\sqrt {2941}}{50}\right ) a_{k +2}+a_{k}-5 a_{k +1}+133 a_{k +2}}{5 \left (25 k^{2}+50 k \left (\frac {21}{50}+\frac {\sqrt {2941}}{50}\right )+25 \left (\frac {21}{50}+\frac {\sqrt {2941}}{50}\right )^{2}+129 k +\frac {9559}{50}+\frac {129 \sqrt {2941}}{50}\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\frac {21}{50}+\frac {\sqrt {2941}}{50} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +\frac {21}{50}+\frac {\sqrt {2941}}{50}}, a_{k +3}=\frac {k^{2} a_{k}-15 k^{2} a_{k +1}+75 k^{2} a_{k +2}+2 k \left (\frac {21}{50}+\frac {\sqrt {2941}}{50}\right ) a_{k}-30 k \left (\frac {21}{50}+\frac {\sqrt {2941}}{50}\right ) a_{k +1}+150 k \left (\frac {21}{50}+\frac {\sqrt {2941}}{50}\right ) a_{k +2}+\left (\frac {21}{50}+\frac {\sqrt {2941}}{50}\right )^{2} a_{k}-15 \left (\frac {21}{50}+\frac {\sqrt {2941}}{50}\right )^{2} a_{k +1}+75 \left (\frac {21}{50}+\frac {\sqrt {2941}}{50}\right )^{2} a_{k +2}-k a_{k}-15 k a_{k +1}+229 k a_{k +2}-\left (\frac {21}{50}+\frac {\sqrt {2941}}{50}\right ) a_{k}-15 \left (\frac {21}{50}+\frac {\sqrt {2941}}{50}\right ) a_{k +1}+229 \left (\frac {21}{50}+\frac {\sqrt {2941}}{50}\right ) a_{k +2}+a_{k}-5 a_{k +1}+133 a_{k +2}}{5 \left (25 k^{2}+50 k \left (\frac {21}{50}+\frac {\sqrt {2941}}{50}\right )+25 \left (\frac {21}{50}+\frac {\sqrt {2941}}{50}\right )^{2}+129 k +\frac {9559}{50}+\frac {129 \sqrt {2941}}{50}\right )}, a_{1}=\frac {a_{0} \left (75 \left (\frac {21}{50}+\frac {\sqrt {2941}}{50}\right )^{2}-\frac {2741}{50}-\frac {71 \sqrt {2941}}{50}\right )}{5 \left (25 \left (\frac {21}{50}+\frac {\sqrt {2941}}{50}\right )^{2}-\frac {441}{50}+\frac {29 \sqrt {2941}}{50}\right )}, a_{2}=\frac {6 a_{0} \left (625 \left (\frac {21}{50}+\frac {\sqrt {2941}}{50}\right )^{4}+50 \left (\frac {21}{50}+\frac {\sqrt {2941}}{50}\right )^{3}-989 \left (\frac {21}{50}+\frac {\sqrt {2941}}{50}\right )^{2}-\frac {497}{25}-\frac {232 \sqrt {2941}}{25}\right )}{25 \left (625 \left (\frac {21}{50}+\frac {\sqrt {2941}}{50}\right )^{4}+2700 \left (\frac {21}{50}+\frac {\sqrt {2941}}{50}\right )^{3}+2591 \left (\frac {21}{50}+\frac {\sqrt {2941}}{50}\right )^{2}-\frac {24696}{25}-\frac {351 \sqrt {2941}}{25}\right )}\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +\frac {21}{50}-\frac {\sqrt {2941}}{50}}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} x^{k +\frac {21}{50}+\frac {\sqrt {2941}}{50}}\right ), a_{k +3}=\frac {k^{2} a_{k}-15 k^{2} a_{k +1}+75 k^{2} a_{k +2}+2 k \left (\frac {21}{50}-\frac {\sqrt {2941}}{50}\right ) a_{k}-30 k \left (\frac {21}{50}-\frac {\sqrt {2941}}{50}\right ) a_{k +1}+150 k \left (\frac {21}{50}-\frac {\sqrt {2941}}{50}\right ) a_{k +2}+\left (\frac {21}{50}-\frac {\sqrt {2941}}{50}\right )^{2} a_{k}-15 \left (\frac {21}{50}-\frac {\sqrt {2941}}{50}\right )^{2} a_{k +1}+75 \left (\frac {21}{50}-\frac {\sqrt {2941}}{50}\right )^{2} a_{k +2}-k a_{k}-15 k a_{k +1}+229 k a_{k +2}-\left (\frac {21}{50}-\frac {\sqrt {2941}}{50}\right ) a_{k}-15 \left (\frac {21}{50}-\frac {\sqrt {2941}}{50}\right ) a_{k +1}+229 \left (\frac {21}{50}-\frac {\sqrt {2941}}{50}\right ) a_{k +2}+a_{k}-5 a_{k +1}+133 a_{k +2}}{5 \left (25 k^{2}+50 k \left (\frac {21}{50}-\frac {\sqrt {2941}}{50}\right )+25 \left (\frac {21}{50}-\frac {\sqrt {2941}}{50}\right )^{2}+129 k +\frac {9559}{50}-\frac {129 \sqrt {2941}}{50}\right )}, a_{1}=\frac {a_{0} \left (75 \left (\frac {21}{50}-\frac {\sqrt {2941}}{50}\right )^{2}-\frac {2741}{50}+\frac {71 \sqrt {2941}}{50}\right )}{5 \left (25 \left (\frac {21}{50}-\frac {\sqrt {2941}}{50}\right )^{2}-\frac {441}{50}-\frac {29 \sqrt {2941}}{50}\right )}, a_{2}=\frac {6 a_{0} \left (625 \left (\frac {21}{50}-\frac {\sqrt {2941}}{50}\right )^{4}+50 \left (\frac {21}{50}-\frac {\sqrt {2941}}{50}\right )^{3}-989 \left (\frac {21}{50}-\frac {\sqrt {2941}}{50}\right )^{2}-\frac {497}{25}+\frac {232 \sqrt {2941}}{25}\right )}{25 \left (625 \left (\frac {21}{50}-\frac {\sqrt {2941}}{50}\right )^{4}+2700 \left (\frac {21}{50}-\frac {\sqrt {2941}}{50}\right )^{3}+2591 \left (\frac {21}{50}-\frac {\sqrt {2941}}{50}\right )^{2}-\frac {24696}{25}+\frac {351 \sqrt {2941}}{25}\right )}, b_{k +3}=\frac {k^{2} b_{k}-15 k^{2} b_{k +1}+75 k^{2} b_{k +2}+2 k \left (\frac {21}{50}+\frac {\sqrt {2941}}{50}\right ) b_{k}-30 k \left (\frac {21}{50}+\frac {\sqrt {2941}}{50}\right ) b_{k +1}+150 k \left (\frac {21}{50}+\frac {\sqrt {2941}}{50}\right ) b_{k +2}+\left (\frac {21}{50}+\frac {\sqrt {2941}}{50}\right )^{2} b_{k}-15 \left (\frac {21}{50}+\frac {\sqrt {2941}}{50}\right )^{2} b_{k +1}+75 \left (\frac {21}{50}+\frac {\sqrt {2941}}{50}\right )^{2} b_{k +2}-k b_{k}-15 k b_{k +1}+229 k b_{k +2}-\left (\frac {21}{50}+\frac {\sqrt {2941}}{50}\right ) b_{k}-15 \left (\frac {21}{50}+\frac {\sqrt {2941}}{50}\right ) b_{k +1}+229 \left (\frac {21}{50}+\frac {\sqrt {2941}}{50}\right ) b_{k +2}+b_{k}-5 b_{k +1}+133 b_{k +2}}{5 \left (25 k^{2}+50 k \left (\frac {21}{50}+\frac {\sqrt {2941}}{50}\right )+25 \left (\frac {21}{50}+\frac {\sqrt {2941}}{50}\right )^{2}+129 k +\frac {9559}{50}+\frac {129 \sqrt {2941}}{50}\right )}, b_{1}=\frac {b_{0} \left (75 \left (\frac {21}{50}+\frac {\sqrt {2941}}{50}\right )^{2}-\frac {2741}{50}-\frac {71 \sqrt {2941}}{50}\right )}{5 \left (25 \left (\frac {21}{50}+\frac {\sqrt {2941}}{50}\right )^{2}-\frac {441}{50}+\frac {29 \sqrt {2941}}{50}\right )}, b_{2}=\frac {6 b_{0} \left (625 \left (\frac {21}{50}+\frac {\sqrt {2941}}{50}\right )^{4}+50 \left (\frac {21}{50}+\frac {\sqrt {2941}}{50}\right )^{3}-989 \left (\frac {21}{50}+\frac {\sqrt {2941}}{50}\right )^{2}-\frac {497}{25}-\frac {232 \sqrt {2941}}{25}\right )}{25 \left (625 \left (\frac {21}{50}+\frac {\sqrt {2941}}{50}\right )^{4}+2700 \left (\frac {21}{50}+\frac {\sqrt {2941}}{50}\right )^{3}+2591 \left (\frac {21}{50}+\frac {\sqrt {2941}}{50}\right )^{2}-\frac {24696}{25}-\frac {351 \sqrt {2941}}{25}\right )}\right ] \end {array} \]
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 -> Kummer -> 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 HeunC ODE, case a = 0, e <> 0, c <> 0 `
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 1179
Order:=6; dsolve(x^2*(x-5)^2*diff(y(x),x$2)+4*x*diff(y(x),x)+(x^2-25)*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = x^{\frac {21}{50}} \left (c_{1} x^{-\frac {\sqrt {2941}}{50}} \left (1+\frac {-1166-4 \sqrt {2941}}{-3125+125 \sqrt {2941}} x -\frac {9}{15625} \frac {879 \sqrt {2941}-79709}{\left (-25+\sqrt {2941}\right ) \left (-50+\sqrt {2941}\right )} x^{2}+\frac {\frac {15291084 \sqrt {2941}}{1953125}-\frac {906742764}{1953125}}{\left (-25+\sqrt {2941}\right ) \left (-50+\sqrt {2941}\right ) \left (-75+\sqrt {2941}\right )} x^{3}-\frac {12}{244140625} \frac {2200649681 \sqrt {2941}-122814219551}{\left (-25+\sqrt {2941}\right ) \left (-50+\sqrt {2941}\right ) \left (-75+\sqrt {2941}\right ) \left (-100+\sqrt {2941}\right )} x^{4}+\frac {\frac {181292058002304 \sqrt {2941}}{152587890625}-\frac {10008934775328384}{152587890625}}{\left (-25+\sqrt {2941}\right ) \left (-50+\sqrt {2941}\right ) \left (-75+\sqrt {2941}\right ) \left (-100+\sqrt {2941}\right ) \left (-125+\sqrt {2941}\right )} x^{5}+\operatorname {O}\left (x^{6}\right )\right )+c_{2} x^{\frac {\sqrt {2941}}{50}} \left (1+\frac {1166-4 \sqrt {2941}}{125 \sqrt {2941}+3125} x +\frac {\frac {7911 \sqrt {2941}}{15625}+\frac {717381}{15625}}{\left (\sqrt {2941}+25\right ) \left (50+\sqrt {2941}\right )} x^{2}+\frac {\frac {15291084 \sqrt {2941}}{1953125}+\frac {906742764}{1953125}}{\left (\sqrt {2941}+25\right ) \left (50+\sqrt {2941}\right ) \left (\sqrt {2941}+75\right )} x^{3}+\frac {\frac {26407796172 \sqrt {2941}}{244140625}+\frac {1473770634612}{244140625}}{\left (\sqrt {2941}+25\right ) \left (50+\sqrt {2941}\right ) \left (\sqrt {2941}+75\right ) \left (100+\sqrt {2941}\right )} x^{4}+\frac {\frac {181292058002304 \sqrt {2941}}{152587890625}+\frac {10008934775328384}{152587890625}}{\left (\sqrt {2941}+25\right ) \left (50+\sqrt {2941}\right ) \left (\sqrt {2941}+75\right ) \left (100+\sqrt {2941}\right ) \left (125+\sqrt {2941}\right )} x^{5}+\operatorname {O}\left (x^{6}\right )\right )\right ) \]
✓ Solution by Mathematica
Time used: 0.008 (sec). Leaf size: 5384
AsymptoticDSolveValue[x^2*(x-5)^2*y''[x]+4*x*y'[x]+(x^2-25)*y[x]==0,y[x],{x,0,5}]
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