Internal problem ID [14804]
Internal file name [OUTPUT/14484_Monday_April_08_2024_06_25_49_AM_84008714/index.tex]
Book: INTRODUCTORY DIFFERENTIAL EQUATIONS. Martha L. Abell, James P. Braselton.
Fourth edition 2014. ElScAe. 2014
Section: Chapter 4. Higher Order Equations. Exercises 4.9, page 215
Problem number: 4.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second order series method. Ordinary point", "second order series method. Taylor series method"
Maple gives the following as the ode type
[[_2nd_order, _with_linear_symmetries]]
\[ \boxed {\left (x^{2}-25\right )^{2} y^{\prime \prime }-\left (x +5\right ) y^{\prime }+10 y=0} \] With the expansion point for the power series method at \(x = 0\).
Solving ode using Taylor series method. This gives review on how the Taylor series method works for solving second order ode.
Let \[ y^{\prime \prime }=f\left ( x,y,y^{\prime }\right ) \] Assuming expansion is at \(x_{0}=0\) (we can always shift the actual expansion point to \(0\) by change of variables) and assuming \(f\left ( x,y,y^{\prime }\right ) \) is analytic at \(x_{0}\) which must be the case for an ordinary point. Let initial conditions be \(y\left ( x_{0}\right ) =y_{0}\) and \(y^{\prime }\left ( x_{0}\right ) =y_{0}^{\prime }\). Using Taylor series gives\begin {align*} y\left ( x\right ) & =y\left ( x_{0}\right ) +\left ( x-x_{0}\right ) y^{\prime }\left ( x_{0}\right ) +\frac {\left ( x-x_{0}\right ) ^{2}}{2}y^{\prime \prime }\left ( x_{0}\right ) +\frac {\left ( x-x_{0}\right ) ^{3}}{3!}y^{\prime \prime \prime }\left ( x_{0}\right ) +\cdots \\ & =y_{0}+xy_{0}^{\prime }+\frac {x^{2}}{2}\left . f\right \vert _{x_{0},y_{0},y_{0}^{\prime }}+\frac {x^{3}}{3!}\left . f^{\prime }\right \vert _{x_{0},y_{0},y_{0}^{\prime }}+\cdots \\ & =y_{0}+xy_{0}^{\prime }+\sum _{n=0}^{\infty }\frac {x^{n+2}}{\left ( n+2\right ) !}\left . \frac {d^{n}f}{dx^{n}}\right \vert _{x_{0},y_{0},y_{0}^{\prime }} \end {align*}
But \begin {align} \frac {df}{dx} & =\frac {\partial f}{\partial x}\frac {dx}{dx}+\frac {\partial f}{\partial y}\frac {dy}{dx}+\frac {\partial f}{\partial y^{\prime }}\frac {dy^{\prime }}{dx}\tag {1}\\ & =\frac {\partial f}{\partial x}+\frac {\partial f}{\partial y}y^{\prime }+\frac {\partial f}{\partial y^{\prime }}y^{\prime \prime }\\ & =\frac {\partial f}{\partial x}+\frac {\partial f}{\partial y}y^{\prime }+\frac {\partial f}{\partial y^{\prime }}f\\ \frac {d^{2}f}{dx^{2}} & =\frac {d}{dx}\left ( \frac {df}{dx}\right ) \nonumber \\ & =\frac {\partial }{\partial x}\left ( \frac {df}{dx}\right ) +\frac {\partial }{\partial y}\left ( \frac {df}{dx}\right ) y^{\prime }+\frac {\partial }{\partial y^{\prime }}\left ( \frac {df}{dx}\right ) f\tag {2}\\ \frac {d^{3}f}{dx^{3}} & =\frac {d}{dx}\left ( \frac {d^{2}f}{dx^{2}}\right ) \nonumber \\ & =\frac {\partial }{\partial x}\left ( \frac {d^{2}f}{dx^{2}}\right ) +\left ( \frac {\partial }{\partial y}\frac {d^{2}f}{dx^{2}}\right ) y^{\prime }+\frac {\partial }{\partial y^{\prime }}\left ( \frac {d^{2}f}{dx^{2}}\right ) f\tag {3}\\ & \vdots \nonumber \end {align}
And so on. Hence if we name \(F_{0}=f\left ( x,y,y^{\prime }\right ) \) then the above can be written as \begin {align} F_{0} & =f\left ( x,y,y^{\prime }\right ) \tag {4}\\ F_{1} & =\frac {df}{dx}\nonumber \\ & =\frac {dF_{0}}{dx}\nonumber \\ & =\frac {\partial f}{\partial x}+\frac {\partial f}{\partial y}y^{\prime }+\frac {\partial f}{\partial y^{\prime }}y^{\prime \prime }\nonumber \\ & =\frac {\partial f}{\partial x}+\frac {\partial f}{\partial y}y^{\prime }+\frac {\partial f}{\partial y^{\prime }}f\tag {5}\\ & =\frac {\partial F_{0}}{\partial x}+\frac {\partial F_{0}}{\partial y}y^{\prime }+\frac {\partial F_{0}}{\partial y^{\prime }}F_{0}\nonumber \\ F_{2} & =\frac {d}{dx}\left ( \frac {d}{dx}f\right ) \nonumber \\ & =\frac {d}{dx}\left ( F_{1}\right ) \nonumber \\ & =\frac {\partial }{\partial x}F_{1}+\left ( \frac {\partial F_{1}}{\partial y}\right ) y^{\prime }+\left ( \frac {\partial F_{1}}{\partial y^{\prime }}\right ) y^{\prime \prime }\nonumber \\ & =\frac {\partial }{\partial x}F_{1}+\left ( \frac {\partial F_{1}}{\partial y}\right ) y^{\prime }+\left ( \frac {\partial F_{1}}{\partial y^{\prime }}\right ) F_{0}\nonumber \\ & \vdots \nonumber \\ F_{n} & =\frac {d}{dx}\left ( F_{n-1}\right ) \nonumber \\ & =\frac {\partial }{\partial x}F_{n-1}+\left ( \frac {\partial F_{n-1}}{\partial y}\right ) y^{\prime }+\left ( \frac {\partial F_{n-1}}{\partial y^{\prime }}\right ) y^{\prime \prime }\nonumber \\ & =\frac {\partial }{\partial x}F_{n-1}+\left ( \frac {\partial F_{n-1}}{\partial y}\right ) y^{\prime }+\left ( \frac {\partial F_{n-1}}{\partial y^{\prime }}\right ) F_{0} \tag {6} \end {align}
Therefore (6) can be used from now on along with \begin {equation} y\left ( x\right ) =y_{0}+xy_{0}^{\prime }+\sum _{n=0}^{\infty }\frac {x^{n+2}}{\left ( n+2\right ) !}\left . F_{n}\right \vert _{x_{0},y_{0},y_{0}^{\prime }} \tag {7} \end {equation} To find \(y\left ( x\right ) \) series solution around \(x=0\). Hence \begin {align*} F_0 &= \frac {y^{\prime } x +5 y^{\prime }-10 y}{x^{4}-50 x^{2}+625}\\ F_1 &= \frac {d F_0}{dx} \\ &= \frac {\partial F_{0}}{\partial x}+ \frac {\partial F_{0}}{\partial y} y^{\prime }+ \frac {\partial F_{0}}{\partial y^{\prime }} F_0 \\ &= \frac {\left (-13 x^{3}+45 x^{2}+326 x -1120\right ) y^{\prime }+40 y \left (x^{2}-5 x -\frac {1}{4}\right )}{\left (x +5\right )^{3} \left (x -5\right )^{4}}\\ F_2 &= \frac {d F_1}{dx} \\ &= \frac {\partial F_{1}}{\partial x}+ \frac {\partial F_{1}}{\partial y} y^{\prime }+ \frac {\partial F_{1}}{\partial y^{\prime }} F_1 \\ &= \frac {\left (92 x^{5}-820 x^{4}-429 x^{3}+20085 x^{2}-46774 x +10380\right ) y^{\prime }-200 y \left (x^{4}-10 x^{3}+29 x^{2}-43 x +\frac {2301}{20}\right )}{\left (x -5\right )^{6} \left (x +5\right )^{4}}\\ F_3 &= \frac {d F_2}{dx} \\ &= \frac {\partial F_{2}}{\partial x}+ \frac {\partial F_{2}}{\partial y} y^{\prime }+ \frac {\partial F_{2}}{\partial y^{\prime }} F_2 \\ &= \frac {\left (-660 x^{7}+9300 x^{6}-31005 x^{5}-111125 x^{4}+952802 x^{3}-2706630 x^{2}+5870576 x -8193620\right ) y^{\prime }+1200 \left (x^{6}-15 x^{5}+\frac {879}{10} x^{4}-\frac {1943}{6} x^{3}+\frac {40713}{40} x^{2}-\frac {6859}{4} x -\frac {4813}{60}\right ) y}{\left (x +5\right )^{5} \left (x -5\right )^{8}}\\ F_4 &= \frac {d F_3}{dx} \\ &= \frac {\partial F_{3}}{\partial x}+ \frac {\partial F_{3}}{\partial y} y^{\prime }+ \frac {\partial F_{3}}{\partial y^{\prime }} F_3 \\ &= \frac {\left (5160 x^{9}-99000 x^{8}+664860 x^{7}-1356700 x^{6}-6090135 x^{5}+61145225 x^{4}-302239320 x^{3}+862218900 x^{2}-1041745124 x +99074380\right ) y^{\prime }-8400 \left (x^{8}-20 x^{7}+\frac {6178}{35} x^{6}-\frac {21811}{21} x^{5}+\frac {999581}{210} x^{4}-\frac {613495}{42} x^{3}+\frac {19811839}{840} x^{2}-\frac {1826309}{84} x +\frac {6201119}{210}\right ) y}{\left (x -5\right )^{10} \left (x +5\right )^{6}} \end {align*}
And so on. Evaluating all the above at initial conditions \(x = 0\) and \(y \left (0\right ) = y \left (0\right )\) and \(y^{\prime }\left (0\right ) = y^{\prime }\left (0\right )\) gives \begin {align*} F_0 &= -\frac {2 y \left (0\right )}{125}+\frac {y^{\prime }\left (0\right )}{125}\\ F_1 &= -\frac {2 y \left (0\right )}{15625}-\frac {224 y^{\prime }\left (0\right )}{15625}\\ F_2 &= -\frac {4602 y \left (0\right )}{1953125}+\frac {2076 y^{\prime }\left (0\right )}{1953125}\\ F_3 &= -\frac {19252 y \left (0\right )}{244140625}-\frac {1638724 y^{\prime }\left (0\right )}{244140625}\\ F_4 &= -\frac {49608952 y \left (0\right )}{30517578125}+\frac {19814876 y^{\prime }\left (0\right )}{30517578125} \end {align*}
Substituting all the above in (7) and simplifying gives the solution as \[ y = \left (1-\frac {1}{125} x^{2}-\frac {1}{46875} x^{3}-\frac {767}{7812500} x^{4}-\frac {4813}{7324218750} x^{5}-\frac {6201119}{2746582031250} x^{6}\right ) y \left (0\right )+\left (x +\frac {1}{250} x^{2}-\frac {112}{46875} x^{3}+\frac {173}{3906250} x^{4}-\frac {409681}{7324218750} x^{5}+\frac {4953719}{5493164062500} x^{6}\right ) y^{\prime }\left (0\right )+O\left (x^{6}\right ) \] Since the expansion point \(x = 0\) is an ordinary, we can also solve this using standard power series The ode is normalized to be \[ y^{\prime \prime } \left (x^{4}-50 x^{2}+625\right )+\left (-x -5\right ) y^{\prime }+10 y = 0 \] Let the solution be represented as power series of the form \[ y = \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n} \] Then \begin {align*} y^{\prime } &= \moverset {\infty }{\munderset {n =1}{\sum }}n a_{n} x^{n -1}\\ y^{\prime \prime } &= \moverset {\infty }{\munderset {n =2}{\sum }}n \left (n -1\right ) a_{n} x^{n -2} \end {align*}
Substituting the above back into the ode gives \begin {align*} \left (\moverset {\infty }{\munderset {n =2}{\sum }}n \left (n -1\right ) a_{n} x^{n -2}\right ) \left (x^{4}-50 x^{2}+625\right )+\left (-x -5\right ) \left (\moverset {\infty }{\munderset {n =1}{\sum }}n a_{n} x^{n -1}\right )+10 \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n}\right ) = 0\tag {1} \end {align*}
Which simplifies to \begin{equation} \tag{2} \left (\moverset {\infty }{\munderset {n =2}{\sum }}n \,x^{n +2} a_{n} \left (n -1\right )\right )+\moverset {\infty }{\munderset {n =2}{\sum }}\left (-50 x^{n} a_{n} n \left (n -1\right )\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}625 n \left (n -1\right ) a_{n} x^{n -2}\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-n a_{n} x^{n}\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-5 n a_{n} x^{n -1}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}10 a_{n} x^{n}\right ) = 0 \end{equation} The next step is to make all powers of \(x\) be \(n\) in each summation term. Going over each summation term above with power of \(x\) in it which is not already \(x^{n}\) and adjusting the power and the corresponding index gives \begin{align*} \moverset {\infty }{\munderset {n =2}{\sum }}n \,x^{n +2} a_{n} \left (n -1\right ) &= \moverset {\infty }{\munderset {n =4}{\sum }}\left (n -2\right ) a_{n -2} \left (n -3\right ) x^{n} \\ \moverset {\infty }{\munderset {n =2}{\sum }}625 n \left (n -1\right ) a_{n} x^{n -2} &= \moverset {\infty }{\munderset {n =0}{\sum }}625 \left (n +2\right ) a_{n +2} \left (n +1\right ) x^{n} \\ \moverset {\infty }{\munderset {n =1}{\sum }}\left (-5 n a_{n} x^{n -1}\right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}\left (-5 \left (n +1\right ) a_{n +1} x^{n}\right ) \\ \end{align*} Substituting all the above in Eq (2) gives the following equation where now all powers of \(x\) are the same and equal to \(n\). \begin{equation} \tag{3} \left (\moverset {\infty }{\munderset {n =4}{\sum }}\left (n -2\right ) a_{n -2} \left (n -3\right ) x^{n}\right )+\moverset {\infty }{\munderset {n =2}{\sum }}\left (-50 x^{n} a_{n} n \left (n -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}625 \left (n +2\right ) a_{n +2} \left (n +1\right ) x^{n}\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-n a_{n} x^{n}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-5 \left (n +1\right ) a_{n +1} x^{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}10 a_{n} x^{n}\right ) = 0 \end{equation} \(n=0\) gives \[ 1250 a_{2}-5 a_{1}+10 a_{0}=0 \] \[ a_{2} = -\frac {a_{0}}{125}+\frac {a_{1}}{250} \] \(n=1\) gives \[ 3750 a_{3}+9 a_{1}-10 a_{2}=0 \] Which after substituting earlier equations, simplifies to \[ a_{3} = -\frac {a_{0}}{46875}-\frac {112 a_{1}}{46875} \] \(n=2\) gives \[ -92 a_{2}+7500 a_{4}-15 a_{3} = 0 \] Which after substituting earlier equations, simplifies to \[ \frac {2301 a_{0}}{3125}-\frac {1038 a_{1}}{3125}+7500 a_{4} = 0 \] Or \[ a_{4} = -\frac {767 a_{0}}{7812500}+\frac {173 a_{1}}{3906250} \] \(n=3\) gives \[ -293 a_{3}+12500 a_{5}-20 a_{4} = 0 \] Which after substituting earlier equations, simplifies to \[ \frac {9626 a_{0}}{1171875}+\frac {819362 a_{1}}{1171875}+12500 a_{5} = 0 \] Or \[ a_{5} = -\frac {4813 a_{0}}{7324218750}-\frac {409681 a_{1}}{7324218750} \] For \(4\le n\), the recurrence equation is \begin{equation} \tag{4} \left (n -2\right ) a_{n -2} \left (n -3\right )-50 n a_{n} \left (n -1\right )+625 \left (n +2\right ) a_{n +2} \left (n +1\right )-n a_{n}-5 \left (n +1\right ) a_{n +1}+10 a_{n} = 0 \end{equation} Solving for \(a_{n +2}\), gives \begin{align*} \tag{5} a_{n +2}&= \frac {50 n^{2} a_{n}-n^{2} a_{n -2}-49 n a_{n}+5 n a_{n -2}+5 n a_{n +1}-10 a_{n}-6 a_{n -2}+5 a_{n +1}}{625 \left (n +2\right ) \left (n +1\right )} \\ &= \frac {\left (50 n^{2}-49 n -10\right ) a_{n}}{625 \left (n +2\right ) \left (n +1\right )}+\frac {\left (-n^{2}+5 n -6\right ) a_{n -2}}{625 \left (n +2\right ) \left (n +1\right )}+\frac {\left (5 n +5\right ) a_{n +1}}{625 \left (n +2\right ) \left (n +1\right )} \\ \end{align*} For \(n = 4\) the recurrence equation gives \[ 2 a_{2}-594 a_{4}+18750 a_{6}-25 a_{5} = 0 \] Which after substituting the earlier terms found becomes \[ a_{6} = -\frac {6201119 a_{0}}{2746582031250}+\frac {4953719 a_{1}}{5493164062500} \] For \(n = 5\) the recurrence equation gives \[ 6 a_{3}-995 a_{5}+26250 a_{7}-30 a_{6} = 0 \] Which after substituting the earlier terms found becomes \[ a_{7} = -\frac {12076457 a_{0}}{534057617187500}-\frac {210023921 a_{1}}{133514404296875} \] And so on. Therefore the solution is \begin {align*} y &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n}\\ &= a_{3} x^{3}+a_{2} x^{2}+a_{1} x +a_{0} + \dots \end {align*}
Substituting the values for \(a_{n}\) found above, the solution becomes \[ y = a_{0}+a_{1} x +\left (-\frac {a_{0}}{125}+\frac {a_{1}}{250}\right ) x^{2}+\left (-\frac {a_{0}}{46875}-\frac {112 a_{1}}{46875}\right ) x^{3}+\left (-\frac {767 a_{0}}{7812500}+\frac {173 a_{1}}{3906250}\right ) x^{4}+\left (-\frac {4813 a_{0}}{7324218750}-\frac {409681 a_{1}}{7324218750}\right ) x^{5}+\dots \] Collecting terms, the solution becomes \begin{equation} \tag{3} y = \left (1-\frac {1}{125} x^{2}-\frac {1}{46875} x^{3}-\frac {767}{7812500} x^{4}-\frac {4813}{7324218750} x^{5}\right ) a_{0}+\left (x +\frac {1}{250} x^{2}-\frac {112}{46875} x^{3}+\frac {173}{3906250} x^{4}-\frac {409681}{7324218750} x^{5}\right ) a_{1}+O\left (x^{6}\right ) \end{equation} At \(x = 0\) the solution above becomes \[ y = \left (1-\frac {1}{125} x^{2}-\frac {1}{46875} x^{3}-\frac {767}{7812500} x^{4}-\frac {4813}{7324218750} x^{5}\right ) c_{1} +\left (x +\frac {1}{250} x^{2}-\frac {112}{46875} x^{3}+\frac {173}{3906250} x^{4}-\frac {409681}{7324218750} x^{5}\right ) c_{2} +O\left (x^{6}\right ) \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \left (1-\frac {1}{125} x^{2}-\frac {1}{46875} x^{3}-\frac {767}{7812500} x^{4}-\frac {4813}{7324218750} x^{5}-\frac {6201119}{2746582031250} x^{6}\right ) y \left (0\right )+\left (x +\frac {1}{250} x^{2}-\frac {112}{46875} x^{3}+\frac {173}{3906250} x^{4}-\frac {409681}{7324218750} x^{5}+\frac {4953719}{5493164062500} x^{6}\right ) y^{\prime }\left (0\right )+O\left (x^{6}\right ) \\ \tag{2} y &= \left (1-\frac {1}{125} x^{2}-\frac {1}{46875} x^{3}-\frac {767}{7812500} x^{4}-\frac {4813}{7324218750} x^{5}\right ) c_{1} +\left (x +\frac {1}{250} x^{2}-\frac {112}{46875} x^{3}+\frac {173}{3906250} x^{4}-\frac {409681}{7324218750} x^{5}\right ) c_{2} +O\left (x^{6}\right ) \\ \end{align*}
Verification of solutions
\[ y = \left (1-\frac {1}{125} x^{2}-\frac {1}{46875} x^{3}-\frac {767}{7812500} x^{4}-\frac {4813}{7324218750} x^{5}-\frac {6201119}{2746582031250} x^{6}\right ) y \left (0\right )+\left (x +\frac {1}{250} x^{2}-\frac {112}{46875} x^{3}+\frac {173}{3906250} x^{4}-\frac {409681}{7324218750} x^{5}+\frac {4953719}{5493164062500} x^{6}\right ) y^{\prime }\left (0\right )+O\left (x^{6}\right ) \] Verified OK.
\[ y = \left (1-\frac {1}{125} x^{2}-\frac {1}{46875} x^{3}-\frac {767}{7812500} x^{4}-\frac {4813}{7324218750} x^{5}\right ) c_{1} +\left (x +\frac {1}{250} x^{2}-\frac {112}{46875} x^{3}+\frac {173}{3906250} x^{4}-\frac {409681}{7324218750} x^{5}\right ) c_{2} +O\left (x^{6}\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime \prime } \left (x^{4}-50 x^{2}+625\right )+\left (-x -5\right ) y^{\prime }+10 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 {10 y}{x^{4}-50 x^{2}+625}+\frac {y^{\prime }}{x^{3}-5 x^{2}-25 x +125} \\ \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 {y^{\prime }}{x^{3}-5 x^{2}-25 x +125}+\frac {10 y}{x^{4}-50 x^{2}+625}=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 {1}{x^{3}-5 x^{2}-25 x +125}, P_{3}\left (x \right )=\frac {10}{x^{4}-50 x^{2}+625}\right ] \\ {} & \circ & \left (x +5\right )\cdot P_{2}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =-5 \\ {} & {} & \left (\left (x +5\right )\cdot P_{2}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}-5}}}=-\frac {1}{100} \\ {} & \circ & \left (x +5\right )^{2}\cdot P_{3}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =-5 \\ {} & {} & \left (\left (x +5\right )^{2}\cdot P_{3}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}-5}}}=\frac {1}{10} \\ {} & \circ & x =-5\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}=-5 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & y^{\prime \prime } \left (x^{3}-5 x^{2}-25 x +125\right ) \left (x^{4}-50 x^{2}+625\right )+\left (-x^{4}+50 x^{2}-625\right ) y^{\prime }+\left (10 x^{3}-50 x^{2}-250 x +1250\right ) y=0 \\ \bullet & {} & \textrm {Change variables using}\hspace {3pt} x =u -5\hspace {3pt}\textrm {so that the regular singular point is at}\hspace {3pt} u =0 \\ {} & {} & \left (u^{7}-40 u^{6}+600 u^{5}-4000 u^{4}+10000 u^{3}\right ) \left (\frac {d^{2}}{d u^{2}}y \left (u \right )\right )+\left (-u^{4}+20 u^{3}-100 u^{2}\right ) \left (\frac {d}{d u}y \left (u \right )\right )+\left (10 u^{3}-200 u^{2}+1000 u \right ) y \left (u \right )=0 \\ \bullet & {} & \textrm {Assume series solution for}\hspace {3pt} y \left (u \right ) \\ {} & {} & y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k +r} \\ \square & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & \circ & \textrm {Convert}\hspace {3pt} u^{m}\cdot y \left (u \right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =1..3 \\ {} & {} & u^{m}\cdot y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k +r +m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k -m \\ {} & {} & u^{m}\cdot y \left (u \right )=\moverset {\infty }{\munderset {k =m}{\sum }}a_{k -m} u^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} u^{m}\cdot \left (\frac {d}{d u}y \left (u \right )\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =2..4 \\ {} & {} & u^{m}\cdot \left (\frac {d}{d u}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) u^{k +r -1+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1-m \\ {} & {} & u^{m}\cdot \left (\frac {d}{d u}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =-1+m}{\sum }}a_{k +1-m} \left (k +1-m +r \right ) u^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} u^{m}\cdot \left (\frac {d^{2}}{d u^{2}}y \left (u \right )\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =3..7 \\ {} & {} & u^{m}\cdot \left (\frac {d^{2}}{d u^{2}}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k +r -1\right ) u^{k +r -2+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2-m \\ {} & {} & u^{m}\cdot \left (\frac {d^{2}}{d u^{2}}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =-2+m}{\sum }}a_{k +2-m} \left (k +2-m +r \right ) \left (k +1-m +r \right ) u^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & 100 a_{0} \left (100 r^{2}-101 r +10\right ) u^{1+r}+\left (100 a_{1} \left (100 r^{2}+99 r +9\right )-20 a_{0} \left (200 r^{2}-201 r +10\right )\right ) u^{2+r}+\left (100 a_{2} \left (100 r^{2}+299 r +208\right )-20 a_{1} \left (200 r^{2}+199 r +9\right )+a_{0} \left (600 r^{2}-601 r +10\right )\right ) u^{3+r}+\left (100 a_{3} \left (100 r^{2}+499 r +607\right )-20 a_{2} \left (200 r^{2}+599 r +408\right )+a_{1} \left (600 r^{2}+599 r +9\right )-40 a_{0} r \left (-1+r \right )\right ) u^{4+r}+\left (\moverset {\infty }{\munderset {k =5}{\sum }}\left (100 a_{k -1} \left (100 \left (k -1\right )^{2}+200 \left (k -1\right ) r +100 r^{2}-101 k +111-101 r \right )-20 a_{k -2} \left (200 \left (k -2\right )^{2}+400 \left (k -2\right ) r +200 r^{2}-201 k +412-201 r \right )+a_{k -3} \left (600 \left (k -3\right )^{2}+1200 \left (k -3\right ) r +600 r^{2}-601 k +1813-601 r \right )-40 a_{k -4} \left (k -4+r \right ) \left (k -5+r \right )+a_{k -5} \left (k -5+r \right ) \left (k -6+r \right )\right ) u^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & 10000 r^{2}-10100 r +1000=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{\frac {101}{200}-\frac {3 \sqrt {689}}{200}, \frac {101}{200}+\frac {3 \sqrt {689}}{200}\right \} \\ \bullet & {} & \textrm {The coefficients of each power of}\hspace {3pt} u \hspace {3pt}\textrm {must be 0}\hspace {3pt} \\ {} & {} & \left [100 a_{1} \left (100 r^{2}+99 r +9\right )-20 a_{0} \left (200 r^{2}-201 r +10\right )=0, 100 a_{2} \left (100 r^{2}+299 r +208\right )-20 a_{1} \left (200 r^{2}+199 r +9\right )+a_{0} \left (600 r^{2}-601 r +10\right )=0, 100 a_{3} \left (100 r^{2}+499 r +607\right )-20 a_{2} \left (200 r^{2}+599 r +408\right )+a_{1} \left (600 r^{2}+599 r +9\right )-40 a_{0} r \left (-1+r \right )=0\right ] \\ \bullet & {} & \textrm {Solve for the dependent coefficient(s)}\hspace {3pt} \\ {} & {} & \left \{a_{1}=\frac {a_{0} \left (200 r^{2}-201 r +10\right )}{5 \left (100 r^{2}+99 r +9\right )}, a_{2}=\frac {a_{0} \left (100000 r^{4}-900 r^{3}-91697 r^{2}+5143 r +270\right )}{100 \left (10000 r^{4}+39800 r^{3}+51301 r^{2}+23283 r +1872\right )}, a_{3}=\frac {a_{0} \left (5000000 r^{6}+14940000 r^{5}+5380400 r^{4}-13226651 r^{3}-8199223 r^{2}+494558 r +45720\right )}{250 \left (1000000 r^{6}+8970000 r^{5}+31060300 r^{4}+52086099 r^{3}+42945124 r^{2}+15066909 r +1136304\right )}\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & \left (a_{k -5}-40 a_{k -4}+600 a_{k -3}-4000 a_{k -2}+10000 a_{k -1}\right ) k^{2}+\left (2 \left (a_{k -5}-40 a_{k -4}+600 a_{k -3}-4000 a_{k -2}+10000 a_{k -1}\right ) r -11 a_{k -5}+360 a_{k -4}-4201 a_{k -3}+20020 a_{k -2}-30100 a_{k -1}\right ) k +\left (a_{k -5}-40 a_{k -4}+600 a_{k -3}-4000 a_{k -2}+10000 a_{k -1}\right ) r^{2}+\left (-11 a_{k -5}+360 a_{k -4}-4201 a_{k -3}+20020 a_{k -2}-30100 a_{k -1}\right ) r +30 a_{k -5}-800 a_{k -4}+7213 a_{k -3}-24240 a_{k -2}+21100 a_{k -1}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +5 \\ {} & {} & \left (a_{k}-40 a_{k +1}+600 a_{k +2}-4000 a_{k +3}+10000 a_{k +4}\right ) \left (k +5\right )^{2}+\left (2 \left (a_{k}-40 a_{k +1}+600 a_{k +2}-4000 a_{k +3}+10000 a_{k +4}\right ) r -11 a_{k}+360 a_{k +1}-4201 a_{k +2}+20020 a_{k +3}-30100 a_{k +4}\right ) \left (k +5\right )+\left (a_{k}-40 a_{k +1}+600 a_{k +2}-4000 a_{k +3}+10000 a_{k +4}\right ) r^{2}+\left (-11 a_{k}+360 a_{k +1}-4201 a_{k +2}+20020 a_{k +3}-30100 a_{k +4}\right ) r +30 a_{k}-800 a_{k +1}+7213 a_{k +2}-24240 a_{k +3}+21100 a_{k +4}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +4}=-\frac {k^{2} a_{k}-40 k^{2} a_{k +1}+600 k^{2} a_{k +2}-4000 k^{2} a_{k +3}+2 k r a_{k}-80 k r a_{k +1}+1200 k r a_{k +2}-8000 k r a_{k +3}+r^{2} a_{k}-40 r^{2} a_{k +1}+600 r^{2} a_{k +2}-4000 r^{2} a_{k +3}-k a_{k}-40 k a_{k +1}+1799 k a_{k +2}-19980 k a_{k +3}-r a_{k}-40 r a_{k +1}+1799 r a_{k +2}-19980 r a_{k +3}+1208 a_{k +2}-24140 a_{k +3}}{100 \left (100 k^{2}+200 k r +100 r^{2}+699 k +699 r +1206\right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {101}{200}-\frac {3 \sqrt {689}}{200} \\ {} & {} & a_{k +4}=-\frac {k^{2} a_{k}-40 k^{2} a_{k +1}+600 k^{2} a_{k +2}-4000 k^{2} a_{k +3}+2 k \left (\frac {101}{200}-\frac {3 \sqrt {689}}{200}\right ) a_{k}-80 k \left (\frac {101}{200}-\frac {3 \sqrt {689}}{200}\right ) a_{k +1}+1200 k \left (\frac {101}{200}-\frac {3 \sqrt {689}}{200}\right ) a_{k +2}-8000 k \left (\frac {101}{200}-\frac {3 \sqrt {689}}{200}\right ) a_{k +3}+\left (\frac {101}{200}-\frac {3 \sqrt {689}}{200}\right )^{2} a_{k}-40 \left (\frac {101}{200}-\frac {3 \sqrt {689}}{200}\right )^{2} a_{k +1}+600 \left (\frac {101}{200}-\frac {3 \sqrt {689}}{200}\right )^{2} a_{k +2}-4000 \left (\frac {101}{200}-\frac {3 \sqrt {689}}{200}\right )^{2} a_{k +3}-k a_{k}-40 k a_{k +1}+1799 k a_{k +2}-19980 k a_{k +3}-\left (\frac {101}{200}-\frac {3 \sqrt {689}}{200}\right ) a_{k}-40 \left (\frac {101}{200}-\frac {3 \sqrt {689}}{200}\right ) a_{k +1}+1799 \left (\frac {101}{200}-\frac {3 \sqrt {689}}{200}\right ) a_{k +2}-19980 \left (\frac {101}{200}-\frac {3 \sqrt {689}}{200}\right ) a_{k +3}+1208 a_{k +2}-24140 a_{k +3}}{100 \left (100 k^{2}+200 k \left (\frac {101}{200}-\frac {3 \sqrt {689}}{200}\right )+100 \left (\frac {101}{200}-\frac {3 \sqrt {689}}{200}\right )^{2}+699 k +\frac {311799}{200}-\frac {2097 \sqrt {689}}{200}\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\frac {101}{200}-\frac {3 \sqrt {689}}{200} \\ {} & {} & \left [y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k +\frac {101}{200}-\frac {3 \sqrt {689}}{200}}, a_{k +4}=-\frac {k^{2} a_{k}-40 k^{2} a_{k +1}+600 k^{2} a_{k +2}-4000 k^{2} a_{k +3}+2 k \left (\frac {101}{200}-\frac {3 \sqrt {689}}{200}\right ) a_{k}-80 k \left (\frac {101}{200}-\frac {3 \sqrt {689}}{200}\right ) a_{k +1}+1200 k \left (\frac {101}{200}-\frac {3 \sqrt {689}}{200}\right ) a_{k +2}-8000 k \left (\frac {101}{200}-\frac {3 \sqrt {689}}{200}\right ) a_{k +3}+\left (\frac {101}{200}-\frac {3 \sqrt {689}}{200}\right )^{2} a_{k}-40 \left (\frac {101}{200}-\frac {3 \sqrt {689}}{200}\right )^{2} a_{k +1}+600 \left (\frac {101}{200}-\frac {3 \sqrt {689}}{200}\right )^{2} a_{k +2}-4000 \left (\frac {101}{200}-\frac {3 \sqrt {689}}{200}\right )^{2} a_{k +3}-k a_{k}-40 k a_{k +1}+1799 k a_{k +2}-19980 k a_{k +3}-\left (\frac {101}{200}-\frac {3 \sqrt {689}}{200}\right ) a_{k}-40 \left (\frac {101}{200}-\frac {3 \sqrt {689}}{200}\right ) a_{k +1}+1799 \left (\frac {101}{200}-\frac {3 \sqrt {689}}{200}\right ) a_{k +2}-19980 \left (\frac {101}{200}-\frac {3 \sqrt {689}}{200}\right ) a_{k +3}+1208 a_{k +2}-24140 a_{k +3}}{100 \left (100 k^{2}+200 k \left (\frac {101}{200}-\frac {3 \sqrt {689}}{200}\right )+100 \left (\frac {101}{200}-\frac {3 \sqrt {689}}{200}\right )^{2}+699 k +\frac {311799}{200}-\frac {2097 \sqrt {689}}{200}\right )}, a_{1}=\frac {a_{0} \left (200 \left (\frac {101}{200}-\frac {3 \sqrt {689}}{200}\right )^{2}-\frac {18301}{200}+\frac {603 \sqrt {689}}{200}\right )}{5 \left (100 \left (\frac {101}{200}-\frac {3 \sqrt {689}}{200}\right )^{2}+\frac {11799}{200}-\frac {297 \sqrt {689}}{200}\right )}, a_{2}=\frac {a_{0} \left (100000 \left (\frac {101}{200}-\frac {3 \sqrt {689}}{200}\right )^{4}-900 \left (\frac {101}{200}-\frac {3 \sqrt {689}}{200}\right )^{3}-91697 \left (\frac {101}{200}-\frac {3 \sqrt {689}}{200}\right )^{2}+\frac {573443}{200}-\frac {15429 \sqrt {689}}{200}\right )}{100 \left (10000 \left (\frac {101}{200}-\frac {3 \sqrt {689}}{200}\right )^{4}+39800 \left (\frac {101}{200}-\frac {3 \sqrt {689}}{200}\right )^{3}+51301 \left (\frac {101}{200}-\frac {3 \sqrt {689}}{200}\right )^{2}+\frac {2725983}{200}-\frac {69849 \sqrt {689}}{200}\right )}, a_{3}=\frac {a_{0} \left (5000000 \left (\frac {101}{200}-\frac {3 \sqrt {689}}{200}\right )^{6}+14940000 \left (\frac {101}{200}-\frac {3 \sqrt {689}}{200}\right )^{5}+5380400 \left (\frac {101}{200}-\frac {3 \sqrt {689}}{200}\right )^{4}-13226651 \left (\frac {101}{200}-\frac {3 \sqrt {689}}{200}\right )^{3}-8199223 \left (\frac {101}{200}-\frac {3 \sqrt {689}}{200}\right )^{2}+\frac {29547179}{100}-\frac {741837 \sqrt {689}}{100}\right )}{250 \left (1000000 \left (\frac {101}{200}-\frac {3 \sqrt {689}}{200}\right )^{6}+8970000 \left (\frac {101}{200}-\frac {3 \sqrt {689}}{200}\right )^{5}+31060300 \left (\frac {101}{200}-\frac {3 \sqrt {689}}{200}\right )^{4}+52086099 \left (\frac {101}{200}-\frac {3 \sqrt {689}}{200}\right )^{3}+42945124 \left (\frac {101}{200}-\frac {3 \sqrt {689}}{200}\right )^{2}+\frac {1749018609}{200}-\frac {45200727 \sqrt {689}}{200}\right )}\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =x +5 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x +5\right )^{k +\frac {101}{200}-\frac {3 \sqrt {689}}{200}}, a_{k +4}=-\frac {k^{2} a_{k}-40 k^{2} a_{k +1}+600 k^{2} a_{k +2}-4000 k^{2} a_{k +3}+2 k \left (\frac {101}{200}-\frac {3 \sqrt {689}}{200}\right ) a_{k}-80 k \left (\frac {101}{200}-\frac {3 \sqrt {689}}{200}\right ) a_{k +1}+1200 k \left (\frac {101}{200}-\frac {3 \sqrt {689}}{200}\right ) a_{k +2}-8000 k \left (\frac {101}{200}-\frac {3 \sqrt {689}}{200}\right ) a_{k +3}+\left (\frac {101}{200}-\frac {3 \sqrt {689}}{200}\right )^{2} a_{k}-40 \left (\frac {101}{200}-\frac {3 \sqrt {689}}{200}\right )^{2} a_{k +1}+600 \left (\frac {101}{200}-\frac {3 \sqrt {689}}{200}\right )^{2} a_{k +2}-4000 \left (\frac {101}{200}-\frac {3 \sqrt {689}}{200}\right )^{2} a_{k +3}-k a_{k}-40 k a_{k +1}+1799 k a_{k +2}-19980 k a_{k +3}-\left (\frac {101}{200}-\frac {3 \sqrt {689}}{200}\right ) a_{k}-40 \left (\frac {101}{200}-\frac {3 \sqrt {689}}{200}\right ) a_{k +1}+1799 \left (\frac {101}{200}-\frac {3 \sqrt {689}}{200}\right ) a_{k +2}-19980 \left (\frac {101}{200}-\frac {3 \sqrt {689}}{200}\right ) a_{k +3}+1208 a_{k +2}-24140 a_{k +3}}{100 \left (100 k^{2}+200 k \left (\frac {101}{200}-\frac {3 \sqrt {689}}{200}\right )+100 \left (\frac {101}{200}-\frac {3 \sqrt {689}}{200}\right )^{2}+699 k +\frac {311799}{200}-\frac {2097 \sqrt {689}}{200}\right )}, a_{1}=\frac {a_{0} \left (200 \left (\frac {101}{200}-\frac {3 \sqrt {689}}{200}\right )^{2}-\frac {18301}{200}+\frac {603 \sqrt {689}}{200}\right )}{5 \left (100 \left (\frac {101}{200}-\frac {3 \sqrt {689}}{200}\right )^{2}+\frac {11799}{200}-\frac {297 \sqrt {689}}{200}\right )}, a_{2}=\frac {a_{0} \left (100000 \left (\frac {101}{200}-\frac {3 \sqrt {689}}{200}\right )^{4}-900 \left (\frac {101}{200}-\frac {3 \sqrt {689}}{200}\right )^{3}-91697 \left (\frac {101}{200}-\frac {3 \sqrt {689}}{200}\right )^{2}+\frac {573443}{200}-\frac {15429 \sqrt {689}}{200}\right )}{100 \left (10000 \left (\frac {101}{200}-\frac {3 \sqrt {689}}{200}\right )^{4}+39800 \left (\frac {101}{200}-\frac {3 \sqrt {689}}{200}\right )^{3}+51301 \left (\frac {101}{200}-\frac {3 \sqrt {689}}{200}\right )^{2}+\frac {2725983}{200}-\frac {69849 \sqrt {689}}{200}\right )}, a_{3}=\frac {a_{0} \left (5000000 \left (\frac {101}{200}-\frac {3 \sqrt {689}}{200}\right )^{6}+14940000 \left (\frac {101}{200}-\frac {3 \sqrt {689}}{200}\right )^{5}+5380400 \left (\frac {101}{200}-\frac {3 \sqrt {689}}{200}\right )^{4}-13226651 \left (\frac {101}{200}-\frac {3 \sqrt {689}}{200}\right )^{3}-8199223 \left (\frac {101}{200}-\frac {3 \sqrt {689}}{200}\right )^{2}+\frac {29547179}{100}-\frac {741837 \sqrt {689}}{100}\right )}{250 \left (1000000 \left (\frac {101}{200}-\frac {3 \sqrt {689}}{200}\right )^{6}+8970000 \left (\frac {101}{200}-\frac {3 \sqrt {689}}{200}\right )^{5}+31060300 \left (\frac {101}{200}-\frac {3 \sqrt {689}}{200}\right )^{4}+52086099 \left (\frac {101}{200}-\frac {3 \sqrt {689}}{200}\right )^{3}+42945124 \left (\frac {101}{200}-\frac {3 \sqrt {689}}{200}\right )^{2}+\frac {1749018609}{200}-\frac {45200727 \sqrt {689}}{200}\right )}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {101}{200}+\frac {3 \sqrt {689}}{200} \\ {} & {} & a_{k +4}=-\frac {k^{2} a_{k}-40 k^{2} a_{k +1}+600 k^{2} a_{k +2}-4000 k^{2} a_{k +3}+2 k \left (\frac {101}{200}+\frac {3 \sqrt {689}}{200}\right ) a_{k}-80 k \left (\frac {101}{200}+\frac {3 \sqrt {689}}{200}\right ) a_{k +1}+1200 k \left (\frac {101}{200}+\frac {3 \sqrt {689}}{200}\right ) a_{k +2}-8000 k \left (\frac {101}{200}+\frac {3 \sqrt {689}}{200}\right ) a_{k +3}+\left (\frac {101}{200}+\frac {3 \sqrt {689}}{200}\right )^{2} a_{k}-40 \left (\frac {101}{200}+\frac {3 \sqrt {689}}{200}\right )^{2} a_{k +1}+600 \left (\frac {101}{200}+\frac {3 \sqrt {689}}{200}\right )^{2} a_{k +2}-4000 \left (\frac {101}{200}+\frac {3 \sqrt {689}}{200}\right )^{2} a_{k +3}-k a_{k}-40 k a_{k +1}+1799 k a_{k +2}-19980 k a_{k +3}-\left (\frac {101}{200}+\frac {3 \sqrt {689}}{200}\right ) a_{k}-40 \left (\frac {101}{200}+\frac {3 \sqrt {689}}{200}\right ) a_{k +1}+1799 \left (\frac {101}{200}+\frac {3 \sqrt {689}}{200}\right ) a_{k +2}-19980 \left (\frac {101}{200}+\frac {3 \sqrt {689}}{200}\right ) a_{k +3}+1208 a_{k +2}-24140 a_{k +3}}{100 \left (100 k^{2}+200 k \left (\frac {101}{200}+\frac {3 \sqrt {689}}{200}\right )+100 \left (\frac {101}{200}+\frac {3 \sqrt {689}}{200}\right )^{2}+699 k +\frac {311799}{200}+\frac {2097 \sqrt {689}}{200}\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\frac {101}{200}+\frac {3 \sqrt {689}}{200} \\ {} & {} & \left [y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k +\frac {101}{200}+\frac {3 \sqrt {689}}{200}}, a_{k +4}=-\frac {k^{2} a_{k}-40 k^{2} a_{k +1}+600 k^{2} a_{k +2}-4000 k^{2} a_{k +3}+2 k \left (\frac {101}{200}+\frac {3 \sqrt {689}}{200}\right ) a_{k}-80 k \left (\frac {101}{200}+\frac {3 \sqrt {689}}{200}\right ) a_{k +1}+1200 k \left (\frac {101}{200}+\frac {3 \sqrt {689}}{200}\right ) a_{k +2}-8000 k \left (\frac {101}{200}+\frac {3 \sqrt {689}}{200}\right ) a_{k +3}+\left (\frac {101}{200}+\frac {3 \sqrt {689}}{200}\right )^{2} a_{k}-40 \left (\frac {101}{200}+\frac {3 \sqrt {689}}{200}\right )^{2} a_{k +1}+600 \left (\frac {101}{200}+\frac {3 \sqrt {689}}{200}\right )^{2} a_{k +2}-4000 \left (\frac {101}{200}+\frac {3 \sqrt {689}}{200}\right )^{2} a_{k +3}-k a_{k}-40 k a_{k +1}+1799 k a_{k +2}-19980 k a_{k +3}-\left (\frac {101}{200}+\frac {3 \sqrt {689}}{200}\right ) a_{k}-40 \left (\frac {101}{200}+\frac {3 \sqrt {689}}{200}\right ) a_{k +1}+1799 \left (\frac {101}{200}+\frac {3 \sqrt {689}}{200}\right ) a_{k +2}-19980 \left (\frac {101}{200}+\frac {3 \sqrt {689}}{200}\right ) a_{k +3}+1208 a_{k +2}-24140 a_{k +3}}{100 \left (100 k^{2}+200 k \left (\frac {101}{200}+\frac {3 \sqrt {689}}{200}\right )+100 \left (\frac {101}{200}+\frac {3 \sqrt {689}}{200}\right )^{2}+699 k +\frac {311799}{200}+\frac {2097 \sqrt {689}}{200}\right )}, a_{1}=\frac {a_{0} \left (200 \left (\frac {101}{200}+\frac {3 \sqrt {689}}{200}\right )^{2}-\frac {18301}{200}-\frac {603 \sqrt {689}}{200}\right )}{5 \left (100 \left (\frac {101}{200}+\frac {3 \sqrt {689}}{200}\right )^{2}+\frac {11799}{200}+\frac {297 \sqrt {689}}{200}\right )}, a_{2}=\frac {a_{0} \left (100000 \left (\frac {101}{200}+\frac {3 \sqrt {689}}{200}\right )^{4}-900 \left (\frac {101}{200}+\frac {3 \sqrt {689}}{200}\right )^{3}-91697 \left (\frac {101}{200}+\frac {3 \sqrt {689}}{200}\right )^{2}+\frac {573443}{200}+\frac {15429 \sqrt {689}}{200}\right )}{100 \left (10000 \left (\frac {101}{200}+\frac {3 \sqrt {689}}{200}\right )^{4}+39800 \left (\frac {101}{200}+\frac {3 \sqrt {689}}{200}\right )^{3}+51301 \left (\frac {101}{200}+\frac {3 \sqrt {689}}{200}\right )^{2}+\frac {2725983}{200}+\frac {69849 \sqrt {689}}{200}\right )}, a_{3}=\frac {a_{0} \left (5000000 \left (\frac {101}{200}+\frac {3 \sqrt {689}}{200}\right )^{6}+14940000 \left (\frac {101}{200}+\frac {3 \sqrt {689}}{200}\right )^{5}+5380400 \left (\frac {101}{200}+\frac {3 \sqrt {689}}{200}\right )^{4}-13226651 \left (\frac {101}{200}+\frac {3 \sqrt {689}}{200}\right )^{3}-8199223 \left (\frac {101}{200}+\frac {3 \sqrt {689}}{200}\right )^{2}+\frac {29547179}{100}+\frac {741837 \sqrt {689}}{100}\right )}{250 \left (1000000 \left (\frac {101}{200}+\frac {3 \sqrt {689}}{200}\right )^{6}+8970000 \left (\frac {101}{200}+\frac {3 \sqrt {689}}{200}\right )^{5}+31060300 \left (\frac {101}{200}+\frac {3 \sqrt {689}}{200}\right )^{4}+52086099 \left (\frac {101}{200}+\frac {3 \sqrt {689}}{200}\right )^{3}+42945124 \left (\frac {101}{200}+\frac {3 \sqrt {689}}{200}\right )^{2}+\frac {1749018609}{200}+\frac {45200727 \sqrt {689}}{200}\right )}\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =x +5 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x +5\right )^{k +\frac {101}{200}+\frac {3 \sqrt {689}}{200}}, a_{k +4}=-\frac {k^{2} a_{k}-40 k^{2} a_{k +1}+600 k^{2} a_{k +2}-4000 k^{2} a_{k +3}+2 k \left (\frac {101}{200}+\frac {3 \sqrt {689}}{200}\right ) a_{k}-80 k \left (\frac {101}{200}+\frac {3 \sqrt {689}}{200}\right ) a_{k +1}+1200 k \left (\frac {101}{200}+\frac {3 \sqrt {689}}{200}\right ) a_{k +2}-8000 k \left (\frac {101}{200}+\frac {3 \sqrt {689}}{200}\right ) a_{k +3}+\left (\frac {101}{200}+\frac {3 \sqrt {689}}{200}\right )^{2} a_{k}-40 \left (\frac {101}{200}+\frac {3 \sqrt {689}}{200}\right )^{2} a_{k +1}+600 \left (\frac {101}{200}+\frac {3 \sqrt {689}}{200}\right )^{2} a_{k +2}-4000 \left (\frac {101}{200}+\frac {3 \sqrt {689}}{200}\right )^{2} a_{k +3}-k a_{k}-40 k a_{k +1}+1799 k a_{k +2}-19980 k a_{k +3}-\left (\frac {101}{200}+\frac {3 \sqrt {689}}{200}\right ) a_{k}-40 \left (\frac {101}{200}+\frac {3 \sqrt {689}}{200}\right ) a_{k +1}+1799 \left (\frac {101}{200}+\frac {3 \sqrt {689}}{200}\right ) a_{k +2}-19980 \left (\frac {101}{200}+\frac {3 \sqrt {689}}{200}\right ) a_{k +3}+1208 a_{k +2}-24140 a_{k +3}}{100 \left (100 k^{2}+200 k \left (\frac {101}{200}+\frac {3 \sqrt {689}}{200}\right )+100 \left (\frac {101}{200}+\frac {3 \sqrt {689}}{200}\right )^{2}+699 k +\frac {311799}{200}+\frac {2097 \sqrt {689}}{200}\right )}, a_{1}=\frac {a_{0} \left (200 \left (\frac {101}{200}+\frac {3 \sqrt {689}}{200}\right )^{2}-\frac {18301}{200}-\frac {603 \sqrt {689}}{200}\right )}{5 \left (100 \left (\frac {101}{200}+\frac {3 \sqrt {689}}{200}\right )^{2}+\frac {11799}{200}+\frac {297 \sqrt {689}}{200}\right )}, a_{2}=\frac {a_{0} \left (100000 \left (\frac {101}{200}+\frac {3 \sqrt {689}}{200}\right )^{4}-900 \left (\frac {101}{200}+\frac {3 \sqrt {689}}{200}\right )^{3}-91697 \left (\frac {101}{200}+\frac {3 \sqrt {689}}{200}\right )^{2}+\frac {573443}{200}+\frac {15429 \sqrt {689}}{200}\right )}{100 \left (10000 \left (\frac {101}{200}+\frac {3 \sqrt {689}}{200}\right )^{4}+39800 \left (\frac {101}{200}+\frac {3 \sqrt {689}}{200}\right )^{3}+51301 \left (\frac {101}{200}+\frac {3 \sqrt {689}}{200}\right )^{2}+\frac {2725983}{200}+\frac {69849 \sqrt {689}}{200}\right )}, a_{3}=\frac {a_{0} \left (5000000 \left (\frac {101}{200}+\frac {3 \sqrt {689}}{200}\right )^{6}+14940000 \left (\frac {101}{200}+\frac {3 \sqrt {689}}{200}\right )^{5}+5380400 \left (\frac {101}{200}+\frac {3 \sqrt {689}}{200}\right )^{4}-13226651 \left (\frac {101}{200}+\frac {3 \sqrt {689}}{200}\right )^{3}-8199223 \left (\frac {101}{200}+\frac {3 \sqrt {689}}{200}\right )^{2}+\frac {29547179}{100}+\frac {741837 \sqrt {689}}{100}\right )}{250 \left (1000000 \left (\frac {101}{200}+\frac {3 \sqrt {689}}{200}\right )^{6}+8970000 \left (\frac {101}{200}+\frac {3 \sqrt {689}}{200}\right )^{5}+31060300 \left (\frac {101}{200}+\frac {3 \sqrt {689}}{200}\right )^{4}+52086099 \left (\frac {101}{200}+\frac {3 \sqrt {689}}{200}\right )^{3}+42945124 \left (\frac {101}{200}+\frac {3 \sqrt {689}}{200}\right )^{2}+\frac {1749018609}{200}+\frac {45200727 \sqrt {689}}{200}\right )}\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x +5\right )^{k +\frac {101}{200}-\frac {3 \sqrt {689}}{200}}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} \left (x +5\right )^{k +\frac {101}{200}+\frac {3 \sqrt {689}}{200}}\right ), a_{k +4}=-\frac {k^{2} a_{k}-40 k^{2} a_{k +1}+600 k^{2} a_{k +2}-4000 k^{2} a_{k +3}+2 k \left (\frac {101}{200}-\frac {3 \sqrt {689}}{200}\right ) a_{k}-80 k \left (\frac {101}{200}-\frac {3 \sqrt {689}}{200}\right ) a_{k +1}+1200 k \left (\frac {101}{200}-\frac {3 \sqrt {689}}{200}\right ) a_{k +2}-8000 k \left (\frac {101}{200}-\frac {3 \sqrt {689}}{200}\right ) a_{k +3}+\left (\frac {101}{200}-\frac {3 \sqrt {689}}{200}\right )^{2} a_{k}-40 \left (\frac {101}{200}-\frac {3 \sqrt {689}}{200}\right )^{2} a_{k +1}+600 \left (\frac {101}{200}-\frac {3 \sqrt {689}}{200}\right )^{2} a_{k +2}-4000 \left (\frac {101}{200}-\frac {3 \sqrt {689}}{200}\right )^{2} a_{k +3}-k a_{k}-40 k a_{k +1}+1799 k a_{k +2}-19980 k a_{k +3}-\left (\frac {101}{200}-\frac {3 \sqrt {689}}{200}\right ) a_{k}-40 \left (\frac {101}{200}-\frac {3 \sqrt {689}}{200}\right ) a_{k +1}+1799 \left (\frac {101}{200}-\frac {3 \sqrt {689}}{200}\right ) a_{k +2}-19980 \left (\frac {101}{200}-\frac {3 \sqrt {689}}{200}\right ) a_{k +3}+1208 a_{k +2}-24140 a_{k +3}}{100 \left (100 k^{2}+200 k \left (\frac {101}{200}-\frac {3 \sqrt {689}}{200}\right )+100 \left (\frac {101}{200}-\frac {3 \sqrt {689}}{200}\right )^{2}+699 k +\frac {311799}{200}-\frac {2097 \sqrt {689}}{200}\right )}, a_{1}=\frac {a_{0} \left (200 \left (\frac {101}{200}-\frac {3 \sqrt {689}}{200}\right )^{2}-\frac {18301}{200}+\frac {603 \sqrt {689}}{200}\right )}{5 \left (100 \left (\frac {101}{200}-\frac {3 \sqrt {689}}{200}\right )^{2}+\frac {11799}{200}-\frac {297 \sqrt {689}}{200}\right )}, a_{2}=\frac {a_{0} \left (100000 \left (\frac {101}{200}-\frac {3 \sqrt {689}}{200}\right )^{4}-900 \left (\frac {101}{200}-\frac {3 \sqrt {689}}{200}\right )^{3}-91697 \left (\frac {101}{200}-\frac {3 \sqrt {689}}{200}\right )^{2}+\frac {573443}{200}-\frac {15429 \sqrt {689}}{200}\right )}{100 \left (10000 \left (\frac {101}{200}-\frac {3 \sqrt {689}}{200}\right )^{4}+39800 \left (\frac {101}{200}-\frac {3 \sqrt {689}}{200}\right )^{3}+51301 \left (\frac {101}{200}-\frac {3 \sqrt {689}}{200}\right )^{2}+\frac {2725983}{200}-\frac {69849 \sqrt {689}}{200}\right )}, a_{3}=\frac {a_{0} \left (5000000 \left (\frac {101}{200}-\frac {3 \sqrt {689}}{200}\right )^{6}+14940000 \left (\frac {101}{200}-\frac {3 \sqrt {689}}{200}\right )^{5}+5380400 \left (\frac {101}{200}-\frac {3 \sqrt {689}}{200}\right )^{4}-13226651 \left (\frac {101}{200}-\frac {3 \sqrt {689}}{200}\right )^{3}-8199223 \left (\frac {101}{200}-\frac {3 \sqrt {689}}{200}\right )^{2}+\frac {29547179}{100}-\frac {741837 \sqrt {689}}{100}\right )}{250 \left (1000000 \left (\frac {101}{200}-\frac {3 \sqrt {689}}{200}\right )^{6}+8970000 \left (\frac {101}{200}-\frac {3 \sqrt {689}}{200}\right )^{5}+31060300 \left (\frac {101}{200}-\frac {3 \sqrt {689}}{200}\right )^{4}+52086099 \left (\frac {101}{200}-\frac {3 \sqrt {689}}{200}\right )^{3}+42945124 \left (\frac {101}{200}-\frac {3 \sqrt {689}}{200}\right )^{2}+\frac {1749018609}{200}-\frac {45200727 \sqrt {689}}{200}\right )}, b_{k +4}=-\frac {k^{2} b_{k}-40 k^{2} b_{k +1}+600 k^{2} b_{k +2}-4000 k^{2} b_{k +3}+2 k \left (\frac {101}{200}+\frac {3 \sqrt {689}}{200}\right ) b_{k}-80 k \left (\frac {101}{200}+\frac {3 \sqrt {689}}{200}\right ) b_{k +1}+1200 k \left (\frac {101}{200}+\frac {3 \sqrt {689}}{200}\right ) b_{k +2}-8000 k \left (\frac {101}{200}+\frac {3 \sqrt {689}}{200}\right ) b_{k +3}+\left (\frac {101}{200}+\frac {3 \sqrt {689}}{200}\right )^{2} b_{k}-40 \left (\frac {101}{200}+\frac {3 \sqrt {689}}{200}\right )^{2} b_{k +1}+600 \left (\frac {101}{200}+\frac {3 \sqrt {689}}{200}\right )^{2} b_{k +2}-4000 \left (\frac {101}{200}+\frac {3 \sqrt {689}}{200}\right )^{2} b_{k +3}-k b_{k}-40 k b_{k +1}+1799 k b_{k +2}-19980 k b_{k +3}-\left (\frac {101}{200}+\frac {3 \sqrt {689}}{200}\right ) b_{k}-40 \left (\frac {101}{200}+\frac {3 \sqrt {689}}{200}\right ) b_{k +1}+1799 \left (\frac {101}{200}+\frac {3 \sqrt {689}}{200}\right ) b_{k +2}-19980 \left (\frac {101}{200}+\frac {3 \sqrt {689}}{200}\right ) b_{k +3}+1208 b_{k +2}-24140 b_{k +3}}{100 \left (100 k^{2}+200 k \left (\frac {101}{200}+\frac {3 \sqrt {689}}{200}\right )+100 \left (\frac {101}{200}+\frac {3 \sqrt {689}}{200}\right )^{2}+699 k +\frac {311799}{200}+\frac {2097 \sqrt {689}}{200}\right )}, b_{1}=\frac {b_{0} \left (200 \left (\frac {101}{200}+\frac {3 \sqrt {689}}{200}\right )^{2}-\frac {18301}{200}-\frac {603 \sqrt {689}}{200}\right )}{5 \left (100 \left (\frac {101}{200}+\frac {3 \sqrt {689}}{200}\right )^{2}+\frac {11799}{200}+\frac {297 \sqrt {689}}{200}\right )}, b_{2}=\frac {b_{0} \left (100000 \left (\frac {101}{200}+\frac {3 \sqrt {689}}{200}\right )^{4}-900 \left (\frac {101}{200}+\frac {3 \sqrt {689}}{200}\right )^{3}-91697 \left (\frac {101}{200}+\frac {3 \sqrt {689}}{200}\right )^{2}+\frac {573443}{200}+\frac {15429 \sqrt {689}}{200}\right )}{100 \left (10000 \left (\frac {101}{200}+\frac {3 \sqrt {689}}{200}\right )^{4}+39800 \left (\frac {101}{200}+\frac {3 \sqrt {689}}{200}\right )^{3}+51301 \left (\frac {101}{200}+\frac {3 \sqrt {689}}{200}\right )^{2}+\frac {2725983}{200}+\frac {69849 \sqrt {689}}{200}\right )}, b_{3}=\frac {b_{0} \left (5000000 \left (\frac {101}{200}+\frac {3 \sqrt {689}}{200}\right )^{6}+14940000 \left (\frac {101}{200}+\frac {3 \sqrt {689}}{200}\right )^{5}+5380400 \left (\frac {101}{200}+\frac {3 \sqrt {689}}{200}\right )^{4}-13226651 \left (\frac {101}{200}+\frac {3 \sqrt {689}}{200}\right )^{3}-8199223 \left (\frac {101}{200}+\frac {3 \sqrt {689}}{200}\right )^{2}+\frac {29547179}{100}+\frac {741837 \sqrt {689}}{100}\right )}{250 \left (1000000 \left (\frac {101}{200}+\frac {3 \sqrt {689}}{200}\right )^{6}+8970000 \left (\frac {101}{200}+\frac {3 \sqrt {689}}{200}\right )^{5}+31060300 \left (\frac {101}{200}+\frac {3 \sqrt {689}}{200}\right )^{4}+52086099 \left (\frac {101}{200}+\frac {3 \sqrt {689}}{200}\right )^{3}+42945124 \left (\frac {101}{200}+\frac {3 \sqrt {689}}{200}\right )^{2}+\frac {1749018609}{200}+\frac {45200727 \sqrt {689}}{200}\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 <- hyper3 successful: received ODE is equivalent to the 1F1 ODE <- Kummer successful <- special function solution successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 54
Order:=6; dsolve((x^2-25)^2*diff(y(x),x$2)-(x+5)*diff(y(x),x)+10*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = \left (1-\frac {1}{125} x^{2}-\frac {1}{46875} x^{3}-\frac {767}{7812500} x^{4}-\frac {4813}{7324218750} x^{5}\right ) y \left (0\right )+\left (x +\frac {1}{250} x^{2}-\frac {112}{46875} x^{3}+\frac {173}{3906250} x^{4}-\frac {409681}{7324218750} x^{5}\right ) D\left (y \right )\left (0\right )+O\left (x^{6}\right ) \]
✓ Solution by Mathematica
Time used: 0.001 (sec). Leaf size: 70
AsymptoticDSolveValue[(x^2-25)^2*y''[x]-(x+5)*y'[x]+10*y[x]==0,y[x],{x,0,5}]
\[ y(x)\to c_1 \left (-\frac {4813 x^5}{7324218750}-\frac {767 x^4}{7812500}-\frac {x^3}{46875}-\frac {x^2}{125}+1\right )+c_2 \left (-\frac {409681 x^5}{7324218750}+\frac {173 x^4}{3906250}-\frac {112 x^3}{46875}+\frac {x^2}{250}+x\right ) \]