problem 14

Internal problem ID [6648]
Internal ﬁle name [OUTPUT/5896_Sunday_June_05_2022_04_00_29_PM_1035037/index.tex]

Book: DIFFERENTIAL EQUATIONS with Boundary Value Problems. DENNIS G. ZILL, WARREN S. WRIGHT, MICHAEL R. CULLEN. Brooks/Cole. Boston, MA. 2013. 8th edition.
Section: CHAPTER 6 SERIES SOLUTIONS OF LINEAR EQUATIONS. CHAPTER 6 IN REVIEW. Page 271
Problem number: 14.
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"

\[ \boxed {\cos \left (x \right ) y^{\prime \prime }+y=0} \] 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}{\cos \left (x \right )}\\ 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 \\ &= -\sec \left (x \right ) \left (y \tan \left (x \right )+y^{\prime }\right )\\ 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 \\ &= -2 \sec \left (x \right ) \left (y^{\prime } \tan \left (x \right )+y \left (\sec \left (x \right )+\frac {1}{2}\right ) \left (\sec \left (x \right )-1\right )\right )\\ 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 \\ &= \left (\left (-6 \sec \left (x \right )^{2}+\sec \left (x \right )+3\right ) y^{\prime }+\tan \left (x \right ) \sec \left (x \right )^{2} y \left (\cos \left (x \right )^{2}+4 \cos \left (x \right )-6\right )\right ) \sec \left (x \right )\\ 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 \\ &= -24 \sec \left (x \right ) \left (-\frac {\sec \left (x \right )^{2} \left (\cos \left (x \right )^{2}+\frac {3 \cos \left (x \right )}{2}-6\right ) \tan \left (x \right ) y^{\prime }}{6}+y \left (\sec \left (x \right )^{4}-\frac {3 \sec \left (x \right )^{3}}{4}-\frac {19 \sec \left (x \right )^{2}}{24}+\frac {11 \sec \left (x \right )}{24}+\frac {1}{24}\right )\right )\\ F_5 &= \frac {d F_4}{dx} \\ &= \frac {\partial F_{4}}{\partial x}+ \frac {\partial F_{4}}{\partial y} y^{\prime }+ \frac {\partial F_{4}}{\partial y^{\prime }} F_4 \\ &= -\sec \left (x \right ) \left (\left (120 \sec \left (x \right )^{4}-36 \sec \left (x \right )^{3}-99 \sec \left (x \right )^{2}+23 \sec \left (x \right )+5\right ) y^{\prime }+\tan \left (x \right ) \sec \left (x \right )^{4} y \left (-1+\cos \left (x \right )\right ) \left (\cos \left (x \right )^{3}+27 \cos \left (x \right )^{2}-24 \cos \left (x \right )-120\right )\right )\\ F_6 &= \frac {d F_5}{dx} \\ &= \frac {\partial F_{5}}{\partial x}+ \frac {\partial F_{5}}{\partial y} y^{\prime }+ \frac {\partial F_{5}}{\partial y^{\prime }} F_5 \\ &= -720 \sec \left (x \right ) \left (\frac {\tan \left (x \right ) \sec \left (x \right )^{4} \left (\cos \left (x \right )^{4}+12 \cos \left (x \right )^{3}-58 \cos \left (x \right )^{2}-40 \cos \left (x \right )+120\right ) y^{\prime }}{120}+y \left (\sec \left (x \right )^{6}-\frac {5 \sec \left (x \right )^{5}}{6}-\frac {16 \sec \left (x \right )^{4}}{15}+\frac {187 \sec \left (x \right )^{3}}{240}+\frac {11 \sec \left (x \right )^{2}}{60}-\frac {19 \sec \left (x \right )}{240}-\frac {1}{720}\right )\right ) \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 &= -y \left (0\right )\\ F_1 &= -y^{\prime }\left (0\right )\\ F_2 &= 0\\ F_3 &= -2 y^{\prime }\left (0\right )\\ F_4 &= y \left (0\right )\\ F_5 &= -13 y^{\prime }\left (0\right )\\ F_6 &= 13 y \left (0\right ) \end {align*}

Substituting all the above in (7) and simplifying gives the solution as \[ y = \left (1-\frac {1}{2} x^{2}+\frac {1}{720} x^{6}+\frac {13}{40320} x^{8}\right ) y \left (0\right )+\left (x -\frac {1}{6} x^{3}-\frac {1}{60} x^{5}-\frac {13}{5040} x^{7}\right ) y^{\prime }\left (0\right )+O\left (x^{8}\right ) \] Since the expansion point \(x = 0\) is an ordinary, we can also solve this using standard power series 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*} \moverset {\infty }{\munderset {n =2}{\sum }}n \left (n -1\right ) a_{n} x^{n -2} = -\frac {\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n}}{\cos \left (x \right )}\tag {1} \end {align*}

Expanding \(\cos \left (x \right )\) as Taylor series around \(x=0\) and keeping only the ﬁrst \(8\) terms gives \begin {align*} \cos \left (x \right ) &= \frac {1}{40320} x^{8}-\frac {1}{720} x^{6}+\frac {1}{24} x^{4}-\frac {1}{2} x^{2}+1 + \dots \\ &= \frac {1}{40320} x^{8}-\frac {1}{720} x^{6}+\frac {1}{24} x^{4}-\frac {1}{2} x^{2}+1 \end {align*}

Hence the ODE in Eq (1) becomes \[ \left (\frac {1}{40320} x^{8}-\frac {1}{720} x^{6}+\frac {1}{24} x^{4}-\frac {1}{2} x^{2}+1\right ) \left (\moverset {\infty }{\munderset {n =2}{\sum }}n \left (n -1\right ) a_{n} x^{n -2}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n}\right ) = 0 \] Expanding the ﬁrst term in (1) gives \[ \frac {x^{8}}{40320}\eslowast \left (\moverset {\infty }{\munderset {n =2}{\sum }}n \left (n -1\right ) a_{n} x^{n -2}\right )-\frac {x^{6}}{720}\eslowast \left (\moverset {\infty }{\munderset {n =2}{\sum }}n \left (n -1\right ) a_{n} x^{n -2}\right )+\frac {x^{4}}{24}\eslowast \left (\moverset {\infty }{\munderset {n =2}{\sum }}n \left (n -1\right ) a_{n} x^{n -2}\right )-\frac {x^{2}}{2}\eslowast \left (\moverset {\infty }{\munderset {n =2}{\sum }}n \left (n -1\right ) a_{n} x^{n -2}\right )+\left (1\eslowast \left (\moverset {\infty }{\munderset {n =2}{\sum }}n \left (n -1\right ) a_{n} x^{n -2}\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n}\right ) = 0 \] Which simpliﬁes to \begin{equation} \tag{2} \left (\moverset {\infty }{\munderset {n =2}{\sum }}\frac {n \,x^{n +6} a_{n} \left (n -1\right )}{40320}\right )+\moverset {\infty }{\munderset {n =2}{\sum }}\left (-\frac {n \,x^{n +4} a_{n} \left (n -1\right )}{720}\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}\frac {n \,x^{n +2} a_{n} \left (n -1\right )}{24}\right )+\moverset {\infty }{\munderset {n =2}{\sum }}\left (-\frac {n a_{n} x^{n} \left (n -1\right )}{2}\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}n \left (n -1\right ) a_{n} x^{n -2}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}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 }}\frac {n \,x^{n +6} a_{n} \left (n -1\right )}{40320} &= \moverset {\infty }{\munderset {n =8}{\sum }}\frac {\left (n -6\right ) a_{n -6} \left (n -7\right ) x^{n}}{40320} \\ \moverset {\infty }{\munderset {n =2}{\sum }}\left (-\frac {n \,x^{n +4} a_{n} \left (n -1\right )}{720}\right ) &= \moverset {\infty }{\munderset {n =6}{\sum }}\left (-\frac {\left (n -4\right ) a_{n -4} \left (n -5\right ) x^{n}}{720}\right ) \\ \moverset {\infty }{\munderset {n =2}{\sum }}\frac {n \,x^{n +2} a_{n} \left (n -1\right )}{24} &= \moverset {\infty }{\munderset {n =4}{\sum }}\frac {\left (n -2\right ) a_{n -2} \left (n -3\right ) x^{n}}{24} \\ \moverset {\infty }{\munderset {n =2}{\sum }}n \left (n -1\right ) a_{n} x^{n -2} &= \moverset {\infty }{\munderset {n =0}{\sum }}\left (n +2\right ) a_{n +2} \left (n +1\right ) x^{n} \\ \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 =8}{\sum }}\frac {\left (n -6\right ) a_{n -6} \left (n -7\right ) x^{n}}{40320}\right )+\moverset {\infty }{\munderset {n =6}{\sum }}\left (-\frac {\left (n -4\right ) a_{n -4} \left (n -5\right ) x^{n}}{720}\right )+\left (\moverset {\infty }{\munderset {n =4}{\sum }}\frac {\left (n -2\right ) a_{n -2} \left (n -3\right ) x^{n}}{24}\right )+\moverset {\infty }{\munderset {n =2}{\sum }}\left (-\frac {n a_{n} x^{n} \left (n -1\right )}{2}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +2\right ) a_{n +2} \left (n +1\right ) x^{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n}\right ) = 0 \end{equation} \(n=0\) gives \[ 2 a_{2}+a_{0}=0 \] \[ a_{2} = -\frac {a_{0}}{2} \] \(n=1\) gives \[ 6 a_{3}+a_{1}=0 \] Which after substituting earlier equations, simpliﬁes to \[ a_{3} = -\frac {a_{1}}{6} \] \(n=3\) gives \[ -2 a_{3}+20 a_{5} = 0 \] Which after substituting earlier equations, simpliﬁes to \[ \frac {a_{1}}{3}+20 a_{5} = 0 \] Or \[ a_{5} = -\frac {a_{1}}{60} \] \(n=4\) gives \[ \frac {a_{2}}{12}-5 a_{4}+30 a_{6} = 0 \] Which after substituting earlier equations, simpliﬁes to \[ -\frac {a_{0}}{24}+30 a_{6} = 0 \] Or \[ a_{6} = \frac {a_{0}}{720} \] \(n=5\) gives \[ \frac {a_{3}}{4}-9 a_{5}+42 a_{7} = 0 \] Which after substituting earlier equations, simpliﬁes to \[ \frac {13 a_{1}}{120}+42 a_{7} = 0 \] Or \[ a_{7} = -\frac {13 a_{1}}{5040} \] \(n=6\) gives \[ -\frac {a_{2}}{360}+\frac {a_{4}}{2}-14 a_{6}+56 a_{8} = 0 \] Which after substituting earlier equations, simpliﬁes to \[ -\frac {13 a_{0}}{720}+56 a_{8} = 0 \] Or \[ a_{8} = \frac {13 a_{0}}{40320} \] \(n=7\) gives \[ -\frac {a_{3}}{120}+\frac {5 a_{5}}{6}-20 a_{7}+72 a_{9} = 0 \] Which after substituting earlier equations, simpliﬁes to \[ \frac {197 a_{1}}{5040}+72 a_{9} = 0 \] Or \[ a_{9} = -\frac {197 a_{1}}{362880} \] For \(8\le n\), the recurrence equation is \begin{equation} \tag{4} \frac {\left (n -6\right ) a_{n -6} \left (n -7\right )}{40320}-\frac {\left (n -4\right ) a_{n -4} \left (n -5\right )}{720}+\frac {\left (n -2\right ) a_{n -2} \left (n -3\right )}{24}-\frac {n a_{n} \left (n -1\right )}{2}+\left (n +2\right ) a_{n +2} \left (n +1\right )+a_{n} = 0 \end{equation} Solving for \(a_{n +2}\), gives \begin{align*} \tag{5} a_{n +2}&= \frac {20160 n^{2} a_{n}-n^{2} a_{n -6}+56 n^{2} a_{n -4}-1680 n^{2} a_{n -2}-20160 n a_{n}+13 n a_{n -6}-504 n a_{n -4}+8400 n a_{n -2}-40320 a_{n}-42 a_{n -6}+1120 a_{n -4}-10080 a_{n -2}}{40320 \left (n +2\right ) \left (n +1\right )} \\ &= \frac {\left (20160 n^{2}-20160 n -40320\right ) a_{n}}{40320 \left (n +2\right ) \left (n +1\right )}+\frac {\left (-n^{2}+13 n -42\right ) a_{n -6}}{40320 \left (n +2\right ) \left (n +1\right )}+\frac {\left (56 n^{2}-504 n +1120\right ) a_{n -4}}{40320 \left (n +2\right ) \left (n +1\right )}+\frac {\left (-1680 n^{2}+8400 n -10080\right ) a_{n -2}}{40320 \left (n +2\right ) \left (n +1\right )} \\ \end{align*} 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 \[ y = a_{0}+a_{1} x -\frac {1}{2} a_{0} x^{2}-\frac {1}{6} a_{1} x^{3}-\frac {1}{60} a_{1} x^{5}+\frac {1}{720} a_{0} x^{6}-\frac {13}{5040} a_{1} x^{7}+\dots \] Collecting terms, the solution becomes \begin{equation} \tag{3} y = \left (1-\frac {1}{2} x^{2}+\frac {1}{720} x^{6}\right ) a_{0}+\left (x -\frac {1}{6} x^{3}-\frac {1}{60} x^{5}-\frac {13}{5040} x^{7}\right ) a_{1}+O\left (x^{8}\right ) \end{equation} At \(x = 0\) the solution above becomes \[ y = \left (1-\frac {1}{2} x^{2}+\frac {1}{720} x^{6}\right ) c_{1} +\left (x -\frac {1}{6} x^{3}-\frac {1}{60} x^{5}-\frac {13}{5040} x^{7}\right ) c_{2} +O\left (x^{8}\right ) \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \left (1-\frac {1}{2} x^{2}+\frac {1}{720} x^{6}+\frac {13}{40320} x^{8}\right ) y \left (0\right )+\left (x -\frac {1}{6} x^{3}-\frac {1}{60} x^{5}-\frac {13}{5040} x^{7}\right ) y^{\prime }\left (0\right )+O\left (x^{8}\right ) \\ \tag{2} y &= \left (1-\frac {1}{2} x^{2}+\frac {1}{720} x^{6}\right ) c_{1} +\left (x -\frac {1}{6} x^{3}-\frac {1}{60} x^{5}-\frac {13}{5040} x^{7}\right ) c_{2} +O\left (x^{8}\right ) \\ \end{align*}

Veriﬁcation of solutions

\[ y = \left (1-\frac {1}{2} x^{2}+\frac {1}{720} x^{6}+\frac {13}{40320} x^{8}\right ) y \left (0\right )+\left (x -\frac {1}{6} x^{3}-\frac {1}{60} x^{5}-\frac {13}{5040} x^{7}\right ) y^{\prime }\left (0\right )+O\left (x^{8}\right ) \] Veriﬁed OK.

\[ y = \left (1-\frac {1}{2} x^{2}+\frac {1}{720} x^{6}\right ) c_{1} +\left (x -\frac {1}{6} x^{3}-\frac {1}{60} x^{5}-\frac {13}{5040} x^{7}\right ) c_{2} +O\left (x^{8}\right ) \] Veriﬁed OK.

Maple trace

`Methods for second order ODEs: 
--- Trying classification methods --- 
trying a symmetry of the form [xi=0, eta=F(x)] 
checking if the LODE is missing y 
-> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius 
-> trying a solution of the form r0(x) * Y + r1(x) * Y where Y = exp(int(r(x), dx)) * 2F1([a1, a2], [b1], f) 
-> Trying changes of variables to rationalize or make the ODE simpler 
   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  HeunG  ODE, case  a <> 0, e <> 0, g <> 0, c = 0 
   Change of variables used: 
      [x = arccos(t)] 
   Linear ODE actually solved: 
      u(t)-t^2*diff(u(t),t)+(-t^3+t)*diff(diff(u(t),t),t) = 0 
<- change of variables successful`

\[ y \left (x \right ) = \left (1-\frac {1}{2} x^{2}+\frac {1}{720} x^{6}\right ) y \left (0\right )+\left (x -\frac {1}{6} x^{3}-\frac {1}{60} x^{5}-\frac {13}{5040} x^{7}\right ) D\left (y \right )\left (0\right )+O\left (x^{8}\right ) \]

\[ y(x)\to c_1 \left (\frac {x^6}{720}-\frac {x^2}{2}+1\right )+c_2 \left (-\frac {13 x^7}{5040}-\frac {x^5}{60}-\frac {x^3}{6}+x\right ) \]

4.6 problem 14