problem 5

Internal problem ID [5023]
Internal ﬁle name [OUTPUT/4516_Sunday_June_05_2022_02_59_57_PM_80071994/index.tex]

Book: Fundamentals of Diﬀerential Equations. By Nagle, Saﬀ and Snider. 9th edition. Boston. Pearson 2018.
Section: Chapter 8, Series solutions of diﬀerential equations. Section 8.4. page 449
Problem number: 5.
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"

\[ \boxed {y^{\prime \prime }-\tan \left (x \right ) y^{\prime }+y=0} \] With the expansion point for the power series method at \(x = 1\).

The ode does not have its expansion point at \(x = 0\), therefore to simplify the computation of power series expansion, change of variable is made on the independent variable to shift the initial conditions and the expasion point back to zero. The new ode is then solved more easily since the expansion point is now at zero. The solution converted back to the original independent variable. Let \[ t = x -1 \] The ode is converted to be in terms of the new independent variable \(t\). This results in \[ \frac {d^{2}}{d t^{2}}y \left (t \right )-\tan \left (t +1\right ) \left (\frac {d}{d t}y \left (t \right )\right )+y \left (t \right ) = 0 \] With its expansion point and initial conditions now at \(t = 0\). The transformed ODE is now solved. 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 &= \tan \left (t +1\right ) \left (\frac {d}{d t}y \left (t \right )\right )-y \left (t \right )\\ F_1 &= \frac {d F_0}{dt} \\ &= \frac {\partial F_{0}}{\partial t}+ \frac {\partial F_{0}}{\partial y} \frac {d}{d t}y \left (t \right )+ \frac {\partial F_{0}}{\partial \frac {d}{d t}y \left (t \right )} F_0 \\ &= -\tan \left (t +1\right ) \left (-2 \tan \left (t +1\right ) \left (\frac {d}{d t}y \left (t \right )\right )+y \left (t \right )\right )\\ F_2 &= \frac {d F_1}{dt} \\ &= \frac {\partial F_{1}}{\partial t}+ \frac {\partial F_{1}}{\partial y} \frac {d}{d t}y \left (t \right )+ \frac {\partial F_{1}}{\partial \frac {d}{d t}y \left (t \right )} F_1 \\ &= \left (6 \tan \left (t +1\right ) \left (\frac {d}{d t}y \left (t \right )\right )-3 y \left (t \right )\right ) \sec \left (t +1\right )^{2}-3 \tan \left (t +1\right ) \left (\frac {d}{d t}y \left (t \right )\right )+2 y \left (t \right )\\ F_3 &= \frac {d F_2}{dt} \\ &= \frac {\partial F_{2}}{\partial t}+ \frac {\partial F_{2}}{\partial y} \frac {d}{d t}y \left (t \right )+ \frac {\partial F_{2}}{\partial \frac {d}{d t}y \left (t \right )} F_2 \\ &= 24 \sec \left (t +1\right )^{4} \left (\frac {d}{d t}y \left (t \right )\right )+\left (-12 \tan \left (t +1\right ) y \left (t \right )-27 \frac {d}{d t}y \left (t \right )\right ) \sec \left (t +1\right )^{2}+3 \tan \left (t +1\right ) y \left (t \right )+5 \frac {d}{d t}y \left (t \right )\\ F_4 &= \frac {d F_3}{dt} \\ &= \frac {\partial F_{3}}{\partial t}+ \frac {\partial F_{3}}{\partial y} \frac {d}{d t}y \left (t \right )+ \frac {\partial F_{3}}{\partial \frac {d}{d t}y \left (t \right )} F_3 \\ &= \left (8 \left (\frac {d}{d t}y \left (t \right )\right ) \left (\cos \left (t +1\right )^{4}-\frac {93 \cos \left (t +1\right )^{2}}{8}+15\right ) \tan \left (t +1\right )-60 y \left (t \right )\right ) \sec \left (t +1\right )^{4}+54 y \left (t \right ) \sec \left (t +1\right )^{2}-5 y \left (t \right ) \end {align*}

And so on. Evaluating all the above at initial conditions \(t = 0\) 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 &= \tan \left (1\right ) y^{\prime }\left (0\right )-y \left (0\right )\\ F_1 &= 2 y^{\prime }\left (0\right ) \tan \left (1\right )^{2}-y \left (0\right ) \tan \left (1\right )\\ F_2 &= 6 \sec \left (1\right )^{2} \tan \left (1\right ) y^{\prime }\left (0\right )-3 \sec \left (1\right )^{2} y \left (0\right )-3 \tan \left (1\right ) y^{\prime }\left (0\right )+2 y \left (0\right )\\ F_3 &= 24 \sec \left (1\right )^{4} y^{\prime }\left (0\right )-12 \sec \left (1\right )^{2} \tan \left (1\right ) y \left (0\right )-27 \sec \left (1\right )^{2} y^{\prime }\left (0\right )+3 y \left (0\right ) \tan \left (1\right )+5 y^{\prime }\left (0\right )\\ F_4 &= 8 \tan \left (1\right ) y^{\prime }\left (0\right )-93 \sec \left (1\right )^{2} \tan \left (1\right ) y^{\prime }\left (0\right )+120 \sec \left (1\right )^{4} \tan \left (1\right ) y^{\prime }\left (0\right )-60 \sec \left (1\right )^{4} y \left (0\right )+54 \sec \left (1\right )^{2} y \left (0\right )-5 y \left (0\right ) \end {align*}

Substituting all the above in (7) and simplifying gives the solution as \[ y \left (t \right ) = \left (1-\frac {t^{2}}{2}-\frac {t^{3} \tan \left (1\right )}{6}-\frac {t^{4} \sec \left (1\right )^{2}}{8}+\frac {t^{4}}{12}-\frac {t^{5} \sec \left (1\right )^{2} \tan \left (1\right )}{10}+\frac {t^{5} \tan \left (1\right )}{40}-\frac {t^{6} \sec \left (1\right )^{4}}{12}+\frac {3 t^{6} \sec \left (1\right )^{2}}{40}-\frac {t^{6}}{144}\right ) y \left (0\right )+\left (t +\frac {t^{2} \tan \left (1\right )}{2}+\frac {t^{3} \tan \left (1\right )^{2}}{3}+\frac {t^{4} \sec \left (1\right )^{2} \tan \left (1\right )}{4}-\frac {t^{4} \tan \left (1\right )}{8}+\frac {t^{5} \sec \left (1\right )^{4}}{5}-\frac {9 t^{5} \sec \left (1\right )^{2}}{40}+\frac {t^{5}}{24}+\frac {t^{6} \tan \left (1\right )}{90}-\frac {31 t^{6} \sec \left (1\right )^{2} \tan \left (1\right )}{240}+\frac {t^{6} \sec \left (1\right )^{4} \tan \left (1\right )}{6}\right ) y^{\prime }\left (0\right )+O\left (t^{6}\right ) \] Since the expansion point \(t = 0\) is an ordinary, we can also solve this using standard power series Let the solution be represented as power series of the form \[ y \left (t \right ) = \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} t^{n} \] Then \begin {align*} \frac {d}{d t}y \left (t \right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}n a_{n} t^{n -1}\\ \frac {d^{2}}{d t^{2}}y \left (t \right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}n \left (n -1\right ) a_{n} t^{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} t^{n -2} = \tan \left (t +1\right ) \left (\moverset {\infty }{\munderset {n =1}{\sum }}n a_{n} t^{n -1}\right )-\left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} t^{n}\right )\tag {1} \end {align*}

Expanding \(-\tan \left (t +1\right )\) as Taylor series around \(t=0\) and keeping only the ﬁrst \(6\) terms gives \begin {align*} -\tan \left (t +1\right ) &= -\tan \left (1\right )+\left (-1-\tan \left (1\right )^{2}\right ) t -\left (1+\tan \left (1\right )^{2}\right ) \tan \left (1\right ) t^{2}+\left (-\frac {1}{3}-\frac {4 \tan \left (1\right )^{2}}{3}-\tan \left (1\right )^{4}\right ) t^{3}+\left (-\frac {5 \tan \left (1\right )^{3}}{3}-\frac {2 \tan \left (1\right )}{3}-\tan \left (1\right )^{5}\right ) t^{4}+\left (-\frac {2}{15}-\frac {17 \tan \left (1\right )^{2}}{15}-2 \tan \left (1\right )^{4}-\tan \left (1\right )^{6}\right ) t^{5}+\left (-\frac {77 \tan \left (1\right )^{3}}{45}-\frac {17 \tan \left (1\right )}{45}-\frac {7 \tan \left (1\right )^{5}}{3}-\tan \left (1\right )^{7}\right ) t^{6} + \dots \\ &= -\tan \left (1\right )+\left (-1-\tan \left (1\right )^{2}\right ) t -\left (1+\tan \left (1\right )^{2}\right ) \tan \left (1\right ) t^{2}+\left (-\frac {1}{3}-\frac {4 \tan \left (1\right )^{2}}{3}-\tan \left (1\right )^{4}\right ) t^{3}+\left (-\frac {5 \tan \left (1\right )^{3}}{3}-\frac {2 \tan \left (1\right )}{3}-\tan \left (1\right )^{5}\right ) t^{4}+\left (-\frac {2}{15}-\frac {17 \tan \left (1\right )^{2}}{15}-2 \tan \left (1\right )^{4}-\tan \left (1\right )^{6}\right ) t^{5}+\left (-\frac {77 \tan \left (1\right )^{3}}{45}-\frac {17 \tan \left (1\right )}{45}-\frac {7 \tan \left (1\right )^{5}}{3}-\tan \left (1\right )^{7}\right ) t^{6} \end {align*}

Hence the ODE in Eq (1) becomes \[ \left (\moverset {\infty }{\munderset {n =2}{\sum }}n \left (n -1\right ) a_{n} t^{n -2}\right )+\left (-\tan \left (1\right )+\left (-1-\tan \left (1\right )^{2}\right ) t -\left (1+\tan \left (1\right )^{2}\right ) \tan \left (1\right ) t^{2}+\left (-\frac {1}{3}-\frac {4 \tan \left (1\right )^{2}}{3}-\tan \left (1\right )^{4}\right ) t^{3}+\left (-\frac {5 \tan \left (1\right )^{3}}{3}-\frac {2 \tan \left (1\right )}{3}-\tan \left (1\right )^{5}\right ) t^{4}+\left (-\frac {2}{15}-\frac {17 \tan \left (1\right )^{2}}{15}-2 \tan \left (1\right )^{4}-\tan \left (1\right )^{6}\right ) t^{5}+\left (-\frac {77 \tan \left (1\right )^{3}}{45}-\frac {17 \tan \left (1\right )}{45}-\frac {7 \tan \left (1\right )^{5}}{3}-\tan \left (1\right )^{7}\right ) t^{6}\right ) \left (\moverset {\infty }{\munderset {n =1}{\sum }}n a_{n} t^{n -1}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} t^{n}\right ) = 0 \] Expanding the second term in (1) gives \[ \left (\moverset {\infty }{\munderset {n =2}{\sum }}n \left (n -1\right ) a_{n} t^{n -2}\right )-\tan \left (1\right )\eslowast \left (\moverset {\infty }{\munderset {n =1}{\sum }}n a_{n} t^{n -1}\right )+\left (-1-\tan \left (1\right )^{2}\right ) t \eslowast \left (\moverset {\infty }{\munderset {n =1}{\sum }}n a_{n} t^{n -1}\right )-\left (1+\tan \left (1\right )^{2}\right ) \tan \left (1\right ) t^{2}\eslowast \left (\moverset {\infty }{\munderset {n =1}{\sum }}n a_{n} t^{n -1}\right )+\left (-\frac {1}{3}-\frac {4 \tan \left (1\right )^{2}}{3}-\tan \left (1\right )^{4}\right ) t^{3}\eslowast \left (\moverset {\infty }{\munderset {n =1}{\sum }}n a_{n} t^{n -1}\right )+\left (-\frac {5 \tan \left (1\right )^{3}}{3}-\frac {2 \tan \left (1\right )}{3}-\tan \left (1\right )^{5}\right ) t^{4}\eslowast \left (\moverset {\infty }{\munderset {n =1}{\sum }}n a_{n} t^{n -1}\right )+\left (-\frac {2}{15}-\frac {17 \tan \left (1\right )^{2}}{15}-2 \tan \left (1\right )^{4}-\tan \left (1\right )^{6}\right ) t^{5}\eslowast \left (\moverset {\infty }{\munderset {n =1}{\sum }}n a_{n} t^{n -1}\right )+\left (-\frac {77 \tan \left (1\right )^{3}}{45}-\frac {17 \tan \left (1\right )}{45}-\frac {7 \tan \left (1\right )^{5}}{3}-\tan \left (1\right )^{7}\right ) t^{6}\eslowast \left (\moverset {\infty }{\munderset {n =1}{\sum }}n a_{n} t^{n -1}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} t^{n}\right ) = 0 \] Which simpliﬁes to \begin{equation} \tag{2} \left (\moverset {\infty }{\munderset {n =2}{\sum }}n \left (n -1\right ) a_{n} t^{n -2}\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-\frac {n \,t^{n +5} a_{n} \tan \left (1\right ) \sec \left (1\right )^{6} \left (123+\cos \left (4\right )-56 \cos \left (2\right )\right )}{180}\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-n \,t^{n +4} \left (\sec \left (1\right )^{4}-\sec \left (1\right )^{2}+\frac {2}{15}\right ) a_{n} \sec \left (1\right )^{2}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}\frac {n \,t^{n +3} a_{n} \tan \left (1\right ) \sec \left (1\right )^{4} \left (-5+\cos \left (2\right )\right )}{6}\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-n \,t^{n +2} \left (\sec \left (1\right )^{2}-\frac {2}{3}\right ) a_{n} \sec \left (1\right )^{2}\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-n \,t^{1+n} a_{n} \tan \left (1\right ) \sec \left (1\right )^{2}\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-n a_{n} t^{n} \sec \left (1\right )^{2}\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-n \,t^{n -1} a_{n} \tan \left (1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} t^{n}\right ) = 0 \end{equation} The next step is to make all powers of \(t\) be \(n\) in each summation term. Going over each summation term above with power of \(t\) in it which is not already \(t^{n}\) and adjusting the power and the corresponding index gives \begin{align*} \moverset {\infty }{\munderset {n =2}{\sum }}n \left (n -1\right ) a_{n} t^{n -2} &= \moverset {\infty }{\munderset {n =0}{\sum }}\left (n +2\right ) a_{n +2} \left (1+n \right ) t^{n} \\ \moverset {\infty }{\munderset {n =1}{\sum }}\left (-\frac {n \,t^{n +5} a_{n} \tan \left (1\right ) \sec \left (1\right )^{6} \left (123+\cos \left (4\right )-56 \cos \left (2\right )\right )}{180}\right ) &= \moverset {\infty }{\munderset {n =6}{\sum }}\left (-\frac {\left (n -5\right ) a_{n -5} \sec \left (1\right )^{6} \left (123+\cos \left (4\right )-56 \cos \left (2\right )\right ) \tan \left (1\right ) t^{n}}{180}\right ) \\ \moverset {\infty }{\munderset {n =1}{\sum }}\left (-n \,t^{n +4} \left (\sec \left (1\right )^{4}-\sec \left (1\right )^{2}+\frac {2}{15}\right ) a_{n} \sec \left (1\right )^{2}\right ) &= \moverset {\infty }{\munderset {n =5}{\sum }}\left (-\left (n -4\right ) a_{n -4} \sec \left (1\right )^{2} \left (\sec \left (1\right )^{4}-\sec \left (1\right )^{2}+\frac {2}{15}\right ) t^{n}\right ) \\ \moverset {\infty }{\munderset {n =1}{\sum }}\frac {n \,t^{n +3} a_{n} \tan \left (1\right ) \sec \left (1\right )^{4} \left (-5+\cos \left (2\right )\right )}{6} &= \moverset {\infty }{\munderset {n =4}{\sum }}\frac {\left (n -3\right ) a_{n -3} \sec \left (1\right )^{4} \left (-5+\cos \left (2\right )\right ) \tan \left (1\right ) t^{n}}{6} \\ \moverset {\infty }{\munderset {n =1}{\sum }}\left (-n \,t^{n +2} \left (\sec \left (1\right )^{2}-\frac {2}{3}\right ) a_{n} \sec \left (1\right )^{2}\right ) &= \moverset {\infty }{\munderset {n =3}{\sum }}\left (-\left (n -2\right ) a_{n -2} \sec \left (1\right )^{2} \left (\sec \left (1\right )^{2}-\frac {2}{3}\right ) t^{n}\right ) \\ \moverset {\infty }{\munderset {n =1}{\sum }}\left (-n \,t^{1+n} a_{n} \tan \left (1\right ) \sec \left (1\right )^{2}\right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}\left (-\left (n -1\right ) a_{n -1} \sec \left (1\right )^{2} \tan \left (1\right ) t^{n}\right ) \\ \moverset {\infty }{\munderset {n =1}{\sum }}\left (-n \,t^{n -1} a_{n} \tan \left (1\right )\right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}\left (-\left (1+n \right ) a_{1+n} \tan \left (1\right ) t^{n}\right ) \\ \end{align*} Substituting all the above in Eq (2) gives the following equation where now all powers of \(t\) are the same and equal to \(n\). \begin{equation} \tag{3} \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +2\right ) a_{n +2} \left (1+n \right ) t^{n}\right )+\moverset {\infty }{\munderset {n =6}{\sum }}\left (-\frac {\left (n -5\right ) a_{n -5} \sec \left (1\right )^{6} \left (123+\cos \left (4\right )-56 \cos \left (2\right )\right ) \tan \left (1\right ) t^{n}}{180}\right )+\moverset {\infty }{\munderset {n =5}{\sum }}\left (-\left (n -4\right ) a_{n -4} \sec \left (1\right )^{2} \left (\sec \left (1\right )^{4}-\sec \left (1\right )^{2}+\frac {2}{15}\right ) t^{n}\right )+\left (\moverset {\infty }{\munderset {n =4}{\sum }}\frac {\left (n -3\right ) a_{n -3} \sec \left (1\right )^{4} \left (-5+\cos \left (2\right )\right ) \tan \left (1\right ) t^{n}}{6}\right )+\moverset {\infty }{\munderset {n =3}{\sum }}\left (-\left (n -2\right ) a_{n -2} \sec \left (1\right )^{2} \left (\sec \left (1\right )^{2}-\frac {2}{3}\right ) t^{n}\right )+\moverset {\infty }{\munderset {n =2}{\sum }}\left (-\left (n -1\right ) a_{n -1} \sec \left (1\right )^{2} \tan \left (1\right ) t^{n}\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-n a_{n} t^{n} \sec \left (1\right )^{2}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-\left (1+n \right ) a_{1+n} \tan \left (1\right ) t^{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} t^{n}\right ) = 0 \end{equation} \(n=0\) gives \[ 2 a_{2}-a_{1} \tan \left (1\right )+a_{0}=0 \] \[ a_{2} = -\frac {a_{0}}{2}+\frac {a_{1} \tan \left (1\right )}{2} \] \(n=1\) gives \[ 6 a_{3}-a_{1} \sec \left (1\right )^{2}-2 a_{2} \tan \left (1\right )+a_{1}=0 \] Which after substituting earlier equations, simpliﬁes to \[ a_{3} = -\frac {\tan \left (1\right ) a_{0}}{6}+\frac {a_{1} \tan \left (1\right )^{2}}{6}+\frac {a_{1} \sec \left (1\right )^{2}}{6}-\frac {a_{1}}{6} \] \(n=2\) gives \[ 12 a_{4}-a_{1} \sec \left (1\right )^{2} \tan \left (1\right )-2 a_{2} \sec \left (1\right )^{2}-3 a_{3} \tan \left (1\right )+a_{2}=0 \] Which after substituting earlier equations, simpliﬁes to \[ a_{4} = -\frac {\sec \left (1\right )^{2} a_{0}}{12}-\frac {\tan \left (1\right )^{2} a_{0}}{24}+\frac {a_{0}}{24}+\frac {5 a_{1} \sec \left (1\right )^{2} \tan \left (1\right )}{24}+\frac {a_{1} \tan \left (1\right )^{3}}{24}-\frac {a_{1} \tan \left (1\right )}{12} \] \(n=3\) gives \[ 20 a_{5}-a_{1} \sec \left (1\right )^{2} \left (\sec \left (1\right )^{2}-\frac {2}{3}\right )-2 a_{2} \sec \left (1\right )^{2} \tan \left (1\right )-3 a_{3} \sec \left (1\right )^{2}-4 a_{4} \tan \left (1\right )+a_{3}=0 \] Which after substituting earlier equations, simpliﬁes to \[ a_{5} = -\frac {11 \sec \left (1\right )^{2} \tan \left (1\right ) a_{0}}{120}-\frac {\tan \left (1\right )^{3} a_{0}}{120}+\frac {\tan \left (1\right ) a_{0}}{60}+\frac {7 \sec \left (1\right )^{2} \tan \left (1\right )^{2} a_{1}}{60}+\frac {3 a_{1} \sec \left (1\right )^{4}}{40}-\frac {a_{1} \sec \left (1\right )^{2}}{15}+\frac {a_{1} \tan \left (1\right )^{4}}{120}-\frac {a_{1} \tan \left (1\right )^{2}}{40}+\frac {a_{1}}{120} \] \(n=4\) gives \[ 30 a_{6}+\frac {a_{1} \sec \left (1\right )^{4} \left (-5+\cos \left (2\right )\right ) \tan \left (1\right )}{6}-2 a_{2} \sec \left (1\right )^{2} \left (\sec \left (1\right )^{2}-\frac {2}{3}\right )-3 a_{3} \sec \left (1\right )^{2} \tan \left (1\right )-4 a_{4} \sec \left (1\right )^{2}-5 a_{5} \tan \left (1\right )+a_{4}=0 \] Which after substituting earlier equations, simpliﬁes to \[ a_{6} = -\frac {a_{0}}{720}+\frac {\tan \left (1\right )^{2} a_{0}}{240}-\frac {2 \sec \left (1\right )^{4} a_{0}}{45}-\frac {\tan \left (1\right )^{4} a_{0}}{720}+\frac {11 \sec \left (1\right )^{2} a_{0}}{360}-\frac {3 \sec \left (1\right )^{2} \tan \left (1\right )^{2} a_{0}}{80}+\frac {a_{1} \tan \left (1\right )}{240}-\frac {a_{1} \tan \left (1\right )^{3}}{180}+\frac {a_{1} \tan \left (1\right )^{5}}{720}-\frac {49 a_{1} \sec \left (1\right )^{2} \tan \left (1\right )}{720}+\frac {17 a_{1} \sec \left (1\right )^{4} \tan \left (1\right )}{144}+\frac {\sec \left (1\right )^{2} \tan \left (1\right )^{3} a_{1}}{24}-\frac {a_{1} \sec \left (1\right )^{4} \tan \left (1\right ) \cos \left (2\right )}{180} \] \(n=5\) gives \[ 42 a_{7}-a_{1} \sec \left (1\right )^{2} \left (\sec \left (1\right )^{4}-\sec \left (1\right )^{2}+\frac {2}{15}\right )+\frac {a_{2} \sec \left (1\right )^{4} \left (-5+\cos \left (2\right )\right ) \tan \left (1\right )}{3}-3 a_{3} \sec \left (1\right )^{2} \left (\sec \left (1\right )^{2}-\frac {2}{3}\right )-4 a_{4} \sec \left (1\right )^{2} \tan \left (1\right )-5 a_{5} \sec \left (1\right )^{2}-6 a_{6} \tan \left (1\right )+a_{5}=0 \] Which after substituting earlier equations, simpliﬁes to \[ a_{7} = \frac {103 \sec \left (1\right )^{2} \tan \left (1\right ) a_{0}}{5040}-\frac {41 \sec \left (1\right )^{4} \tan \left (1\right ) a_{0}}{720}-\frac {13 \sec \left (1\right )^{2} \tan \left (1\right )^{3} a_{0}}{1260}-\frac {\tan \left (1\right ) a_{0}}{1680}+\frac {\tan \left (1\right )^{3} a_{0}}{1260}-\frac {\tan \left (1\right )^{5} a_{0}}{5040}+\frac {\sec \left (1\right )^{4} \tan \left (1\right ) a_{0} \cos \left (2\right )}{252}+\frac {23 a_{1} \sec \left (1\right )^{2}}{1680}-\frac {a_{1}}{5040}-\frac {79 \sec \left (1\right )^{2} \tan \left (1\right )^{2} a_{1}}{2520}+\frac {83 \sec \left (1\right )^{4} \tan \left (1\right )^{2} a_{1}}{1008}+\frac {11 \sec \left (1\right )^{2} \tan \left (1\right )^{4} a_{1}}{1008}-\frac {269 a_{1} \sec \left (1\right )^{4}}{5040}+\frac {5 a_{1} \sec \left (1\right )^{6}}{112}+\frac {a_{1} \tan \left (1\right )^{2}}{840}-\frac {a_{1} \tan \left (1\right )^{4}}{1008}+\frac {a_{1} \tan \left (1\right )^{6}}{5040}-\frac {\sec \left (1\right )^{4} \tan \left (1\right )^{2} a_{1} \cos \left (2\right )}{210} \] For \(6\le n\), the recurrence equation is \begin{equation} \tag{4} \left (n +2\right ) a_{n +2} \left (1+n \right )-\frac {\left (n -5\right ) a_{n -5} \sec \left (1\right )^{6} \left (123+\cos \left (4\right )-56 \cos \left (2\right )\right ) \tan \left (1\right )}{180}-\left (n -4\right ) a_{n -4} \sec \left (1\right )^{2} \left (\sec \left (1\right )^{4}-\sec \left (1\right )^{2}+\frac {2}{15}\right )+\frac {\left (n -3\right ) a_{n -3} \sec \left (1\right )^{4} \left (-5+\cos \left (2\right )\right ) \tan \left (1\right )}{6}-\left (n -2\right ) a_{n -2} \sec \left (1\right )^{2} \left (\sec \left (1\right )^{2}-\frac {2}{3}\right )-\left (n -1\right ) a_{n -1} \sec \left (1\right )^{2} \tan \left (1\right )-n a_{n} \sec \left (1\right )^{2}-\left (1+n \right ) a_{1+n} \tan \left (1\right )+a_{n} = 0 \end{equation} Solving for \(a_{n +2}\), gives \begin{align*} \tag{5} a_{n +2}&= \frac {180 a_{n -4} \sec \left (1\right )^{6} n -180 a_{n -4} \sec \left (1\right )^{4} n +24 a_{n -4} \sec \left (1\right )^{2} n +180 a_{n -2} \sec \left (1\right )^{4} n -120 a_{n -2} \sec \left (1\right )^{2} n +180 a_{1+n} \tan \left (1\right ) n -180 a_{n -1} \sec \left (1\right )^{2} \tan \left (1\right )-615 a_{n -5} \sec \left (1\right )^{6} \tan \left (1\right )-450 a_{n -3} \sec \left (1\right )^{4} \tan \left (1\right )+180 a_{n -1} \sec \left (1\right )^{2} \tan \left (1\right ) n -5 \tan \left (1\right ) \cos \left (4\right ) \sec \left (1\right )^{6} a_{n -5}+280 \tan \left (1\right ) \cos \left (2\right ) \sec \left (1\right )^{6} a_{n -5}+123 \tan \left (1\right ) \sec \left (1\right )^{6} n a_{n -5}+90 \tan \left (1\right ) \cos \left (2\right ) \sec \left (1\right )^{4} a_{n -3}+150 \tan \left (1\right ) \sec \left (1\right )^{4} n a_{n -3}+180 n a_{n} \sec \left (1\right )^{2}+\tan \left (1\right ) \cos \left (4\right ) \sec \left (1\right )^{6} n a_{n -5}-56 \tan \left (1\right ) \cos \left (2\right ) \sec \left (1\right )^{6} n a_{n -5}-30 \tan \left (1\right ) \cos \left (2\right ) \sec \left (1\right )^{4} n a_{n -3}-180 a_{n}+180 a_{1+n} \tan \left (1\right )+240 a_{n -2} \sec \left (1\right )^{2}-96 a_{n -4} \sec \left (1\right )^{2}-720 a_{n -4} \sec \left (1\right )^{6}+720 a_{n -4} \sec \left (1\right )^{4}-360 a_{n -2} \sec \left (1\right )^{4}}{180 \left (n +2\right ) \left (1+n \right )} \\ &= \frac {\left (180 n \sec \left (1\right )^{2}-180\right ) a_{n}}{180 \left (n +2\right ) \left (1+n \right )}+\frac {\left (180 \tan \left (1\right ) n +180 \tan \left (1\right )\right ) a_{1+n}}{180 \left (n +2\right ) \left (1+n \right )}+\frac {\left (\tan \left (1\right ) \cos \left (4\right ) \sec \left (1\right )^{6} n -56 \tan \left (1\right ) \cos \left (2\right ) \sec \left (1\right )^{6} n -5 \tan \left (1\right ) \sec \left (1\right )^{6} \cos \left (4\right )+280 \tan \left (1\right ) \sec \left (1\right )^{6} \cos \left (2\right )+123 \tan \left (1\right ) \sec \left (1\right )^{6} n -615 \tan \left (1\right ) \sec \left (1\right )^{6}\right ) a_{n -5}}{180 \left (n +2\right ) \left (1+n \right )}+\frac {\left (180 \sec \left (1\right )^{6} n -720 \sec \left (1\right )^{6}-180 n \sec \left (1\right )^{4}+720 \sec \left (1\right )^{4}+24 n \sec \left (1\right )^{2}-96 \sec \left (1\right )^{2}\right ) a_{n -4}}{180 \left (n +2\right ) \left (1+n \right )}+\frac {\left (-30 \tan \left (1\right ) \cos \left (2\right ) \sec \left (1\right )^{4} n +90 \sec \left (1\right )^{4} \tan \left (1\right ) \cos \left (2\right )+150 \tan \left (1\right ) \sec \left (1\right )^{4} n -450 \sec \left (1\right )^{4} \tan \left (1\right )\right ) a_{n -3}}{180 \left (n +2\right ) \left (1+n \right )}+\frac {\left (180 n \sec \left (1\right )^{4}-360 \sec \left (1\right )^{4}-120 n \sec \left (1\right )^{2}+240 \sec \left (1\right )^{2}\right ) a_{n -2}}{180 \left (n +2\right ) \left (1+n \right )}+\frac {\left (180 \tan \left (1\right ) \sec \left (1\right )^{2} n -180 \sec \left (1\right )^{2} \tan \left (1\right )\right ) a_{n -1}}{180 \left (n +2\right ) \left (1+n \right )} \\ \end{align*} And so on. Therefore the solution is \begin {align*} y \left (t \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} t^{n}\\ &= a_{3} t^{3}+a_{2} t^{2}+a_{1} t +a_{0} + \dots \end {align*}

Substituting the values for \(a_{n}\) found above, the solution becomes \[ y \left (t \right ) = a_{0}+a_{1} t +\left (-\frac {a_{0}}{2}+\frac {a_{1} \tan \left (1\right )}{2}\right ) t^{2}+\left (-\frac {\tan \left (1\right ) a_{0}}{6}+\frac {a_{1} \tan \left (1\right )^{2}}{6}+\frac {a_{1} \sec \left (1\right )^{2}}{6}-\frac {a_{1}}{6}\right ) t^{3}+\left (-\frac {\sec \left (1\right )^{2} a_{0}}{12}-\frac {\tan \left (1\right )^{2} a_{0}}{24}+\frac {a_{0}}{24}+\frac {5 a_{1} \sec \left (1\right )^{2} \tan \left (1\right )}{24}+\frac {a_{1} \tan \left (1\right )^{3}}{24}-\frac {a_{1} \tan \left (1\right )}{12}\right ) t^{4}+\left (-\frac {11 \sec \left (1\right )^{2} \tan \left (1\right ) a_{0}}{120}-\frac {\tan \left (1\right )^{3} a_{0}}{120}+\frac {\tan \left (1\right ) a_{0}}{60}+\frac {7 \sec \left (1\right )^{2} \tan \left (1\right )^{2} a_{1}}{60}+\frac {3 a_{1} \sec \left (1\right )^{4}}{40}-\frac {a_{1} \sec \left (1\right )^{2}}{15}+\frac {a_{1} \tan \left (1\right )^{4}}{120}-\frac {a_{1} \tan \left (1\right )^{2}}{40}+\frac {a_{1}}{120}\right ) t^{5}+\dots \] Collecting terms, the solution becomes \begin{equation} \tag{3} y \left (t \right ) = \left (1-\frac {t^{2}}{2}-\frac {t^{3} \tan \left (1\right )}{6}+\left (-\frac {\sec \left (1\right )^{2}}{12}-\frac {\tan \left (1\right )^{2}}{24}+\frac {1}{24}\right ) t^{4}+\left (-\frac {11 \sec \left (1\right )^{2} \tan \left (1\right )}{120}-\frac {\tan \left (1\right )^{3}}{120}+\frac {\tan \left (1\right )}{60}\right ) t^{5}\right ) a_{0}+\left (t +\frac {t^{2} \tan \left (1\right )}{2}+\left (\frac {\tan \left (1\right )^{2}}{6}+\frac {\sec \left (1\right )^{2}}{6}-\frac {1}{6}\right ) t^{3}+\left (\frac {5 \sec \left (1\right )^{2} \tan \left (1\right )}{24}+\frac {\tan \left (1\right )^{3}}{24}-\frac {\tan \left (1\right )}{12}\right ) t^{4}+\left (\frac {7 \sec \left (1\right )^{2} \tan \left (1\right )^{2}}{60}+\frac {3 \sec \left (1\right )^{4}}{40}-\frac {\sec \left (1\right )^{2}}{15}+\frac {\tan \left (1\right )^{4}}{120}-\frac {\tan \left (1\right )^{2}}{40}+\frac {1}{120}\right ) t^{5}\right ) a_{1}+O\left (t^{6}\right ) \end{equation} At \(t = 0\) the solution above becomes \[ y \left (t \right ) = \left (1-\frac {t^{2}}{2}-\frac {t^{3} \tan \left (1\right )}{6}+\left (-\frac {\sec \left (1\right )^{2}}{12}-\frac {\tan \left (1\right )^{2}}{24}+\frac {1}{24}\right ) t^{4}+\left (-\frac {11 \sec \left (1\right )^{2} \tan \left (1\right )}{120}-\frac {\tan \left (1\right )^{3}}{120}+\frac {\tan \left (1\right )}{60}\right ) t^{5}\right ) c_{1} +\left (t +\frac {t^{2} \tan \left (1\right )}{2}+\left (\frac {\tan \left (1\right )^{2}}{6}+\frac {\sec \left (1\right )^{2}}{6}-\frac {1}{6}\right ) t^{3}+\left (\frac {5 \sec \left (1\right )^{2} \tan \left (1\right )}{24}+\frac {\tan \left (1\right )^{3}}{24}-\frac {\tan \left (1\right )}{12}\right ) t^{4}+\left (\frac {7 \sec \left (1\right )^{2} \tan \left (1\right )^{2}}{60}+\frac {3 \sec \left (1\right )^{4}}{40}-\frac {\sec \left (1\right )^{2}}{15}+\frac {\tan \left (1\right )^{4}}{120}-\frac {\tan \left (1\right )^{2}}{40}+\frac {1}{120}\right ) t^{5}\right ) c_{2} +O\left (t^{6}\right ) \] Replacing \(t\) in the above with the original independent variable \(xs\)using \(t = x -1\) results in \[ y = \left (1-\frac {\left (x -1\right )^{2}}{2}-\frac {\left (x -1\right )^{3} \tan \left (1\right )}{6}-\frac {\left (x -1\right )^{4} \sec \left (1\right )^{2}}{8}+\frac {\left (x -1\right )^{4}}{12}-\frac {\left (x -1\right )^{5} \sec \left (1\right )^{2} \tan \left (1\right )}{10}+\frac {\left (x -1\right )^{5} \tan \left (1\right )}{40}-\frac {\left (x -1\right )^{6} \sec \left (1\right )^{4}}{12}+\frac {3 \left (x -1\right )^{6} \sec \left (1\right )^{2}}{40}-\frac {\left (x -1\right )^{6}}{144}\right ) y \left (1\right )+\left (x -1+\frac {\left (x -1\right )^{2} \tan \left (1\right )}{2}+\frac {\left (x -1\right )^{3} \tan \left (1\right )^{2}}{3}+\frac {\left (x -1\right )^{4} \sec \left (1\right )^{2} \tan \left (1\right )}{4}-\frac {\left (x -1\right )^{4} \tan \left (1\right )}{8}+\frac {\left (x -1\right )^{5} \sec \left (1\right )^{4}}{5}-\frac {9 \left (x -1\right )^{5} \sec \left (1\right )^{2}}{40}+\frac {\left (x -1\right )^{5}}{24}+\frac {\left (x -1\right )^{6} \tan \left (1\right )}{90}-\frac {31 \left (x -1\right )^{6} \sec \left (1\right )^{2} \tan \left (1\right )}{240}+\frac {\left (x -1\right )^{6} \sec \left (1\right )^{4} \tan \left (1\right )}{6}\right ) y^{\prime }\left (1\right )+O\left (\left (x -1\right )^{6}\right ) \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \left (1-\frac {\left (x -1\right )^{2}}{2}-\frac {\left (x -1\right )^{3} \tan \left (1\right )}{6}-\frac {\left (x -1\right )^{4} \sec \left (1\right )^{2}}{8}+\frac {\left (x -1\right )^{4}}{12}-\frac {\left (x -1\right )^{5} \sec \left (1\right )^{2} \tan \left (1\right )}{10}+\frac {\left (x -1\right )^{5} \tan \left (1\right )}{40}-\frac {\left (x -1\right )^{6} \sec \left (1\right )^{4}}{12}+\frac {3 \left (x -1\right )^{6} \sec \left (1\right )^{2}}{40}-\frac {\left (x -1\right )^{6}}{144}\right ) y \left (1\right )+\left (x -1+\frac {\left (x -1\right )^{2} \tan \left (1\right )}{2}+\frac {\left (x -1\right )^{3} \tan \left (1\right )^{2}}{3}+\frac {\left (x -1\right )^{4} \sec \left (1\right )^{2} \tan \left (1\right )}{4}-\frac {\left (x -1\right )^{4} \tan \left (1\right )}{8}+\frac {\left (x -1\right )^{5} \sec \left (1\right )^{4}}{5}-\frac {9 \left (x -1\right )^{5} \sec \left (1\right )^{2}}{40}+\frac {\left (x -1\right )^{5}}{24}+\frac {\left (x -1\right )^{6} \tan \left (1\right )}{90}-\frac {31 \left (x -1\right )^{6} \sec \left (1\right )^{2} \tan \left (1\right )}{240}+\frac {\left (x -1\right )^{6} \sec \left (1\right )^{4} \tan \left (1\right )}{6}\right ) y^{\prime }\left (1\right )+O\left (\left (x -1\right )^{6}\right ) \\ \end{align*}

Veriﬁcation of solutions

\[ y = \left (1-\frac {\left (x -1\right )^{2}}{2}-\frac {\left (x -1\right )^{3} \tan \left (1\right )}{6}-\frac {\left (x -1\right )^{4} \sec \left (1\right )^{2}}{8}+\frac {\left (x -1\right )^{4}}{12}-\frac {\left (x -1\right )^{5} \sec \left (1\right )^{2} \tan \left (1\right )}{10}+\frac {\left (x -1\right )^{5} \tan \left (1\right )}{40}-\frac {\left (x -1\right )^{6} \sec \left (1\right )^{4}}{12}+\frac {3 \left (x -1\right )^{6} \sec \left (1\right )^{2}}{40}-\frac {\left (x -1\right )^{6}}{144}\right ) y \left (1\right )+\left (x -1+\frac {\left (x -1\right )^{2} \tan \left (1\right )}{2}+\frac {\left (x -1\right )^{3} \tan \left (1\right )^{2}}{3}+\frac {\left (x -1\right )^{4} \sec \left (1\right )^{2} \tan \left (1\right )}{4}-\frac {\left (x -1\right )^{4} \tan \left (1\right )}{8}+\frac {\left (x -1\right )^{5} \sec \left (1\right )^{4}}{5}-\frac {9 \left (x -1\right )^{5} \sec \left (1\right )^{2}}{40}+\frac {\left (x -1\right )^{5}}{24}+\frac {\left (x -1\right )^{6} \tan \left (1\right )}{90}-\frac {31 \left (x -1\right )^{6} \sec \left (1\right )^{2} \tan \left (1\right )}{240}+\frac {\left (x -1\right )^{6} \sec \left (1\right )^{4} \tan \left (1\right )}{6}\right ) y^{\prime }\left (1\right )+O\left (\left (x -1\right )^{6}\right ) \] 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 
      <- Legendre successful 
   <- special function solution successful 
   Change of variables used: 
      [x = arcsin(t)] 
   Linear ODE actually solved: 
      u(t)-2*t*diff(u(t),t)+(-t^2+1)*diff(diff(u(t),t),t) = 0 
<- change of variables successful`

\[ y \left (x \right ) = \left (1-\frac {\left (x -1\right )^{2}}{2}-\frac {\tan \left (1\right ) \left (x -1\right )^{3}}{6}+\left (\frac {1}{12}-\frac {\sec \left (1\right )^{2}}{8}\right ) \left (x -1\right )^{4}+\frac {\tan \left (1\right ) \left (1-4 \sec \left (1\right )^{2}\right ) \left (x -1\right )^{5}}{40}\right ) y \left (1\right )+\left (x -1+\frac {\tan \left (1\right ) \left (x -1\right )^{2}}{2}+\frac {\tan \left (1\right )^{2} \left (x -1\right )^{3}}{3}+\frac {\tan \left (1\right ) \left (2 \sec \left (1\right )^{2}-1\right ) \left (x -1\right )^{4}}{8}+\frac {\left (5-27 \sec \left (1\right )^{2}+24 \sec \left (1\right )^{4}\right ) \left (x -1\right )^{5}}{120}\right ) D\left (y \right )\left (1\right )+O\left (x^{6}\right ) \]

\[ y(x)\to c_1 \left (\frac {1}{24} (x-1)^4-\frac {1}{2} (x-1)^2+\frac {1}{20} (x-1)^5 \left (-\tan ^3(1)-\tan (1)\right )-\frac {1}{120} (x-1)^5 \tan ^3(1)-\frac {1}{40} (x-1)^5 \tan (1) \left (1+\tan ^2(1)\right )+\frac {1}{60} (x-1)^5 \tan (1) \left (-1-\tan ^2(1)\right )+\frac {1}{12} (x-1)^4 \left (-1-\tan ^2(1)\right )-\frac {1}{24} (x-1)^4 \tan ^2(1)+\frac {1}{60} (x-1)^5 \tan (1)-\frac {1}{6} (x-1)^3 \tan (1)+1\right )+c_2 \left (\frac {1}{120} (x-1)^5-\frac {1}{6} (x-1)^3+x+\frac {1}{120} (x-1)^5 \tan ^4(1)-\frac {1}{15} (x-1)^5 \tan (1) \left (-\tan ^3(1)-\tan (1)\right )-\frac {1}{12} (x-1)^4 \left (-\tan ^3(1)-\tan (1)\right )+\frac {1}{24} (x-1)^4 \tan ^3(1)-\frac {1}{40} (x-1)^5 \left (-1-\tan ^2(1)\right ) \left (1+\tan ^2(1)\right )+\frac {1}{40} (x-1)^5 \tan ^2(1) \left (1+\tan ^2(1)\right )-\frac {1}{40} (x-1)^5 \left (1+\tan ^2(1)\right )-\frac {1}{40} (x-1)^5 \tan ^2(1) \left (-1-\tan ^2(1)\right )+\frac {1}{120} (x-1)^5 \left (-1-\tan ^2(1)\right )-\frac {1}{40} (x-1)^5 \tan ^2(1)-\frac {1}{8} (x-1)^4 \tan (1) \left (-1-\tan ^2(1)\right )-\frac {1}{6} (x-1)^3 \left (-1-\tan ^2(1)\right )+\frac {1}{6} (x-1)^3 \tan ^2(1)-\frac {1}{60} (x-1)^5 \left (-1-3 \tan ^4(1)-4 \tan ^2(1)\right )-\frac {1}{12} (x-1)^4 \tan (1)+\frac {1}{2} (x-1)^2 \tan (1)-1\right ) \]

6.5 problem 5