23.6 problem 729

Internal problem ID [15472]
Internal file name [OUTPUT/15473_Wednesday_May_08_2024_04_01_05_PM_36847634/index.tex]

Book: A book of problems in ordinary differential equations. M.L. KRASNOV, A.L. KISELYOV, G.I. MARKARENKO. MIR, MOSCOW. 1983
Section: Chapter 2 (Higher order ODE’s). Section 18.1 Integration of differential equation in series. Power series. Exercises page 171
Problem number: 729.
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, _linear, _nonhomogeneous]]

\[ \boxed {x y^{\prime \prime }+\sin \left (x \right ) y=x} \] With initial conditions \begin {align*} [y \left (\pi \right ) = 1, y^{\prime }\left (\pi \right ) = 0] \end {align*}

With the expansion point for the power series method at \(x = \pi \).

23.6.1 Existence and uniqueness analysis

This is a linear ODE. In canonical form it is written as \begin {align*} y^{\prime \prime } + p(x)y^{\prime } + q(x) y &= F \end {align*}

Where here \begin {align*} p(x) &=0\\ q(x) &=\frac {\sin \left (x \right )}{x}\\ F &=1 \end {align*}

Hence the ode is \begin {align*} y^{\prime \prime }+\frac {\sin \left (x \right ) y}{x} = 1 \end {align*}

The domain of \(p(x)=0\) is \[ \{-\infty

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 \[ t = x -\pi \] The ode is converted to be in terms of the new independent variable \(t\). This results in \[ \left (t +\pi \right ) \left (\frac {d^{2}}{d t^{2}}y \left (t \right )\right )-\sin \left (t \right ) y \left (t \right ) = t +\pi \] With its expansion point and initial conditions now at \(t = 0\). With initial conditions now becoming \begin {align*} y(0) &= 1\\ y'(0) &= 0 \end {align*}

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 &= \frac {\sin \left (t \right ) y \left (t \right )+t +\pi }{t +\pi }\\ 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 \\ &= \frac {\sin \left (t \right ) \left (t +\pi \right ) \left (\frac {d}{d t}y \left (t \right )\right )+\left (-\sin \left (t \right )+\cos \left (t \right ) \left (t +\pi \right )\right ) y \left (t \right )}{\left (t +\pi \right )^{2}}\\ 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 \\ &= \frac {2 \left (-\sin \left (t \right )+\cos \left (t \right ) \left (t +\pi \right )\right ) \left (t +\pi \right ) \left (\frac {d}{d t}y \left (t \right )\right )+y \left (t \right ) \left (t +\pi \right ) \sin \left (t \right )^{2}+\left (\left (-\pi ^{2}-2 \pi t -t^{2}+2\right ) y \left (t \right )+\left (t +\pi \right )^{2}\right ) \sin \left (t \right )-2 y \left (t \right ) \left (t +\pi \right ) \cos \left (t \right )}{\left (t +\pi \right )^{3}}\\ 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 \\ &= \frac {-3 \left (\frac {\left (-t -\pi \right ) \sin \left (t \right )^{2}}{3}+\left (\pi ^{2}+2 \pi t +t^{2}-2\right ) \sin \left (t \right )+2 \cos \left (t \right ) \left (t +\pi \right )\right ) \left (t +\pi \right ) \left (\frac {d}{d t}y \left (t \right )\right )-4 y \left (t \right ) \left (t +\pi \right ) \sin \left (t \right )^{2}+\left (4 y \left (t \right ) \left (t +\pi \right )^{2} \cos \left (t \right )+3 \left (\pi ^{2}+2 \pi t +t^{2}-2\right ) y \left (t \right )-3 \left (t +\pi \right )^{2}\right ) \sin \left (t \right )-\left (\left (\pi ^{2}+2 \pi t +t^{2}-6\right ) y \left (t \right )-3 \left (t +\pi \right )^{2}\right ) \left (t +\pi \right ) \cos \left (t \right )}{\left (t +\pi \right )^{4}}\\ 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 \\ &= \frac {-4 \left (\frac {3 \left (t +\pi \right ) \sin \left (t \right )^{2}}{2}+3 \left (2-\frac {\left (t +\pi \right )^{2} \cos \left (t \right )}{2}-t^{2}-2 \pi t -\pi ^{2}\right ) \sin \left (t \right )+\cos \left (t \right ) \left (t +\pi \right ) \left (\pi ^{2}+2 \pi t +t^{2}-6\right )\right ) \left (t +\pi \right ) \left (\frac {d}{d t}y \left (t \right )\right )+y \left (t \right ) \left (t +\pi \right )^{2} \sin \left (t \right )^{3}-7 \left (\left (t^{2}+2 \pi t +\pi ^{2}-\frac {18}{7}\right ) y \left (t \right )-\frac {\left (t +\pi \right )^{2}}{7}\right ) \left (t +\pi \right ) \sin \left (t \right )^{2}+\left (\left (24-22 \left (t +\pi \right )^{2} \cos \left (t \right )+\pi ^{4}+4 \pi ^{3} t +6 \left (t^{2}-2\right ) \pi ^{2}+4 \left (t^{3}-6 t \right ) \pi +t^{4}-12 t^{2}\right ) y \left (t \right )-6 \left (\pi ^{2}+2 \pi t +t^{2}-2\right ) \left (t +\pi \right )^{2}\right ) \sin \left (t \right )+4 \left (\left (\left (t +\pi \right )^{2} \cos \left (t \right )+t^{2}+2 \pi t +\pi ^{2}-6\right ) y \left (t \right )-3 \left (t +\pi \right )^{2}\right ) \left (t +\pi \right ) \cos \left (t \right )}{\left (t +\pi \right )^{5}} \end {align*}

And so on. Evaluating all the above at initial conditions \(t = 0\) and \(y \left (0\right ) = 1\) and \(y^{\prime }\left (0\right ) = 0\) gives \begin {align*} F_0 &= 1\\ F_1 &= \frac {1}{\pi }\\ F_2 &= -\frac {2}{\pi ^{2}}\\ F_3 &= \frac {2 \pi ^{2}+6}{\pi ^{3}}\\ F_4 &= \frac {-4 \pi ^{2}-24}{\pi ^{4}} \end {align*}

Substituting all the above in (7) and simplifying gives the solution as \[ y \left (t \right ) = 1+\frac {t^{2}}{2}+\frac {t^{3}}{6 \pi }-\frac {t^{4}}{12 \pi ^{2}}+\frac {t^{5}}{60 \pi }+\frac {t^{5}}{20 \pi ^{3}}-\frac {t^{6}}{180 \pi ^{2}}-\frac {t^{6}}{30 \pi ^{4}}+O\left (t^{6}\right ) \] \[ y \left (t \right ) = 1+\frac {t^{2}}{2}+\frac {t^{3}}{6 \pi }-\frac {t^{4}}{12 \pi ^{2}}+\frac {t^{5}}{60 \pi }+\frac {t^{5}}{20 \pi ^{3}}-\frac {t^{6}}{180 \pi ^{2}}-\frac {t^{6}}{30 \pi ^{4}}+O\left (t^{6}\right ) \] Since the expansion point \(t = 0\) is an ordinary, we can also solve this using standard power series The ode is normalized to be \[ \left (t +\pi \right ) \left (\frac {d^{2}}{d t^{2}}y \left (t \right )\right )-\sin \left (t \right ) y \left (t \right ) = t +\pi \] 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*} \left (t +\pi \right ) \left (\moverset {\infty }{\munderset {n =2}{\sum }}n \left (n -1\right ) a_{n} t^{n -2}\right )-\sin \left (t \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} t^{n}\right ) = t +\pi \tag {1} \end {align*}

Expanding \(-\sin \left (t \right )\) as Taylor series around \(t=0\) and keeping only the first \(6\) terms gives \begin {align*} -\sin \left (t \right ) &= -t +\frac {1}{6} t^{3}-\frac {1}{120} t^{5}+\frac {1}{5040} t^{7} + \dots \\ &= -t +\frac {1}{6} t^{3}-\frac {1}{120} t^{5}+\frac {1}{5040} t^{7} \end {align*}

Expanding \(t +\pi \) as Taylor series around \(t=0\) and keeping only the first \(6\) terms gives \begin {align*} t +\pi &= t +\pi + \dots \\ &= t +\pi \end {align*}

Hence the ODE in Eq (1) becomes \[ \left (t +\pi \right ) \left (\moverset {\infty }{\munderset {n =2}{\sum }}n \left (n -1\right ) a_{n} t^{n -2}\right )+\left (-t +\frac {1}{6} t^{3}-\frac {1}{120} t^{5}+\frac {1}{5040} t^{7}\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} t^{n}\right ) = t +\pi \] Expanding the second term in (1) gives \[ \left (t +\pi \right ) \left (\moverset {\infty }{\munderset {n =2}{\sum }}n \left (n -1\right ) a_{n} t^{n -2}\right )-t \eslowast \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} t^{n}\right )+\frac {t^{3}}{6}\eslowast \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} t^{n}\right )-\frac {t^{5}}{120}\eslowast \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} t^{n}\right )+\frac {t^{7}}{5040}\eslowast \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} t^{n}\right ) = t +\pi \] Which simplifies to \begin{equation} \tag{2} \left (\moverset {\infty }{\munderset {n =2}{\sum }}n \,t^{n -1} a_{n} \left (n -1\right )\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}n \pi \,t^{n -2} a_{n} \left (n -1\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-t^{1+n} a_{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {t^{n +3} a_{n}}{6}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {t^{n +5} a_{n}}{120}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {t^{n +7} a_{n}}{5040}\right ) = t +\pi \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 \,t^{n -1} a_{n} \left (n -1\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (1+n \right ) a_{1+n} n \,t^{n} \\ \moverset {\infty }{\munderset {n =2}{\sum }}n \pi \,t^{n -2} a_{n} \left (n -1\right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}\pi \left (n +2\right ) a_{n +2} \left (1+n \right ) t^{n} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-t^{1+n} a_{n}\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-a_{n -1} t^{n}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\frac {t^{n +3} a_{n}}{6} &= \moverset {\infty }{\munderset {n =3}{\sum }}\frac {a_{n -3} t^{n}}{6} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {t^{n +5} a_{n}}{120}\right ) &= \moverset {\infty }{\munderset {n =5}{\sum }}\left (-\frac {a_{n -5} t^{n}}{120}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\frac {t^{n +7} a_{n}}{5040} &= \moverset {\infty }{\munderset {n =7}{\sum }}\frac {a_{n -7} t^{n}}{5040} \\ \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 =1}{\sum }}\left (1+n \right ) a_{1+n} n \,t^{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\pi \left (n +2\right ) a_{n +2} \left (1+n \right ) t^{n}\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-a_{n -1} t^{n}\right )+\left (\moverset {\infty }{\munderset {n =3}{\sum }}\frac {a_{n -3} t^{n}}{6}\right )+\moverset {\infty }{\munderset {n =5}{\sum }}\left (-\frac {a_{n -5} t^{n}}{120}\right )+\left (\moverset {\infty }{\munderset {n =7}{\sum }}\frac {a_{n -7} t^{n}}{5040}\right ) = t +\pi \end{equation} \(n=0\) gives \begin{align*} \left (2 \pi a_{2}\right ) 1 &= \pi \\ 2 \pi a_{2} &= \pi \\ \end{align*} Or \[ a_{2} = {\frac {1}{2}} \] \(n=1\) gives \begin{align*} \left (6 \pi a_{3}-a_{0}+2 a_{2}\right ) t &= t \\ 6 \pi a_{3}-a_{0}+2 a_{2} &= 1 \\ \end{align*} Which after substituting earlier equations, simplifies to \[ a_{3} = \frac {a_{0}}{6 \pi } \] \(n=2\) gives \[ 12 \pi a_{4}-a_{1}+6 a_{3}=0 \] Which after substituting earlier equations, simplifies to \[ a_{4} = \frac {\pi a_{1}-a_{0}}{12 \pi ^{2}} \] \(n=3\) gives \[ 20 \pi a_{5}-a_{2}+12 a_{4}+\frac {a_{0}}{6}=0 \] Which after substituting earlier equations, simplifies to \[ a_{5} = -\frac {\pi ^{2} a_{0}-3 \pi ^{2}+6 \pi a_{1}-6 a_{0}}{120 \pi ^{3}} \] \(n=4\) gives \[ 30 \pi a_{6}-a_{3}+20 a_{5}+\frac {a_{1}}{6}=0 \] Which after substituting earlier equations, simplifies to \[ a_{6} = -\frac {\pi ^{3} a_{1}-2 \pi ^{2} a_{0}+3 \pi ^{2}-6 \pi a_{1}+6 a_{0}}{180 \pi ^{4}} \] \(n=5\) gives \[ 42 \pi a_{7}-a_{4}+30 a_{6}+\frac {a_{2}}{6}-\frac {a_{0}}{120}=0 \] Which after substituting earlier equations, simplifies to \[ a_{7} = \frac {\pi ^{4} a_{0}-10 \pi ^{4}+30 \pi ^{3} a_{1}-50 \pi ^{2} a_{0}+60 \pi ^{2}-120 \pi a_{1}+120 a_{0}}{5040 \pi ^{5}} \] For \(7\le n\), the recurrence equation is \begin{equation} \tag{4} \left (\left (1+n \right ) a_{1+n} n +\pi \left (n +2\right ) a_{n +2} \left (1+n \right )-a_{n -1}+\frac {a_{n -3}}{6}-\frac {a_{n -5}}{120}+\frac {a_{n -7}}{5040}\right ) t^{n} = t +\pi \end{equation} 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 \[ y \left (t \right ) = a_{0}+a_{1} t +\frac {t^{2}}{2}+\frac {a_{0} t^{3}}{6 \pi }+\frac {\left (\pi a_{1}-a_{0}\right ) t^{4}}{12 \pi ^{2}}-\frac {\left (\pi ^{2} a_{0}-3 \pi ^{2}+6 \pi a_{1}-6 a_{0}\right ) t^{5}}{120 \pi ^{3}}+\dots \] Collecting terms, the solution becomes \begin{equation} \tag{3} y \left (t \right ) = \left (1+\frac {t^{3}}{6 \pi }-\frac {t^{4}}{12 \pi ^{2}}-\frac {\left (\pi ^{2}-6\right ) t^{5}}{120 \pi ^{3}}\right ) a_{0}+\left (t +\frac {t^{4}}{12 \pi }-\frac {t^{5}}{20 \pi ^{2}}\right ) a_{1}+\frac {t^{2}}{2}+\frac {t^{5}}{40 \pi }+O\left (t^{6}\right ) \end{equation} At \(t = 0\) the solution above becomes \[ y \left (t \right ) = \left (1+\frac {t^{3}}{6 \pi }-\frac {t^{4}}{12 \pi ^{2}}-\frac {\left (\pi ^{2}-6\right ) t^{5}}{120 \pi ^{3}}\right ) c_{1} +\left (t +\frac {t^{4}}{12 \pi }-\frac {t^{5}}{20 \pi ^{2}}\right ) c_{2} +\frac {t^{2}}{2}+\frac {t^{5}}{40 \pi }+O\left (t^{6}\right ) \] \[ y \left (t \right ) = 1+\frac {t^{3}}{6 \pi }-\frac {t^{4}}{12 \pi ^{2}}+\frac {t^{5}}{60 \pi }+\frac {t^{5}}{20 \pi ^{3}}+\frac {t^{2}}{2}+O\left (t^{6}\right ) \] Replacing \(t\) in the above with the original independent variable \(xs\)using \(t = x -\pi \) results in \[ y = 1+\frac {\left (x -\pi \right )^{2}}{2}+\frac {\left (x -\pi \right )^{3}}{6 \pi }-\frac {\left (x -\pi \right )^{4}}{12 \pi ^{2}}+\frac {\left (x -\pi \right )^{5}}{60 \pi }+\frac {\left (x -\pi \right )^{5}}{20 \pi ^{3}}-\frac {\left (x -\pi \right )^{6}}{180 \pi ^{2}}-\frac {\left (x -\pi \right )^{6}}{30 \pi ^{4}}+O\left (\left (x -\pi \right )^{6}\right ) \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= 1+\frac {\left (x -\pi \right )^{2}}{2}+\frac {\left (x -\pi \right )^{3}}{6 \pi }-\frac {\left (x -\pi \right )^{4}}{12 \pi ^{2}}+\frac {\left (x -\pi \right )^{5}}{60 \pi }+\frac {\left (x -\pi \right )^{5}}{20 \pi ^{3}}-\frac {\left (x -\pi \right )^{6}}{180 \pi ^{2}}-\frac {\left (x -\pi \right )^{6}}{30 \pi ^{4}}+O\left (\left (x -\pi \right )^{6}\right ) \\ \end{align*}

Verification of solutions

\[ y = 1+\frac {\left (x -\pi \right )^{2}}{2}+\frac {\left (x -\pi \right )^{3}}{6 \pi }-\frac {\left (x -\pi \right )^{4}}{12 \pi ^{2}}+\frac {\left (x -\pi \right )^{5}}{60 \pi }+\frac {\left (x -\pi \right )^{5}}{20 \pi ^{3}}-\frac {\left (x -\pi \right )^{6}}{180 \pi ^{2}}-\frac {\left (x -\pi \right )^{6}}{30 \pi ^{4}}+O\left (\left (x -\pi \right )^{6}\right ) \] Verified OK.

Maple trace

`Methods for second order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
trying high order exact linear fully integrable 
trying differential order: 2; linear nonhomogeneous with symmetry [0,1] 
trying a double symmetry of the form [xi=0, eta=F(x)] 
trying symmetries linear in x and y(x) 
-> Try solving first the homogeneous part of the ODE 
   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 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 a symmetry of the form [xi=0, eta=F(x)] 
         trying 2nd order exact linear 
         trying symmetries linear in x and y(x) 
         trying to convert to a linear ODE with constant coefficients 
         -> trying with_periodic_functions in the coefficients 
            --- Trying Lie symmetry methods, 2nd order --- 
            `, `-> Computing symmetries using: way = 5 
      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 a symmetry of the form [xi=0, eta=F(x)] 
         trying 2nd order exact linear 
         trying symmetries linear in x and y(x) 
         trying to convert to a linear ODE with constant coefficients 
   <- unable to find a useful change of variables 
      trying symmetries linear in x and y(x) 
      trying to convert to a linear ODE with constant coefficients 
      trying to convert to an ODE of Bessel type 
      -> trying reduction of order to Riccati 
         trying Riccati sub-methods: 
            trying Riccati_symmetries 
            -> trying a symmetry pattern of the form [F(x)*G(y), 0] 
            -> trying a symmetry pattern of the form [0, F(x)*G(y)] 
            -> trying a symmetry pattern of the form [F(x),G(x)*y+H(x)] 
      -> trying with_periodic_functions in the coefficients 
         --- Trying Lie symmetry methods, 2nd order --- 
         `, `-> Computing symmetries using: way = 5`[0, y]
 

✓ Solution by Maple

Time used: 0.015 (sec). Leaf size: 35

Order:=6; 
dsolve([x*diff(y(x),x$2)+y(x)*sin(x)=x,y(Pi) = 1, D(y)(Pi) = 0],y(x),type='series',x=Pi);
 

\[ y \left (x \right ) = 1+\frac {1}{2} \left (-\pi +x \right )^{2}+\frac {1}{6} \frac {1}{\pi } \left (-\pi +x \right )^{3}-\frac {1}{12} \frac {1}{\pi ^{2}} \left (-\pi +x \right )^{4}+\frac {1}{60} \frac {\pi ^{2}+3}{\pi ^{3}} \left (-\pi +x \right )^{5}+\operatorname {O}\left (\left (-\pi +x \right )^{6}\right ) \]

✓ Solution by Mathematica

Time used: 0.025 (sec). Leaf size: 75

AsymptoticDSolveValue[{x*y''[x]+Sin[x]*y[x]==x,{y[Pi]==1,y'[Pi]==0}},y[x],{x,Pi,5}]
 

\[ y(x)\to \frac {1}{60} \left (\frac {3}{2 \pi }-\frac {\pi ^2-6}{2 \pi ^3}\right ) (x-\pi )^5-\frac {(x-\pi )^4}{12 \pi ^2}+\frac {(x-\pi )^3}{6 \pi }+\frac {1}{2} (x-\pi )^2+1 \]